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concepts/mechanics/u1/.gitkeep
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concepts/mechanics/u1/.gitkeep
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113
concepts/mechanics/u1/m1-1-scalars-vectors.tex
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concepts/mechanics/u1/m1-1-scalars-vectors.tex
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\subsection{Scalars, Vectors, and Components}
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This subsection introduces the language used throughout Unit 1 for quantities that have magnitude only and for quantities that also have direction.
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\dfn{Scalars, vectors, and component form}{A \textbf{scalar} quantity is specified completely by a magnitude. Examples include mass, time, temperature, energy, and speed.
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A \textbf{vector} quantity is specified by both a magnitude and a direction. Examples include displacement, velocity, acceleration, and force.
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In a chosen Cartesian coordinate system, let $x$, $y$, and $z$ denote coordinates along the $x$-, $y$-, and $z$-axes, and let $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$ denote unit vectors along those axes. If $\vec{v}$ is a vector with scalar components $v_x$, $v_y$, and $v_z$, then its unit-vector decomposition is
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\[
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\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}.
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\]
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In two dimensions,
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\[
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\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath}.
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\]
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The numbers $v_x$, $v_y$, and $v_z$ are \textbf{components} of $\vec{v}$; they are scalars and can be positive, negative, or zero. The magnitude of $\vec{v}$ is written $|\vec{v}|$.}
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\nt{Speed is a scalar, but velocity is a vector. A component such as $v_x$ is a scalar, not a vector by itself. Also, the magnitude $|\vec{v}|$ is not the same thing as a component: $|\vec{v}|\geq 0$, while a component can be negative if the vector points partly in a negative coordinate direction. Likewise, distance is a scalar, while displacement $\Delta \vec{r}$ is a vector. Later in this unit, $\Delta \vec{r}$, $\vec{v}$, and $\vec{a}$ will all be handled with the same component ideas.}
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\mprop{Operational rules in components}{Let
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\[
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\vec{u} = u_x\hat{\imath} + u_y\hat{\jmath} + u_z\hat{k}
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\qquad\text{and}\qquad
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\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}
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\]
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be vectors written in the same Cartesian coordinate system.
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\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
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\item Vector addition and subtraction are done component-by-component:
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\[
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\vec{u} + \vec{v} = (u_x+v_x)\hat{\imath} + (u_y+v_y)\hat{\jmath} + (u_z+v_z)\hat{k},
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\]
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\[
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\vec{u} - \vec{v} = (u_x-v_x)\hat{\imath} + (u_y-v_y)\hat{\jmath} + (u_z-v_z)\hat{k}.
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\]
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\item In Cartesian coordinates, the magnitude of a vector comes from its components:
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\[
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|\vec{v}| = \sqrt{v_x^2+v_y^2}
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\qquad\text{in 2D,}
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\]
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\[
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|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2}
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\qquad\text{in 3D.}
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\]
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\item In two dimensions, if $\theta$ is the direction angle of $\vec{v}$ measured from the positive $x$-axis, then
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\[
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\tan\theta = \frac{v_y}{v_x},
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\]
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provided $v_x\neq 0$. The signs of $v_x$ and $v_y$ determine the correct quadrant for $\theta$. If $v_x=0$, then the vector points straight along the positive or negative $y$-axis.
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\end{enumerate}}
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\qs{Worked example}{In an $xy$ coordinate system, the positive $x$-axis points east and the positive $y$-axis points north. A student first moves with displacement
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\[
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\Delta \vec{r}_1=(3.0\hat{\imath}+4.0\hat{\jmath})\,\text{m}
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\]
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and then moves with displacement
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\[
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\Delta \vec{r}_2=(-1.0\hat{\imath}+2.0\hat{\jmath})\,\text{m}.
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\]
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Let $\Delta \vec{r}_{\text{tot}}$ denote the total displacement, and let $d$ denote the total distance traveled.
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\begin{enumerate}[label=(\alph*)]
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\item Identify whether $\Delta \vec{r}_{\text{tot}}$, the component $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$, and $d$ are scalars or vectors.
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\item Find $\Delta \vec{r}_{\text{tot}}$ in component form.
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\item Find $|\Delta \vec{r}_{\text{tot}}|$.
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\item Let $\theta$ be the direction of $\Delta \vec{r}_{\text{tot}}$ measured counterclockwise from the positive $x$-axis. Find $\theta$.
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\end{enumerate}}
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\sol For part (a), $\Delta \vec{r}_{\text{tot}}$ is a vector because it has both magnitude and direction. The quantity $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$ is a scalar because it is one component of a vector. The quantity $d$ is also a scalar because distance gives only a path length.
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For part (b), add the two displacements by components:
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\[
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\Delta \vec{r}_{\text{tot}}=\Delta \vec{r}_1+\Delta \vec{r}_2.
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\]
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Therefore,
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\[
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\Delta \vec{r}_{\text{tot}}=(3.0-1.0)\hat{\imath}+(4.0+2.0)\hat{\jmath}.
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\]
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So,
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\[
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\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m}.
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\]
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For part (c), use the magnitude formula in two dimensions:
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\[
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|\Delta \vec{r}_{\text{tot}}|=\sqrt{(2.0\,\text{m})^2+(6.0\,\text{m})^2}.
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\]
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Thus,
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\[
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|\Delta \vec{r}_{\text{tot}}|=\sqrt{40}\,\text{m}=6.32\,\text{m}.
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\]
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For part (d), use the component ratio. Since both components of $\Delta \vec{r}_{\text{tot}}$ are positive, the vector lies in the first quadrant. Then
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\[
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\tan\theta=\frac{6.0}{2.0}=3.0.
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\]
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So,
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\[
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\theta=\tan^{-1}(3.0)=71.6^\circ.
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\]
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Therefore, the total displacement is
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\[
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\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m},
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\]
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with magnitude $6.32\,\text{m}$ and direction $71.6^\circ$ counterclockwise from the positive $x$-axis.
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As a useful comparison, the total distance traveled is
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\[
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d=|\Delta \vec{r}_1|+|\Delta \vec{r}_2|=5.0\,\text{m}+\sqrt{5}\,\text{m}=7.24\,\text{m},
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\]
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which is a scalar and is not equal to the magnitude of the total displacement.
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93
concepts/mechanics/u1/m1-2-position-displacement.tex
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concepts/mechanics/u1/m1-2-position-displacement.tex
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\subsection{Position, Displacement, Distance, and Reference Frames}
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\dfn{Reference frame, position, displacement, and distance}{A \emph{reference frame} is a choice of origin $O$, coordinate axes $x$, $y$, and $z$, and a clock for measuring time $t$. In AP kinematics, we describe motion in an inertial reference frame.
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The \emph{position vector} of an object at time $t$ is
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\[
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\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
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\]
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where $x(t)$, $y(t)$, and $z(t)$ are the object's coordinates in the chosen frame.
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If the object is at initial position $\vec{r}_i$ and later at final position $\vec{r}_f$, the \emph{displacement} over that time interval is the vector
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\[
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\Delta \vec{r}=\vec{r}_f-\vec{r}_i.
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\]
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The \emph{distance traveled}, denoted by $d$, is the total length of the path actually followed. Distance is a scalar, while displacement is a vector.}
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\nt{The sign of a coordinate such as $x$ depends on the chosen axis direction, and the value of a position such as $\vec{r}(t)$ depends on the chosen origin. Thus position is frame-dependent. Displacement $\Delta \vec{r}$ compares two positions in the same frame, so changing the origin alone changes $\vec{r}_i$ and $\vec{r}_f$ but not their difference. Distance $d$ is not the same as $\lvert \Delta \vec{r} \rvert$ in general: $d$ is the total path length, while $\lvert \Delta \vec{r} \rvert$ is the straight-line separation between the initial and final positions. If an object turns around or follows a curved path, then $d>\lvert \Delta \vec{r} \rvert$.}
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\mprop{Component formulas and the straight-line special case}{If the initial position is
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\[
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\vec{r}_i=x_i\,\hat{\imath}+y_i\,\hat{\jmath}+z_i\,\hat{k}
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\]
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and the final position is
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\[
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\vec{r}_f=x_f\,\hat{\imath}+y_f\,\hat{\jmath}+z_f\,\hat{k},
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\]
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then the displacement is
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\[
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\Delta \vec{r}=(x_f-x_i)\,\hat{\imath}+(y_f-y_i)\,\hat{\jmath}+(z_f-z_i)\,\hat{k}.
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\]
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Its magnitude is
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\[
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\lvert \Delta \vec{r} \rvert=\sqrt{(x_f-x_i)^2+(y_f-y_i)^2+(z_f-z_i)^2}.
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\]
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If the object moves along a straight path without changing direction, then the distance traveled equals the magnitude of the displacement:
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\[
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d=\lvert \Delta \vec{r} \rvert.
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\]
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For any other path, the distance satisfies
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\[
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d\ge \lvert \Delta \vec{r} \rvert.
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\]}
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\qs{Worked example}{In a laboratory reference frame, the origin is marked on the floor, the $x$-axis points east, and the $y$-axis points north. A robot starts at the position
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\[
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\vec{r}_i=(2\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
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\]
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It then moves $5\,\mathrm{m}$ east, then $3\,\mathrm{m}$ south, and then $2\,\mathrm{m}$ west. Find the final position $\vec{r}_f$, the displacement $\Delta \vec{r}$, the magnitude $\lvert \Delta \vec{r} \rvert$, and the total distance traveled $d$.}
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\sol The initial position is
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\[
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\vec{r}_i=(2\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
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\]
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After the first motion, the robot moves $5\,\mathrm{m}$ east, so its position becomes
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\[
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(7\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
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\]
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After the second motion, the robot moves $3\,\mathrm{m}$ south, so its position becomes
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\[
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(7\,\mathrm{m})\,\hat{\imath}+(-2\,\mathrm{m})\,\hat{\jmath}.
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\]
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After the third motion, the robot moves $2\,\mathrm{m}$ west, so the final position is
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\[
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\vec{r}_f=(5\,\mathrm{m})\,\hat{\imath}+(-2\,\mathrm{m})\,\hat{\jmath}.
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\]
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Therefore the displacement is
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\[
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\Delta \vec{r}=\vec{r}_f-\vec{r}_i=\bigl[(5-2)\,\mathrm{m}\bigr]\hat{\imath}+\bigl[(-2)-1\bigr]\mathrm{m}\,\hat{\jmath}.
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\]
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So
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\[
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\Delta \vec{r}=(3\,\mathrm{m})\,\hat{\imath}+(-3\,\mathrm{m})\,\hat{\jmath}.
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\]
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Its magnitude is
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\[
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\lvert \Delta \vec{r} \rvert=\sqrt{(3\,\mathrm{m})^2+(-3\,\mathrm{m})^2}=\sqrt{18}\,\mathrm{m}=3\sqrt{2}\,\mathrm{m}.
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\]
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The total distance traveled is the sum of the three path segments:
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\[
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d=5\,\mathrm{m}+3\,\mathrm{m}+2\,\mathrm{m}=10\,\mathrm{m}.
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\]
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Thus,
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\[
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\vec{r}_f=(5\,\mathrm{m})\,\hat{\imath}-(2\,\mathrm{m})\,\hat{\jmath},
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\qquad
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\Delta \vec{r}=(3\,\mathrm{m})\,\hat{\imath}-(3\,\mathrm{m})\,\hat{\jmath},
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\]
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\[
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\lvert \Delta \vec{r} \rvert=3\sqrt{2}\,\mathrm{m},
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\qquad
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d=10\,\mathrm{m}.
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\]
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133
concepts/mechanics/u1/m1-3-velocity-acceleration.tex
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concepts/mechanics/u1/m1-3-velocity-acceleration.tex
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\subsection{Velocity and Acceleration as Derivatives}
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This subsection describes motion locally: starting from the position vector $\vec{r}(t)$, velocity and acceleration are defined by derivatives at an instant. Later sections will reverse these local definitions with definite integrals to recover displacement and changes in velocity over a time interval.
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\dfn{Average and instantaneous velocity and acceleration}{Let $t$ denote time, let $\Delta t$ denote a nonzero time interval, and let $\vec{r}(t)$ denote the position vector of a particle.
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The \emph{average velocity} from time $t$ to time $t+\Delta t$ is
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\[
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\vec{v}_{\mathrm{avg}}=\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}.
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\]
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If the limit exists, the \emph{instantaneous velocity} at time $t$ is the vector
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\[
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\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
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\]
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Let $\vec{v}(t)$ denote the instantaneous velocity. Then the \emph{average acceleration} from time $t$ to time $t+\Delta t$ is
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\[
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\vec{a}_{\mathrm{avg}}=\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}.
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\]
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If the limit exists, the \emph{instantaneous acceleration} at time $t$ is the vector
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\[
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\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
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\]
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The \emph{speed} at time $t$ is the scalar magnitude $|\vec{v}(t)|$.}
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\thm{Derivative relations in vector and component form}{Let
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\[
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\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
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\]
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where $x(t)$, $y(t)$, and $z(t)$ are coordinate functions.
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Then the velocity vector is
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\[
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\vec{v}(t)=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
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\]
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If $v_x(t)$, $v_y(t)$, and $v_z(t)$ denote the components of $\vec{v}(t)$, then
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\[
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v_x=\frac{dx}{dt},\qquad v_y=\frac{dy}{dt},\qquad v_z=\frac{dz}{dt}.
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\]
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The acceleration vector is
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\[
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\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{dt^2}=\frac{dv_x}{dt}\,\hat{\imath}+\frac{dv_y}{dt}\,\hat{\jmath}+\frac{dv_z}{dt}\,\hat{k}.
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\]
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If $a_x(t)$, $a_y(t)$, and $a_z(t)$ denote the components of $\vec{a}(t)$, then
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\[
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a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2},\qquad a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt^2},\qquad a_z=\frac{dv_z}{dt}=\frac{d^2z}{dt^2}.
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\]
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In two-dimensional motion, the same formulas hold with the $z$-terms omitted.}
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\pf{Why these formulas are true}{By definition,
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\[
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\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
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\]
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If
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\[
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\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
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\]
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then differentiating component-by-component gives
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\[
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\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
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\]
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Likewise,
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\[
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\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
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\]
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Differentiating the velocity components gives
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\[
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a_x=\frac{dv_x}{dt},\qquad a_y=\frac{dv_y}{dt},\qquad a_z=\frac{dv_z}{dt},
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\]
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and substituting $v_x=dx/dt$, $v_y=dy/dt$, and $v_z=dz/dt$ gives
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\[
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a_x=\frac{d^2x}{dt^2},\qquad a_y=\frac{d^2y}{dt^2},\qquad a_z=\frac{d^2z}{dt^2}.
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\]}
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\cor{One-dimensional motion and a speed caution}{If motion is confined to the $x$-axis, so that
|
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\[
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||||
\vec{r}(t)=x(t)\,\hat{\imath},
|
||||
\]
|
||||
then
|
||||
\[
|
||||
\vec{v}(t)=v_x(t)\,\hat{\imath}\qquad\text{and}\qquad \vec{a}(t)=a_x(t)\,\hat{\imath},
|
||||
\]
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||||
with
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||||
\[
|
||||
v_x=\frac{dx}{dt},\qquad a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2}.
|
||||
\]
|
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However, in two or three dimensions, constant speed $|\vec{v}|$ does not by itself imply zero acceleration, because the direction of $\vec{v}$ can change even when its magnitude stays the same.}
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|
||||
\qs{Worked example}{Let $t$ denote time in seconds. A particle moves in the $xy$-plane with position vector
|
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\[
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||||
\vec{r}(t)=\bigl(t^2-4t\bigr)\hat{\imath}+\bigl(3t-t^2\bigr)\hat{\jmath}\;\mathrm{m}.
|
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\]
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Let $x(t)=t^2-4t$ and let $y(t)=3t-t^2$, where $x(t)$ and $y(t)$ are measured in meters.
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|
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Find $\vec{v}(t)$ and $\vec{a}(t)$. Then find $\vec{v}(1.0\,\mathrm{s})$, $\vec{a}(1.0\,\mathrm{s})$, and the speed at $t=1.0\,\mathrm{s}$. Interpret the signs of the velocity components at $t=1.0\,\mathrm{s}$.}
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\sol From
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\[
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x(t)=t^2-4t\qquad\text{and}\qquad y(t)=3t-t^2,
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\]
|
||||
the component formulas for velocity give
|
||||
\[
|
||||
v_x=\frac{dx}{dt}=2t-4\qquad\text{and}\qquad v_y=\frac{dy}{dt}=3-2t.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{v}(t)=\bigl(2t-4\bigr)\hat{\imath}+\bigl(3-2t\bigr)\hat{\jmath}\;\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Differentiate again to find the acceleration components:
|
||||
\[
|
||||
a_x=\frac{dv_x}{dt}=2\qquad\text{and}\qquad a_y=\frac{dv_y}{dt}=-2.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{a}(t)=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
At $t=1.0\,\mathrm{s}$,
|
||||
\[
|
||||
\vec{v}(1.0\,\mathrm{s})=\bigl(2(1.0)-4\bigr)\hat{\imath}+\bigl(3-2(1.0)\bigr)\hat{\jmath}=-2\hat{\imath}+\hat{\jmath}\;\mathrm{m/s},
|
||||
\]
|
||||
and
|
||||
\[
|
||||
\vec{a}(1.0\,\mathrm{s})=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
The speed at $t=1.0\,\mathrm{s}$ is the magnitude of the velocity vector:
|
||||
\[
|
||||
|\vec{v}(1.0\,\mathrm{s})|=\sqrt{(-2)^2+(1)^2}\;\mathrm{m/s}=\sqrt{5}\;\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Because $v_x(1.0\,\mathrm{s})=-2\,\mathrm{m/s}$, the particle is moving in the negative $x$-direction at that instant. Because $v_y(1.0\,\mathrm{s})=1\,\mathrm{m/s}$, the particle is moving in the positive $y$-direction at that instant. So at $t=1.0\,\mathrm{s}$ the particle is moving left and upward, with speed $\sqrt{5}\,\mathrm{m/s}$.
|
||||
137
concepts/mechanics/u1/m1-4-motion-graphs.tex
Normal file
137
concepts/mechanics/u1/m1-4-motion-graphs.tex
Normal file
@@ -0,0 +1,137 @@
|
||||
\subsection{Motion Graphs, Slopes, and Signed Areas}
|
||||
|
||||
This subsection connects the local language of derivatives to the global language of accumulation. In one-dimensional motion, or when motion is analyzed along the $x$-axis, AP problems often ask you to translate among a verbal description, a graph such as $x(t)$, $v_x(t)$, or $a_x(t)$, and equations relating slope and area.
|
||||
|
||||
\dfn{Common motion graphs and the meaning of slope}{Let $t$ denote time. Let $x(t)$ denote the position coordinate along the $x$-axis, let $v_x(t)$ denote the $x$-component of velocity, and let $a_x(t)$ denote the $x$-component of acceleration.
|
||||
|
||||
An $x(t)$ graph shows position as a function of time. A $v_x(t)$ graph shows velocity as a function of time. An $a_x(t)$ graph shows acceleration as a function of time.
|
||||
|
||||
If $t_1$ and $t_2$ are two times with $t_2>t_1$, then the \emph{average slope} of a graph of a quantity $q(t)$ over the interval from $t_1$ to $t_2$ is
|
||||
\[
|
||||
\frac{q(t_2)-q(t_1)}{t_2-t_1}.
|
||||
\]
|
||||
This is the slope of the secant line through the two points on the graph.
|
||||
|
||||
The \emph{slope at a point} is the slope of the tangent line at that time. For motion graphs, the slope at a point gives an instantaneous rate of change, while the average slope over an interval gives an average rate of change over that interval.}
|
||||
|
||||
\nt{Do not confuse a graph's \emph{value} with its \emph{slope}. On an $x(t)$ graph, a point high above the axis means large position, not large velocity. On a $v_x(t)$ graph, a point at $v_x=0$ means zero velocity at that instant, while a horizontal tangent means zero acceleration at that instant. Likewise, zero slope is not the same as zero value. Also, signed area under a $v_x(t)$ graph gives displacement, not total distance traveled. If velocity changes sign, distance in one dimension is found from $\int |v_x|\,dt$, so areas below the axis must be counted with positive magnitude when finding distance.}
|
||||
|
||||
\mprop{Operational graph rules for one-dimensional motion}{Let $t_1$ and $t_2$ be times with $t_2>t_1$. Let $\Delta x=x(t_2)-x(t_1)$ and let $\Delta v_x=v_x(t_2)-v_x(t_1)$.
|
||||
|
||||
For position and velocity graphs,
|
||||
\[
|
||||
\text{slope of }x(t)\text{ at time }t=v_x(t),
|
||||
\]
|
||||
so the average slope of the position graph over $[t_1,t_2]$ is the average velocity:
|
||||
\[
|
||||
\frac{x(t_2)-x(t_1)}{t_2-t_1}=\frac{\Delta x}{t_2-t_1}.
|
||||
\]
|
||||
|
||||
For velocity and acceleration graphs,
|
||||
\[
|
||||
\text{slope of }v_x(t)\text{ at time }t=a_x(t),
|
||||
\]
|
||||
so the average slope of the velocity graph over $[t_1,t_2]$ is the average acceleration:
|
||||
\[
|
||||
\frac{v_x(t_2)-v_x(t_1)}{t_2-t_1}=\frac{\Delta v_x}{t_2-t_1}.
|
||||
\]
|
||||
|
||||
The signed area under the velocity graph from $t_1$ to $t_2$ gives displacement:
|
||||
\[
|
||||
\Delta x=\int_{t_1}^{t_2} v_x(t)\,dt.
|
||||
\]
|
||||
Area above the time axis contributes positively, and area below the time axis contributes negatively.
|
||||
|
||||
The signed area under the acceleration graph from $t_1$ to $t_2$ gives change in velocity:
|
||||
\[
|
||||
\Delta v_x=\int_{t_1}^{t_2} a_x(t)\,dt.
|
||||
\]
|
||||
|
||||
In one dimension, the total distance traveled from $t_1$ to $t_2$ is
|
||||
\[
|
||||
\text{distance}=\int_{t_1}^{t_2} |v_x(t)|\,dt,
|
||||
\]
|
||||
which equals the total unsigned area between the $v_x(t)$ graph and the time axis.}
|
||||
|
||||
\qs{Worked example}{A particle moves along the $x$-axis. Its velocity graph $v_x(t)$ is described as follows.
|
||||
|
||||
From $t=0$ to $t=2.0\,\mathrm{s}$, the velocity increases linearly from $0$ to $4.0\,\mathrm{m/s}$. From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the velocity is constant at $4.0\,\mathrm{m/s}$. From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the velocity decreases linearly from $4.0\,\mathrm{m/s}$ to $-2.0\,\mathrm{m/s}$.
|
||||
|
||||
Find the acceleration on each time interval, the displacement from $t=0$ to $t=7.0\,\mathrm{s}$, the total distance traveled from $t=0$ to $t=7.0\,\mathrm{s}$, and the average velocity over the full $7.0\,\mathrm{s}$ interval. State when the particle moves in the negative $x$-direction.}
|
||||
|
||||
\sol Let $a_x$ denote the slope of the velocity graph.
|
||||
|
||||
From $t=0$ to $t=2.0\,\mathrm{s}$,
|
||||
\[
|
||||
a_x=\frac{4.0-0}{2.0-0}=2.0\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the graph is horizontal, so
|
||||
\[
|
||||
a_x=0.
|
||||
\]
|
||||
|
||||
From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$,
|
||||
\[
|
||||
a_x=\frac{-2.0-4.0}{7.0-5.0}=-3.0\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Now find displacement from the signed area under the $v_x(t)$ graph.
|
||||
|
||||
From $t=0$ to $t=2.0\,\mathrm{s}$, the area is a triangle with base $2.0\,\mathrm{s}$ and height $4.0\,\mathrm{m/s}$:
|
||||
\[
|
||||
\Delta x_1=\frac{1}{2}(2.0)(4.0)=4.0\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the area is a rectangle:
|
||||
\[
|
||||
\Delta x_2=(3.0)(4.0)=12.0\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the signed area is a trapezoid:
|
||||
\[
|
||||
\Delta x_3=\frac{4.0+(-2.0)}{2}(2.0)=2.0\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Therefore the total displacement is
|
||||
\[
|
||||
\Delta x=\Delta x_1+\Delta x_2+\Delta x_3=4.0+12.0+2.0=18.0\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For total distance, any part of the velocity graph below the axis must be counted positively. The velocity becomes zero during the last interval, so first find that time. Starting from $v_x=4.0\,\mathrm{m/s}$ at $t=5.0\,\mathrm{s}$ with slope $-3.0\,\mathrm{m/s^2}$,
|
||||
\[
|
||||
0=4.0+(-3.0)(t-5.0).
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
t-5.0=\frac{4.0}{3.0},
|
||||
\qquad
|
||||
t=\frac{19}{3}\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
From $t=5.0\,\mathrm{s}$ to $t=19/3\,\mathrm{s}$, the graph is above the axis, giving a triangle of area
|
||||
\[
|
||||
A_+=\frac{1}{2}\left(\frac{4}{3}\right)(4.0)=\frac{8}{3}\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
From $t=19/3\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the graph is below the axis, giving a triangle with signed area $-\frac{2}{3}\,\mathrm{m}$ and magnitude
|
||||
\[
|
||||
A_-=\frac{1}{2}\left(\frac{2}{3}\right)(2.0)=\frac{2}{3}\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Thus the total distance traveled is
|
||||
\[
|
||||
d=4.0+12.0+\frac{8}{3}+\frac{2}{3}=16.0+\frac{10}{3}=\frac{58}{3}\,\mathrm{m}\approx 19.3\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
The average velocity over the full interval is displacement divided by the total elapsed time of $7.0\,\mathrm{s}$:
|
||||
\[
|
||||
v_{x,\mathrm{avg}}=\frac{18.0\,\mathrm{m}}{7.0\,\mathrm{s}}\approx 2.57\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
The particle moves in the negative $x$-direction when $v_x<0$, which occurs after the graph crosses the axis. Therefore it moves in the negative $x$-direction for
|
||||
\[
|
||||
\frac{19}{3}\,\mathrm{s}<t\le 7.0\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
So the interval accelerations are $2.0\,\mathrm{m/s^2}$, $0$, and $-3.0\,\mathrm{m/s^2}$; the displacement is $18.0\,\mathrm{m}$; the total distance is $58/3\,\mathrm{m}$; and the average velocity is about $2.57\,\mathrm{m/s}$.
|
||||
147
concepts/mechanics/u1/m1-5-constant-acceleration.tex
Normal file
147
concepts/mechanics/u1/m1-5-constant-acceleration.tex
Normal file
@@ -0,0 +1,147 @@
|
||||
\subsection{Constant-Acceleration Motion and Free Fall}
|
||||
|
||||
This subsection treats motion over a time interval during which an acceleration component is constant. The standard kinematic formulas are not separate facts; they are the integrated consequences of the local derivative relation between acceleration and velocity.
|
||||
|
||||
\dfn{Constant acceleration and the free-fall approximation}{Let $t$ denote time measured from a chosen initial instant $t=0$. Let $x(t)$ and $y(t)$ denote position components, let $v_x(t)=dx/dt$ and $v_y(t)=dy/dt$ denote velocity components, and let $a_x(t)=dv_x/dt$ and $a_y(t)=dv_y/dt$ denote acceleration components.
|
||||
|
||||
Motion has \emph{constant acceleration in the $x$-direction} if there is a constant scalar $a_x$ such that $a_x(t)=a_x$ throughout the time interval of interest. Likewise, motion has \emph{constant acceleration in the $y$-direction} if there is a constant scalar $a_y$ such that $a_y(t)=a_y$ throughout the interval.
|
||||
|
||||
Near Earth's surface, and neglecting air resistance, \emph{free fall} is motion with constant vertical acceleration of magnitude $g\approx 9.8\,\mathrm{m/s^2}$. If the positive $y$-axis is chosen upward, then $a_y=-g$. If the positive $y$-axis is chosen downward, then $a_y=+g$.}
|
||||
|
||||
\thm{Integrated kinematics in component form}{Let $t$ denote elapsed time from the initial instant $t=0$. Let $x_0=x(0)$ and $v_{x0}=v_x(0)$, and let $x=x(t)$ denote the later $x$-coordinate at time $t$. If $a_x$ is constant, then
|
||||
\[
|
||||
v_x=v_{x0}+a_x t,
|
||||
\]
|
||||
\[
|
||||
x=x_0+v_{x0}t+\tfrac12 a_x t^2,
|
||||
\]
|
||||
and
|
||||
\[
|
||||
v_x^2=v_{x0}^2+2a_x(x-x_0).
|
||||
\]
|
||||
|
||||
Likewise, let $y_0=y(0)$ and $v_{y0}=v_y(0)$, and let $y=y(t)$ denote the later $y$-coordinate at time $t$. If $a_y$ is constant, then
|
||||
\[
|
||||
v_y=v_{y0}+a_y t,
|
||||
\]
|
||||
\[
|
||||
y=y_0+v_{y0}t+\tfrac12 a_y t^2,
|
||||
\]
|
||||
and
|
||||
\[
|
||||
v_y^2=v_{y0}^2+2a_y(y-y_0).
|
||||
\]
|
||||
If both $a_x$ and $a_y$ are constant, then each component evolves independently according to these formulas.}
|
||||
|
||||
\pf{Derivation from a constant acceleration component}{Assume first that $a_x(t)=a_x$ is constant. By the local definition of acceleration,
|
||||
\[
|
||||
\frac{dv_x}{dt}=a_x.
|
||||
\]
|
||||
Integrate from the initial time $0$ to a later time $t$:
|
||||
\[
|
||||
\int_0^t \frac{dv_x}{dt'}\,dt'=\int_0^t a_x\,dt'.
|
||||
\]
|
||||
The left side is $v_x-v_{x0}$, so
|
||||
\[
|
||||
v_x-v_{x0}=a_x t,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
v_x=v_{x0}+a_x t.
|
||||
\]
|
||||
|
||||
Now use $dx/dt=v_x$ and substitute the result just found:
|
||||
\[
|
||||
\frac{dx}{dt}=v_{x0}+a_x t.
|
||||
\]
|
||||
Integrating from $0$ to $t$ gives
|
||||
\[
|
||||
\int_0^t \frac{dx}{dt'}\,dt'=\int_0^t \bigl(v_{x0}+a_x t'\bigr)\,dt'.
|
||||
\]
|
||||
The left side is $x-x_0$, so
|
||||
\[
|
||||
x-x_0=v_{x0}t+\tfrac12 a_x t^2,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
x=x_0+v_{x0}t+\tfrac12 a_x t^2.
|
||||
\]
|
||||
|
||||
To eliminate time, apply the chain rule:
|
||||
\[
|
||||
a_x=\frac{dv_x}{dt}=\frac{dv_x}{dx}\frac{dx}{dt}=v_x\frac{dv_x}{dx}.
|
||||
\]
|
||||
Integrate from $x_0$ to $x$ and from $v_{x0}$ to $v_x$:
|
||||
\[
|
||||
\int_{v_{x0}}^{v_x} v\,dv=\int_{x_0}^{x} a_x\,dx'.
|
||||
\]
|
||||
This gives
|
||||
\[
|
||||
\tfrac12\bigl(v_x^2-v_{x0}^2\bigr)=a_x(x-x_0),
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_x^2=v_{x0}^2+2a_x(x-x_0).
|
||||
\]
|
||||
The $y$-component formulas follow in exactly the same way after replacing $x$ by $y$, $v_x$ by $v_y$, and $a_x$ by $a_y$.}
|
||||
|
||||
\cor{Vertical free-fall formulas and a sign caution}{Choose the positive $y$-axis upward. Let $y_0=y(0)$, let $v_{y0}=v_y(0)$, and let $y=y(t)$ be the height at time $t$. For free fall near Earth with negligible air resistance, $a_y=-g$, where $g>0$. Therefore,
|
||||
\[
|
||||
v_y=v_{y0}-gt,
|
||||
\]
|
||||
\[
|
||||
y=y_0+v_{y0}t-\tfrac12 gt^2,
|
||||
\]
|
||||
and
|
||||
\[
|
||||
v_y^2=v_{y0}^2-2g(y-y_0).
|
||||
\]
|
||||
Negative acceleration does not automatically mean an object is slowing down. An object slows down only when its velocity and acceleration point in opposite directions. Thus, in free fall with $a_y=-g$, an object moving upward has decreasing speed, but an object moving downward has increasing speed.}
|
||||
|
||||
\qs{Worked example}{Choose the positive $y$-axis upward, and let $y$ denote height above the ground. A ball is thrown straight upward from a balcony. Let the initial time be $t=0$, let the initial height be $y_0=24.0\,\mathrm{m}$, let the initial vertical velocity be $v_{y0}=+12.0\,\mathrm{m/s}$, and let the constant vertical acceleration be $a_y=-9.8\,\mathrm{m/s^2}$. Neglect air resistance.
|
||||
|
||||
Find the time when the ball hits the ground and the vertical velocity just before impact.}
|
||||
|
||||
\sol The ball hits the ground when its height is $y=0$. Using
|
||||
\[
|
||||
y=y_0+v_{y0}t+\tfrac12 a_y t^2,
|
||||
\]
|
||||
we substitute the stated values:
|
||||
\[
|
||||
0=24.0\,\mathrm{m}+(12.0\,\mathrm{m/s})t+\tfrac12(-9.8\,\mathrm{m/s^2})t^2.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
0=24.0+12.0t-4.9t^2.
|
||||
\]
|
||||
Rewriting into standard quadratic form gives
|
||||
\[
|
||||
4.9t^2-12.0t-24.0=0.
|
||||
\]
|
||||
Using the quadratic formula,
|
||||
\[
|
||||
t=\frac{12.0\pm\sqrt{(-12.0)^2-4(4.9)(-24.0)}}{2(4.9)}
|
||||
=\frac{12.0\pm\sqrt{614.4}}{9.8}.
|
||||
\]
|
||||
This gives two mathematical roots,
|
||||
\[
|
||||
t\approx -1.30\,\mathrm{s}\qquad\text{or}\qquad t\approx 3.75\,\mathrm{s}.
|
||||
\]
|
||||
The negative time does not fit the physical situation after the throw, so the ball hits the ground at
|
||||
\[
|
||||
t\approx 3.75\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
Now find the vertical velocity at that time from
|
||||
\[
|
||||
v_y=v_{y0}+a_y t.
|
||||
\]
|
||||
Substituting the known values and the physical root gives
|
||||
\[
|
||||
v_y=(12.0\,\mathrm{m/s})+(-9.8\,\mathrm{m/s^2})(3.75\,\mathrm{s})\approx -24.8\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore, just before impact, the ball's vertical velocity is
|
||||
\[
|
||||
v_y\approx -24.8\,\mathrm{m/s},
|
||||
\]
|
||||
which means the ball is moving downward with speed $24.8\,\mathrm{m/s}$. The sign is negative because the positive axis was chosen upward. During the descent, both $v_y$ and $a_y$ are negative, so the ball speeds up even though the acceleration is negative.
|
||||
135
concepts/mechanics/u1/m1-6-relative-projectile.tex
Normal file
135
concepts/mechanics/u1/m1-6-relative-projectile.tex
Normal file
@@ -0,0 +1,135 @@
|
||||
\subsection{Relative Motion and Projectile Motion}
|
||||
|
||||
This subsection combines two central AP kinematics ideas in an inertial frame: relative velocity between different observers, and two-dimensional projectile motion with negligible air resistance. In both settings, vectors are handled component-by-component, and careful notation keeps track of what is being measured.
|
||||
|
||||
\dfn{Relative velocity and the projectile-motion setup}{Let $A$ and $B$ denote moving objects, and let $E$ denote an inertial reference frame such as Earth. If $\vec{v}_{A/E}$ denotes the velocity of object $A$ measured in frame $E$, then the \emph{relative velocity of $A$ with respect to $B$} is the velocity vector of $A$ as measured in the frame moving with $B$, written $\vec{v}_{A/B}$.
|
||||
|
||||
For projectile motion, choose Cartesian axes before writing equations. Let $t$ denote time after launch, let $x(t)$ denote the horizontal coordinate, and let $y(t)$ denote the vertical coordinate measured upward. Let $x_0=x(0)$ and $y_0=y(0)$ denote the initial coordinates, let $v_x(t)$ and $v_y(t)$ denote the velocity components, let $\vec{v}(t)=v_x(t)\hat{\imath}+v_y(t)\hat{\jmath}$ denote the velocity vector, and let $v_{0x}=v_x(0)$ and $v_{0y}=v_y(0)$ denote the initial velocity components. In the AP model, air resistance is neglected and the only acceleration is gravity, so the acceleration vector is
|
||||
\[
|
||||
\vec{a}=-g\hat{\jmath},
|
||||
\]
|
||||
where $g$ denotes the positive magnitude of the gravitational acceleration. Thus the component equations are
|
||||
\[
|
||||
\frac{d^2x}{dt^2}=0,
|
||||
\qquad
|
||||
\frac{d^2y}{dt^2}=-g.
|
||||
\]}
|
||||
|
||||
\thm{Relative-velocity addition and projectile component formulas}{Let $A$, $B$, and $E$ denote objects or frames in classical mechanics. Then the relative-velocity addition law is
|
||||
\[
|
||||
\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}.
|
||||
\]
|
||||
Equivalently,
|
||||
\[
|
||||
\vec{v}_{A/B}=\vec{v}_{A/E}-\vec{v}_{B/E}.
|
||||
\]
|
||||
These equations are vector equations, so they may be applied component-by-component in any chosen axes.
|
||||
|
||||
For projectile motion with the setup in the definition,
|
||||
\[
|
||||
\frac{d^2x}{dt^2}=0,
|
||||
\qquad
|
||||
\frac{d^2y}{dt^2}=-g.
|
||||
\]
|
||||
Integrating once gives the velocity components
|
||||
\[
|
||||
v_x(t)=v_{0x},
|
||||
\qquad
|
||||
v_y(t)=v_{0y}-gt.
|
||||
\]
|
||||
Integrating again gives the position components
|
||||
\[
|
||||
x(t)=x_0+v_{0x}t,
|
||||
\qquad
|
||||
y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2.
|
||||
\]
|
||||
If $v_0$ denotes the initial speed and $\theta$ denotes the launch angle measured above the positive $x$-axis, then
|
||||
\[
|
||||
v_{0x}=v_0\cos\theta,
|
||||
\qquad
|
||||
v_{0y}=v_0\sin\theta.
|
||||
\]
|
||||
The horizontal and vertical motions are independent in the sense that each component has its own equation, but they are linked by the same time variable $t$.}
|
||||
|
||||
\ex{Illustrative relative-motion example}{Let $A$ denote a student walking on a moving walkway, let $B$ denote the walkway, and let $E$ denote the ground frame. Suppose
|
||||
\[
|
||||
\vec{v}_{A/B}=3.0\hat{\jmath}\,\mathrm{m/s}
|
||||
\qquad\text{and}\qquad
|
||||
\vec{v}_{B/E}=4.0\hat{\imath}\,\mathrm{m/s}.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}=4.0\hat{\imath}+3.0\hat{\jmath}\,\mathrm{m/s}.
|
||||
\]
|
||||
So the student moves relative to the ground with speed
|
||||
\[
|
||||
|\vec{v}_{A/E}|=\sqrt{4.0^2+3.0^2}\,\mathrm{m/s}=5.0\,\mathrm{m/s}.
|
||||
\]}
|
||||
|
||||
\nt{In $\vec{v}_{A/B}$, the first label tells \emph{whose} velocity is being described, and the second label tells \emph{which frame} measures it. Reversing the labels changes the meaning. In projectile motion, choose axes first so that the signs of $v_{0x}$, $v_{0y}$, and $-g$ are clear. The horizontal and vertical motions must be evaluated at the same time $t$; they are not separate motions with separate clocks. Common mistakes include adding speeds instead of velocity vectors in relative-motion problems, forgetting that $v_x$ stays constant only when air resistance is neglected, and setting $v_y=0$ for the entire flight instead of only at the top of the path.}
|
||||
|
||||
\qs{Worked example}{A ball is launched from the top of a platform. Let $t$ denote time in seconds after launch. Choose the $x$-axis horizontal and the $y$-axis vertical upward. Let $x(t)$ and $y(t)$ denote the coordinates of the ball in meters. At $t=0$, let $x_0=0$, let $y_0=30.0\,\mathrm{m}$, let $v_{0x}=12.0\,\mathrm{m/s}$, and let $v_{0y}=5.0\,\mathrm{m/s}$. Let $g=10.0\,\mathrm{m/s^2}$.
|
||||
|
||||
Find the time when the ball hits the ground, where the ground is given by $y=0$. Then find the horizontal distance traveled and the velocity vector just before impact.}
|
||||
|
||||
\sol From the projectile component formulas,
|
||||
\[
|
||||
x(t)=x_0+v_{0x}t=12.0t,
|
||||
\]
|
||||
and
|
||||
\[
|
||||
y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2=30.0+5.0t-5.0t^2.
|
||||
\]
|
||||
|
||||
The ball hits the ground when $y=0$, so solve
|
||||
\[
|
||||
30.0+5.0t-5.0t^2=0.
|
||||
\]
|
||||
Divide by $5.0$:
|
||||
\[
|
||||
6+t-t^2=0.
|
||||
\]
|
||||
Rearrange:
|
||||
\[
|
||||
t^2-t-6=0.
|
||||
\]
|
||||
Factor:
|
||||
\[
|
||||
(t-3)(t+2)=0.
|
||||
\]
|
||||
Thus the two algebraic solutions are $t=3.0\,\mathrm{s}$ and $t=-2.0\,\mathrm{s}$. The negative time is not physically relevant after launch, so the impact time is
|
||||
\[
|
||||
t=3.0\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
The horizontal distance traveled is the $x$-coordinate at this time:
|
||||
\[
|
||||
x(3.0)=12.0(3.0)\,\mathrm{m}=36.0\,\mathrm{m}.
|
||||
\]
|
||||
So the ball lands $36.0\,\mathrm{m}$ horizontally from the launch point.
|
||||
|
||||
Next find the velocity components. The horizontal component is constant:
|
||||
\[
|
||||
v_x(t)=v_{0x}=12.0\,\mathrm{m/s}.
|
||||
\]
|
||||
The vertical component is
|
||||
\[
|
||||
v_y(t)=v_{0y}-gt=5.0-10.0t.
|
||||
\]
|
||||
At impact,
|
||||
\[
|
||||
v_y(3.0)=5.0-10.0(3.0)\,\mathrm{m/s}=-25.0\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore the velocity vector just before impact is
|
||||
\[
|
||||
\vec{v}(3.0\,\mathrm{s})=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
|
||||
\]
|
||||
Its negative $y$-component shows that the ball is moving downward at impact. If the impact speed is also desired, then
|
||||
\[
|
||||
|\vec{v}(3.0\,\mathrm{s})|=\sqrt{12.0^2+(-25.0)^2}\,\mathrm{m/s}=\sqrt{769}\,\mathrm{m/s}\approx 27.7\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Thus the ball hits the ground after $3.0\,\mathrm{s}$, lands $36.0\,\mathrm{m}$ from the launch point, and has impact velocity
|
||||
\[
|
||||
\vec{v}=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
|
||||
\]
|
||||
0
concepts/mechanics/u2/.gitkeep
Normal file
0
concepts/mechanics/u2/.gitkeep
Normal file
133
concepts/mechanics/u2/m2-1-newton-laws.tex
Normal file
133
concepts/mechanics/u2/m2-1-newton-laws.tex
Normal file
@@ -0,0 +1,133 @@
|
||||
\subsection{System Choice, Free-Body Diagrams, and Newton's Laws}
|
||||
|
||||
This subsection gives the standard AP mechanics workflow: choose a system, identify the interactions on that system, draw a free-body diagram, choose axes, and then apply Newton's laws in an inertial frame.
|
||||
|
||||
\dfn{System choice, interactions, free-body diagrams, and net external force}{Let $S$ denote a chosen \textbf{system}. In this subsection, $S$ will usually be a single body treated as a particle or rigid object.
|
||||
|
||||
An \textbf{interaction} is a physical influence between the system $S$ and something in the environment, such as gravity from Earth or contact with a surface, rope, or another object. Each interaction can exert a force on $S$.
|
||||
|
||||
A \textbf{free-body diagram} for $S$ is a diagram that shows only the forces acting \emph{on} $S$. If a force is exerted by the environment on $S$, that force belongs on the free-body diagram. Forces exerted by $S$ on other objects do not belong on the free-body diagram of $S$.
|
||||
|
||||
Let $\vec{F}_{\mathrm{ext}}$ denote the net external force on $S$. If the external forces acting on $S$ are $\vec{F}_1$, $\vec{F}_2$, $\dots$, and $\vec{F}_n$, then
|
||||
\[
|
||||
\vec{F}_{\mathrm{ext}}=\sum_{i=1}^n \vec{F}_i=\vec{F}_1+\vec{F}_2+\cdots+\vec{F}_n.
|
||||
\]
|
||||
This is a vector sum. After axes are chosen, one may resolve a force into components for calculation, but those components are not additional physical forces to be added to the free-body diagram.}
|
||||
|
||||
\thm{Newton's laws in AP-usable form}{Work in an inertial reference frame. Let $m$ denote the mass of the chosen body, let $\vec{v}$ denote its velocity, let $\vec{a}$ denote its acceleration, and let $\vec{F}_{\mathrm{ext}}$ denote the vector sum of all external forces on that body.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item \textbf{Newton I.} If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then $\vec{a}=\vec{0}$. Thus the body is either at rest or moves with constant velocity.
|
||||
|
||||
\item \textbf{Newton II.} For the chosen body,
|
||||
\[
|
||||
\vec{F}_{\mathrm{ext}}=m\vec{a}.
|
||||
\]
|
||||
This is the main working law for AP mechanics.
|
||||
|
||||
\item \textbf{Newton III.} If body $A$ exerts a force $\vec{F}_{A\to B}$ on body $B$, then body $B$ exerts a force $\vec{F}_{B\to A}$ on body $A$ such that
|
||||
\[
|
||||
\vec{F}_{B\to A}=-\vec{F}_{A\to B}.
|
||||
\]
|
||||
These two forces act on different bodies, so they do not cancel on a single free-body diagram.
|
||||
\end{enumerate}}
|
||||
|
||||
\pf{Why the vector law becomes component equations}{Choose Cartesian axes with unit vectors $\hat{\imath}$ and $\hat{\jmath}$. Let the acceleration be
|
||||
\[
|
||||
\vec{a}=a_x\hat{\imath}+a_y\hat{\jmath},
|
||||
\]
|
||||
where $a_x$ and $a_y$ are scalar components. Let the net external force be
|
||||
\[
|
||||
\vec{F}_{\mathrm{ext}}=(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath},
|
||||
\]
|
||||
where $\sum F_x$ and $\sum F_y$ are the scalar sums of force components along the chosen axes.
|
||||
|
||||
Substitute these into Newton II:
|
||||
\[
|
||||
(\sum F_x)\hat{\imath}+(\sum F_y)\hat{\jmath}=m a_x\hat{\imath}+m a_y\hat{\jmath}.
|
||||
\]
|
||||
Because the unit vectors $\hat{\imath}$ and $\hat{\jmath}$ are independent, the corresponding scalar components must match. Therefore,
|
||||
\[
|
||||
\sum F_x=ma_x,
|
||||
\qquad
|
||||
\sum F_y=ma_y.
|
||||
\]
|
||||
In practice, one first draws only the actual forces on the free-body diagram, then chooses axes, and only then resolves forces into components if that makes the equations simpler.}
|
||||
|
||||
\cor{Zero net force and equilibrium}{If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then Newton II gives $\vec{a}=\vec{0}$. This does \emph{not} mean the velocity must be zero. A body can have zero net force while moving with a nonzero constant velocity. A special case is equilibrium: if a body is initially at rest and $\vec{F}_{\mathrm{ext}}=\vec{0}$, then it remains at rest.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=5.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field near Earth. Choose the system to be the block.
|
||||
|
||||
Find the acceleration of the block and the magnitude of the normal force exerted by the incline on the block.}
|
||||
|
||||
\sol The system is the block. The interactions on the block are the gravitational interaction with Earth and the contact interaction with the incline. Therefore the free-body diagram contains only two forces: the weight $\vec{W}$ exerted by Earth on the block and the normal force $\vec{N}$ exerted by the incline on the block. Let $N$ denote the magnitude of $\vec{N}$.
|
||||
|
||||
Choose axes so that the $x$-axis is parallel to the incline and positive down the incline, and the $y$-axis is perpendicular to the incline and positive away from the surface. Let $a_x$ and $a_y$ denote the acceleration components in these directions.
|
||||
|
||||
The block stays in contact with the plane, so there is no acceleration perpendicular to the surface. Thus
|
||||
\[
|
||||
a_y=0.
|
||||
\]
|
||||
Resolve the weight into components relative to the chosen axes. The component of $\vec{W}$ parallel to the incline has magnitude
|
||||
\[
|
||||
W_x=mg\sin\theta,
|
||||
\]
|
||||
and the component of $\vec{W}$ perpendicular to the incline has magnitude
|
||||
\[
|
||||
W_y=mg\cos\theta.
|
||||
\]
|
||||
These are components of the single force $\vec{W}$; they are not extra forces on the free-body diagram.
|
||||
|
||||
Now apply Newton II by components.
|
||||
|
||||
Along the incline,
|
||||
\[
|
||||
\sum F_x=ma_x.
|
||||
\]
|
||||
The only force component along the incline is $mg\sin\theta$ in the positive $x$-direction, so
|
||||
\[
|
||||
mg\sin\theta=ma_x.
|
||||
\]
|
||||
Cancel $m$:
|
||||
\[
|
||||
a_x=g\sin\theta.
|
||||
\]
|
||||
Substitute the stated values:
|
||||
\[
|
||||
a_x=(9.8\,\mathrm{m/s^2})\sin 30^\circ=(9.8\,\mathrm{m/s^2})(0.50)=4.9\,\mathrm{m/s^2}.
|
||||
\]
|
||||
So the block accelerates at
|
||||
\[
|
||||
4.9\,\mathrm{m/s^2}
|
||||
\]
|
||||
down the incline.
|
||||
|
||||
Perpendicular to the incline,
|
||||
\[
|
||||
\sum F_y=ma_y.
|
||||
\]
|
||||
The positive $y$-direction is away from the surface, so the normal force is positive and the perpendicular component of the weight is negative. Since $a_y=0$,
|
||||
\[
|
||||
N-mg\cos\theta=0.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
N=mg\cos\theta.
|
||||
\]
|
||||
Substitute the stated values:
|
||||
\[
|
||||
N=(5.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.
|
||||
\]
|
||||
Using $\cos 30^\circ\approx 0.866$ gives
|
||||
\[
|
||||
N\approx (49.0\,\mathrm{N})(0.866)=42.4\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Therefore, the block's acceleration is
|
||||
\[
|
||||
4.9\,\mathrm{m/s^2}
|
||||
\]
|
||||
down the incline, and the magnitude of the normal force is
|
||||
\[
|
||||
42.4\,\mathrm{N}.
|
||||
\]
|
||||
121
concepts/mechanics/u2/m2-2-center-of-mass.tex
Normal file
121
concepts/mechanics/u2/m2-2-center-of-mass.tex
Normal file
@@ -0,0 +1,121 @@
|
||||
\subsection{Center of Mass and Translational Motion of Systems}
|
||||
|
||||
This subsection extends Newton's second law from a single particle to a system of particles. The key idea is that the overall translational motion of the system is described by its center of mass, even when the particles exert complicated internal forces on one another.
|
||||
|
||||
\dfn{Center of mass for a system}{Consider $N$ particles labeled by an index $i=1,2,\dots,N$. Let $m_i$ denote the mass of particle $i$, let $\vec{r}_i$ denote the position vector of particle $i$, and let
|
||||
\[
|
||||
M=\sum_{i=1}^N m_i
|
||||
\]
|
||||
denote the total mass of the system. The \emph{center of mass} is the vector
|
||||
\[
|
||||
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{r}_i.
|
||||
\]
|
||||
Its velocity and acceleration are
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{d\vec{r}_{\mathrm{cm}}}{dt},
|
||||
\qquad
|
||||
\vec{a}_{\mathrm{cm}}=\frac{d\vec{v}_{\mathrm{cm}}}{dt}.
|
||||
\]
|
||||
For a continuous mass distribution, let $dm$ denote a differential mass element located at position vector $\vec{r}$. Then the continuous analog is
|
||||
\[
|
||||
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\int \vec{r}\,dm.
|
||||
\]}
|
||||
|
||||
\thm{Newton's second law for a system}{For the system above, let $\vec{F}_{\mathrm{ext},i}$ denote the net external force on particle $i$. Let
|
||||
\[
|
||||
\sum \vec{F}_{\mathrm{ext}}=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}
|
||||
\]
|
||||
denote the net external force on the whole system. Then
|
||||
\[
|
||||
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
|
||||
\]
|
||||
Therefore the center of mass moves as if all the system mass $M$ were concentrated at the center of mass and acted on by the net external force.}
|
||||
|
||||
\nt{Internal forces can change the separations, shape, or rotation of the parts of a system, but they do not change the motion of the center of mass. When the internal forces are added over the whole system, they cancel in equal-and-opposite pairs. As a result, only external forces determine $\vec{a}_{\mathrm{cm}}$. In particular, if $\sum \vec{F}_{\mathrm{ext}}=\vec{0}$, then $\vec{a}_{\mathrm{cm}}=\vec{0}$, so the center of mass moves with constant velocity even if an explosion or collision makes the individual parts fly apart.}
|
||||
|
||||
\pf{Derivation by summing Newton's second law over the particles}{For each particle $i$, let $\vec{a}_i$ denote its acceleration, and let $\vec{F}_{\mathrm{int},i}$ denote the net internal force on that particle from the other particles in the system. Newton's second law for particle $i$ gives
|
||||
\[
|
||||
m_i\vec{a}_i=\vec{F}_{\mathrm{ext},i}+\vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
Now sum over all particles:
|
||||
\[
|
||||
\sum_{i=1}^N m_i\vec{a}_i=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}+\sum_{i=1}^N \vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
By Newton's third law, the internal forces cancel in pairs, so
|
||||
\[
|
||||
\sum_{i=1}^N \vec{F}_{\mathrm{int},i}=\vec{0}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\sum_{i=1}^N m_i\vec{a}_i=\sum \vec{F}_{\mathrm{ext}}.
|
||||
\]
|
||||
From the definition
|
||||
\[
|
||||
\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{r}_i,
|
||||
\]
|
||||
differentiate twice with respect to time:
|
||||
\[
|
||||
\vec{a}_{\mathrm{cm}}=\frac{1}{M}\sum_{i=1}^N m_i\vec{a}_i.
|
||||
\]
|
||||
Multiplying by $M$ gives
|
||||
\[
|
||||
M\vec{a}_{\mathrm{cm}}=\sum_{i=1}^N m_i\vec{a}_i.
|
||||
\]
|
||||
Substitute this into the previous result to obtain
|
||||
\[
|
||||
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
|
||||
\]}
|
||||
|
||||
\qs{Worked example}{Two carts move on a horizontal frictionless track. Cart 1 has mass $m_1=2.0\,\mathrm{kg}$ and cart 2 has mass $m_2=3.0\,\mathrm{kg}$. At the instant of interest, cart 1 is at position coordinate $x_1=0.40\,\mathrm{m}$ and cart 2 is at position coordinate $x_2=1.60\,\mathrm{m}$. The carts interact with each other through a light compressed spring between them, so the spring forces are internal to the two-cart system. A student pulls on cart 1 so that the net external force on the two-cart system is a constant horizontal force of magnitude $10.0\,\mathrm{N}$ to the right. At that instant the center of mass is moving to the right with speed $v_{\mathrm{cm},0}=1.2\,\mathrm{m/s}$.
|
||||
|
||||
Find the $x$-coordinate of the center of mass, the acceleration of the center of mass, and the speed of the center of mass $t=2.0\,\mathrm{s}$ later.}
|
||||
|
||||
\sol For one-dimensional motion along the $x$-axis, let $x_{\mathrm{cm}}$ denote the $x$-coordinate of $\vec{r}_{\mathrm{cm}}$. Also let
|
||||
\[
|
||||
M=m_1+m_2=2.0\,\mathrm{kg}+3.0\,\mathrm{kg}=5.0\,\mathrm{kg}.
|
||||
\]
|
||||
Then the center-of-mass coordinate is
|
||||
\[
|
||||
x_{\mathrm{cm}}=\frac{m_1x_1+m_2x_2}{M},
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
x_{\mathrm{cm}}=\frac{(2.0\,\mathrm{kg})(0.40\,\mathrm{m})+(3.0\,\mathrm{kg})(1.60\,\mathrm{m})}{5.0\,\mathrm{kg}}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
x_{\mathrm{cm}}=\frac{0.80+4.80}{5.0}\,\mathrm{m}=1.12\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Now use Newton's second law for the system:
|
||||
\[
|
||||
\sum \vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}.
|
||||
\]
|
||||
The net external force has magnitude $10.0\,\mathrm{N}$ to the right, so the center-of-mass acceleration has magnitude
|
||||
\[
|
||||
a_{\mathrm{cm}}=\frac{10.0\,\mathrm{N}}{5.0\,\mathrm{kg}}=2.0\,\mathrm{m/s^2}
|
||||
\]
|
||||
to the right. Equivalently, $\vec{a}_{\mathrm{cm}}$ points in the positive $x$-direction with magnitude $2.0\,\mathrm{m/s^2}$.
|
||||
|
||||
Because the external force is constant, the center of mass has constant acceleration during the $2.0\,\mathrm{s}$ interval. Let $v_{\mathrm{cm}}$ denote the center-of-mass speed at the later time. Then
|
||||
\[
|
||||
v_{\mathrm{cm}}=v_{\mathrm{cm},0}+a_{\mathrm{cm}}t.
|
||||
\]
|
||||
Substituting gives
|
||||
\[
|
||||
v_{\mathrm{cm}}=1.2\,\mathrm{m/s}+(2.0\,\mathrm{m/s^2})(2.0\,\mathrm{s})=5.2\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
So the center of mass is at
|
||||
\[
|
||||
x_{\mathrm{cm}}=1.12\,\mathrm{m},
|
||||
\]
|
||||
its acceleration has magnitude
|
||||
\[
|
||||
a_{\mathrm{cm}}=2.0\,\mathrm{m/s^2},
|
||||
\]
|
||||
with $\vec{a}_{\mathrm{cm}}$ pointing in the positive $x$-direction, and its speed $2.0\,\mathrm{s}$ later is
|
||||
\[
|
||||
v_{\mathrm{cm}}=5.2\,\mathrm{m/s}.
|
||||
\]
|
||||
Even if the spring pushes the carts apart strongly, that spring force is internal to the chosen system. It can change the individual motions of the carts, but it does not change these center-of-mass results.
|
||||
101
concepts/mechanics/u2/m2-3-gravitation.tex
Normal file
101
concepts/mechanics/u2/m2-3-gravitation.tex
Normal file
@@ -0,0 +1,101 @@
|
||||
\subsection{Gravitation and the Inverse-Square Field}
|
||||
|
||||
This subsection treats gravity from the local field viewpoint. A source mass creates a gravitational field $\vec{g}(\vec{r})$, and a test mass placed at position $\vec{r}$ experiences a gravitational force determined by that field.
|
||||
|
||||
\dfn{Gravitational field and the source--test-mass viewpoint}{Let $\vec{r}$ denote the position vector of a field point, and let a test mass $m$ be placed at that point. If $\vec{F}_g(\vec{r})$ denotes the gravitational force on the test mass due to some source mass distribution, then the \emph{gravitational field} at $\vec{r}$ is defined by
|
||||
\[
|
||||
\vec{g}(\vec{r})=\frac{\vec{F}_g(\vec{r})}{m}.
|
||||
\]
|
||||
Thus the field is the gravitational force per unit mass. Equivalently, any test mass $m$ placed at that point satisfies
|
||||
\[
|
||||
\vec{F}_g=m\vec{g}.
|
||||
\]
|
||||
The SI units of $\vec{g}$ are $\mathrm{N/kg}$, which are equivalent to $\mathrm{m/s^2}$. The test mass is assumed small enough that it does not significantly change the source field.}
|
||||
|
||||
\thm{Newton's law of gravitation and the inverse-square field}{Let $G$ denote the universal gravitational constant. Let a point mass or a spherically symmetric source have total mass $M$ and center at the origin. Let $\vec{r}$ denote the position vector of a field point, let $r=|\vec{r}|$ denote its distance from the center, and let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector. If the source is spherically symmetric, assume the field point lies outside the source. If a test mass $m$ is placed at that point, then the gravitational force on the test mass is
|
||||
\[
|
||||
\vec{F}_g(\vec{r})=-\frac{GMm}{r^2}\hat{r}.
|
||||
\]
|
||||
Therefore the gravitational field produced by the source is
|
||||
\[
|
||||
\vec{g}(\vec{r})=-\frac{GM}{r^2}\hat{r}.
|
||||
\]
|
||||
Its magnitude is
|
||||
\[
|
||||
g(r)=\frac{GM}{r^2}.
|
||||
\]
|
||||
The negative sign shows that the field points toward the source, so gravity is attractive.}
|
||||
|
||||
\pf{Connecting the force law to the field law}{Let $M$ denote the source mass, let $m$ denote the test mass, and let $r$ denote the separation between their centers. Newton's law of gravitation states that the magnitude of the gravitational force is
|
||||
\[
|
||||
F_g=\frac{GMm}{r^2}.
|
||||
\]
|
||||
Because gravity is attractive, the force on the test mass points toward the source. Since $\hat{r}$ points radially outward, the force direction is $-\hat{r}$. Therefore,
|
||||
\[
|
||||
\vec{F}_g(\vec{r})=-\frac{GMm}{r^2}\hat{r}.
|
||||
\]
|
||||
Now apply the definition of gravitational field:
|
||||
\[
|
||||
\vec{g}(\vec{r})=\frac{\vec{F}_g(\vec{r})}{m}=-\frac{GM}{r^2}\hat{r}.
|
||||
\]
|
||||
Thus the field depends on the source mass $M$ and the location $\vec{r}$, but not on the test mass $m$.}
|
||||
|
||||
\cor{Near-Earth constant-$g$ approximation}{Let Earth have mass $M_E$ and radius $R_E$. Let a point above Earth's surface have altitude $h$, and let $r=R_E+h$ denote its distance from Earth's center. Then the gravitational field is
|
||||
\[
|
||||
\vec{g}(\vec{r})=-\frac{GM_E}{(R_E+h)^2}\hat{r}.
|
||||
\]
|
||||
If $h\ll R_E$, then $R_E+h\approx R_E$, so the field magnitude is approximately constant:
|
||||
\[
|
||||
g(r)\approx \frac{GM_E}{R_E^2}=g.
|
||||
\]
|
||||
Over a small region near Earth's surface, the radial direction changes very little. If the positive $y$-axis is chosen upward, then locally
|
||||
\[
|
||||
\vec{g}\approx -g\hat{\jmath},
|
||||
\]
|
||||
which is the constant-acceleration free-fall model used near Earth's surface.}
|
||||
|
||||
\qs{Worked example}{Let Earth have mass $M_E=5.97\times 10^{24}\,\mathrm{kg}$, let Earth have radius $R_E=6.37\times 10^6\,\mathrm{m}$, and let the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}$. A spacecraft has mass $m=500\,\mathrm{kg}$ and is at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let $r=R_E+h$ denote the spacecraft's distance from Earth's center, let $\hat{r}$ denote the outward radial unit vector, and let $g_0=9.80\,\mathrm{m/s^2}$ denote the near-surface gravitational field magnitude.
|
||||
|
||||
Find the gravitational field magnitude $g(r)$ at the spacecraft, the gravitational force $\vec{F}_g$ on the spacecraft, and the percent by which the near-surface value $g_0$ overestimates the true field magnitude at that altitude.}
|
||||
|
||||
\sol First compute the distance from Earth's center:
|
||||
\[
|
||||
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
|
||||
\]
|
||||
The inverse-square field magnitude is
|
||||
\[
|
||||
g(r)=\frac{GM_E}{r^2}.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
g(r)=\frac{\left(6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}\right)\left(5.97\times 10^{24}\,\mathrm{kg}\right)}{\left(6.77\times 10^6\,\mathrm{m}\right)^2}.
|
||||
\]
|
||||
This gives
|
||||
\[
|
||||
g(r)\approx 8.69\,\mathrm{m/s^2}.
|
||||
\]
|
||||
So the gravitational field at the spacecraft has magnitude $8.69\,\mathrm{m/s^2}$ and points toward Earth's center.
|
||||
|
||||
Now use $\vec{F}_g=m\vec{g}$. Since $\vec{g}=-(8.69\,\mathrm{m/s^2})\hat{r}$ at that location,
|
||||
\[
|
||||
\vec{F}_g=m\vec{g}=\left(500\,\mathrm{kg}\right)\left[-(8.69\,\mathrm{m/s^2})\hat{r}\right].
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{F}_g\approx -4.34\times 10^3\hat{r}\,\mathrm{N}.
|
||||
\]
|
||||
Its magnitude is
|
||||
\[
|
||||
F_g\approx 4.34\times 10^3\,\mathrm{N},
|
||||
\]
|
||||
and the negative sign means the force points toward Earth's center.
|
||||
|
||||
Finally, compare this with the near-surface value $g_0=9.80\,\mathrm{m/s^2}$. The amount of overestimate is
|
||||
\[
|
||||
g_0-g(r)=9.80-8.69=1.11\,\mathrm{m/s^2}.
|
||||
\]
|
||||
Thus the percent overestimate is
|
||||
\[
|
||||
\frac{g_0-g(r)}{g(r)}\times 100\% = \frac{1.11}{8.69}\times 100\% \approx 12.8\%.
|
||||
\]
|
||||
So at an altitude of $4.00\times 10^5\,\mathrm{m}$, the true gravitational field magnitude is about $8.69\,\mathrm{m/s^2}$, the spacecraft's gravitational force is about $4.34\times 10^3\,\mathrm{N}$ toward Earth, and the near-surface constant-$g$ model overestimates the field by about $12.8\%$.
|
||||
182
concepts/mechanics/u2/m2-4-normal-tension.tex
Normal file
182
concepts/mechanics/u2/m2-4-normal-tension.tex
Normal file
@@ -0,0 +1,182 @@
|
||||
\subsection{Normal Force, Tension, and Constrained Motion}
|
||||
|
||||
This subsection treats normal force and tension as contact forces set by constraints. In AP mechanics, a surface constrains motion perpendicular to itself, and an ideal string constrains connected objects to have linked motions.
|
||||
|
||||
\dfn{Normal force, tension, and mechanical constraints}{Let a body be in contact with a surface whose outward unit normal is $\hat{n}$. The \emph{normal force} on the body is the contact force exerted by the surface perpendicular to the surface, so
|
||||
\[
|
||||
\vec{N}=N\hat{n},
|
||||
\]
|
||||
where $N\ge 0$ is its magnitude.
|
||||
|
||||
Let a body be attached to a taut string, rope, or cable whose direction at the body is given by the unit vector $\hat{t}$. The \emph{tension force} exerted by that connector acts along the connector, so
|
||||
\[
|
||||
\vec{T}=T\hat{t},
|
||||
\]
|
||||
where $T\ge 0$ is its magnitude. In the ideal AP model, the string is massless and inextensible, and any pulley is massless and frictionless.
|
||||
|
||||
A \emph{constraint} is a geometric restriction on motion. Contact with a rigid surface constrains motion perpendicular to the surface, and an inextensible string constrains connected bodies to move so that their accelerations along the string are related. Therefore normal force and tension are found from the constraint together with Newton's second law, not chosen independently.}
|
||||
|
||||
\nt{The symbols $N$ and $T$ do not automatically mean $mg$. The equality $N=mg$ holds only in special cases such as a body on a horizontal surface with no vertical acceleration and no other vertical forces. Likewise, $T=mg$ holds only for special cases such as a hanging mass in equilibrium or moving with constant velocity. On an incline, $N$ is often less than $mg$. In an accelerating elevator, $N$ can be greater than or less than $mg$. In a connected multi-body system, $T$ is usually set by the common acceleration of the system, so it is generally not equal to the weight of either mass.}
|
||||
|
||||
\ex{Short example: normal force in an accelerating elevator}{Choose the positive $y$-axis upward. A student of mass $m=60.0\,\mathrm{kg}$ stands on a scale in an elevator that accelerates upward with magnitude $a=2.0\,\mathrm{m/s^2}$. Let $N$ denote the magnitude of the scale's normal force on the student, and let $g=9.8\,\mathrm{m/s^2}$.
|
||||
|
||||
Newton's second law in the vertical direction gives
|
||||
\[
|
||||
N-mg=ma.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
N=m(g+a)=(60.0\,\mathrm{kg})(11.8\,\mathrm{m/s^2})=708\,\mathrm{N}.
|
||||
\]
|
||||
So here $N>mg$. The normal force is determined by the acceleration constraint, not by a rule that it must equal the weight.}
|
||||
|
||||
\mprop{Operational rules for frictionless contact and ideal strings}{Let $\vec{a}$ denote the acceleration of a body in contact with a frictionless surface, let $\hat{n}$ denote a unit vector perpendicular to the surface, and let $a_{\perp}=\vec{a}\cdot \hat{n}$ denote the acceleration component perpendicular to the surface. Then Newton's second law in the perpendicular direction is
|
||||
\[
|
||||
\sum F_{\perp}=ma_{\perp}.
|
||||
\]
|
||||
If the contact surface is a fixed plane and the body remains in contact without leaving that plane, then $a_{\perp}=0$, so the perpendicular force equation determines $N$. For a plane at angle $\theta$ to the horizontal, if the only force with a perpendicular component besides $\vec{N}$ is the weight, then
|
||||
\[
|
||||
N=mg\cos\theta.
|
||||
\]
|
||||
But if another force has a perpendicular component, it must also be included, so $N$ is not automatically $mg\cos\theta$.
|
||||
|
||||
Let two or more bodies be connected by one ideal string over massless, frictionless pulleys. Then the tension magnitude is the same everywhere in that string:
|
||||
\[
|
||||
T_1=T_2=\cdots=T.
|
||||
\]
|
||||
If $a_1$ and $a_2$ denote acceleration components of two connected bodies measured along their allowed directions of motion, then the inextensible-string constraint gives equal magnitudes:
|
||||
\[
|
||||
|a_1|=|a_2|.
|
||||
\]
|
||||
With a consistent choice of positive directions, this often becomes a signed relation such as
|
||||
\[
|
||||
a_1=a_2
|
||||
\qquad\text{or}\qquad
|
||||
a_1=-a_2.
|
||||
\]
|
||||
Then write one Newton's second law equation for each body and solve those equations together with the constraint relation.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m_1=3.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. The block is connected by a light inextensible string that passes over a massless frictionless pulley to a hanging block of mass $m_2=2.0\,\mathrm{kg}$. Let $g=9.8\,\mathrm{m/s^2}$.
|
||||
|
||||
Find the magnitude and direction of the acceleration of the system, the tension $T$ in the string, and the magnitude of the normal force $N$ exerted on $m_1$ by the incline.}
|
||||
|
||||
\sol Choose the system to be both blocks, but write Newton's second-law equations separately for each block.
|
||||
|
||||
For block $m_1$, choose the positive $x$-axis up the incline and the positive $y$-axis perpendicular to the incline, away from the surface. Let $a$ denote the common acceleration magnitude. If $m_2$ moves downward, then $m_1$ moves up the incline with the same acceleration magnitude because the string is ideal and inextensible.
|
||||
|
||||
For block $m_2$, choose the positive direction downward so that both bodies have acceleration component $+a$ along their chosen directions.
|
||||
|
||||
Now identify the forces.
|
||||
|
||||
On $m_1$, the forces are the weight $\vec{W}_1$, the normal force $\vec{N}$, and the tension $\vec{T}$. Resolve the weight into components relative to the incline:
|
||||
\[
|
||||
W_{1,\parallel}=m_1g\sin\theta,
|
||||
\qquad
|
||||
W_{1,\perp}=m_1g\cos\theta.
|
||||
\]
|
||||
|
||||
On $m_2$, the forces are its weight $\vec{W}_2$ downward and the tension $\vec{T}$ upward.
|
||||
|
||||
First check the likely direction of motion. The hanging weight has magnitude
|
||||
\[
|
||||
m_2g=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=19.6\,\mathrm{N},
|
||||
\]
|
||||
while the component of $m_1$'s weight down the incline has magnitude
|
||||
\[
|
||||
m_1g\sin\theta=(3.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin 30^\circ=14.7\,\mathrm{N}.
|
||||
\]
|
||||
Since $19.6\,\mathrm{N}>14.7\,\mathrm{N}$, the system accelerates with $m_2$ downward and $m_1$ up the incline, consistent with the chosen positive directions.
|
||||
|
||||
Apply Newton's second law to $m_1$ along the incline:
|
||||
\[
|
||||
\sum F_{\parallel}=m_1a.
|
||||
\]
|
||||
The positive direction is up the incline, so
|
||||
\[
|
||||
T-m_1g\sin\theta=m_1a.
|
||||
\]
|
||||
|
||||
Apply Newton's second law to $m_1$ perpendicular to the incline:
|
||||
\[
|
||||
\sum F_{\perp}=m_1a_{\perp}.
|
||||
\]
|
||||
Because the block stays in contact with the incline, $a_{\perp}=0$. Therefore,
|
||||
\[
|
||||
N-m_1g\cos\theta=0,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
N=m_1g\cos\theta.
|
||||
\]
|
||||
|
||||
Apply Newton's second law to $m_2$ in the downward positive direction:
|
||||
\[
|
||||
\sum F=m_2a.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
m_2g-T=m_2a.
|
||||
\]
|
||||
|
||||
Now solve the two equations containing $a$ and $T$:
|
||||
\[
|
||||
T-m_1g\sin\theta=m_1a,
|
||||
\]
|
||||
\[
|
||||
m_2g-T=m_2a.
|
||||
\]
|
||||
Add them to eliminate $T$:
|
||||
\[
|
||||
m_2g-m_1g\sin\theta=(m_1+m_2)a.
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
a=\frac{m_2g-m_1g\sin\theta}{m_1+m_2}.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
a=\frac{19.6-14.7}{3.0+2.0}\,\mathrm{m/s^2}
|
||||
=\frac{4.9}{5.0}\,\mathrm{m/s^2}
|
||||
=0.98\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Now find the tension from either block's equation. Using block $m_2$,
|
||||
\[
|
||||
m_2g-T=m_2a,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
T=m_2g-m_2a.
|
||||
\]
|
||||
Substitute:
|
||||
\[
|
||||
T=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})-(2.0\,\mathrm{kg})(0.98\,\mathrm{m/s^2})
|
||||
=19.6\,\mathrm{N}-1.96\,\mathrm{N}
|
||||
=17.64\,\mathrm{N}.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
T\approx 17.6\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Finally, compute the normal force:
|
||||
\[
|
||||
N=m_1g\cos\theta=(3.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.
|
||||
\]
|
||||
Using $\cos 30^\circ\approx 0.866$ gives
|
||||
\[
|
||||
N\approx (29.4\,\mathrm{N})(0.866)=25.5\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Therefore the system accelerates with magnitude
|
||||
\[
|
||||
0.98\,\mathrm{m/s^2},
|
||||
\]
|
||||
with $m_2$ moving downward and $m_1$ moving up the incline, the string tension is
|
||||
\[
|
||||
T\approx 17.6\,\mathrm{N},
|
||||
\]
|
||||
and the normal force on the incline block is
|
||||
\[
|
||||
N\approx 25.5\,\mathrm{N}.
|
||||
\]
|
||||
141
concepts/mechanics/u2/m2-5-friction.tex
Normal file
141
concepts/mechanics/u2/m2-5-friction.tex
Normal file
@@ -0,0 +1,141 @@
|
||||
\subsection{Static and Kinetic Friction}
|
||||
|
||||
This subsection gives the AP dry-friction model for a body in contact with a surface. Friction is a contact force parallel to the surface, tied to the tendency for slipping or to actual slipping, while the normal force is perpendicular to the surface.
|
||||
|
||||
\dfn{Static friction, kinetic friction, and coefficients of friction}{Let a body be in contact with a surface. Let $\vec{N}$ denote the normal force exerted by the surface on the body, let $N=|\vec{N}|$ denote its magnitude, and let $\vec{f}$ denote the friction force exerted by the surface on the body.
|
||||
|
||||
The friction force acts parallel to the contact surface and opposes the relative motion or the tendency of relative motion between the surfaces.
|
||||
|
||||
If the surfaces are not slipping relative to each other, the friction is called \emph{static friction}. Let $\vec{f}_s$ denote the static-friction force and let $f_s=|\vec{f}_s|$ denote its magnitude. Then static friction can adjust in magnitude up to a maximum value:
|
||||
\[
|
||||
f_s\le \mu_s N,
|
||||
\]
|
||||
where $\mu_s$ is the coefficient of static friction.
|
||||
|
||||
If the surfaces are sliding relative to each other, the friction is called \emph{kinetic friction}. Let $\vec{f}_k$ denote the kinetic-friction force and let $f_k=|\vec{f}_k|$ denote its magnitude. In the idealized AP dry-friction model,
|
||||
\[
|
||||
f_k=\mu_k N,
|
||||
\]
|
||||
where $\mu_k$ is the coefficient of kinetic friction.
|
||||
|
||||
The coefficients $\mu_s$ and $\mu_k$ are dimensionless constants for the pair of surfaces in the chosen model.}
|
||||
|
||||
\nt{Static friction is \emph{not} always equal to $\mu_s N$. The quantity $\mu_s N$ is the \emph{maximum possible} static-friction magnitude. The actual static-friction magnitude is whatever value is required by Newton's second law to prevent slipping, as long as that required value does not exceed $\mu_s N$. Only at the threshold of slipping does $f_s=\mu_s N$.}
|
||||
|
||||
\ex{Illustrative example}{Let a block rest on a horizontal table. Let $\vec{P}=(3.0\,\mathrm{N})\hat{\imath}$ denote a horizontal applied force on the block, and let the maximum possible static-friction magnitude be $\mu_s N=5.0\,\mathrm{N}$.
|
||||
|
||||
Because the required friction to prevent motion is only $3.0\,\mathrm{N}$, the block remains at rest and the actual static-friction force is
|
||||
\[
|
||||
\vec{f}_s=-(3.0\,\mathrm{N})\hat{\imath},
|
||||
\]
|
||||
not $-(5.0\,\mathrm{N})\hat{\imath}$. If the applied-force magnitude were increased beyond $5.0\,\mathrm{N}$, static friction could no longer hold the block at rest and slipping would begin.}
|
||||
|
||||
\mprop{Operational laws and direction rules}{Let $\vec{N}$ denote the normal force on a body, let $N=|\vec{N}|$, and let a tangent axis be chosen along the contact surface.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item If there is no slipping at the contact, then the friction is static. Its magnitude must satisfy
|
||||
\[
|
||||
f_s\le \mu_s N.
|
||||
\]
|
||||
Its direction is opposite the direction the body would move \emph{relative to the surface} if friction were absent.
|
||||
|
||||
\item If the body slides relative to the surface, then the friction is kinetic. Its magnitude is
|
||||
\[
|
||||
f_k=\mu_k N,
|
||||
\]
|
||||
and its direction is opposite the relative velocity of the sliding surfaces.
|
||||
|
||||
\item Friction and the normal force are different parts of the same contact interaction: $\vec{N}$ is perpendicular to the surface, while $\vec{f}$ is parallel to the surface.
|
||||
|
||||
\item On an incline that makes an angle $\theta$ with the horizontal, if the only forces perpendicular to the surface are the normal force and the perpendicular component of the weight, then
|
||||
\[
|
||||
N=mg\cos\theta.
|
||||
\]
|
||||
Thus the friction magnitudes are often written as
|
||||
\[
|
||||
f_{s,\max}=\mu_s mg\cos\theta,
|
||||
\qquad
|
||||
f_k=\mu_k mg\cos\theta.
|
||||
\]
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=6.0\,\mathrm{kg}$ is released from rest on a rough incline that makes an angle $\theta=35^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field. Let the coefficient of static friction be $\mu_s=0.40$ and the coefficient of kinetic friction be $\mu_k=0.30$.
|
||||
|
||||
Determine whether the block remains at rest or starts to slide. If it slides, find the magnitude and direction of its acceleration.}
|
||||
|
||||
\sol Choose axes so that the $x$-axis is parallel to the incline and positive down the incline, and the $y$-axis is perpendicular to the incline and positive away from the surface. Let $a_x$ and $a_y$ denote the acceleration components in these directions.
|
||||
|
||||
The forces on the block are the weight $\vec{W}$, the normal force $\vec{N}$ exerted by the incline, and a friction force $\vec{f}$ exerted by the incline.
|
||||
|
||||
Because the block remains on the surface, there is no acceleration perpendicular to the incline, so
|
||||
\[
|
||||
a_y=0.
|
||||
\]
|
||||
Resolve the weight into components relative to the chosen axes. The component parallel to the incline has magnitude
|
||||
\[
|
||||
W_x=mg\sin\theta,
|
||||
\]
|
||||
and the component perpendicular to the incline has magnitude
|
||||
\[
|
||||
W_y=mg\cos\theta.
|
||||
\]
|
||||
|
||||
Apply Newton's second law perpendicular to the incline:
|
||||
\[
|
||||
\sum F_y=ma_y.
|
||||
\]
|
||||
Since the positive $y$-direction is away from the surface,
|
||||
\[
|
||||
N-mg\cos\theta=0.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
N=mg\cos\theta.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
N=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 35^\circ\approx 48.2\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Now check whether static friction can prevent motion. The component of the weight down the incline is
|
||||
\[
|
||||
mg\sin\theta=(6.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\sin 35^\circ\approx 33.7\,\mathrm{N}.
|
||||
\]
|
||||
The maximum possible static-friction magnitude is
|
||||
\[
|
||||
f_{s,\max}=\mu_s N=(0.40)(48.2\,\mathrm{N})\approx 19.3\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
To keep the block at rest, static friction would need to balance the $33.7\,\mathrm{N}$ downslope component of the weight by acting upslope with magnitude $33.7\,\mathrm{N}$. But
|
||||
\[
|
||||
33.7\,\mathrm{N}>19.3\,\mathrm{N}.
|
||||
\]
|
||||
So static friction is not large enough to hold the block at rest. The block starts to slide down the incline.
|
||||
|
||||
Once the block is sliding, the friction is kinetic and points up the incline. Its magnitude is
|
||||
\[
|
||||
f_k=\mu_k N=(0.30)(48.2\,\mathrm{N})\approx 14.5\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Now apply Newton's second law parallel to the incline:
|
||||
\[
|
||||
\sum F_x=ma_x.
|
||||
\]
|
||||
Taking down the incline as positive, the component of the weight is positive and the kinetic friction is negative, so
|
||||
\[
|
||||
mg\sin\theta-f_k=ma_x.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
33.7\,\mathrm{N}-14.5\,\mathrm{N}=(6.0\,\mathrm{kg})a_x.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
a_x=\frac{19.2\,\mathrm{N}}{6.0\,\mathrm{kg}}\approx 3.2\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Therefore, the block does not remain at rest. It slides down the incline with acceleration
|
||||
\[
|
||||
3.2\,\mathrm{m/s^2}
|
||||
\]
|
||||
down the incline.
|
||||
159
concepts/mechanics/u2/m2-6-springs.tex
Normal file
159
concepts/mechanics/u2/m2-6-springs.tex
Normal file
@@ -0,0 +1,159 @@
|
||||
\subsection{Hooke's Law and Spring Models}
|
||||
|
||||
This subsection uses the local displacement-from-equilibrium viewpoint for spring forces. That viewpoint makes the restoring nature of the force and the modeling of combined springs especially clear.
|
||||
|
||||
\dfn{Spring constant and displacement from equilibrium}{Let an ideal spring act along a line with positive direction given by the unit vector $\hat{u}$. Let $x$ denote the signed displacement of the attached object from the spring's equilibrium position, measured along that line, so $x>0$ means displacement in the $+\hat{u}$ direction and $x<0$ means displacement in the opposite direction.
|
||||
|
||||
Let $k>0$ denote the \emph{spring constant}. It measures the stiffness of the spring: a larger $k$ means a larger restoring force for the same displacement. The SI units of $k$ are $\mathrm{N/m}$.}
|
||||
|
||||
\thm{Hooke's law and equivalent spring constants}{Let $x$ denote the signed displacement from equilibrium for an ideal spring with spring constant $k$, measured along the spring's line of action with unit vector $\hat{u}$. Then the spring force on the attached object is
|
||||
\[
|
||||
\vec{F}_s=-kx\hat{u}.
|
||||
\]
|
||||
In one-dimensional scalar form,
|
||||
\[
|
||||
F_s=-kx.
|
||||
\]
|
||||
The negative sign means the spring force opposes the displacement from equilibrium.
|
||||
|
||||
If several ideal springs are modeled by one equivalent spring with constant $k_{\mathrm{eq}}$, then:
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item \textbf{Parallel:} if all springs undergo the same displacement $x$, then
|
||||
\[
|
||||
k_{\mathrm{eq}}=k_1+k_2+\cdots+k_n.
|
||||
\]
|
||||
|
||||
\item \textbf{Series:} if all springs carry the same force magnitude, then
|
||||
\[
|
||||
\frac{1}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}.
|
||||
\]
|
||||
\end{enumerate}}
|
||||
|
||||
\pf{Why the sign and combination rules are correct}{If $x>0$, the object is displaced in the $+\hat{u}$ direction, so the restoring spring force must point in the $-\hat{u}$ direction. If $x<0$, the restoring force must point in the $+\hat{u}$ direction. Both cases are captured by the single vector formula
|
||||
\[
|
||||
\vec{F}_s=-kx\hat{u}.
|
||||
\]
|
||||
|
||||
For springs in parallel, each spring has the same displacement $x$, so the forces add:
|
||||
\[
|
||||
\vec{F}_{\mathrm{tot}}=-(k_1+k_2+\cdots+k_n)x\hat{u}.
|
||||
\]
|
||||
Therefore $k_{\mathrm{eq}}=k_1+k_2+\cdots+k_n$.
|
||||
|
||||
For springs in series, let $F$ denote the common force magnitude through each spring, and let $x_i$ denote the magnitude of the displacement of spring $i$. Since $F=k_ix_i$, we have
|
||||
\[
|
||||
x_i=\frac{F}{k_i}.
|
||||
\]
|
||||
The total displacement magnitude is then
|
||||
\[
|
||||
x=x_1+x_2+\cdots+x_n=F\left(\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}\right).
|
||||
\]
|
||||
If the combination is replaced by one equivalent spring, then $F=k_{\mathrm{eq}}x$, so
|
||||
\[
|
||||
\frac{1}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}.
|
||||
\]}
|
||||
|
||||
\cor{Vertical equilibrium shift and the local spring equation}{Let a mass $m$ hang from a vertical spring of spring constant $k$. Choose downward as positive. Let $y$ denote the downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium value of $y$, and let
|
||||
\[
|
||||
x=y-y_{\mathrm{eq}}
|
||||
\]
|
||||
denote the displacement from equilibrium.
|
||||
|
||||
At static equilibrium,
|
||||
\[
|
||||
mg-ky_{\mathrm{eq}}=0,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
y_{\mathrm{eq}}=\frac{mg}{k}.
|
||||
\]
|
||||
Then the net force on the mass is
|
||||
\[
|
||||
F_{\mathrm{net}}=mg-ky=mg-k(y_{\mathrm{eq}}+x)=-kx.
|
||||
\]
|
||||
Thus, when motion is measured from equilibrium, gravity has already been accounted for, and the net force again has Hooke form.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=0.50\,\mathrm{kg}$ hangs at rest from a vertical spring with spring constant $k=200\,\mathrm{N/m}$. Choose downward as positive. Let $y$ denote the block's downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium displacement, and let $x=y-y_{\mathrm{eq}}$ denote the displacement from equilibrium.
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find $y_{\mathrm{eq}}$.
|
||||
\item The block is pulled downward $0.030\,\mathrm{m}$ from equilibrium and released from rest. Find the spring force and the net force at the instant of release.
|
||||
\item Find the acceleration at the instant of release.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol At equilibrium, the acceleration is zero, so the net force is zero:
|
||||
\[
|
||||
mg-ky_{\mathrm{eq}}=0.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
y_{\mathrm{eq}}=\frac{mg}{k}=\frac{(0.50\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{200\,\mathrm{N/m}}=0.0245\,\mathrm{m}.
|
||||
\]
|
||||
So the equilibrium position is
|
||||
\[
|
||||
2.45\times 10^{-2}\,\mathrm{m}
|
||||
\]
|
||||
below the unstretched length.
|
||||
|
||||
At the instant of release, the block is displaced
|
||||
\[
|
||||
x=+0.030\,\mathrm{m}
|
||||
\]
|
||||
from equilibrium, because downward is positive. The actual spring displacement from the unstretched length is
|
||||
\[
|
||||
y=y_{\mathrm{eq}}+x=0.0245\,\mathrm{m}+0.030\,\mathrm{m}=0.0545\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
The spring force along the chosen vertical axis is
|
||||
\[
|
||||
F_s=-ky=-(200\,\mathrm{N/m})(0.0545\,\mathrm{m})=-10.9\,\mathrm{N}.
|
||||
\]
|
||||
The negative sign means the spring force is upward. Its magnitude is therefore
|
||||
\[
|
||||
10.9\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
The weight is
|
||||
\[
|
||||
mg=(0.50\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=4.90\,\mathrm{N}
|
||||
\]
|
||||
in the positive direction. Hence the net force is
|
||||
\[
|
||||
F_{\mathrm{net}}=mg-ky=4.90\,\mathrm{N}-10.9\,\mathrm{N}=-6.0\,\mathrm{N}.
|
||||
\]
|
||||
Equivalently, using the local displacement-from-equilibrium form,
|
||||
\[
|
||||
F_{\mathrm{net}}=-kx=-(200\,\mathrm{N/m})(0.030\,\mathrm{m})=-6.0\,\mathrm{N},
|
||||
\]
|
||||
which agrees.
|
||||
|
||||
Now apply Newton's second law:
|
||||
\[
|
||||
F_{\mathrm{net}}=ma.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
a=\frac{F_{\mathrm{net}}}{m}=\frac{-6.0\,\mathrm{N}}{0.50\,\mathrm{kg}}=-12\,\mathrm{m/s^2}.
|
||||
\]
|
||||
The negative sign means the acceleration is upward, so the block's acceleration at release has magnitude
|
||||
\[
|
||||
12\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
y_{\mathrm{eq}}=0.0245\,\mathrm{m},
|
||||
\]
|
||||
the spring force at release is
|
||||
\[
|
||||
10.9\,\mathrm{N}
|
||||
\]
|
||||
upward, the net force is
|
||||
\[
|
||||
6.0\,\mathrm{N}
|
||||
\]
|
||||
upward, and the acceleration is
|
||||
\[
|
||||
12\,\mathrm{m/s^2}
|
||||
\]
|
||||
upward.
|
||||
201
concepts/mechanics/u2/m2-7-drag-terminal.tex
Normal file
201
concepts/mechanics/u2/m2-7-drag-terminal.tex
Normal file
@@ -0,0 +1,201 @@
|
||||
\subsection{Drag Forces and Terminal Velocity}
|
||||
|
||||
This subsection uses the AP linear-drag model, in which the resistive force depends on the object's instantaneous velocity. The local-first viewpoint is Newton's second law with a velocity-dependent force.
|
||||
|
||||
\dfn{Linear drag and terminal velocity}{Let an object move through a fluid with velocity $\vec{v}$ relative to the fluid, and let $b>0$ denote the linear-drag coefficient. In the linear-drag model, the drag force exerted by the fluid on the object is
|
||||
\[
|
||||
\vec{F}_D=-b\vec{v}.
|
||||
\]
|
||||
The negative sign means that the drag force always points opposite the velocity.
|
||||
|
||||
If an object falls vertically through the fluid and eventually moves with constant velocity, that steady velocity is called the \emph{terminal velocity}. At terminal velocity, the net force is zero, so the acceleration is zero.}
|
||||
|
||||
\thm{Vertical-fall ODE and terminal-speed result}{Choose the vertical axis positive downward. Let $v(t)$ denote the downward velocity of an object of mass $m$ at time $t$, let $g$ denote the gravitational field strength, and let $b>0$ denote the linear-drag coefficient. Then the vertical equation of motion is
|
||||
\[
|
||||
m\frac{dv}{dt}=mg-bv.
|
||||
\]
|
||||
The terminal speed $v_T$ is the steady-state value obtained by setting the net force equal to zero:
|
||||
\[
|
||||
mg-bv_T=0,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_T=\frac{mg}{b}.
|
||||
\]
|
||||
|
||||
If the initial velocity is $v(0)=v_0$, then
|
||||
\[
|
||||
v(t)=v_T+(v_0-v_T)e^{-bt/m}.
|
||||
\]
|
||||
In particular, if the object is released from rest, then
|
||||
\[
|
||||
v(t)=v_T\left(1-e^{-bt/m}\right).
|
||||
\]}
|
||||
|
||||
\pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction:
|
||||
\[
|
||||
m\frac{dv}{dt}=mg-bv.
|
||||
\]
|
||||
Define the terminal speed by
|
||||
\[
|
||||
v_T=\frac{mg}{b}.
|
||||
\]
|
||||
Then the differential equation becomes
|
||||
\[
|
||||
\frac{dv}{dt}=\frac{b}{m}(v_T-v).
|
||||
\]
|
||||
Separate variables:
|
||||
\[
|
||||
\frac{dv}{v_T-v}=\frac{b}{m}\,dt.
|
||||
\]
|
||||
Integrating gives
|
||||
\[
|
||||
-\ln|v_T-v|=\frac{b}{m}t+C.
|
||||
\]
|
||||
Therefore
|
||||
\[
|
||||
v_T-v=Ce^{-bt/m}
|
||||
\]
|
||||
for some constant $C$, so
|
||||
\[
|
||||
v=v_T-Ce^{-bt/m}.
|
||||
\]
|
||||
Now use the initial condition $v(0)=v_0$:
|
||||
\[
|
||||
v_0=v_T-C.
|
||||
\]
|
||||
Thus $C=v_T-v_0$, and
|
||||
\[
|
||||
v(t)=v_T-(v_T-v_0)e^{-bt/m}=v_T+(v_0-v_T)e^{-bt/m}.
|
||||
\]
|
||||
If $v_0=0$, this reduces to
|
||||
\[
|
||||
v(t)=v_T\left(1-e^{-bt/m}\right).
|
||||
\]
|
||||
As $t\to\infty$, the exponential term approaches $0$, so $v(t)\to v_T$.}
|
||||
|
||||
\ex{Illustrative example}{Choose downward as positive. A ball of mass $m=0.20\,\mathrm{kg}$ falls through air with linear drag coefficient $b=0.50\,\mathrm{N\cdot s/m}$. Find the terminal speed and the acceleration when the downward speed is $v=2.0\,\mathrm{m/s}$.
|
||||
|
||||
The terminal speed is
|
||||
\[
|
||||
v_T=\frac{mg}{b}=\frac{(0.20\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{0.50\,\mathrm{N\cdot s/m}}=3.92\,\mathrm{m/s}.
|
||||
\]
|
||||
When $v=2.0\,\mathrm{m/s}$, Newton's second law gives
|
||||
\[
|
||||
m\frac{dv}{dt}=mg-bv,
|
||||
\]
|
||||
so the acceleration is
|
||||
\[
|
||||
a=g-\frac{b}{m}v=9.8-\frac{0.50}{0.20}(2.0)=4.8\,\mathrm{m/s^2}.
|
||||
\]
|
||||
Since this value is positive in the downward-positive coordinate system, the ball is still accelerating downward.}
|
||||
|
||||
\qs{Worked example}{A small package of mass $m=0.40\,\mathrm{kg}$ is dropped from rest and falls vertically through air. Choose downward as positive. Let $v(t)$ denote the package's downward velocity at time $t$, let $g=9.8\,\mathrm{m/s^2}$, and let the drag force be modeled by
|
||||
\[
|
||||
\vec{F}_D=-b\vec{v}
|
||||
\]
|
||||
with $b=0.80\,\mathrm{N\cdot s/m}$.
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Write the differential equation for $v(t)$.
|
||||
\item Find the terminal speed.
|
||||
\item Find an explicit formula for $v(t)$.
|
||||
\item Find the package's velocity and acceleration at $t=1.0\,\mathrm{s}$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol The forces on the package are its weight $\vec{W}$ downward and the drag force $\vec{F}_D$ upward because the package is moving downward. Since downward is chosen as positive, the scalar force equation is
|
||||
\[
|
||||
mg-bv=m\frac{dv}{dt}.
|
||||
\]
|
||||
|
||||
For part (a), the differential equation is therefore
|
||||
\[
|
||||
m\frac{dv}{dt}=mg-bv.
|
||||
\]
|
||||
Substitute $m=0.40\,\mathrm{kg}$ and $b=0.80\,\mathrm{N\cdot s/m}$:
|
||||
\[
|
||||
(0.40)\frac{dv}{dt}=(0.40)(9.8)-0.80v.
|
||||
\]
|
||||
So an equivalent form is
|
||||
\[
|
||||
\frac{dv}{dt}=9.8-2.0v.
|
||||
\]
|
||||
|
||||
For part (b), terminal speed occurs when the acceleration is zero, so
|
||||
\[
|
||||
\frac{dv}{dt}=0.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
mg-bv_T=0,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
v_T=\frac{mg}{b}=\frac{(0.40)(9.8)}{0.80}=4.9\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
For part (c), because the package is dropped from rest, the initial condition is
|
||||
\[
|
||||
v(0)=0.
|
||||
\]
|
||||
Using the linear-drag result for release from rest,
|
||||
\[
|
||||
v(t)=v_T\left(1-e^{-bt/m}\right).
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
v(t)=4.9\left(1-e^{-(0.80/0.40)t}\right).
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
For part (d), evaluate this at $t=1.0\,\mathrm{s}$:
|
||||
\[
|
||||
v(1.0)=4.9\left(1-e^{-2.0}\right)\,\mathrm{m/s}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
e^{-2.0}\approx 0.135,
|
||||
\]
|
||||
we get
|
||||
\[
|
||||
v(1.0)\approx 4.9(0.865)=4.24\,\mathrm{m/s}.
|
||||
\]
|
||||
So after $1.0\,\mathrm{s}$ the package is moving downward at about
|
||||
\[
|
||||
4.24\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now find the acceleration. From the differential equation,
|
||||
\[
|
||||
a=\frac{dv}{dt}=9.8-2.0v.
|
||||
\]
|
||||
At $t=1.0\,\mathrm{s}$,
|
||||
\[
|
||||
a=9.8-2.0(4.24)=1.32\,\mathrm{m/s^2}.
|
||||
\]
|
||||
This is positive in the downward-positive coordinate system, so the acceleration is still downward.
|
||||
|
||||
Therefore the differential equation is
|
||||
\[
|
||||
\frac{dv}{dt}=9.8-2.0v,
|
||||
\]
|
||||
the terminal speed is
|
||||
\[
|
||||
4.9\,\mathrm{m/s},
|
||||
\]
|
||||
the velocity function is
|
||||
\[
|
||||
v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s},
|
||||
\]
|
||||
and at $t=1.0\,\mathrm{s}$ the package has velocity
|
||||
\[
|
||||
4.24\,\mathrm{m/s}
|
||||
\]
|
||||
downward and acceleration
|
||||
\[
|
||||
1.32\,\mathrm{m/s^2}
|
||||
\]
|
||||
downward.
|
||||
140
concepts/mechanics/u2/m2-8-circular-orbital.tex
Normal file
140
concepts/mechanics/u2/m2-8-circular-orbital.tex
Normal file
@@ -0,0 +1,140 @@
|
||||
\subsection{Circular Motion and Orbital Dynamics}
|
||||
|
||||
This subsection resolves motion on a circle into radial and tangential parts. In AP mechanics, the key idea is that even when the speed stays constant, the velocity direction changes, so there is still an inward acceleration. In a circular orbit, gravity supplies that inward acceleration.
|
||||
|
||||
\dfn{Radial and tangential directions for circular motion}{Let a particle move on a circle of radius $r$ centered at a fixed point. Let $\vec{r}$ denote the particle's position vector from the center, let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector, let $\hat{t}$ denote the unit vector tangent to the path in the direction of motion, let $v$ denote the speed, and let $\vec{a}$ denote the acceleration.
|
||||
|
||||
The acceleration can be decomposed into radial and tangential parts:
|
||||
\[
|
||||
\vec{a}=a_r\hat{r}+a_t\hat{t}.
|
||||
\]
|
||||
The tangential component $a_t$ changes the speed, while the radial component changes the direction of the velocity. For circular motion, the radial acceleration points toward the center, so its direction is $-\hat{r}$.}
|
||||
|
||||
\thm{Centripetal acceleration and circular-orbit speed}{Let a particle of mass $m$ move on a circle of radius $r$ with speed $v$. Let $\hat{r}$ denote the outward radial unit vector and let $\hat{t}$ denote the tangential unit vector in the direction of motion. Then the acceleration is
|
||||
\[
|
||||
\vec{a}=-\frac{v^2}{r}\hat{r}+\frac{dv}{dt}\hat{t}.
|
||||
\]
|
||||
In particular, for uniform circular motion, $dv/dt=0$, so
|
||||
\[
|
||||
\vec{a}=-\frac{v^2}{r}\hat{r},
|
||||
\qquad
|
||||
a_r=\frac{v^2}{r}\text{ inward}.
|
||||
\]
|
||||
|
||||
Now let the same particle be in a circular orbit around a spherically symmetric body of mass $M$, with orbital radius $r$. If gravity is the only significant radial force, then
|
||||
\[
|
||||
\frac{GMm}{r^2}=\frac{mv^2}{r},
|
||||
\]
|
||||
so the orbital speed is
|
||||
\[
|
||||
v=\sqrt{\frac{GM}{r}}.
|
||||
\]}
|
||||
|
||||
\pf{Short derivation from radial force balance}{Let $F_r$ denote the net inward radial force on the particle. For circular motion of radius $r$ and speed $v$, the required inward acceleration has magnitude $v^2/r$, so Newton's second law in the radial direction gives
|
||||
\[
|
||||
F_r=m\frac{v^2}{r}.
|
||||
\]
|
||||
Because inward is the $-\hat{r}$ direction,
|
||||
\[
|
||||
\vec{a}_r=-\frac{v^2}{r}\hat{r}.
|
||||
\]
|
||||
If the speed changes, the tangential component is
|
||||
\[
|
||||
\vec{a}_t=\frac{dv}{dt}\hat{t}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{a}=\vec{a}_r+\vec{a}_t=-\frac{v^2}{r}\hat{r}+\frac{dv}{dt}\hat{t}.
|
||||
\]
|
||||
For a circular orbit, gravity provides the entire inward force, so
|
||||
\[
|
||||
F_r=\frac{GMm}{r^2}=m\frac{v^2}{r}.
|
||||
\]
|
||||
Canceling $m$ and solving for $v$ gives
|
||||
\[
|
||||
v=\sqrt{\frac{GM}{r}}.
|
||||
\]}
|
||||
|
||||
\cor{Circular-orbit period}{Let $T$ denote the period of a circular orbit of radius $r$ around a spherically symmetric body of mass $M$. Since one orbit has circumference $2\pi r$,
|
||||
\[
|
||||
v=\frac{2\pi r}{T}.
|
||||
\]
|
||||
Combine this with
|
||||
\[
|
||||
v=\sqrt{\frac{GM}{r}}
|
||||
\]
|
||||
to obtain
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{r^3}{GM}}.
|
||||
\]
|
||||
Near Earth's surface, if $r\approx R_E$ and $g=GM/R_E^2$, this becomes
|
||||
\[
|
||||
T\approx 2\pi\sqrt{\frac{R_E}{g}}.
|
||||
\]}
|
||||
|
||||
\qs{Worked example}{A satellite moves in a circular orbit at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let Earth have mass $M_E=5.97\times 10^{24}\,\mathrm{kg}$, radius $R_E=6.37\times 10^6\,\mathrm{m}$, and let the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}$.
|
||||
|
||||
Find the satellite's orbital radius $r$, orbital speed $v$, orbital period $T$, and centripetal acceleration magnitude $a_r$.}
|
||||
|
||||
\sol First compute the orbital radius from Earth's center:
|
||||
\[
|
||||
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For a circular orbit, gravity supplies the centripetal force, so
|
||||
\[
|
||||
\frac{GM_E m}{r^2}=\frac{mv^2}{r}.
|
||||
\]
|
||||
Cancel $m$ and solve for $v$:
|
||||
\[
|
||||
v=\sqrt{\frac{GM_E}{r}}.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
v=\sqrt{\frac{\left(6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}\right)\left(5.97\times 10^{24}\,\mathrm{kg}\right)}{6.77\times 10^6\,\mathrm{m}}}.
|
||||
\]
|
||||
This gives
|
||||
\[
|
||||
v\approx 7.67\times 10^3\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now find the period from
|
||||
\[
|
||||
T=\frac{2\pi r}{v}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
T=\frac{2\pi\left(6.77\times 10^6\,\mathrm{m}\right)}{7.67\times 10^3\,\mathrm{m/s}}
|
||||
\approx 5.55\times 10^3\,\mathrm{s}.
|
||||
\]
|
||||
In minutes,
|
||||
\[
|
||||
T\approx \frac{5.55\times 10^3\,\mathrm{s}}{60}\approx 92.4\,\mathrm{min}.
|
||||
\]
|
||||
|
||||
Finally, the centripetal acceleration magnitude is
|
||||
\[
|
||||
a_r=\frac{v^2}{r}.
|
||||
\]
|
||||
Using the speed just found,
|
||||
\[
|
||||
a_r=\frac{\left(7.67\times 10^3\,\mathrm{m/s}\right)^2}{6.77\times 10^6\,\mathrm{m}}
|
||||
\approx 8.69\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Therefore the satellite's orbital radius is
|
||||
\[
|
||||
r=6.77\times 10^6\,\mathrm{m},
|
||||
\]
|
||||
its orbital speed is
|
||||
\[
|
||||
v\approx 7.67\times 10^3\,\mathrm{m/s},
|
||||
\]
|
||||
its orbital period is
|
||||
\[
|
||||
T\approx 5.55\times 10^3\,\mathrm{s}\approx 92.4\,\mathrm{min},
|
||||
\]
|
||||
and its centripetal acceleration magnitude is
|
||||
\[
|
||||
a_r\approx 8.69\,\mathrm{m/s^2}.
|
||||
\]
|
||||
0
concepts/mechanics/u3/.gitkeep
Normal file
0
concepts/mechanics/u3/.gitkeep
Normal file
99
concepts/mechanics/u3/m3-1-work.tex
Normal file
99
concepts/mechanics/u3/m3-1-work.tex
Normal file
@@ -0,0 +1,99 @@
|
||||
\subsection{Work as a Line Integral}
|
||||
|
||||
This subsection develops work from the local relation between force and an infinitesimal displacement. In AP mechanics, this gives a clean calculus-based way to handle variable forces and to track when a force helps, opposes, or does no work on the motion.
|
||||
|
||||
\dfn{Infinitesimal work and total work along a path}{Let a particle move along a path $C$. Let $d\vec{r}$ denote an infinitesimal displacement vector of the particle, let $ds=|d\vec{r}|$ denote the corresponding infinitesimal path length, and let $\vec{F}$ denote the force acting on the particle at that location.
|
||||
|
||||
The \emph{infinitesimal work} done by the force during that displacement is
|
||||
\[
|
||||
dW=\vec{F}\cdot d\vec{r}.
|
||||
\]
|
||||
|
||||
If $\hat{t}$ denotes the unit tangent to the path and $F_{\parallel}=\vec{F}\cdot \hat{t}$ denotes the component of the force parallel to the motion, then
|
||||
\[
|
||||
dW=F_{\parallel}\,ds.
|
||||
\]
|
||||
|
||||
Therefore the \emph{total work} done by the force as the particle moves along the path $C$ is
|
||||
\[
|
||||
W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds.
|
||||
\]}
|
||||
|
||||
\thm{Line-integral form of work and the constant-force special case}{Let a particle move from an initial point to a final point along a path $C$ under a force $\vec{F}$. Then the work done by that force is
|
||||
\[
|
||||
W=\int_C \vec{F}\cdot d\vec{r}.
|
||||
\]
|
||||
Only the component of $\vec{F}$ parallel to the instantaneous displacement contributes, so equivalently
|
||||
\[
|
||||
W=\int_C F_{\parallel}\,ds.
|
||||
\]
|
||||
|
||||
If the force is constant in magnitude and direction, then
|
||||
\[
|
||||
W=\vec{F}\cdot \Delta \vec{r},
|
||||
\]
|
||||
where $\Delta \vec{r}$ is the total displacement from the initial point to the final point. In particular, if the motion is along a straight line parallel to the constant force, then
|
||||
\[
|
||||
W=F\,\Delta s.
|
||||
\]}
|
||||
|
||||
\nt{Work is positive when the force has a component in the same direction as the displacement, negative when that component is opposite the displacement, and zero when the force is perpendicular to the displacement. For a general force, the value of $W=\int_C \vec{F}\cdot d\vec{r}$ can depend on the path $C$, not just on the endpoints. In AP problems, this often appears when the force changes with position or when different paths make the parallel component $F_{\parallel}$ different. Typical zero-work cases include a normal force on motion along a surface or a centripetal force in uniform circular motion, because those forces are perpendicular to the instantaneous displacement.}
|
||||
|
||||
\pf{Why the line integral gives total work}{Break the path into many small displacement vectors $\Delta \vec{r}_1,\Delta \vec{r}_2,\dots,\Delta \vec{r}_n$. Over each small segment, the work is approximately
|
||||
\[
|
||||
\Delta W_k\approx \vec{F}_k\cdot \Delta \vec{r}_k.
|
||||
\]
|
||||
Summing over the path gives
|
||||
\[
|
||||
W\approx \sum_{k=1}^n \vec{F}_k\cdot \Delta \vec{r}_k.
|
||||
\]
|
||||
In the limit as the segments become infinitesimal, this Riemann sum becomes
|
||||
\[
|
||||
W=\int_C \vec{F}\cdot d\vec{r}.
|
||||
\]
|
||||
If $\vec{F}$ is constant, then it can be taken outside the integral:
|
||||
\[
|
||||
W=\vec{F}\cdot \int_C d\vec{r}=\vec{F}\cdot \Delta \vec{r}.
|
||||
\]}
|
||||
|
||||
\qs{Worked example}{A cart moves along a straight horizontal track from $x_i=0$ to $x_f=5.0\,\mathrm{m}$. Let $x$ denote the cart's position coordinate, and let the applied force on the cart be
|
||||
\[
|
||||
\vec{F}(x)=(6.0-2.0x)\hat{\imath}\,\mathrm{N}.
|
||||
\]
|
||||
Find the work done by this force on the cart over the interval from $x=0$ to $x=5.0\,\mathrm{m}$. State where the force does negative work.}
|
||||
|
||||
\sol Because the motion is along the $x$-axis, the displacement element is
|
||||
\[
|
||||
d\vec{r}=dx\,\hat{\imath}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
dW=\vec{F}\cdot d\vec{r}=[(6.0-2.0x)\hat{\imath}]\cdot (dx\,\hat{\imath})=(6.0-2.0x)\,dx.
|
||||
\]
|
||||
So the total work is
|
||||
\[
|
||||
W=\int_0^{5.0} (6.0-2.0x)\,dx.
|
||||
\]
|
||||
Evaluate the integral:
|
||||
\[
|
||||
W=\left[6.0x-x^2\right]_0^{5.0}.
|
||||
\]
|
||||
Substitute the limits:
|
||||
\[
|
||||
W=(6.0)(5.0)-(5.0)^2-\left[(6.0)(0)-0^2\right]=30.0-25.0=5.0\,\mathrm{J}.
|
||||
\]
|
||||
Thus the force does
|
||||
\[
|
||||
5.0\,\mathrm{J}
|
||||
\]
|
||||
of net work on the cart.
|
||||
|
||||
To identify where the force does negative work, find where the force component along the motion becomes negative:
|
||||
\[
|
||||
6.0-2.0x<0.
|
||||
\]
|
||||
This occurs when
|
||||
\[
|
||||
x>3.0\,\mathrm{m}.
|
||||
\]
|
||||
So from $x=3.0\,\mathrm{m}$ to $x=5.0\,\mathrm{m}$, the force points opposite the displacement and does negative work. From $x=0$ to $x=3.0\,\mathrm{m}$, it does positive work.
|
||||
130
concepts/mechanics/u3/m3-2-work-energy.tex
Normal file
130
concepts/mechanics/u3/m3-2-work-energy.tex
Normal file
@@ -0,0 +1,130 @@
|
||||
\subsection{Kinetic Energy and the Work-Energy Theorem}
|
||||
|
||||
This subsection introduces kinetic energy as the energy of motion and the work-energy theorem as the main AP bridge from force and displacement to speed without solving for time explicitly.
|
||||
|
||||
\dfn{Kinetic energy and net work}{Let $m$ denote the mass of a particle or body, let $\vec{v}$ denote its velocity, and let $v=|\vec{v}|$ denote its speed. The \emph{kinetic energy} of the body is
|
||||
\[
|
||||
K=\tfrac12 mv^2.
|
||||
\]
|
||||
|
||||
If forces $\vec{F}_1$, $\vec{F}_2$, $\dots$, and $\vec{F}_n$ act on the body while it undergoes an infinitesimal displacement $d\vec{r}$, then the differential work done by force $i$ is
|
||||
\[
|
||||
dW_i=\vec{F}_i\cdot d\vec{r}.
|
||||
\]
|
||||
Let the net force be
|
||||
\[
|
||||
\vec{F}_{\text{net}}=\sum_{i=1}^n \vec{F}_i.
|
||||
\]
|
||||
Then the \emph{net work} is the sum of the works done by all forces:
|
||||
\[
|
||||
dW_{\text{net}}=\sum_{i=1}^n dW_i=\vec{F}_{\text{net}}\cdot d\vec{r}.
|
||||
\]
|
||||
Over a finite motion,
|
||||
\[
|
||||
W_{\text{net}}=\int \vec{F}_{\text{net}}\cdot d\vec{r}=\sum_{i=1}^n W_i.
|
||||
\]
|
||||
The SI unit of both work and kinetic energy is the joule, where $1\,\mathrm{J}=1\,\mathrm{N\,m}$.}
|
||||
|
||||
\thm{Work-energy theorem}{Let $m$ denote the mass of a body, let $\vec{v}$ denote its velocity, let $v=|\vec{v}|$ denote its speed, and let $d\vec{r}$ denote an infinitesimal displacement of the body. Then
|
||||
\[
|
||||
dK=\vec{F}_{\text{net}}\cdot d\vec{r}.
|
||||
\]
|
||||
Integrating from an initial state to a final state gives the work-energy theorem:
|
||||
\[
|
||||
W_{\text{net}}=\Delta K=K_f-K_i=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
|
||||
\]
|
||||
Thus the net work done on a body equals the change in its kinetic energy.}
|
||||
|
||||
\nt{The work-energy theorem is often more efficient than combining Newton's second law with kinematics when the problem asks for a speed after a known displacement or after a known amount of work. It avoids solving for time and often avoids solving for acceleration explicitly. In AP mechanics, \emph{net work} means the algebraic sum of the work done by all forces on the chosen system. A force parallel to the displacement does positive work, a force opposite the displacement does negative work, and a force perpendicular to the displacement does zero work.}
|
||||
|
||||
\pf{Short derivation from Newton II}{Let $m$ denote the constant mass of the body, let $\vec{v}$ denote its velocity, and let $d\vec{r}$ denote its infinitesimal displacement. Start with Newton's second law,
|
||||
\[
|
||||
\vec{F}_{\text{net}}=m\frac{d\vec{v}}{dt}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
d\vec{r}=\vec{v}\,dt,
|
||||
\]
|
||||
dot both sides of Newton's second law with $d\vec{r}$:
|
||||
\[
|
||||
\vec{F}_{\text{net}}\cdot d\vec{r}=m\frac{d\vec{v}}{dt}\cdot (\vec{v}\,dt)=m\vec{v}\cdot d\vec{v}.
|
||||
\]
|
||||
Now use
|
||||
\[
|
||||
v^2=\vec{v}\cdot \vec{v},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
d(v^2)=2\vec{v}\cdot d\vec{v}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{F}_{\text{net}}\cdot d\vec{r}=m\vec{v}\cdot d\vec{v}=d\!\left(\tfrac12 mv^2\right)=dK.
|
||||
\]
|
||||
Integrating gives
|
||||
\[
|
||||
W_{\text{net}}=\int \vec{F}_{\text{net}}\cdot d\vec{r}=\int dK=K_f-K_i=\Delta K.
|
||||
\]}
|
||||
|
||||
\qs{Worked example}{Choose the positive $x$-axis to the right. A crate of mass $m=4.0\,\mathrm{kg}$ moves to the right on a horizontal floor with initial speed $v_i=3.0\,\mathrm{m/s}$. A constant applied force of magnitude $F_A=20.0\,\mathrm{N}$ acts to the right while the crate moves a horizontal distance $\Delta x=6.0\,\mathrm{m}$. Kinetic friction of magnitude $f_k=4.0\,\mathrm{N}$ acts to the left. The normal force and the weight act vertically.
|
||||
|
||||
Find the crate's final speed $v_f$.}
|
||||
|
||||
\sol Use the work-energy theorem:
|
||||
\[
|
||||
W_{\text{net}}=\Delta K=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
|
||||
\]
|
||||
|
||||
Compute the work done by each force.
|
||||
|
||||
The applied force is parallel to the displacement, so its work is positive:
|
||||
\[
|
||||
W_A=F_A\Delta x=(20.0\,\mathrm{N})(6.0\,\mathrm{m})=120\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
The friction force is opposite the displacement, so its work is negative:
|
||||
\[
|
||||
W_f=-f_k\Delta x=-(4.0\,\mathrm{N})(6.0\,\mathrm{m})=-24\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
The normal force and the weight are perpendicular to the horizontal displacement, so each does zero work:
|
||||
\[
|
||||
W_N=0,
|
||||
\qquad
|
||||
W_g=0.
|
||||
\]
|
||||
|
||||
Therefore the net work is
|
||||
\[
|
||||
W_{\text{net}}=W_A+W_f+W_N+W_g=120\,\mathrm{J}-24\,\mathrm{J}=96\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Now find the initial kinetic energy:
|
||||
\[
|
||||
K_i=\tfrac12 mv_i^2=\tfrac12 (4.0\,\mathrm{kg})(3.0\,\mathrm{m/s})^2=18\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
So the final kinetic energy is
|
||||
\[
|
||||
K_f=K_i+W_{\text{net}}=18\,\mathrm{J}+96\,\mathrm{J}=114\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Use $K_f=\tfrac12 mv_f^2$:
|
||||
\[
|
||||
\tfrac12 (4.0\,\mathrm{kg})v_f^2=114\,\mathrm{J}.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
2.0\,v_f^2=114,
|
||||
\qquad
|
||||
v_f^2=57,
|
||||
\qquad
|
||||
v_f=\sqrt{57}\,\mathrm{m/s}\approx 7.5\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Therefore the crate's final speed is
|
||||
\[
|
||||
v_f\approx 7.5\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
This method is shorter than solving for the acceleration and then using a kinematics equation, because the theorem connects net work directly to the change in speed.
|
||||
148
concepts/mechanics/u3/m3-3-potential-energy.tex
Normal file
148
concepts/mechanics/u3/m3-3-potential-energy.tex
Normal file
@@ -0,0 +1,148 @@
|
||||
\subsection{Conservative Forces and Potential Energy}
|
||||
|
||||
This subsection introduces conservative forces through path-independent work and uses that idea to define potential energy differences for an interacting system.
|
||||
|
||||
\dfn{Conservative force and potential energy difference}{Let $\vec{F}_c$ denote a force associated with some interaction, let $i$ and $f$ denote initial and final positions, and let $C$ denote a path from $i$ to $f$. The force $\vec{F}_c$ is called \emph{conservative} if the work
|
||||
\[
|
||||
W_c(i\to f)=\int_C \vec{F}_c\cdot d\vec{r}
|
||||
\]
|
||||
depends only on the endpoints $i$ and $f$, not on the path $C$.
|
||||
|
||||
For a conservative force, the corresponding \emph{potential energy difference} of the interacting system is defined by
|
||||
\[
|
||||
\Delta U=U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}.
|
||||
\]
|
||||
Thus the work done by the conservative force is
|
||||
\[
|
||||
W_c=-\Delta U.
|
||||
\]}
|
||||
|
||||
\thm{Equivalent conservative-force relations}{Let $\vec{F}_c$ denote a conservative force and let $d\vec{r}$ denote an infinitesimal displacement. Then the following relations hold:
|
||||
\[
|
||||
\oint \vec{F}_c\cdot d\vec{r}=0,
|
||||
\]
|
||||
so the work done by $\vec{F}_c$ around any closed loop is zero.
|
||||
|
||||
Equivalently, for any two paths $C_1$ and $C_2$ connecting the same endpoints,
|
||||
\[
|
||||
\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
|
||||
\]
|
||||
|
||||
The local potential-energy relation is
|
||||
\[
|
||||
dU=-\vec{F}_c\cdot d\vec{r}.
|
||||
\]
|
||||
For one-dimensional motion along the $x$-axis,
|
||||
\[
|
||||
F_x=-\frac{dU}{dx}.
|
||||
\]
|
||||
More generally, one may write lightly
|
||||
\[
|
||||
\vec{F}_c=-\nabla U.
|
||||
\]}
|
||||
|
||||
\nt{Potential energy is a property of a \emph{system}, not of a single isolated object. For example, gravitational potential energy belongs to the Earth-object system, and spring potential energy belongs to the block-spring system. A conservative force can transfer energy between kinetic and potential forms without making the potential difference depend on the path. By contrast, nonconservative forces such as kinetic friction and air resistance have path-dependent work, so a single-valued potential-energy function for that interaction is not defined in this AP sense.}
|
||||
|
||||
\pf{Why zero closed-loop work gives a well-defined $\Delta U$}{Assume that for every closed path,
|
||||
\[
|
||||
\oint \vec{F}_c\cdot d\vec{r}=0.
|
||||
\]
|
||||
Take two paths $C_1$ and $C_2$ from the same initial point $i$ to the same final point $f$. Traverse $C_1$ from $i$ to $f$ and then traverse $C_2$ backward from $f$ to $i$. This makes a closed loop, so
|
||||
\[
|
||||
\int_{C_1} \vec{F}_c\cdot d\vec{r}+\int_{f\to i\text{ on }C_2} \vec{F}_c\cdot d\vec{r}=0.
|
||||
\]
|
||||
Reversing the limits changes the sign of the second integral, giving
|
||||
\[
|
||||
\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
|
||||
\]
|
||||
So the work depends only on the endpoints. Therefore the quantity
|
||||
\[
|
||||
U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}
|
||||
\]
|
||||
is path independent and is a well-defined potential-energy difference.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=0.50\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $k=200\,\mathrm{N/m}$ on a frictionless track. Let $x$ denote the block's displacement from the spring's equilibrium position, with positive $x$ to the right. Initially the block is held at rest at $x_i=+0.15\,\mathrm{m}$ and then released. Find:
|
||||
|
||||
\begin{enumerate}
|
||||
\item the change in spring potential energy $\Delta U_s$ as the block moves to $x_f=0$,
|
||||
\item the work done by the spring during that motion, and
|
||||
\item the block's speed $v_f$ when it passes through equilibrium.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol For an ideal spring, the spring force is
|
||||
\[
|
||||
\vec{F}_s=-kx\,\hat{\imath}.
|
||||
\]
|
||||
Since the motion is one-dimensional, the potential-energy relation gives
|
||||
\[
|
||||
F_x=-\frac{dU_s}{dx}.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
-kx=-\frac{dU_s}{dx}
|
||||
\qquad \Rightarrow \qquad
|
||||
\frac{dU_s}{dx}=kx.
|
||||
\]
|
||||
Integrate with respect to $x$:
|
||||
\[
|
||||
U_s(x)=\int kx\,dx=\tfrac12 kx^2+C.
|
||||
\]
|
||||
Choose the usual reference $U_s=0$ at $x=0$, so $C=0$ and
|
||||
\[
|
||||
U_s(x)=\tfrac12 kx^2.
|
||||
\]
|
||||
|
||||
At the initial position,
|
||||
\[
|
||||
U_{s,i}=\tfrac12 (200)(0.15)^2=100(0.0225)=2.25\,\mathrm{J}.
|
||||
\]
|
||||
At the final position $x_f=0$,
|
||||
\[
|
||||
U_{s,f}=\tfrac12 (200)(0)^2=0.
|
||||
\]
|
||||
Therefore the change in spring potential energy is
|
||||
\[
|
||||
\Delta U_s=U_{s,f}-U_{s,i}=0-2.25\,\mathrm{J}=-2.25\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Because the spring force is conservative,
|
||||
\[
|
||||
W_s=-\Delta U_s=+2.25\,\mathrm{J}.
|
||||
\]
|
||||
So the spring does positive work on the block as the spring relaxes toward equilibrium.
|
||||
|
||||
The track is frictionless, so the spring is the only force doing work on the block in the horizontal direction. Thus
|
||||
\[
|
||||
W_{\text{net}}=\Delta K.
|
||||
\]
|
||||
Since the block starts from rest,
|
||||
\[
|
||||
K_i=0,
|
||||
\qquad
|
||||
K_f=W_{\text{net}}=2.25\,\mathrm{J}.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
\tfrac12 mv_f^2=2.25.
|
||||
\]
|
||||
Substitute $m=0.50\,\mathrm{kg}$:
|
||||
\[
|
||||
\tfrac12 (0.50)v_f^2=2.25
|
||||
\qquad \Rightarrow \qquad
|
||||
0.25v_f^2=2.25
|
||||
\qquad \Rightarrow \qquad
|
||||
v_f^2=9.0.
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
v_f=3.0\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\Delta U_s=-2.25\,\mathrm{J},
|
||||
\qquad
|
||||
W_s=+2.25\,\mathrm{J},
|
||||
\qquad
|
||||
v_f=3.0\,\mathrm{m/s}.
|
||||
\]
|
||||
129
concepts/mechanics/u3/m3-4-energy-conservation.tex
Normal file
129
concepts/mechanics/u3/m3-4-energy-conservation.tex
Normal file
@@ -0,0 +1,129 @@
|
||||
\subsection{Mechanical Energy Conservation}
|
||||
|
||||
This subsection packages work and potential energy into an energy-accounting method. In AP mechanics, the key step is to choose a system first, then decide which interactions are represented by potential energy and which must be tracked as nonconservative work.
|
||||
|
||||
\dfn{Mechanical energy and nonconservative work}{Let a chosen system move between an initial state and a final state. Let $K$ denote the total kinetic energy of the system, and let $U$ denote the total potential energy associated with all conservative interactions included in the system, such as gravitational and spring interactions. The \emph{mechanical energy} of the system is
|
||||
\[
|
||||
E_{\mathrm{mech}}=K+U.
|
||||
\]
|
||||
|
||||
Let $W_{\mathrm{nc}}$ denote the total work done on the system by forces or processes that are not represented by a potential-energy function in the chosen model. With this sign convention,
|
||||
\[
|
||||
W_{\mathrm{nc}}>0 \text{ increases } E_{\mathrm{mech}},
|
||||
\qquad
|
||||
W_{\mathrm{nc}}<0 \text{ decreases } E_{\mathrm{mech}}.
|
||||
\]
|
||||
Typical examples include kinetic friction, air drag, and an external applied force not absorbed into $U$.}
|
||||
|
||||
\thm{Mechanical energy equation}{Let $K_i$ and $U_i$ denote the initial kinetic and potential energies of a chosen system, and let $K_f$ and $U_f$ denote the corresponding final quantities. If $W_{\mathrm{nc}}$ is the total nonconservative work done on the system, then
|
||||
\[
|
||||
\Delta E_{\mathrm{mech}}=\Delta(K+U)=W_{\mathrm{nc}}.
|
||||
\]
|
||||
Equivalently,
|
||||
\[
|
||||
K_i+U_i+W_{\mathrm{nc}}=K_f+U_f.
|
||||
\]
|
||||
If the motion is governed only by conservative forces already accounted for in $U$, then $W_{\mathrm{nc}}=0$ and mechanical energy is conserved:
|
||||
\[
|
||||
\Delta(K+U)=0,
|
||||
\qquad
|
||||
K_i+U_i=K_f+U_f.
|
||||
\]}
|
||||
|
||||
\nt{Choose the system before writing any energy equation. If the system is \emph{object + Earth}, then gravitational potential energy belongs in $U$ and gravity should not also be counted as separate work. If the system is \emph{object + spring}, then spring potential energy belongs in $U$. If the system is \emph{object + Earth + spring}, then both $U_g$ and $U_s$ belong in $U$. Mechanical energy is conserved only when no nonconservative work changes $K+U$ for that chosen system. When friction, drag, or an external agent transfers energy into or out of the system, use
|
||||
\[
|
||||
\Delta(K+U)=W_{\mathrm{nc}}
|
||||
\]
|
||||
instead of setting $\Delta(K+U)$ equal to zero. Total energy is still conserved overall; it is specifically \emph{mechanical} energy that may change.}
|
||||
|
||||
\pf{Derivation from the work-energy theorem}{Let $W_{\mathrm{net}}$ denote the net work done on the chosen system. By the work-energy theorem,
|
||||
\[
|
||||
\Delta K=W_{\mathrm{net}}.
|
||||
\]
|
||||
Split the net work into conservative and nonconservative parts:
|
||||
\[
|
||||
W_{\mathrm{net}}=W_c+W_{\mathrm{nc}}.
|
||||
\]
|
||||
For the conservative forces represented by the potential energy $U$,
|
||||
\[
|
||||
W_c=-\Delta U.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\Delta K=-\Delta U+W_{\mathrm{nc}}.
|
||||
\]
|
||||
Rearranging gives
|
||||
\[
|
||||
\Delta K+\Delta U=W_{\mathrm{nc}},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\Delta(K+U)=W_{\mathrm{nc}}.
|
||||
\]
|
||||
If $W_{\mathrm{nc}}=0$, then
|
||||
\[
|
||||
\Delta(K+U)=0,
|
||||
\]
|
||||
which is the conservation of mechanical energy.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=2.0\,\mathrm{kg}$ is released from rest at a height $h=1.20\,\mathrm{m}$ above the bottom of a ramp. The ramp is frictionless. After reaching the bottom, the block crosses a rough horizontal surface of length $d=0.80\,\mathrm{m}$ with coefficient of kinetic friction $\mu_k=0.25$. The block then compresses a horizontal spring of spring constant $k=400\,\mathrm{N/m}$ on a frictionless section of track and momentarily comes to rest at maximum compression. Let $x$ denote the maximum spring compression.
|
||||
|
||||
Find $x$.}
|
||||
|
||||
\sol Choose the system to be \emph{block + Earth + spring}. Then gravity and the spring are accounted for through potential energy, and the only nonconservative work is the work done by kinetic friction on the rough horizontal section.
|
||||
|
||||
Take the initial state to be the release point and the final state to be the instant of maximum compression. Let the gravitational potential energy be zero at the bottom of the ramp, and let the spring potential energy be zero when the spring is uncompressed.
|
||||
|
||||
Use the mechanical energy equation
|
||||
\[
|
||||
K_i+U_i+W_{\mathrm{nc}}=K_f+U_f.
|
||||
\]
|
||||
|
||||
At the initial state, the block is released from rest, so
|
||||
\[
|
||||
K_i=0.
|
||||
\]
|
||||
Its gravitational potential energy is
|
||||
\[
|
||||
U_i=mgh=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})(1.20\,\mathrm{m})=23.52\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
At the final state, the block momentarily stops at maximum compression, so
|
||||
\[
|
||||
K_f=0.
|
||||
\]
|
||||
Its gravitational potential energy is zero because it is at the bottom level, and its spring potential energy is
|
||||
\[
|
||||
U_f=\tfrac12 kx^2.
|
||||
\]
|
||||
|
||||
Now compute the nonconservative work. On the rough horizontal section, the kinetic friction force has magnitude
|
||||
\[
|
||||
f_k=\mu_k mg=(0.25)(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=4.9\,\mathrm{N}.
|
||||
\]
|
||||
Because friction opposes the motion over the distance $d=0.80\,\mathrm{m}$, the work done by friction is
|
||||
\[
|
||||
W_{\mathrm{nc}}=-f_k d=-(4.9\,\mathrm{N})(0.80\,\mathrm{m})=-3.92\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Substitute into the energy equation:
|
||||
\[
|
||||
0+23.52\,\mathrm{J}-3.92\,\mathrm{J}=0+\tfrac12 (400\,\mathrm{N/m})x^2.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
19.60\,\mathrm{J}=200x^2.
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
x^2=0.0980,
|
||||
\qquad
|
||||
x=\sqrt{0.0980}\,\mathrm{m}\approx 0.313\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Therefore the maximum spring compression is
|
||||
\[
|
||||
x\approx 0.31\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
If the rough section were absent, then $W_{\mathrm{nc}}=0$ and mechanical energy would be conserved exactly. Here the negative friction work reduces the mechanical energy before the block reaches the spring.
|
||||
112
concepts/mechanics/u3/m3-5-power.tex
Normal file
112
concepts/mechanics/u3/m3-5-power.tex
Normal file
@@ -0,0 +1,112 @@
|
||||
\subsection{Power and Instantaneous Power}
|
||||
|
||||
This subsection introduces power as the rate at which work is done. In AP mechanics, the key idea is local: the instantaneous power delivered by a force depends on the component of that force along the motion, so the sign of the power tells whether the force is adding energy to the object or removing it.
|
||||
|
||||
\dfn{Average and instantaneous power}{Let $\Delta W$ denote the work done by a force during a time interval $\Delta t>0$. The \emph{average power} over that interval is
|
||||
\[
|
||||
P_{\text{avg}}=\frac{\Delta W}{\Delta t}.
|
||||
\]
|
||||
The \emph{instantaneous power} at time $t$ is the limit of the average power over shorter and shorter time intervals:
|
||||
\[
|
||||
P=\lim_{\Delta t\to 0}\frac{\Delta W}{\Delta t}=\frac{dW}{dt}.
|
||||
\]
|
||||
Power is a scalar quantity. Its SI unit is the watt:
|
||||
\[
|
||||
1\,\mathrm{W}=1\,\mathrm{J/s}.
|
||||
\]
|
||||
If $P>0$, the force is doing positive work and transferring energy to the object. If $P<0$, the force is doing negative work and removing mechanical energy from the object.}
|
||||
|
||||
\thm{Mechanical power from force and velocity}{Let $\vec{F}$ denote the force acting on a particle, let $\vec{v}=d\vec{r}/dt$ denote the particle's velocity, let $F=|\vec{F}|$ denote the magnitude of the force, let $v=|\vec{v}|$ denote the speed, let $\theta$ denote the angle between $\vec{F}$ and $\vec{v}$, and let $F_{\parallel}$ denote the component of the force parallel to the motion. Then the instantaneous power delivered by the force is
|
||||
\[
|
||||
P=\vec{F}\cdot \vec{v}=Fv\cos\theta=F_{\parallel}v.
|
||||
\]
|
||||
Therefore only the component of the force along the motion contributes to power. If $\theta<90^\circ$, then $P>0$; if $\theta>90^\circ$, then $P<0$; and if $\theta=90^\circ$, then $P=0$. In one-dimensional motion along the $x$-axis,
|
||||
\[
|
||||
P=F_xv_x.
|
||||
\]}
|
||||
|
||||
\pf{Short derivation from $P=dW/dt$}{Let $d\vec{r}$ denote the particle's infinitesimal displacement. From the local definition of work,
|
||||
\[
|
||||
dW=\vec{F}\cdot d\vec{r}.
|
||||
\]
|
||||
Divide by $dt$ to obtain the instantaneous rate at which work is done:
|
||||
\[
|
||||
P=\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{r}}{dt}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
\frac{d\vec{r}}{dt}=\vec{v},
|
||||
\]
|
||||
it follows that
|
||||
\[
|
||||
P=\vec{F}\cdot \vec{v}.
|
||||
\]
|
||||
If $\theta$ is the angle between $\vec{F}$ and $\vec{v}$, then the dot-product form gives
|
||||
\[
|
||||
P=Fv\cos\theta=F_{\parallel}v.
|
||||
\]}
|
||||
|
||||
\cor{Perpendicular forces do zero instantaneous power}{If a force is perpendicular to the velocity at a given instant, then
|
||||
\[
|
||||
\vec{F}\cdot \vec{v}=0,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
P=0.
|
||||
\]
|
||||
Thus a force can change the direction of motion without transferring energy through work at that instant. Common AP examples are a normal force on frictionless motion along a surface and the centripetal force in uniform circular motion.}
|
||||
|
||||
\qs{Worked example}{A student pulls a sled across level snow with a rope of tension magnitude $T=85\,\mathrm{N}$ at an angle $\theta=25^\circ$ above the horizontal. The sled moves horizontally with constant speed $v=1.8\,\mathrm{m/s}$ for a time interval $\Delta t=40\,\mathrm{s}$. Let $\vec{T}$ denote the tension force and let $\vec{v}$ denote the sled's velocity.
|
||||
|
||||
(a) Find the instantaneous power delivered by the tension.
|
||||
|
||||
(b) Find the work done by the tension during the $40\,\mathrm{s}$ interval.
|
||||
|
||||
(c) Find the average power delivered by the tension over that interval.}
|
||||
|
||||
\sol Because the sled's velocity is horizontal, only the horizontal component of the tension contributes to the power and the work. The component of the tension along the motion is
|
||||
\[
|
||||
T_{\parallel}=T\cos\theta=(85\,\mathrm{N})\cos 25^\circ\approx 77.0\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
For part (a), the instantaneous power is
|
||||
\[
|
||||
P=\vec{T}\cdot \vec{v}=Tv\cos\theta.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
P=(85\,\mathrm{N})(1.8\,\mathrm{m/s})\cos 25^\circ\approx 1.39\times 10^2\,\mathrm{W}.
|
||||
\]
|
||||
So the instantaneous power delivered by the tension is
|
||||
\[
|
||||
P\approx 1.4\times 10^2\,\mathrm{W}.
|
||||
\]
|
||||
The power is positive because the tension has a component in the same direction as the motion.
|
||||
|
||||
For part (b), the sled's horizontal displacement during the interval is
|
||||
\[
|
||||
\Delta x=v\Delta t=(1.8\,\mathrm{m/s})(40\,\mathrm{s})=72\,\mathrm{m}.
|
||||
\]
|
||||
The work done by the tension is
|
||||
\[
|
||||
W=T\Delta x\cos\theta.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
W=(85\,\mathrm{N})(72\,\mathrm{m})\cos 25^\circ\approx 5.55\times 10^3\,\mathrm{J}.
|
||||
\]
|
||||
So the tension does
|
||||
\[
|
||||
W\approx 5.6\times 10^3\,\mathrm{J}
|
||||
\]
|
||||
of positive work on the sled. The vertical component of the tension does no work because the displacement is horizontal.
|
||||
|
||||
For part (c), the average power is
|
||||
\[
|
||||
P_{\text{avg}}=\frac{\Delta W}{\Delta t}=\frac{5.55\times 10^3\,\mathrm{J}}{40\,\mathrm{s}}\approx 1.39\times 10^2\,\mathrm{W}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
P_{\text{avg}}\approx 1.4\times 10^2\,\mathrm{W}.
|
||||
\]
|
||||
This matches the instantaneous power because the force magnitude, the angle, and the speed are all constant throughout the motion.
|
||||
0
concepts/mechanics/u4/.gitkeep
Normal file
0
concepts/mechanics/u4/.gitkeep
Normal file
185
concepts/mechanics/u4/m4-1-linear-momentum.tex
Normal file
185
concepts/mechanics/u4/m4-1-linear-momentum.tex
Normal file
@@ -0,0 +1,185 @@
|
||||
\subsection{Linear Momentum}
|
||||
|
||||
This subsection introduces linear momentum as the vector state variable for translational motion. In AP mechanics, momentum is the quantity that naturally leads into impulse and conservation ideas.
|
||||
|
||||
\dfn{Particle and system momentum}{Let $m$ denote the mass of a particle, let $\vec{v}$ denote its velocity measured in a chosen inertial reference frame, and let $\vec{p}$ denote its linear momentum. The \emph{linear momentum} of the particle is
|
||||
\[
|
||||
\vec{p}=m\vec{v}.
|
||||
\]
|
||||
Because $m$ is a scalar and $\vec{v}$ is a vector, $\vec{p}$ is a vector in the same direction as $\vec{v}$. Its SI unit is
|
||||
\[
|
||||
1\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
For a system of $N$ particles labeled by $i=1,2,\dots,N$, let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its momentum. The total linear momentum of the system is
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
|
||||
\]}
|
||||
|
||||
\nt{Momentum is not the same thing as speed. Speed is the scalar $v=|\vec{v}|$, while momentum is the vector $\vec{p}=m\vec{v}$. Two objects can have the same speed but different momenta if their masses are different or if they move in different directions. Momentum also depends on the chosen reference frame because velocity does: an object at rest in one frame has $\vec{p}=\vec{0}$ in that frame, but it can have nonzero momentum in another frame. In one-dimensional motion, a negative momentum component means motion in the negative coordinate direction; it does not mean negative mass or negative speed.}
|
||||
|
||||
\mprop{Component and system relations}{Let
|
||||
\[
|
||||
\vec{v}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat{k}
|
||||
\]
|
||||
be the velocity of a particle of mass $m$, and let
|
||||
\[
|
||||
\vec{p}=p_x\hat{\imath}+p_y\hat{\jmath}+p_z\hat{k}
|
||||
\]
|
||||
be its momentum. For a system of particles, let
|
||||
\[
|
||||
M=\sum_{i=1}^N m_i
|
||||
\]
|
||||
denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item Momentum components are found component-by-component from the velocity:
|
||||
\[
|
||||
p_x=mv_x,
|
||||
\qquad
|
||||
p_y=mv_y,
|
||||
\qquad
|
||||
p_z=mv_z.
|
||||
\]
|
||||
In one-dimensional motion along the $x$-axis,
|
||||
\[
|
||||
p_x=mv_x.
|
||||
\]
|
||||
|
||||
\item For a single particle, the momentum magnitude is
|
||||
\[
|
||||
p=|\vec{p}|=m|\vec{v}|=mv.
|
||||
\]
|
||||
In two dimensions,
|
||||
\[
|
||||
p=\sqrt{p_x^2+p_y^2},
|
||||
\]
|
||||
and in three dimensions,
|
||||
\[
|
||||
p=\sqrt{p_x^2+p_y^2+p_z^2}.
|
||||
\]
|
||||
|
||||
\item System momentum is the vector sum of the particle momenta:
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i.
|
||||
\]
|
||||
Therefore its components also add:
|
||||
\[
|
||||
p_{\mathrm{sys},x}=\sum_{i=1}^N p_{i,x},
|
||||
\qquad
|
||||
p_{\mathrm{sys},y}=\sum_{i=1}^N p_{i,y},
|
||||
\qquad
|
||||
p_{\mathrm{sys},z}=\sum_{i=1}^N p_{i,z}.
|
||||
\]
|
||||
Equivalently,
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}}.
|
||||
\]
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{In a laboratory frame, two pucks slide on nearly frictionless ice. Puck 1 has mass $m_1=2.0\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_1=(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Puck 2 has mass $m_2=1.0\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Let $\vec{p}_1$ and $\vec{p}_2$ denote the individual momenta, let $\vec{p}_{\mathrm{sys}}$ denote the total momentum, let $M$ denote the total mass, and let $\theta$ denote the direction of $\vec{p}_{\mathrm{sys}}$ measured counterclockwise from the positive $x$-axis.
|
||||
|
||||
Find $\vec{p}_1$, $\vec{p}_2$, $\vec{p}_{\mathrm{sys}}$, the magnitude $|\vec{p}_{\mathrm{sys}}|$, the direction $\theta$, and the center-of-mass velocity $\vec{v}_{\mathrm{cm}}$.}
|
||||
|
||||
\sol Use $\vec{p}=m\vec{v}$ for each puck.
|
||||
|
||||
For puck 1,
|
||||
\[
|
||||
\vec{p}_1=m_1\vec{v}_1=(2.0\,\mathrm{kg})(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
For puck 2,
|
||||
\[
|
||||
\vec{p}_2=m_2\vec{v}_2=(1.0\,\mathrm{kg})(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Now add the momenta component-by-component:
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=\vec{p}_1+\vec{p}_2.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=(3.0-1.0)\hat{\imath}+(1.0+4.0)\hat{\jmath}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Its magnitude is
|
||||
\[
|
||||
|\vec{p}_{\mathrm{sys}}|=\sqrt{(2.0)^2+(5.0)^2}\,\mathrm{kg\cdot m/s}=\sqrt{29}\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
Numerically,
|
||||
\[
|
||||
|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
To find the direction, use the component ratio. Since both components of $\vec{p}_{\mathrm{sys}}$ are positive, the vector lies in the first quadrant. Thus,
|
||||
\[
|
||||
\tan\theta=\frac{p_{\mathrm{sys},y}}{p_{\mathrm{sys},x}}=\frac{5.0}{2.0}=2.5.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\theta=\tan^{-1}(2.5)\approx 68.2^\circ.
|
||||
\]
|
||||
|
||||
Now find the center-of-mass velocity. The total mass is
|
||||
\[
|
||||
M=m_1+m_2=2.0\,\mathrm{kg}+1.0\,\mathrm{kg}=3.0\,\mathrm{kg}.
|
||||
\]
|
||||
Using
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}},
|
||||
\]
|
||||
we get
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{\vec{p}_{\mathrm{sys}}}{M}=\frac{(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{3.0\,\mathrm{kg}}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\left(\frac{2.0}{3.0}\hat{\imath}+\frac{5.0}{3.0}\hat{\jmath}\right)\,\mathrm{m/s}
|
||||
\]
|
||||
or numerically,
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
So the individual and system momenta are
|
||||
\[
|
||||
\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
|
||||
\qquad
|
||||
\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
|
||||
\]
|
||||
\[
|
||||
\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
|
||||
\]
|
||||
with magnitude
|
||||
\[
|
||||
|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s},
|
||||
\]
|
||||
direction
|
||||
\[
|
||||
\theta\approx 68.2^\circ,
|
||||
\]
|
||||
and center-of-mass velocity
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
This example shows why momentum must be handled as a vector: the total momentum is found by adding components, not by adding speeds.
|
||||
122
concepts/mechanics/u4/m4-2-impulse.tex
Normal file
122
concepts/mechanics/u4/m4-2-impulse.tex
Normal file
@@ -0,0 +1,122 @@
|
||||
\subsection{Impulse and Momentum Transfer}
|
||||
|
||||
This subsection connects the local momentum law $\vec{F}_{\text{net}}=d\vec{p}/dt$ to finite-time interactions such as hits, kicks, and collisions, where a large force acts for a short time and changes momentum by a measurable amount.
|
||||
|
||||
\dfn{Momentum change and impulse}{Let a body of constant mass $m$ have velocity $\vec{v}$. Its linear momentum is
|
||||
\[
|
||||
\vec{p}=m\vec{v}.
|
||||
\]
|
||||
Let $\vec{p}_i$ and $\vec{p}_f$ denote the initial and final momenta over some time interval from $t_i$ to $t_f$. The change in momentum is
|
||||
\[
|
||||
\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
|
||||
\]
|
||||
Let $\vec{F}_{\text{net}}(t)$ denote the net external force on the body during that interval. The impulse delivered to the body is
|
||||
\[
|
||||
\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt.
|
||||
\]
|
||||
If the net force is constant, then this reduces to
|
||||
\[
|
||||
\vec{J}=\vec{F}_{\text{net}}\Delta t,
|
||||
\qquad
|
||||
\Delta t=t_f-t_i.
|
||||
\]
|
||||
The SI unit of impulse is $\mathrm{N\,s}$, which is equivalent to $\mathrm{kg\,m/s}$.}
|
||||
|
||||
\thm{Impulse-momentum theorem}{Let $\vec{p}(t)$ denote the momentum of a body and let $\vec{F}_{\text{net}}(t)$ denote the net external force on it. Over any time interval from $t_i$ to $t_f$,
|
||||
\[
|
||||
\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
|
||||
\]
|
||||
Thus the net impulse on a body equals its change in momentum. In one dimension along the $x$-axis, let $J_x$, $F_{\text{net},x}$, and $p_x$ denote the corresponding $x$-components. Then
|
||||
\[
|
||||
J_x=\int_{t_i}^{t_f} F_{\text{net},x}\,dt=\Delta p_x.
|
||||
\]}
|
||||
|
||||
\nt{Impulse is a vector, so its direction is the direction of $\Delta \vec{p}$. A body can be moving in one direction while the impulse points in the opposite direction if the interaction slows or reverses the motion. Impulse depends on both force and time: a large force acting briefly can produce the same impulse as a smaller force acting longer. If an average net force $\vec{F}_{\text{avg}}$ over a time interval $\Delta t$ is defined so that it has the same effect as the actual time-varying force, then
|
||||
\[
|
||||
\vec{J}=\vec{F}_{\text{avg}}\Delta t.
|
||||
\]
|
||||
On a force-versus-time graph, the signed area under the curve gives impulse. In component form, the signed area under an $F_x(t)$ graph gives $J_x=\Delta p_x$.}
|
||||
|
||||
\pf{Short derivation from $d\vec{p}/dt$}{Start with the local momentum law
|
||||
\[
|
||||
\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt}.
|
||||
\]
|
||||
Multiply by $dt$:
|
||||
\[
|
||||
\vec{F}_{\text{net}}\,dt=d\vec{p}.
|
||||
\]
|
||||
Now integrate from $t_i$ to $t_f$:
|
||||
\[
|
||||
\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\int_{t_i}^{t_f} d\vec{p}.
|
||||
\]
|
||||
The left side is the impulse $\vec{J}$, and the right side is the total change in momentum:
|
||||
\[
|
||||
\vec{J}=\vec{p}_f-\vec{p}_i=\Delta \vec{p}.
|
||||
\]
|
||||
This is the impulse-momentum theorem.}
|
||||
|
||||
\qs{Worked example}{Choose the positive $x$-axis to the right. A tennis ball of mass $m=0.150\,\mathrm{kg}$ moves horizontally with initial velocity
|
||||
\[
|
||||
\vec{v}_i=-18.0\,\hat{\imath}\,\mathrm{m/s}.
|
||||
\]
|
||||
Let $F_x(t)$ denote the net horizontal force on the ball during contact. The force is directed to the right and has the following force-versus-time graph: it increases linearly from $0$ at $t=0$ to $900\,\mathrm{N}$ at $t=4.0\,\mathrm{ms}$, then decreases linearly back to $0$ at $t=8.0\,\mathrm{ms}$.
|
||||
|
||||
Find (a) the impulse delivered to the ball, (b) the ball's final momentum, (c) the ball's final velocity, and (d) the average net force during contact.}
|
||||
|
||||
\sol Let $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$ denote the contact time. Because the force is entirely in the $+x$ direction, the impulse is the area under the triangular $F_x(t)$ graph in the positive direction:
|
||||
\[
|
||||
\vec{J}=\left(\tfrac12 \right)(\Delta t)(900\,\mathrm{N})\,\hat{\imath}.
|
||||
\]
|
||||
Substitute $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$:
|
||||
\[
|
||||
\vec{J}=\left(\tfrac12 \right)(8.0\times 10^{-3}\,\mathrm{s})(900\,\mathrm{N})\,\hat{\imath}=3.6\,\hat{\imath}\,\mathrm{N\,s}.
|
||||
\]
|
||||
Using $1\,\mathrm{N\,s}=1\,\mathrm{kg\,m/s}$,
|
||||
\[
|
||||
\vec{J}=3.6\,\hat{\imath}\,\mathrm{kg\,m/s}.
|
||||
\]
|
||||
|
||||
Now compute the initial momentum:
|
||||
\[
|
||||
\vec{p}_i=m\vec{v}_i=(0.150\,\mathrm{kg})(-18.0\,\hat{\imath}\,\mathrm{m/s})=-2.70\,\hat{\imath}\,\mathrm{kg\,m/s}.
|
||||
\]
|
||||
From the impulse-momentum theorem,
|
||||
\[
|
||||
\vec{J}=\Delta \vec{p}=\vec{p}_f-\vec{p}_i,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\vec{p}_f=\vec{p}_i+\vec{J}.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
\vec{p}_f=(-2.70+3.60)\,\hat{\imath}\,\mathrm{kg\,m/s}=0.90\,\hat{\imath}\,\mathrm{kg\,m/s}.
|
||||
\]
|
||||
|
||||
Then the final velocity is
|
||||
\[
|
||||
\vec{v}_f=\frac{\vec{p}_f}{m}=\frac{0.90\,\hat{\imath}\,\mathrm{kg\,m/s}}{0.150\,\mathrm{kg}}=6.0\,\hat{\imath}\,\mathrm{m/s}.
|
||||
\]
|
||||
The positive sign shows that the ball leaves moving to the right.
|
||||
|
||||
For the average net force, use
|
||||
\[
|
||||
\vec{J}=\vec{F}_{\text{avg}}\Delta t.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{F}_{\text{avg}}=\frac{\vec{J}}{\Delta t}=\frac{3.6\,\hat{\imath}\,\mathrm{N\,s}}{8.0\times 10^{-3}\,\mathrm{s}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\vec{J}=3.6\,\hat{\imath}\,\mathrm{N\,s},
|
||||
\qquad
|
||||
\vec{p}_f=0.90\,\hat{\imath}\,\mathrm{kg\,m/s},
|
||||
\]
|
||||
\[
|
||||
\vec{v}_f=6.0\,\hat{\imath}\,\mathrm{m/s},
|
||||
\qquad
|
||||
\vec{F}_{\text{avg}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
|
||||
\]
|
||||
The graph interpretation is essential here: the triangular area under $F_x(t)$ gives the impulse directly, and that impulse determines the momentum transfer.
|
||||
200
concepts/mechanics/u4/m4-3-momentum-conservation.tex
Normal file
200
concepts/mechanics/u4/m4-3-momentum-conservation.tex
Normal file
@@ -0,0 +1,200 @@
|
||||
\subsection{Conservation of Momentum for Systems}
|
||||
|
||||
This subsection treats momentum conservation as a statement about a chosen system: internal forces can transfer momentum between parts of the system, but if the net external impulse is zero, the total momentum stays constant.
|
||||
|
||||
\dfn{System momentum and momentum-isolated systems}{Consider a system of $N$ objects labeled by an index $i=1,2,\dots,N$. Let $m_i$ denote the mass of object $i$, let $\vec{v}_i$ denote its velocity, and let
|
||||
\[
|
||||
\vec{p}_i=m_i\vec{v}_i
|
||||
\]
|
||||
denote its momentum. The total momentum of the system is the vector sum
|
||||
\[
|
||||
\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
|
||||
\]
|
||||
|
||||
For an interval from time $t_i$ to time $t_f$, let $\sum \vec{F}_{\mathrm{ext}}$ denote the net external force on the whole system, and let
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt
|
||||
\]
|
||||
denote the net external impulse on the system.
|
||||
|
||||
A system is called \emph{closed} over that interval if the set of objects in the system does not change during the interaction. A closed system is \emph{isolated for momentum} over that interval if the net external impulse is zero or negligible:
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\vec{0}.
|
||||
\]}
|
||||
|
||||
\thm{Conservation of momentum for a system}{For a closed system,
|
||||
\[
|
||||
\Delta \vec{P}=\vec{P}_f-\vec{P}_i=\vec{J}_{\mathrm{ext}}.
|
||||
\]
|
||||
Therefore, if the net external impulse on the system is zero,
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\vec{0},
|
||||
\]
|
||||
then the total momentum is conserved:
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i.
|
||||
\]
|
||||
In component form, this means each Cartesian component is conserved separately, such as
|
||||
\[
|
||||
P_{x,f}=P_{x,i}
|
||||
\qquad\text{and}\qquad
|
||||
P_{y,f}=P_{y,i}
|
||||
\]
|
||||
when the external impulse is zero.}
|
||||
|
||||
\nt{The most important step is choosing the system boundary correctly. If two objects collide and both objects are included in the system, then the contact forces between them are internal forces and cancel in the total momentum balance. Forces from outside the system, such as a large friction force from the floor or a push from a person, are external and can change the system momentum. In two-dimensional problems, write momentum conservation separately in the $x$- and $y$-directions and solve the resulting scalar equations. Also, momentum conservation does \emph{not} require kinetic energy to be conserved: in an inelastic collision, the total momentum can stay constant even though the kinetic energy changes.}
|
||||
|
||||
\pf{Derivation from Newton's laws}{For each object $i$, let $\vec{F}_{\mathrm{ext},i}$ denote the net external force on that object, and let $\vec{F}_{\mathrm{int},i}$ denote the net internal force on it from the other objects in the system. Then Newton's second law gives
|
||||
\[
|
||||
\frac{d\vec{p}_i}{dt}=\vec{F}_{\mathrm{ext},i}+\vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
Now sum over all objects:
|
||||
\[
|
||||
\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}+\sum_{i=1}^N \vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\frac{d\vec{P}}{dt},
|
||||
\]
|
||||
and the internal forces cancel in equal-and-opposite pairs by Newton's third law,
|
||||
\[
|
||||
\sum_{i=1}^N \vec{F}_{\mathrm{int},i}=\vec{0},
|
||||
\]
|
||||
we obtain
|
||||
\[
|
||||
\frac{d\vec{P}}{dt}=\sum \vec{F}_{\mathrm{ext}}.
|
||||
\]
|
||||
Integrate from $t_i$ to $t_f$:
|
||||
\[
|
||||
\vec{P}_f-\vec{P}_i=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt=\vec{J}_{\mathrm{ext}}.
|
||||
\]
|
||||
If $\vec{J}_{\mathrm{ext}}=\vec{0}$, then $\vec{P}_f=\vec{P}_i$, so total momentum is conserved.}
|
||||
|
||||
\qs{Worked example}{On a frictionless horizontal air table, puck $A$ has mass $m_A=0.20\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{A,i}=(6.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Puck $B$ has mass $m_B=0.30\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{B,i}=(4.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The pucks collide and stick together. Let
|
||||
\[
|
||||
\vec{v}_f=v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}
|
||||
\]
|
||||
denote their common final velocity.
|
||||
|
||||
Find $\vec{v}_f$, its magnitude and direction, and determine whether kinetic energy is conserved.}
|
||||
|
||||
\sol Choose the system to be \emph{puck $A$ + puck $B$}. During the short collision, the table is frictionless, so the net external horizontal impulse on this two-puck system is zero. Therefore,
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
Because this is a two-dimensional problem, conserve momentum separately in the $x$- and $y$-directions.
|
||||
|
||||
First write the initial momentum components.
|
||||
|
||||
For puck $A$,
|
||||
\[
|
||||
\vec{p}_{A,i}=m_A\vec{v}_{A,i}=(0.20\,\mathrm{kg})(6.0\hat{\imath}\,\mathrm{m/s})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
For puck $B$,
|
||||
\[
|
||||
\vec{p}_{B,i}=m_B\vec{v}_{B,i}=(0.30\,\mathrm{kg})(4.0\hat{\jmath}\,\mathrm{m/s})=(1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
So the total initial momentum is
|
||||
\[
|
||||
\vec{P}_i=(1.2\hat{\imath}+1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
After the collision, the pucks stick together, so the total mass is
|
||||
\[
|
||||
M=m_A+m_B=0.20\,\mathrm{kg}+0.30\,\mathrm{kg}=0.50\,\mathrm{kg}.
|
||||
\]
|
||||
Their final momentum is
|
||||
\[
|
||||
\vec{P}_f=M\vec{v}_f=(0.50\,\mathrm{kg})(v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}).
|
||||
\]
|
||||
|
||||
Now conserve momentum by components.
|
||||
|
||||
In the $x$-direction,
|
||||
\[
|
||||
P_{x,i}=P_{x,f}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,x},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_{f,x}=2.4\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
In the $y$-direction,
|
||||
\[
|
||||
P_{y,i}=P_{y,f}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,y},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_{f,y}=2.4\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Therefore, the common final velocity is
|
||||
\[
|
||||
\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Its magnitude is
|
||||
\[
|
||||
|\vec{v}_f|=\sqrt{(2.4\,\mathrm{m/s})^2+(2.4\,\mathrm{m/s})^2}=3.39\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Let $\theta$ denote the direction of $\vec{v}_f$ measured counterclockwise from the positive $x$-axis. Then
|
||||
\[
|
||||
\tan\theta=\frac{v_{f,y}}{v_{f,x}}=\frac{2.4}{2.4}=1,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\theta=45^\circ.
|
||||
\]
|
||||
|
||||
Now check the kinetic energy.
|
||||
|
||||
Initially,
|
||||
\[
|
||||
K_i=\tfrac12 m_A|\vec{v}_{A,i}|^2+\tfrac12 m_B|\vec{v}_{B,i}|^2
|
||||
=\tfrac12(0.20)(6.0)^2+\tfrac12(0.30)(4.0)^2=6.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Finally,
|
||||
\[
|
||||
K_f=\tfrac12 M|\vec{v}_f|^2=\tfrac12(0.50)(3.39)^2\approx 2.88\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Since
|
||||
\[
|
||||
K_f\neq K_i,
|
||||
\]
|
||||
kinetic energy is not conserved. That is expected because the pucks stick together, so the collision is inelastic.
|
||||
|
||||
Thus the final velocity is
|
||||
\[
|
||||
\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s},
|
||||
\]
|
||||
with magnitude
|
||||
\[
|
||||
|\vec{v}_f|=3.39\,\mathrm{m/s},
|
||||
\]
|
||||
directed
|
||||
\[
|
||||
45^\circ
|
||||
\]
|
||||
above the positive $x$-axis, while momentum is conserved but kinetic energy is not.
|
||||
178
concepts/mechanics/u4/m4-4-collisions.tex
Normal file
178
concepts/mechanics/u4/m4-4-collisions.tex
Normal file
@@ -0,0 +1,178 @@
|
||||
\subsection{Elastic, Inelastic, and Perfectly Inelastic Collisions}
|
||||
|
||||
This subsection classifies collisions by what happens to the system's kinetic energy. In AP mechanics, the first step is still to apply momentum conservation to an isolated system; only elastic collisions add a kinetic-energy conservation equation.
|
||||
|
||||
\dfn{Elastic, inelastic, and perfectly inelastic collisions}{Consider two objects of masses $m_1$ and $m_2$. Let $\vec{v}_{1,i}$ and $\vec{v}_{2,i}$ denote their velocities just before a collision, and let $\vec{v}_{1,f}$ and $\vec{v}_{2,f}$ denote their velocities just after it. Let the total kinetic energy before and after the collision be
|
||||
\[
|
||||
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
\]
|
||||
and
|
||||
\[
|
||||
K_f=\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
|
||||
\]
|
||||
|
||||
A collision is called \emph{elastic} if the total kinetic energy is unchanged, so
|
||||
\[
|
||||
K_f=K_i.
|
||||
\]
|
||||
It is called \emph{inelastic} if the total kinetic energy changes, so
|
||||
\[
|
||||
K_f\ne K_i.
|
||||
\]
|
||||
In the usual AP collision problems without stored energy being released during impact, an inelastic collision has $K_f<K_i$.
|
||||
|
||||
A collision is \emph{perfectly inelastic} if the objects stick together after the collision, so they share one final velocity:
|
||||
\[
|
||||
\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f.
|
||||
\]
|
||||
Every perfectly inelastic collision is inelastic.}
|
||||
|
||||
\nt{Do not decide whether momentum is conserved by asking whether the collision is elastic. Those are different ideas. Momentum conservation depends on the net external impulse on the chosen system. If the system is isolated during the collision, then total momentum is conserved for elastic, inelastic, and perfectly inelastic collisions alike. Kinetic energy supplies an \emph{extra} condition only in the elastic case. In two-dimensional AP problems, conserve momentum separately in the $x$- and $y$-directions.}
|
||||
|
||||
\ex{Illustrative example}{On a frictionless track, cart 1 has mass $m_1=0.40\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{1,i}=(3.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Cart 2 has mass $m_2=0.20\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{2,i}=\vec{0}.
|
||||
\]
|
||||
After the collision, the carts move together with common velocity
|
||||
\[
|
||||
\vec{v}_f=(2.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Because the carts move together after impact, the collision is perfectly inelastic. The initial kinetic energy is
|
||||
\[
|
||||
K_i=\tfrac12(0.40)(3.0)^2=1.8\,\mathrm{J},
|
||||
\]
|
||||
while the final kinetic energy is
|
||||
\[
|
||||
K_f=\tfrac12(0.60)(2.0)^2=1.2\,\mathrm{J}.
|
||||
\]
|
||||
Since $K_f<K_i$, the collision is inelastic, as expected. Momentum is still conserved because
|
||||
\[
|
||||
\vec{P}_i=(0.40)(3.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}
|
||||
\]
|
||||
and
|
||||
\[
|
||||
\vec{P}_f=(0.60)(2.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
|
||||
\]}
|
||||
|
||||
\mprop{Practical relations for isolated collisions}{Consider two objects that form an isolated system during a short collision. Let
|
||||
\[
|
||||
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}
|
||||
\]
|
||||
denote the total initial momentum and let
|
||||
\[
|
||||
\vec{P}_f=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}
|
||||
\]
|
||||
denote the total final momentum.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item For any isolated collision,
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}.
|
||||
\]
|
||||
|
||||
\item In two dimensions, let $P_{x,i}$ and $P_{y,i}$ denote the initial momentum components, and let $P_{x,f}$ and $P_{y,f}$ denote the final momentum components. Then write two separate component equations:
|
||||
\[
|
||||
P_{x,i}=P_{x,f},
|
||||
\qquad
|
||||
P_{y,i}=P_{y,f}.
|
||||
\]
|
||||
|
||||
\item If the collision is perfectly inelastic, then $\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f$, so momentum conservation gives the shared final velocity directly:
|
||||
\[
|
||||
\vec{v}_f=\frac{m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}}{m_1+m_2}.
|
||||
\]
|
||||
|
||||
\item If the collision is elastic, then in addition to momentum conservation,
|
||||
\[
|
||||
\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
=
|
||||
\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
|
||||
\]
|
||||
For a one-dimensional elastic collision, this is equivalent to the relative-speed relation
|
||||
\[
|
||||
v_{1,i}-v_{2,i}=-(v_{1,f}-v_{2,f}),
|
||||
\]
|
||||
where each $v$ is an $x$-component.
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{Choose the positive $x$-axis to the right. On a frictionless track, cart 1 has mass $m_1=1.0\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{1,i}=(4.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Cart 2 has mass $m_2=3.0\,\mathrm{kg}$ and is initially at rest, so
|
||||
\[
|
||||
\vec{v}_{2,i}=\vec{0}.
|
||||
\]
|
||||
After the collision, the carts lock together and move with common final velocity $\vec{v}_f$.
|
||||
|
||||
Find $\vec{v}_f$, determine whether the collision is elastic, inelastic, or perfectly inelastic, and calculate the change in kinetic energy $\Delta K=K_f-K_i$.}
|
||||
|
||||
\sol Because the carts lock together, this is a perfectly inelastic collision by definition. During the short collision the track is frictionless, so the two-cart system is isolated horizontally. Therefore total momentum is conserved:
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
The initial momentum is
|
||||
\[
|
||||
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
\vec{P}_i=(1.0\,\mathrm{kg})(4.0\hat{\imath}\,\mathrm{m/s})+(3.0\,\mathrm{kg})(\vec{0})
|
||||
=4.0\hat{\imath}\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
After the collision the carts move together, so their total mass is
|
||||
\[
|
||||
m_1+m_2=4.0\,\mathrm{kg}
|
||||
\]
|
||||
and their final momentum is
|
||||
\[
|
||||
\vec{P}_f=(m_1+m_2)\vec{v}_f.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
4.0\hat{\imath}\,\mathrm{kg\cdot m/s}=(4.0\,\mathrm{kg})\vec{v}_f,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now compare the kinetic energies.
|
||||
|
||||
Initially,
|
||||
\[
|
||||
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
=\tfrac12(1.0)(4.0)^2+\tfrac12(3.0)(0)^2=8.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Finally,
|
||||
\[
|
||||
K_f=\tfrac12 (m_1+m_2)|\vec{v}_f|^2
|
||||
=\tfrac12(4.0)(1.0)^2=2.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
\Delta K=K_f-K_i=2.0\,\mathrm{J}-8.0\,\mathrm{J}=-6.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
The negative sign means $6.0\,\mathrm{J}$ of kinetic energy was transformed into other forms of energy during the collision. Since the carts stick together and the kinetic energy decreases, the collision is inelastic, more specifically perfectly inelastic.
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s},
|
||||
\qquad
|
||||
\Delta K=-6.0\,\mathrm{J},
|
||||
\]
|
||||
and the collision is perfectly inelastic rather than elastic.
|
||||
244
concepts/mechanics/u4/m4-5-recoil-explosions.tex
Normal file
244
concepts/mechanics/u4/m4-5-recoil-explosions.tex
Normal file
@@ -0,0 +1,244 @@
|
||||
\subsection{Recoil, Explosions, and the Center-of-Mass Viewpoint}
|
||||
|
||||
This subsection treats recoil and explosions as momentum-redistribution processes. In AP mechanics, internal energy may be released during the interaction, but if the net external impulse on the chosen system is zero, the total momentum and the center-of-mass motion do not change.
|
||||
|
||||
\dfn{Recoil/explosion interactions and the center-of-mass viewpoint}{Consider a closed system of particles labeled by $i=1,2,\dots,N$. Let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, let
|
||||
\[
|
||||
\vec{p}_i=m_i\vec{v}_i
|
||||
\]
|
||||
denote its momentum, and let
|
||||
\[
|
||||
\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i
|
||||
\]
|
||||
denote the total momentum. Let
|
||||
\[
|
||||
M=\sum_{i=1}^N m_i
|
||||
\]
|
||||
denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity. Then
|
||||
\[
|
||||
\vec{P}=M\vec{v}_{\mathrm{cm}}.
|
||||
\]
|
||||
|
||||
A \emph{recoil} or \emph{explosion} interaction is a short internal interaction in which parts of the system push apart or are driven apart by released internal energy. If the net external impulse on the system during the interaction is zero or negligible, then
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i,
|
||||
\]
|
||||
so equivalently
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=\vec{v}_{\mathrm{cm},i}.
|
||||
\]
|
||||
If the system is initially at rest, then $\vec{P}_i=\vec{0}$ and $\vec{v}_{\mathrm{cm}}=\vec{0}$ both before and after the interaction.}
|
||||
|
||||
\nt{In recoil and explosion problems, the internal forces between the parts of the system can be very large, but they occur in equal-and-opposite pairs and therefore only redistribute momentum within the system. They can change the individual velocities and can increase the total kinetic energy if internal energy is released, but they do not change the total momentum of an isolated system. From the center-of-mass viewpoint, the whole event is just internal rearrangement: the center of mass continues to remain at rest or to move with constant velocity if the external impulse is zero.}
|
||||
|
||||
\ex{Illustrative example}{A student of mass $m_s=60.0\,\mathrm{kg}$ stands at rest on a frictionless skateboard and throws a backpack of mass $m_b=5.0\,\mathrm{kg}$ horizontally backward with velocity
|
||||
\[
|
||||
\vec{v}_b=(-8.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Let $\vec{v}_s=v_s\hat{\imath}$ denote the student's recoil velocity after the throw.
|
||||
|
||||
Because the student-backpack system starts at rest and the external horizontal impulse is negligible,
|
||||
\[
|
||||
\vec{P}_f=\vec{0}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
m_s\vec{v}_s+m_b\vec{v}_b=\vec{0}.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
(60.0\,\mathrm{kg})v_s\hat{\imath}+(5.0\,\mathrm{kg})(-8.0\hat{\imath}\,\mathrm{m/s})=\vec{0}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
60.0v_s-40.0=0,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
v_s=0.667\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{v}_s=(0.667\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The student recoils forward while the backpack moves backward, and the total momentum remains zero.}
|
||||
|
||||
\mprop{Useful recoil and center-of-mass relations}{Let a system of total mass $M$ have velocity $\vec{V}_0$ just before a recoil or explosion event. Let the final pieces have masses $m_1,m_2,\dots$ and velocities $\vec{v}_1,\vec{v}_2,\dots$. If the net external impulse during the short interaction is zero or negligible, then:
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item Total momentum is conserved:
|
||||
\[
|
||||
\sum \vec{p}_{f}=\sum \vec{p}_{i},
|
||||
\qquad
|
||||
\sum m_k\vec{v}_k=M\vec{V}_0.
|
||||
\]
|
||||
|
||||
\item The center-of-mass velocity stays constant:
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M}=\vec{V}_0.
|
||||
\]
|
||||
If the system starts from rest, then $\vec{v}_{\mathrm{cm}}=\vec{0}$ and
|
||||
\[
|
||||
\sum m_k\vec{v}_k=\vec{0}.
|
||||
\]
|
||||
|
||||
\item For a two-piece explosion from rest,
|
||||
\[
|
||||
m_1\vec{v}_1+m_2\vec{v}_2=\vec{0},
|
||||
\]
|
||||
so the two final momenta are equal in magnitude and opposite in direction:
|
||||
\[
|
||||
m_1\vec{v}_1=-m_2\vec{v}_2.
|
||||
\]
|
||||
|
||||
\item In two dimensions, conserve momentum component-by-component:
|
||||
\[
|
||||
\sum p_{x,f}=\sum p_{x,i},
|
||||
\qquad
|
||||
\sum p_{y,f}=\sum p_{y,i}.
|
||||
\]
|
||||
This is usually the most direct way to find unknown fragment velocities.
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{A firework shell of total mass $M=4.0\,\mathrm{kg}$ is moving with velocity
|
||||
\[
|
||||
\vec{V}_0=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}
|
||||
\]
|
||||
just before it explodes into two fragments. Fragment 1 has mass $m_1=1.5\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_1=(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}
|
||||
\]
|
||||
immediately after the explosion. Fragment 2 has mass $m_2=2.5\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_2=v_{2,x}\hat{\imath}+v_{2,y}\hat{\jmath}.
|
||||
\]
|
||||
Assume the net external impulse during the explosion is negligible.
|
||||
|
||||
Find $\vec{v}_2$, find the center-of-mass velocity after the explosion, and determine whether the explosion could have increased the total kinetic energy.}
|
||||
|
||||
\sol Because the explosion is brief and the net external impulse is negligible, the shell-plus-fragments system conserves momentum:
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
First compute the initial total momentum. Since the shell has total mass $M=4.0\,\mathrm{kg}$ and velocity $\vec{V}_0$ just before the explosion,
|
||||
\[
|
||||
\vec{P}_i=M\vec{V}_0.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
\vec{P}_i=(4.0\,\mathrm{kg})(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Next compute the momentum of fragment 1:
|
||||
\[
|
||||
\vec{p}_1=m_1\vec{v}_1=(1.5\,\mathrm{kg})(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{p}_1=(27.0\hat{\imath}+3.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Let $\vec{p}_2=m_2\vec{v}_2$ denote the momentum of fragment 2. Since
|
||||
\[
|
||||
\vec{P}_f=\vec{p}_1+\vec{p}_2,
|
||||
\]
|
||||
momentum conservation gives
|
||||
\[
|
||||
\vec{p}_2=\vec{P}_i-\vec{p}_1.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\vec{p}_2=(40.0\hat{\imath}+24.0\hat{\jmath})-(27.0\hat{\imath}+3.0\hat{\jmath}).
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{p}_2=(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Now divide by $m_2=2.5\,\mathrm{kg}$ to find the velocity of fragment 2:
|
||||
\[
|
||||
\vec{v}_2=\frac{\vec{p}_2}{m_2}=\frac{(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{2.5\,\mathrm{kg}}.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now find the center-of-mass velocity after the explosion. The total momentum after the explosion is still
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
Using
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M},
|
||||
\]
|
||||
we get
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=\frac{\vec{P}_f}{M}=\frac{(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{4.0\,\mathrm{kg}}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
This is exactly the same as the pre-explosion velocity $\vec{V}_0$, which is the center-of-mass viewpoint: the fragments fly apart, but the center of mass continues with the same velocity because the external impulse is negligible.
|
||||
|
||||
Finally, check whether the total kinetic energy could have increased.
|
||||
|
||||
Before the explosion,
|
||||
\[
|
||||
K_i=\tfrac12 M|\vec{V}_0|^2=\tfrac12 (4.0)\left[(10.0)^2+(6.0)^2\right].
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
K_i=2.0(136)=272\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
After the explosion,
|
||||
\[
|
||||
K_f=\tfrac12 m_1|\vec{v}_1|^2+\tfrac12 m_2|\vec{v}_2|^2.
|
||||
\]
|
||||
For fragment 1,
|
||||
\[
|
||||
|\vec{v}_1|^2=(18.0)^2+(2.0)^2=328,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
K_1=\tfrac12 (1.5)(328)=246\,\mathrm{J}.
|
||||
\]
|
||||
For fragment 2,
|
||||
\[
|
||||
|\vec{v}_2|^2=(5.2)^2+(8.4)^2=27.04+70.56=97.60,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
K_2=\tfrac12 (2.5)(97.60)=122\,\mathrm{J}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
K_f=246\,\mathrm{J}+122\,\mathrm{J}=368\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Since
|
||||
\[
|
||||
K_f>K_i,
|
||||
\]
|
||||
the total kinetic energy increased by
|
||||
\[
|
||||
\Delta K=368\,\mathrm{J}-272\,\mathrm{J}=96\,\mathrm{J}.
|
||||
\]
|
||||
That is possible because the explosion released internal energy. Momentum stayed constant because the system was isolated during the brief explosion, but kinetic energy did not have to remain constant.
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s},
|
||||
\]
|
||||
and
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The explosion changed the fragment velocities and increased the total kinetic energy, but it did not change the total momentum or the motion of the center of mass.
|
||||
0
concepts/mechanics/u5/.gitkeep
Normal file
0
concepts/mechanics/u5/.gitkeep
Normal file
140
concepts/mechanics/u5/m5-1-angular-kinematics.tex
Normal file
140
concepts/mechanics/u5/m5-1-angular-kinematics.tex
Normal file
@@ -0,0 +1,140 @@
|
||||
\subsection{Angular Position, Velocity, and Acceleration}
|
||||
|
||||
This subsection describes rigid-body rotation about a fixed axis using signed angular variables. Once a positive sense of rotation is chosen, the formulas parallel one-dimensional kinematics.
|
||||
|
||||
\dfn{Angular kinematic variables for fixed-axis rotation}{Consider a rigid body rotating about a fixed axis. Choose a positive direction along the axis with unit vector $\hat{k}$ so that positive rotation is counterclockwise by the right-hand rule. Let $\theta(t)$ denote the signed angular position of the body, let $\omega(t)$ denote its angular velocity, and let $\alpha(t)$ denote its angular acceleration. Then
|
||||
\[
|
||||
\omega=\frac{d\theta}{dt},
|
||||
\qquad
|
||||
\alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}.
|
||||
\]
|
||||
Over a time interval $\Delta t$, let $\theta_i$ and $\theta_f$ denote the initial and final angular positions, and let $\omega_i$ and $\omega_f$ denote the initial and final angular velocities. Then
|
||||
\[
|
||||
\omega_{\mathrm{avg}}=\frac{\Delta\theta}{\Delta t},
|
||||
\qquad
|
||||
\alpha_{\mathrm{avg}}=\frac{\Delta\omega}{\Delta t},
|
||||
\]
|
||||
where $\Delta\theta=\theta_f-\theta_i$ and $\Delta\omega=\omega_f-\omega_i$. In vector form for fixed-axis rotation,
|
||||
\[
|
||||
\vec{\omega}=\omega\hat{k},
|
||||
\qquad
|
||||
\vec{\alpha}=\alpha\hat{k}.
|
||||
\]
|
||||
A positive value means rotation or change in rotation in the chosen positive sense, and a negative value means the opposite sense.}
|
||||
|
||||
\nt{Angular position is measured in radians, with $2\pi\,\mathrm{rad}=1$ revolution. Because a radian is a ratio of lengths, it is technically dimensionless, but radians should still be written in angular answers to make the meaning clear. The angular variables $\theta$, $\omega$, and $\alpha$ describe the entire rigid body, whereas the linear quantities of an individual point on the body depend on its distance $r$ from the axis. Also, $\theta$ is a signed coordinate, not a distance, so it can be negative and can exceed $2\pi$ after multiple turns.}
|
||||
|
||||
\mprop{Core fixed-axis relations}{Let $t$ denote elapsed time from an initial instant $t=0$. Let $\theta_0=\theta(0)$ and $\omega_0=\omega(0)$. Let $\theta=\theta(t)$ and $\omega=\omega(t)$ at a later time $t$. For a point on the rigid body at perpendicular distance $r$ from the axis, let $s$ denote its arc length from the chosen reference line, let $v_t$ denote its tangential velocity component in the positive tangential direction, and let $a_t$ denote its tangential acceleration component in that direction.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item Angular and linear quantities are related by
|
||||
\[
|
||||
s=r\theta,
|
||||
\qquad
|
||||
v_t=r\omega,
|
||||
\qquad
|
||||
a_t=r\alpha.
|
||||
\]
|
||||
Thus all points on a rigid body have the same $\theta$, $\omega$, and $\alpha$, but points farther from the axis have larger $|s|$, $|v_t|$, and $|a_t|$.
|
||||
|
||||
\item If $\alpha$ is constant over the interval, then the angular kinematic equations are
|
||||
\[
|
||||
\omega=\omega_0+\alpha t,
|
||||
\]
|
||||
\[
|
||||
\theta=\theta_0+\omega_0 t+\tfrac12 \alpha t^2,
|
||||
\]
|
||||
\[
|
||||
\omega^2=\omega_0^2+2\alpha(\theta-\theta_0).
|
||||
\]
|
||||
|
||||
\item For constant $\alpha$, the average angular velocity over the interval is
|
||||
\[
|
||||
\omega_{\mathrm{avg}}=\frac{\omega_0+\omega}{2},
|
||||
\]
|
||||
so the angular displacement can also be written as
|
||||
\[
|
||||
\Delta\theta=\omega_{\mathrm{avg}}t=\frac{\omega_0+\omega}{2}\,t.
|
||||
\]
|
||||
These equations are the exact rotational analogs of the one-dimensional constant-acceleration formulas.
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{A bicycle wheel rotates about a fixed axle. Choose counterclockwise as positive. At the instant the brakes are applied, let $t=0$, let the wheel's angular position be $\theta_0=0$, and let its angular velocity be $\omega_0=+18.0\,\mathrm{rad/s}$. While braking, the wheel has constant angular acceleration $\alpha=-3.00\,\mathrm{rad/s^2}$ until it stops. The wheel radius is $r=0.340\,\mathrm{m}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the time required to stop,
|
||||
\item the angular displacement before stopping,
|
||||
\item the number of revolutions made while stopping, and
|
||||
\item the initial tangential speed of a point on the rim and the tangential acceleration of the rim during braking.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol First use constant angular acceleration to find the stopping time. At the instant the wheel stops, $\omega=0$. From
|
||||
\[
|
||||
\omega=\omega_0+\alpha t,
|
||||
\]
|
||||
we have
|
||||
\[
|
||||
0=18.0\,\mathrm{rad/s}+(-3.00\,\mathrm{rad/s^2})t.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
t=\frac{18.0\,\mathrm{rad/s}}{3.00\,\mathrm{rad/s^2}}=6.00\,\mathrm{s}.
|
||||
\]
|
||||
Therefore the wheel stops after
|
||||
\[
|
||||
t=6.00\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
Now find the angular displacement. Using
|
||||
\[
|
||||
\theta=\theta_0+\omega_0 t+\tfrac12 \alpha t^2,
|
||||
\]
|
||||
with $\theta_0=0$, $t=6.00\,\mathrm{s}$, $\omega_0=18.0\,\mathrm{rad/s}$, and $\alpha=-3.00\,\mathrm{rad/s^2}$,
|
||||
\[
|
||||
\theta=(0)+(18.0)(6.00)+\tfrac12(-3.00)(6.00)^2.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
\theta=108-54=54.0\,\mathrm{rad}.
|
||||
\]
|
||||
Since $\theta_0=0$, the angular displacement is
|
||||
\[
|
||||
\Delta\theta=54.0\,\mathrm{rad}.
|
||||
\]
|
||||
|
||||
Convert this to revolutions using $2\pi\,\mathrm{rad}=1$ revolution:
|
||||
\[
|
||||
N=\frac{\Delta\theta}{2\pi}=\frac{54.0}{2\pi}\approx 8.59.
|
||||
\]
|
||||
So the wheel makes
|
||||
\[
|
||||
N\approx 8.59\text{ revolutions}
|
||||
\]
|
||||
before stopping.
|
||||
|
||||
For the rim's initial tangential speed,
|
||||
\[
|
||||
v_t=r\omega_0=(0.340\,\mathrm{m})(18.0\,\mathrm{rad/s})=6.12\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
For the tangential acceleration,
|
||||
\[
|
||||
a_t=r\alpha=(0.340\,\mathrm{m})(-3.00\,\mathrm{rad/s^2})=-1.02\,\mathrm{m/s^2}.
|
||||
\]
|
||||
So the tangential acceleration is opposite the positive tangential direction, consistent with the wheel slowing down.
|
||||
|
||||
The results are
|
||||
\[
|
||||
t=6.00\,\mathrm{s},
|
||||
\qquad
|
||||
\Delta\theta=54.0\,\mathrm{rad},
|
||||
\qquad
|
||||
N\approx 8.59,
|
||||
\]
|
||||
\[
|
||||
v_{t,0}=6.12\,\mathrm{m/s},
|
||||
\qquad
|
||||
a_t=-1.02\,\mathrm{m/s^2}.
|
||||
\]
|
||||
Because $\omega_0>0$ and $\alpha<0$, the brake torque reduces the wheel's counterclockwise rotation rate until the angular velocity reaches zero.
|
||||
176
concepts/mechanics/u5/m5-2-linear-rotational-link.tex
Normal file
176
concepts/mechanics/u5/m5-2-linear-rotational-link.tex
Normal file
@@ -0,0 +1,176 @@
|
||||
\subsection{Linear and Rotational Kinematic Correspondence}
|
||||
|
||||
This subsection connects the angular variables of a rigid body in fixed-axis rotation to the linear motion of any specific point on that body. Once a positive sense of rotation is chosen, the tangential quantities behave like signed one-dimensional variables along the circular path, while the radial acceleration always points inward.
|
||||
|
||||
\dfn{Arc length, tangential speed, and radial/tangential acceleration for a rotating point}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $P$ be a point on the body at constant perpendicular distance $r$ from the axis, and let $\vec{r}=r\hat{r}$ be the position vector from the axis to $P$. Choose the positive tangential direction $\hat{t}$ to be the direction of positive rotation. Let $\theta(t)$ denote the signed angular position of the body, let $\omega=d\theta/dt$ denote its angular velocity, and let $\alpha=d\omega/dt$ denote its angular acceleration.
|
||||
|
||||
The signed arc-length coordinate of $P$ along its circular path is
|
||||
\[
|
||||
s=r\theta.
|
||||
\]
|
||||
The tangential velocity component and tangential acceleration component of $P$ are
|
||||
\[
|
||||
v_t=\frac{ds}{dt},
|
||||
\qquad
|
||||
a_t=\frac{dv_t}{dt}.
|
||||
\]
|
||||
Let $a_r\ge 0$ denote the magnitude of the radial component of the acceleration. Then the acceleration vector of $P$ can be written as
|
||||
\[
|
||||
\vec{a}=-a_r\hat{r}+a_t\hat{t},
|
||||
\]
|
||||
because the radial part points toward the axis.}
|
||||
|
||||
\thm{Linear/rotational correspondence for fixed-axis rotation}{For the point $P$ above at radius $r$,
|
||||
\[
|
||||
s=r\theta,
|
||||
\qquad
|
||||
v_t=r\omega,
|
||||
\qquad
|
||||
a_t=r\alpha,
|
||||
\qquad
|
||||
a_r=\frac{v_t^2}{r}=r\omega^2.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
\vec{v}=v_t\hat{t},
|
||||
\qquad
|
||||
\vec{a}=-r\omega^2\hat{r}+r\alpha\hat{t}.
|
||||
\]
|
||||
If $\vec{\omega}=\omega\hat{k}$, then the tangential-velocity relation may also be written as
|
||||
\[
|
||||
\vec{v}=\vec{\omega}\times \vec{r}.
|
||||
\]
|
||||
The first three formulas are the direct linear analogs of angular position, angular velocity, and angular acceleration for a point on the rotating body, while $a_r$ gives the inward acceleration required to keep the point on a circular path.}
|
||||
|
||||
\pf{Short derivation from $s=r\theta$}{Because the point stays a fixed distance $r$ from the axis, its arc-length coordinate satisfies
|
||||
\[
|
||||
s=r\theta.
|
||||
\]
|
||||
Differentiate with respect to time:
|
||||
\[
|
||||
v_t=\frac{ds}{dt}=r\frac{d\theta}{dt}=r\omega.
|
||||
\]
|
||||
Differentiate again:
|
||||
\[
|
||||
a_t=\frac{dv_t}{dt}=r\frac{d\omega}{dt}=r\alpha.
|
||||
\]
|
||||
For the radial part, the point is moving instantaneously on a circle of radius $r$ with speed $|v_t|$, so the inward radial acceleration has magnitude
|
||||
\[
|
||||
a_r=\frac{v_t^2}{r}.
|
||||
\]
|
||||
Substitute $v_t=r\omega$:
|
||||
\[
|
||||
a_r=\frac{(r\omega)^2}{r}=r\omega^2.
|
||||
\]
|
||||
Therefore
|
||||
\[
|
||||
s=r\theta,
|
||||
\qquad
|
||||
v_t=r\omega,
|
||||
\qquad
|
||||
a_t=r\alpha,
|
||||
\qquad
|
||||
a_r=r\omega^2.
|
||||
\]}
|
||||
|
||||
\cor{Same $\omega$, different linear motion at different radii}{Let two points $P_1$ and $P_2$ lie on the same rigid body at radii $r_1$ and $r_2$ from the same fixed axis. At any instant they share the same angular quantities $\theta$, $\omega$, and $\alpha$, but their linear quantities scale with radius:
|
||||
\[
|
||||
\frac{s_2}{s_1}=\frac{r_2}{r_1},
|
||||
\qquad
|
||||
\frac{v_{t,2}}{v_{t,1}}=\frac{r_2}{r_1},
|
||||
\qquad
|
||||
\frac{a_{t,2}}{a_{t,1}}=\frac{r_2}{r_1},
|
||||
\qquad
|
||||
\frac{a_{r,2}}{a_{r,1}}=\frac{r_2}{r_1}.
|
||||
\]
|
||||
So points farther from the axis move faster, have larger tangential acceleration for the same $\alpha$, and require larger inward acceleration for the same $\omega$.}
|
||||
|
||||
\qs{Worked example}{A rigid wheel rotates counterclockwise about a fixed axle. At the instant of interest, the wheel has angular position $\theta=1.20\,\mathrm{rad}$, angular velocity $\omega=6.0\,\mathrm{rad/s}$, and angular acceleration $\alpha=-2.0\,\mathrm{rad/s^2}$. Point $A$ is painted on the wheel at radius $r_A=0.050\,\mathrm{m}$, and point $B$ is painted at radius $r_B=0.150\,\mathrm{m}$.
|
||||
|
||||
For each point, find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the signed arc-length coordinate $s$,
|
||||
\item the tangential velocity component $v_t$,
|
||||
\item the tangential acceleration component $a_t$, and
|
||||
\item the radial acceleration magnitude $a_r$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Choose the positive tangential direction to be counterclockwise, since the wheel's rotation is positive in that sense. Use
|
||||
\[
|
||||
s=r\theta,
|
||||
\qquad
|
||||
v_t=r\omega,
|
||||
\qquad
|
||||
a_t=r\alpha,
|
||||
\qquad
|
||||
a_r=r\omega^2.
|
||||
\]
|
||||
|
||||
For point $A$, $r_A=0.050\,\mathrm{m}$.
|
||||
|
||||
Its arc-length coordinate is
|
||||
\[
|
||||
s_A=r_A\theta=(0.050\,\mathrm{m})(1.20)=0.060\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Its tangential velocity component is
|
||||
\[
|
||||
v_{t,A}=r_A\omega=(0.050\,\mathrm{m})(6.0\,\mathrm{rad/s})=0.30\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Its tangential acceleration component is
|
||||
\[
|
||||
a_{t,A}=r_A\alpha=(0.050\,\mathrm{m})(-2.0\,\mathrm{rad/s^2})=-0.10\,\mathrm{m/s^2}.
|
||||
\]
|
||||
The negative sign means the tangential acceleration is opposite the positive tangential direction at that instant.
|
||||
|
||||
Its radial acceleration magnitude is
|
||||
\[
|
||||
a_{r,A}=r_A\omega^2=(0.050\,\mathrm{m})(6.0\,\mathrm{rad/s})^2
|
||||
=(0.050)(36)=1.8\,\mathrm{m/s^2}.
|
||||
\]
|
||||
This radial part points inward, toward the axle.
|
||||
|
||||
For point $B$, $r_B=0.150\,\mathrm{m}$.
|
||||
|
||||
Its arc-length coordinate is
|
||||
\[
|
||||
s_B=r_B\theta=(0.150\,\mathrm{m})(1.20)=0.180\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
Its tangential velocity component is
|
||||
\[
|
||||
v_{t,B}=r_B\omega=(0.150\,\mathrm{m})(6.0\,\mathrm{rad/s})=0.90\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Its tangential acceleration component is
|
||||
\[
|
||||
a_{t,B}=r_B\alpha=(0.150\,\mathrm{m})(-2.0\,\mathrm{rad/s^2})=-0.30\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Its radial acceleration magnitude is
|
||||
\[
|
||||
a_{r,B}=r_B\omega^2=(0.150\,\mathrm{m})(6.0\,\mathrm{rad/s})^2
|
||||
=(0.150)(36)=5.4\,\mathrm{m/s^2}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
s_A=0.060\,\mathrm{m},
|
||||
\qquad
|
||||
v_{t,A}=0.30\,\mathrm{m/s},
|
||||
\qquad
|
||||
a_{t,A}=-0.10\,\mathrm{m/s^2},
|
||||
\qquad
|
||||
a_{r,A}=1.8\,\mathrm{m/s^2},
|
||||
\]
|
||||
\[
|
||||
s_B=0.180\,\mathrm{m},
|
||||
\qquad
|
||||
v_{t,B}=0.90\,\mathrm{m/s},
|
||||
\qquad
|
||||
a_{t,B}=-0.30\,\mathrm{m/s^2},
|
||||
\qquad
|
||||
a_{r,B}=5.4\,\mathrm{m/s^2}.
|
||||
\]
|
||||
Point $B$, which is three times farther from the axis than point $A$, has three times the arc length, tangential speed, tangential-acceleration component, and radial-acceleration magnitude.
|
||||
115
concepts/mechanics/u5/m5-3-torque.tex
Normal file
115
concepts/mechanics/u5/m5-3-torque.tex
Normal file
@@ -0,0 +1,115 @@
|
||||
\subsection{Torque and Lever Arm}
|
||||
|
||||
This subsection introduces torque as the rotational effect of a force about a chosen point or axis. In AP mechanics, the key computational idea is that only the part of the force with a nonzero lever arm contributes to the turning effect.
|
||||
|
||||
\dfn{Torque vector, lever arm, and fixed-axis sign convention}{Let $O$ denote the pivot or reference point. Let $\vec{r}$ denote the position vector from $O$ to the point where a force $\vec{F}$ is applied. The \emph{torque vector} of $\vec{F}$ about $O$ is the vector quantity that measures the tendency of the force to cause rotation about $O$.
|
||||
|
||||
Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$. The \emph{lever arm} $\ell$ is the perpendicular distance from the pivot to the line of action of the force, so
|
||||
\[
|
||||
\ell=r\sin\phi,
|
||||
\]
|
||||
where $r=|\vec{r}|$.
|
||||
|
||||
For a fixed-axis problem, choose an axis with unit vector $\hat{k}$ and declare a positive sense of rotation by the right-hand rule. The corresponding signed scalar torque is
|
||||
\[
|
||||
\tau=(\vec{r}\times \vec{F})\cdot \hat{k}.
|
||||
\]
|
||||
If the force tends to rotate the object in the chosen positive sense, then $\tau>0$; if it tends to rotate the object in the opposite sense, then $\tau<0$.}
|
||||
|
||||
\thm{Torque formulas}{Let $\vec{r}$ denote the position vector from the pivot to the point of application of a force $\vec{F}$. Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$, let $r=|\vec{r}|$, let $F=|\vec{F}|$, and let $\ell$ denote the lever arm. Then the torque vector is
|
||||
\[
|
||||
\vec{\tau}=\vec{r}\times \vec{F},
|
||||
\]
|
||||
and its magnitude is
|
||||
\[
|
||||
|\vec{\tau}|=rF\sin\phi=F\ell.
|
||||
\]
|
||||
For rotation about a chosen fixed axis with unit vector $\hat{k}$,
|
||||
\[
|
||||
\tau=(\vec{r}\times \vec{F})\cdot \hat{k},
|
||||
\]
|
||||
so the scalar $\tau$ is positive or negative according to the declared sign convention.}
|
||||
|
||||
\pf{Why $rF\sin\phi$ equals $F\ell$}{From the cross-product magnitude formula,
|
||||
\[
|
||||
|\vec{\tau}|=|\vec{r}\times \vec{F}|=rF\sin\phi.
|
||||
\]
|
||||
But by geometry, the lever arm is the perpendicular distance from the pivot to the line of action of the force, so
|
||||
\[
|
||||
\ell=r\sin\phi.
|
||||
\]
|
||||
Substituting this into the previous expression gives
|
||||
\[
|
||||
|\vec{\tau}|=rF\sin\phi=F\ell.
|
||||
\]
|
||||
Thus torque can be found either from the perpendicular component of the force or from the full force multiplied by the lever arm.}
|
||||
|
||||
\cor{If the line of action passes through the pivot, the torque is zero}{If the line of action of $\vec{F}$ passes through the pivot, then the perpendicular distance from the pivot to that line is $\ell=0$. Therefore
|
||||
\[
|
||||
|\vec{\tau}|=F\ell=0.
|
||||
\]
|
||||
Equivalently, in this case $\phi=0$ or $\phi=\pi$, so $rF\sin\phi=0$. Thus a force directed exactly through the pivot can change the net force on an object without producing any torque about that pivot.}
|
||||
|
||||
\qs{Worked example}{A door rotates about a vertical hinge through its left edge. View the door from above and choose counterclockwise rotation as positive. Let $r=0.90\,\mathrm{m}$ be the distance from the hinge to the point where the force is applied at the outer edge. A student pushes with force magnitude $F=40\,\mathrm{N}$. The force lies in the horizontal plane and makes an angle $\phi=30^\circ$ with the door, so it tends to rotate the door counterclockwise.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the lever arm $\ell$,
|
||||
\item the signed torque $\tau$ about the hinge, and
|
||||
\item the magnitude of a force applied perpendicular to the door at the same point that would produce the same torque.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Because the position vector from the hinge to the point of application lies along the door, the given angle $\phi=30^\circ$ is the angle between $\vec{r}$ and $\vec{F}$.
|
||||
|
||||
For part (a), the lever arm is
|
||||
\[
|
||||
\ell=r\sin\phi=(0.90\,\mathrm{m})\sin 30^\circ.
|
||||
\]
|
||||
Since $\sin 30^\circ=0.50$,
|
||||
\[
|
||||
\ell=(0.90)(0.50)=0.45\,\mathrm{m}.
|
||||
\]
|
||||
So the lever arm is
|
||||
\[
|
||||
\ell=0.45\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For part (b), the torque magnitude is
|
||||
\[
|
||||
|\tau|=rF\sin\phi=F\ell.
|
||||
\]
|
||||
Using either form,
|
||||
\[
|
||||
|\tau|=(0.90\,\mathrm{m})(40\,\mathrm{N})\sin 30^\circ=(40\,\mathrm{N})(0.45\,\mathrm{m}).
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
|\tau|=18\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
Because the force tends to rotate the door counterclockwise and counterclockwise was chosen as positive,
|
||||
\[
|
||||
\tau=+18\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
|
||||
For part (c), let $F_{\perp}$ denote the force magnitude that would act perpendicular to the door at the same point. A perpendicular force has lever arm equal to the full distance $r=0.90\,\mathrm{m}$, so its torque magnitude is
|
||||
\[
|
||||
|\tau|=rF_{\perp}.
|
||||
\]
|
||||
Set this equal to the required $18\,\mathrm{N\cdot m}$:
|
||||
\[
|
||||
18\,\mathrm{N\cdot m}=(0.90\,\mathrm{m})F_{\perp}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
F_{\perp}=\frac{18}{0.90}=20\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
The results are
|
||||
\[
|
||||
\ell=0.45\,\mathrm{m},
|
||||
\qquad
|
||||
\tau=+18\,\mathrm{N\cdot m},
|
||||
\qquad
|
||||
F_{\perp}=20\,\mathrm{N}.
|
||||
\]
|
||||
This example shows that the same torque can be produced either by a larger force with a shorter lever arm or by a smaller force applied more effectively.
|
||||
126
concepts/mechanics/u5/m5-4-moment-of-inertia.tex
Normal file
126
concepts/mechanics/u5/m5-4-moment-of-inertia.tex
Normal file
@@ -0,0 +1,126 @@
|
||||
\subsection{Moment of Inertia and Mass Distribution}
|
||||
|
||||
This subsection introduces the rotational analog of mass for fixed-axis motion. In AP mechanics, the key idea is that both the amount of mass and how far that mass lies from the chosen axis determine how strongly an object resists angular acceleration.
|
||||
|
||||
\dfn{Moment of inertia for discrete and continuous mass distributions}{Consider a rigid body about a chosen fixed axis. For a discrete collection of particles, let particle $i$ have mass $m_i$, let $\vec{r}_i$ denote its position vector relative to a point on the axis, and let $r_{\perp,i}$ denote its perpendicular distance to the axis. The \emph{moment of inertia} of the body about that axis is
|
||||
\[
|
||||
I=\sum_i m_i r_{\perp,i}^2.
|
||||
\]
|
||||
For a continuous mass distribution, let $dm$ denote a small mass element and let $r_\perp$ denote that element's perpendicular distance to the same axis. Then
|
||||
\[
|
||||
I=\int r_\perp^2\,dm.
|
||||
\]
|
||||
For a single point mass,
|
||||
\[
|
||||
I=mr_\perp^2.
|
||||
\]
|
||||
The SI unit of moment of inertia is $\mathrm{kg\cdot m^2}$. Because the distance to the axis is squared, mass farther from the axis contributes much more strongly to $I$.}
|
||||
|
||||
\thm{Key fixed-axis relations and axis dependence}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $I$ denote the moment of inertia about that same axis. Then
|
||||
\[
|
||||
\tau_{\mathrm{net}}=I\alpha,
|
||||
\qquad
|
||||
\vec{\tau}_{\mathrm{net}}=I\vec{\alpha}.
|
||||
\]
|
||||
Thus, for the same net torque, a larger moment of inertia gives a smaller angular acceleration.
|
||||
|
||||
Now let $M$ denote the total mass of the body, let $I_{\mathrm{cm}}$ denote the moment of inertia about an axis through the center of mass, and let $d$ denote the perpendicular distance from that axis to a second axis parallel to it. Then the parallel-axis theorem states
|
||||
\[
|
||||
I=I_{\mathrm{cm}}+Md^2.
|
||||
\]
|
||||
Therefore moment of inertia depends on the chosen axis as well as on the mass distribution. Moving the axis farther from the center of mass increases $I$.}
|
||||
|
||||
\ex{Illustrative example}{A light turntable carries two small clay balls, each of mass $m=0.40\,\mathrm{kg}$. In arrangement A, each ball is at distance $r_A=0.10\,\mathrm{m}$ from the axis. In arrangement B, each ball is at distance $r_B=0.20\,\mathrm{m}$ from the axis.
|
||||
|
||||
For arrangement A,
|
||||
\[
|
||||
I_A=2mr_A^2=2(0.40)(0.10)^2=8.0\times 10^{-3}\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
For arrangement B,
|
||||
\[
|
||||
I_B=2mr_B^2=2(0.40)(0.20)^2=3.2\times 10^{-2}\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
So doubling the distance from the axis makes the moment of inertia four times as large, even though the total mass is unchanged.}
|
||||
|
||||
\nt{Equal mass does not guarantee equal rotational response. Two objects can have the same total mass but different moments of inertia if their mass is distributed differently or if the axis is changed. In fixed-axis rotation, the object with larger $I$ has smaller angular acceleration for the same net torque.}
|
||||
|
||||
\qs{Worked example}{A light rigid rod of length $L=0.80\,\mathrm{m}$ lies on a horizontal frictionless table and can rotate about a vertical axle. A small mass $m_1=0.50\,\mathrm{kg}$ is attached at the left end, and a small mass $m_2=1.50\,\mathrm{kg}$ is attached at the right end. First, the axle passes through the center of the rod, so each mass is at distance $r=0.40\,\mathrm{m}$ from the axis. A string pulls perpendicularly on the right end with force magnitude $F=3.0\,\mathrm{N}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the moment of inertia $I$ about the center axle,
|
||||
\item the torque magnitude $\tau$ due to the pull,
|
||||
\item the angular acceleration magnitude $\alpha$, and
|
||||
\item the new moment of inertia and angular acceleration if the axle is moved to the left end of the rod while the same perpendicular force is still applied at the right end.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol For parts (a)--(c), the axis passes through the center of the rod, so each mass is $r=0.40\,\mathrm{m}$ from the axis. Because the rod is light, treat its mass as negligible and include only the two point masses.
|
||||
|
||||
For part (a), the moment of inertia is
|
||||
\[
|
||||
I=\sum m_i r_i^2=m_1r^2+m_2r^2.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
I=(0.50)(0.40)^2+(1.50)(0.40)^2.
|
||||
\]
|
||||
Since $(0.40)^2=0.16$,
|
||||
\[
|
||||
I=(0.50)(0.16)+(1.50)(0.16)=0.08+0.24=0.32\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
So the moment of inertia about the center axle is
|
||||
\[
|
||||
I=0.32\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
|
||||
For part (b), the pull is perpendicular to the rod, so the torque magnitude is
|
||||
\[
|
||||
\tau=rF.
|
||||
\]
|
||||
Here the force is applied at the right end, which is $0.40\,\mathrm{m}$ from the center axle. Therefore,
|
||||
\[
|
||||
\tau=(0.40\,\mathrm{m})(3.0\,\mathrm{N})=1.2\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
|
||||
For part (c), use the rotational form of Newton's second law:
|
||||
\[
|
||||
\tau=I\alpha.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
\alpha=\frac{\tau}{I}=\frac{1.2}{0.32}=3.75\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
So for the center axle,
|
||||
\[
|
||||
\alpha=3.75\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
|
||||
For part (d), move the axle to the left end. Then the left mass is on the axis, so its distance is $r_1=0$, and the right mass is $r_2=L=0.80\,\mathrm{m}$ from the axis. The new moment of inertia is
|
||||
\[
|
||||
I' = m_1r_1^2+m_2r_2^2.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
I'=(0.50)(0)^2+(1.50)(0.80)^2.
|
||||
\]
|
||||
Since $(0.80)^2=0.64$,
|
||||
\[
|
||||
I'=0+(1.50)(0.64)=0.96\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
|
||||
Now the same perpendicular force is applied at the right end, which is $0.80\,\mathrm{m}$ from the new axis, so the torque magnitude is
|
||||
\[
|
||||
\tau'=(0.80\,\mathrm{m})(3.0\,\mathrm{N})=2.4\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
\alpha'=\frac{\tau'}{I'}=\frac{2.4}{0.96}=2.50\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
|
||||
Therefore, with the axle at the left end,
|
||||
\[
|
||||
I'=0.96\,\mathrm{kg\cdot m^2},
|
||||
\qquad
|
||||
\alpha'=2.50\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
Even though the torque becomes larger, the angular acceleration becomes smaller because much more of the mass is farther from the axis, so the moment of inertia increases substantially.
|
||||
141
concepts/mechanics/u5/m5-5-rotational-equilibrium.tex
Normal file
141
concepts/mechanics/u5/m5-5-rotational-equilibrium.tex
Normal file
@@ -0,0 +1,141 @@
|
||||
\subsection{Rotational Equilibrium}
|
||||
|
||||
This subsection treats the equilibrium case of fixed-axis rotational dynamics. In AP statics, a rigid body remains at rest only when both the translational and rotational balances are satisfied.
|
||||
|
||||
\dfn{Rotational equilibrium and static equilibrium}{Consider a rigid body about a chosen fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $\vec{F}_{\mathrm{net}}$ denote the net external force on the body.
|
||||
|
||||
The body is in \emph{rotational equilibrium} if
|
||||
\[
|
||||
\vec{\alpha}=\vec{0},
|
||||
\]
|
||||
so its angular velocity is constant. In the common AP statics case, the body is at rest and is also in \emph{static equilibrium}, meaning that both its translational and rotational motion remain unchanged:
|
||||
\[
|
||||
\vec{F}_{\mathrm{net}}=\vec{0},
|
||||
\qquad
|
||||
\vec{\alpha}=\vec{0}.
|
||||
\]
|
||||
Thus rotational equilibrium is the $\alpha=0$ special case of rotational dynamics, and static equilibrium is the at-rest case in which there is also no translational acceleration.}
|
||||
|
||||
\thm{Equilibrium conditions for a rigid body}{Let $\vec{F}_{\mathrm{net}}$ denote the net external force on a rigid body, and let $\vec{\tau}_{\mathrm{net},O}$ denote the net external torque about a chosen point $O$. For static equilibrium in an inertial frame,
|
||||
\[
|
||||
\sum \vec{F}=\vec{0},
|
||||
\qquad
|
||||
\sum \vec{\tau}_O=\vec{0}.
|
||||
\]
|
||||
For a planar fixed-axis problem, choose a positive sense of rotation about the axis. Then the equivalent scalar conditions are
|
||||
\[
|
||||
\sum F_x=0,
|
||||
\qquad
|
||||
\sum F_y=0,
|
||||
\qquad
|
||||
\sum \tau_O=0,
|
||||
\]
|
||||
where positive and negative torques are assigned by the declared sign convention. Force balance enforces translational equilibrium, and torque balance enforces rotational equilibrium.}
|
||||
|
||||
\nt{In equilibrium problems, the pivot can be chosen wherever it is most convenient. A smart choice is often a point through which one or more unknown forces act, because those forces then contribute zero torque about that point. After using $\sum \tau_O=0$ to solve for a remaining unknown such as a tension, use $\sum \vec{F}=\vec{0}$ to find the support-force components. If $\sum \vec{F}=\vec{0}$, then torque balance about one point is equivalent to torque balance about any other point, so changing the pivot changes the algebra, not the physics. For fixed-axis AP problems, signed scalar torques are fully acceptable once a sign convention such as counterclockwise positive has been declared.}
|
||||
|
||||
\pf{Short explanation from $\tau_{\mathrm{net}}=I\alpha$}{Let $I$ denote the moment of inertia of the rigid body about the chosen fixed axis. Fixed-axis rotational dynamics gives
|
||||
\[
|
||||
\tau_{\mathrm{net}}=I\alpha.
|
||||
\]
|
||||
If the body is in rotational equilibrium, then $\alpha=0$, so
|
||||
\[
|
||||
\tau_{\mathrm{net}}=0.
|
||||
\]
|
||||
Now let $m$ denote the total mass of the body, and let $\vec{a}_{\mathrm{cm}}$ denote the acceleration of its center of mass. For statics the body also has no translational acceleration, so Newton's second law gives
|
||||
\[
|
||||
\sum \vec{F}=m\vec{a}_{\mathrm{cm}}=\vec{0}.
|
||||
\]
|
||||
Therefore a rigid body at rest must satisfy both torque balance and force balance. Conversely, if both balances hold, the body has neither angular acceleration nor translational acceleration, so an initially resting body remains in static equilibrium.}
|
||||
|
||||
\qs{Worked example}{A uniform horizontal beam has length $L=4.0\,\mathrm{m}$ and weight $W_b=200\,\mathrm{N}$. It is hinged to a wall at its left end. A light cable is attached to the right end of the beam and makes an angle $\theta=30^\circ$ above the beam. A sign of weight $W_s=300\,\mathrm{N}$ hangs from the beam at a point $x_s=3.0\,\mathrm{m}$ from the hinge. Let $T$ denote the cable tension. Let the hinge force on the beam be $\vec{H}=H_x\hat{i}+H_y\hat{j}$, where $\hat{i}$ points horizontally to the right and $\hat{j}$ points vertically upward. Choose counterclockwise torque as positive.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the cable tension $T$,
|
||||
\item the horizontal component $H_x$, and
|
||||
\item the vertical component $H_y$ of the hinge force.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Draw the beam's free-body diagram. The external forces on the beam are the hinge force $\vec{H}$ at the left end, the cable force $\vec{T}$ at the right end, the beam's weight $\vec{W}_b$ downward at its center, and the sign's weight $\vec{W}_s$ downward at $x_s=3.0\,\mathrm{m}$.
|
||||
|
||||
Choose the hinge as the pivot. Then the unknown hinge force produces zero torque about that point, which is why this pivot choice is efficient.
|
||||
|
||||
The beam's center is at
|
||||
\[
|
||||
\frac{L}{2}=2.0\,\mathrm{m}
|
||||
\]
|
||||
from the hinge. The horizontal component of the cable force acts along the beam, so its line of action passes through the hinge and it produces zero torque about the hinge. Thus only the cable's vertical component contributes to the torque balance:
|
||||
\[
|
||||
\sum \tau_{\mathrm{hinge}}=0.
|
||||
\]
|
||||
Using counterclockwise as positive,
|
||||
\[
|
||||
(T\sin\theta)L-W_b\left(\frac{L}{2}\right)-W_sx_s=0.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
(T\sin 30^\circ)(4.0\,\mathrm{m})-(200\,\mathrm{N})(2.0\,\mathrm{m})-(300\,\mathrm{N})(3.0\,\mathrm{m})=0.
|
||||
\]
|
||||
Since $\sin 30^\circ=0.50$,
|
||||
\[
|
||||
(0.50T)(4.0)-400-900=0.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
2.0T-1300=0,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
T=650\,\mathrm{N}.
|
||||
\]
|
||||
|
||||
Now apply force balance in the horizontal direction:
|
||||
\[
|
||||
\sum F_x=0.
|
||||
\]
|
||||
The cable pulls the beam toward the wall, so its horizontal component is to the left. Therefore,
|
||||
\[
|
||||
H_x-T\cos 30^\circ=0.
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
H_x=T\cos 30^\circ=(650\,\mathrm{N})(0.866)\approx 5.63\times 10^2\,\mathrm{N}.
|
||||
\]
|
||||
So the horizontal hinge-force component is
|
||||
\[
|
||||
H_x\approx 563\,\mathrm{N}
|
||||
\]
|
||||
to the right.
|
||||
|
||||
Now apply force balance in the vertical direction:
|
||||
\[
|
||||
\sum F_y=0.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
H_y+T\sin 30^\circ-W_b-W_s=0.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
H_y+(650\,\mathrm{N})(0.50)-200\,\mathrm{N}-300\,\mathrm{N}=0.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
H_y+325-500=0,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
H_y=175\,\mathrm{N}.
|
||||
\]
|
||||
Thus the vertical hinge-force component is upward.
|
||||
|
||||
The final answers are
|
||||
\[
|
||||
T=650\,\mathrm{N},
|
||||
\qquad
|
||||
H_x\approx 563\,\mathrm{N}\text{ to the right},
|
||||
\qquad
|
||||
H_y=175\,\mathrm{N}\text{ upward}.
|
||||
\]
|
||||
This is the standard statics strategy: pair $\sum \vec{F}=\vec{0}$ with $\sum \tau=0$, choose a pivot that eliminates unknown torque contributions, solve the torque equation first, and then use force balance to determine the support forces.
|
||||
145
concepts/mechanics/u5/m5-6-rotation-dynamics.tex
Normal file
145
concepts/mechanics/u5/m5-6-rotation-dynamics.tex
Normal file
@@ -0,0 +1,145 @@
|
||||
\subsection{Newton's Second Law for Rotation}
|
||||
|
||||
This subsection gives the fixed-axis rotational analog of $\sum F=ma$. In AP mechanics, the central idea is that the net external torque about a chosen axis determines the angular acceleration, with the moment of inertia setting how strongly the body resists that change in rotation.
|
||||
|
||||
\dfn{Net torque about a fixed axis and the role of rotational inertia}{Consider a rigid body that can rotate about a fixed axis with unit vector $\hat{k}$. Choose a positive sense of rotation by the right-hand rule. For an external force $\vec{F}_i$ applied at position $\vec{r}_i$ relative to a point on the axis, the signed scalar torque about the axis is
|
||||
\[
|
||||
\tau_i=(\vec{r}_i\times \vec{F}_i)\cdot \hat{k}.
|
||||
\]
|
||||
The net torque about the axis is the sum of the signed torques:
|
||||
\[
|
||||
\sum \tau = \sum_i \tau_i.
|
||||
\]
|
||||
|
||||
Let $I$ denote the moment of inertia of the body about that same axis, and let $\alpha$ denote the signed angular acceleration. The quantity $I$ measures the rotational inertia of the body: for the same net torque, a larger $I$ gives a smaller $\alpha$. Thus $I$ plays the rotational role that mass plays in translational motion.}
|
||||
|
||||
\thm{Newton's second law for fixed-axis rotation}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the moment of inertia about that axis, let $\alpha$ denote the signed angular acceleration, and let $\sum \tau$ denote the net external torque about the axis using the declared sign convention. Then
|
||||
\[
|
||||
\sum \tau = I\alpha.
|
||||
\]
|
||||
Equivalently, if $\vec{\tau}_{\mathrm{net}}$ and $\vec{\alpha}$ both lie along the fixed axis, then
|
||||
\[
|
||||
\vec{\tau}_{\mathrm{net}}=I\vec{\alpha}.
|
||||
\]
|
||||
In AP fixed-axis problems, the signed scalar form is usually the most convenient.}
|
||||
|
||||
\pf{Short derivation from point-mass contributions}{Model the rigid body as many particles. Let particle $i$ have mass $m_i$ and perpendicular distance $r_{\perp,i}$ from the axis. Because the body is rigid, every particle has the same angular acceleration $\alpha$, so the tangential acceleration of particle $i$ is
|
||||
\[
|
||||
a_{t,i}=r_{\perp,i}\alpha.
|
||||
\]
|
||||
Its tangential force component therefore satisfies
|
||||
\[
|
||||
F_{t,i}=m_i a_{t,i}=m_i r_{\perp,i}\alpha.
|
||||
\]
|
||||
Only this tangential component contributes to the torque about the axis, so the signed torque from particle $i$ is
|
||||
\[
|
||||
\tau_i=r_{\perp,i}F_{t,i}=m_i r_{\perp,i}^2\alpha.
|
||||
\]
|
||||
Summing over all particles gives
|
||||
\[
|
||||
\sum \tau = \sum_i m_i r_{\perp,i}^2\alpha = \left(\sum_i m_i r_{\perp,i}^2\right)\alpha.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
I=\sum_i m_i r_{\perp,i}^2,
|
||||
\]
|
||||
it follows that
|
||||
\[
|
||||
\sum \tau=I\alpha.
|
||||
\]
|
||||
The same conclusion holds for a continuous rigid body by replacing the sum with an integral.}
|
||||
|
||||
\cor{Constant net torque gives constant angular acceleration}{If a rigid body rotates about a fixed axis with constant moment of inertia $I$ and constant net external torque $\sum \tau$, then
|
||||
\[
|
||||
\alpha=\frac{\sum \tau}{I}
|
||||
\]
|
||||
is constant. Let $\omega_0$ denote the angular velocity and let $\theta_0$ denote the angular position at $t=0$. Then the constant-angular-acceleration kinematic equations apply:
|
||||
\[
|
||||
\omega=\omega_0+\alpha t,
|
||||
\qquad
|
||||
\theta=\theta_0+\omega_0 t+\tfrac12 \alpha t^2.
|
||||
\]
|
||||
So under a constant net torque, the angular velocity changes linearly in time and the angular position changes quadratically in time.}
|
||||
|
||||
\qs{Worked example}{A wheel rotates about a fixed axle. Choose counterclockwise rotation as positive. The wheel has radius $R=0.25\,\mathrm{m}$ and moment of inertia $I=0.50\,\mathrm{kg\cdot m^2}$ about the axle. A tangential force of magnitude $F_1=18\,\mathrm{N}$ is applied at the rim in the counterclockwise direction. At the same time, a second tangential force of magnitude $F_2=10\,\mathrm{N}$ is applied at the rim in the clockwise direction. In addition, friction exerts a constant torque of magnitude $\tau_f=0.50\,\mathrm{N\cdot m}$ in the clockwise direction. The wheel starts from rest at $t=0$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the net torque on the wheel,
|
||||
\item the angular acceleration $\alpha$, and
|
||||
\item the angular speed and angular displacement after $4.0\,\mathrm{s}$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Use the declared sign convention: counterclockwise torques are positive and clockwise torques are negative.
|
||||
|
||||
The torque due to the first force is
|
||||
\[
|
||||
\tau_1=+RF_1=(0.25\,\mathrm{m})(18\,\mathrm{N})=+4.5\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
The torque due to the second force is
|
||||
\[
|
||||
\tau_2=-RF_2=(0.25\,\mathrm{m})(10\,\mathrm{N})=-2.5\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
The friction torque is clockwise, so
|
||||
\[
|
||||
\tau_f=-0.50\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
|
||||
For part (a), the net torque is
|
||||
\[
|
||||
\sum \tau = \tau_1+\tau_2+\tau_f.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
\sum \tau = 4.5-2.5-0.50=1.5\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
So the net torque is
|
||||
\[
|
||||
\sum \tau = +1.5\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
The positive sign means the wheel accelerates counterclockwise.
|
||||
|
||||
For part (b), apply Newton's second law for rotation:
|
||||
\[
|
||||
\sum \tau = I\alpha.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\alpha=\frac{\sum \tau}{I}=\frac{1.5\,\mathrm{N\cdot m}}{0.50\,\mathrm{kg\cdot m^2}}=3.0\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
So the angular acceleration is
|
||||
\[
|
||||
\alpha=+3.0\,\mathrm{rad/s^2}.
|
||||
\]
|
||||
|
||||
For part (c), the net torque is constant, so the angular acceleration is constant. Since the wheel starts from rest, $\omega_0=0$ and take $\theta_0=0$.
|
||||
|
||||
First find the angular speed after $t=4.0\,\mathrm{s}$:
|
||||
\[
|
||||
\omega=\omega_0+\alpha t=0+(3.0\,\mathrm{rad/s^2})(4.0\,\mathrm{s})=12\,\mathrm{rad/s}.
|
||||
\]
|
||||
|
||||
Now find the angular displacement:
|
||||
\[
|
||||
\theta=\theta_0+\omega_0 t+\tfrac12 \alpha t^2.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
\theta=0+0+\tfrac12(3.0\,\mathrm{rad/s^2})(4.0\,\mathrm{s})^2.
|
||||
\]
|
||||
Since $(4.0)^2=16$,
|
||||
\[
|
||||
\theta=\tfrac12(3.0)(16)=24\,\mathrm{rad}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
\sum \tau=+1.5\,\mathrm{N\cdot m},
|
||||
\qquad
|
||||
\alpha=+3.0\,\mathrm{rad/s^2},
|
||||
\qquad
|
||||
\omega=12\,\mathrm{rad/s},
|
||||
\qquad
|
||||
\theta=24\,\mathrm{rad}.
|
||||
\]
|
||||
This example shows the rotational analog of $\sum F=ma$: once the net torque and the moment of inertia are known, the angular acceleration follows directly.
|
||||
0
concepts/mechanics/u6/.gitkeep
Normal file
0
concepts/mechanics/u6/.gitkeep
Normal file
152
concepts/mechanics/u6/m6-1-rotational-energy.tex
Normal file
152
concepts/mechanics/u6/m6-1-rotational-energy.tex
Normal file
@@ -0,0 +1,152 @@
|
||||
\subsection{Rotational Kinetic Energy and Work by Torque}
|
||||
|
||||
This subsection introduces the energy description of fixed-axis rotation. In AP mechanics, the key idea is that a net torque acting through an angular displacement changes a rigid body's rotational kinetic energy in the same way that a net force acting through a displacement changes translational kinetic energy.
|
||||
|
||||
\dfn{Rotational kinetic energy and incremental work by torque}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the body's moment of inertia about that axis, let $\vec{\omega}=\omega\hat{k}$ denote its angular velocity, and let $\omega=|\vec{\omega}|$ denote the angular speed. The \emph{rotational kinetic energy} of the rigid body is
|
||||
\[
|
||||
K_{\mathrm{rot}}=\tfrac12 I\omega^2.
|
||||
\]
|
||||
|
||||
Now let the body undergo an infinitesimal angular displacement $d\theta$ about the same axis, with the positive direction chosen consistently with $\hat{k}$. Let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis. Then the incremental work done by the net torque is
|
||||
\[
|
||||
dW=\tau_{\mathrm{net}}\,d\theta.
|
||||
\]
|
||||
If $\tau_{\mathrm{net}}$ is constant over a finite angular displacement $\Delta\theta$, then
|
||||
\[
|
||||
W=\tau_{\mathrm{net}}\Delta\theta.
|
||||
\]
|
||||
The SI unit of both rotational kinetic energy and work is the joule, where $1\,\mathrm{J}=1\,\mathrm{N\cdot m}$.}
|
||||
|
||||
\thm{Rotational work-energy relation for fixed-axis motion}{Consider a rigid body rotating about a fixed axis with moment of inertia $I$. Let $\omega_i$ and $\omega_f$ denote the initial and final angular speeds, let $\theta_i$ and $\theta_f$ denote the corresponding angular positions, and let $\tau_{\mathrm{net}}$ denote the signed net external torque about the axis. Then the net rotational work from the initial state to the final state is
|
||||
\[
|
||||
W_{\mathrm{net}}=\int_{\theta_i}^{\theta_f} \tau_{\mathrm{net}}\,d\theta,
|
||||
\]
|
||||
and the rotational work-energy theorem is
|
||||
\[
|
||||
W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}=K_{\mathrm{rot},f}-K_{\mathrm{rot},i}
|
||||
=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
|
||||
\]
|
||||
Thus a positive net torque doing positive work increases rotational kinetic energy, while negative net work decreases it.}
|
||||
|
||||
\nt{This result is the rotational analog of the translational work-energy theorem $W_{\mathrm{net}}=\Delta K$ with $K=\tfrac12 mv^2$. The correspondence is
|
||||
\[
|
||||
\vec{F}_{\mathrm{net}}\leftrightarrow \vec{\tau}_{\mathrm{net}},
|
||||
\qquad
|
||||
x\leftrightarrow \theta,
|
||||
\qquad
|
||||
v\leftrightarrow \omega,
|
||||
\qquad
|
||||
\tfrac12 mv^2\leftrightarrow \tfrac12 I\omega^2.
|
||||
\]
|
||||
In this subsection, assume a rigid body rotating about one fixed axis so that every part of the body shares the same angular displacement $\theta$ and angular speed $\omega$, and so that $I$ stays constant about that axis. The sign convention for $\tau_{\mathrm{net}}$ must match the sign convention for $\theta$.}
|
||||
|
||||
\pf{Why $K_{\mathrm{rot}}=\tfrac12 I\omega^2$ and $W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}$}{Model the rigid body as particles labeled by an index $i$. Let particle $i$ have mass $m_i$ and perpendicular distance $r_{\perp,i}$ from the fixed axis. Because the body is rigid, each particle has speed
|
||||
\[
|
||||
v_i=r_{\perp,i}\omega.
|
||||
\]
|
||||
Therefore the total kinetic energy is
|
||||
\[
|
||||
K_{\mathrm{rot}}=\sum_i \tfrac12 m_i v_i^2
|
||||
=\sum_i \tfrac12 m_i(r_{\perp,i}\omega)^2
|
||||
=\tfrac12 \omega^2\sum_i m_i r_{\perp,i}^2.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
I=\sum_i m_i r_{\perp,i}^2,
|
||||
\]
|
||||
it follows that
|
||||
\[
|
||||
K_{\mathrm{rot}}=\tfrac12 I\omega^2.
|
||||
\]
|
||||
|
||||
For the work-energy relation, start with the fixed-axis rotational form of Newton's second law,
|
||||
\[
|
||||
\tau_{\mathrm{net}}=I\alpha,
|
||||
\]
|
||||
where $\alpha=d\omega/dt$. Multiply both sides by $d\theta$:
|
||||
\[
|
||||
\tau_{\mathrm{net}}\,d\theta=I\alpha\,d\theta.
|
||||
\]
|
||||
Using $\omega=d\theta/dt$, we have $d\theta=\omega\,dt$, so
|
||||
\[
|
||||
I\alpha\,d\theta=I\frac{d\omega}{dt}(\omega\,dt)=I\omega\,d\omega.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
dW=\tau_{\mathrm{net}}\,d\theta=I\omega\,d\omega=d\!\left(\tfrac12 I\omega^2\right)=dK_{\mathrm{rot}}.
|
||||
\]
|
||||
Integrating from the initial state to the final state gives
|
||||
\[
|
||||
W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}.
|
||||
\]}
|
||||
|
||||
\qs{Worked example}{A wheel of radius $R=0.25\,\mathrm{m}$ rotates about a frictionless fixed axle. Its moment of inertia about the axle is $I=1.0\,\mathrm{kg\cdot m^2}$. A light string is wrapped around the rim, and a student pulls tangentially on the string with constant force magnitude $F=12\,\mathrm{N}$. The string does not slip, and a length $s=2.0\,\mathrm{m}$ of string unwinds. Initially the wheel is already spinning in the same direction as the applied torque with angular speed $\omega_i=4.0\,\mathrm{rad/s}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the torque magnitude $\tau$ applied by the string,
|
||||
\item the angular displacement $\Delta\theta$ during the pull,
|
||||
\item the work done on the wheel by the torque, and
|
||||
\item the final angular speed $\omega_f$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Because the pull is tangential to the rim, the lever arm equals the radius $R$. Therefore the applied torque magnitude is
|
||||
\[
|
||||
\tau=RF=(0.25\,\mathrm{m})(12\,\mathrm{N})=3.0\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
|
||||
Since the string does not slip, the unwound length equals the arc length at the rim:
|
||||
\[
|
||||
s=R\Delta\theta.
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
\Delta\theta=\frac{s}{R}=\frac{2.0\,\mathrm{m}}{0.25\,\mathrm{m}}=8.0\,\mathrm{rad}.
|
||||
\]
|
||||
|
||||
The torque is constant and acts in the direction of rotation, so the work done on the wheel is
|
||||
\[
|
||||
W=\tau\Delta\theta=(3.0\,\mathrm{N\cdot m})(8.0\,\mathrm{rad})=24\,\mathrm{J}.
|
||||
\]
|
||||
This also agrees with $W=Fs=(12\,\mathrm{N})(2.0\,\mathrm{m})=24\,\mathrm{J}$.
|
||||
|
||||
Now apply the rotational work-energy theorem:
|
||||
\[
|
||||
W=\Delta K_{\mathrm{rot}}=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
|
||||
\]
|
||||
Substitute the known values:
|
||||
\[
|
||||
24=\tfrac12 (1.0)\omega_f^2-\tfrac12 (1.0)(4.0)^2.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
\tfrac12 (1.0)(4.0)^2=8,
|
||||
\]
|
||||
we get
|
||||
\[
|
||||
24=\tfrac12 \omega_f^2-8.
|
||||
\]
|
||||
Add $8$ to both sides:
|
||||
\[
|
||||
32=\tfrac12 \omega_f^2.
|
||||
\]
|
||||
Multiply by $2$:
|
||||
\[
|
||||
\omega_f^2=64.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\omega_f=8.0\,\mathrm{rad/s}.
|
||||
\]
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\tau=3.0\,\mathrm{N\cdot m},
|
||||
\qquad
|
||||
\Delta\theta=8.0\,\mathrm{rad},
|
||||
\qquad
|
||||
W=24\,\mathrm{J},
|
||||
\qquad
|
||||
\omega_f=8.0\,\mathrm{rad/s}.
|
||||
\]
|
||||
The key idea is that the applied torque does positive work, so the wheel's rotational kinetic energy increases.
|
||||
133
concepts/mechanics/u6/m6-2-angular-momentum.tex
Normal file
133
concepts/mechanics/u6/m6-2-angular-momentum.tex
Normal file
@@ -0,0 +1,133 @@
|
||||
\subsection{Angular Momentum and Angular Impulse}
|
||||
|
||||
This subsection connects torque to rotational motion in the same way that linear impulse connects force to linear momentum. The key AP idea is that a net external torque acting for a time interval changes angular momentum about a chosen point.
|
||||
|
||||
\dfn{Angular momentum and angular impulse}{Let $O$ denote a chosen reference point. Let a particle of mass $m$ have position vector $\vec{r}$ measured from $O$, velocity $\vec{v}$, and linear momentum $\vec{p}=m\vec{v}$. The angular momentum of the particle about $O$ is
|
||||
\[
|
||||
\vec{L}_O=\vec{r}\times \vec{p}.
|
||||
\]
|
||||
For a system of particles, the total angular momentum about $O$ is
|
||||
\[
|
||||
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
|
||||
\]
|
||||
Let $\vec{\tau}_{\mathrm{ext},O}(t)$ denote the net external torque about the same point $O$ over a time interval from $t_i$ to $t_f$. The angular impulse about $O$ over that interval is
|
||||
\[
|
||||
\vec{J}_{\tau,O}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
|
||||
\]
|
||||
Its SI unit is $\mathrm{N\cdot m\cdot s}$, which is equivalent to $\mathrm{kg\cdot m^2/s}$. For a rigid body rotating about a fixed axis with moment of inertia $I$ about that axis and angular velocity $\vec{\omega}$, the angular momentum simplifies to
|
||||
\[
|
||||
\vec{L}=I\vec{\omega}.
|
||||
\]
|
||||
}
|
||||
|
||||
\thm{Torque-angular momentum relation and angular impulse theorem}{Let $O$ denote a chosen reference point. Let $\vec{L}_O(t)$ be the total angular momentum of a particle or system about $O$, and let $\vec{\tau}_{\mathrm{ext},O}(t)$ be the net external torque about $O$. Then
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
|
||||
\]
|
||||
Integrating from $t_i$ to $t_f$ gives
|
||||
\[
|
||||
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{J}_{\tau,O}.
|
||||
\]
|
||||
For fixed-axis rotation along a chosen axis with unit vector $\hat{k}$, let $\vec{L}=L\hat{k}$ and let $\vec{\tau}_{\mathrm{ext}}=\tau_{\mathrm{ext}}\hat{k}$. Then
|
||||
\[
|
||||
\frac{dL}{dt}=\tau_{\mathrm{ext}},
|
||||
\qquad
|
||||
\Delta L=\int_{t_i}^{t_f} \tau_{\mathrm{ext}}\,dt,
|
||||
\]
|
||||
and if $\tau_{\mathrm{ext}}$ is constant,
|
||||
\[
|
||||
\Delta L=\tau_{\mathrm{ext}}\Delta t.
|
||||
\]
|
||||
}
|
||||
|
||||
\nt{Angular momentum and torque depend on the chosen point $O$. The same object can have different $\vec{L}_O$ and $\vec{\tau}_{\mathrm{ext},O}$ when a different origin is chosen, so use the same reference point consistently throughout a problem. The direction of $\vec{L}$ and $\vec{\tau}$ is set by the right-hand rule and is perpendicular to the plane of the relevant cross product. In many AP fixed-axis problems, this vector bookkeeping reduces to signed scalars along one axis: counterclockwise may be chosen positive and clockwise negative. The simplification $\vec{L}=I\vec{\omega}$ is valid for rigid rotation about that fixed axis with the stated moment of inertia about the same axis.}
|
||||
|
||||
\pf{Short derivation from $\vec{L}=\vec{r}\times \vec{p}$}{For one particle about point $O$,
|
||||
\[
|
||||
\vec{L}_O=\vec{r}\times \vec{p}.
|
||||
\]
|
||||
Differentiate with respect to time:
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r}\times \frac{d\vec{p}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times \vec{F}_{\mathrm{net}}.
|
||||
\]
|
||||
Because $\vec{v}\times \vec{v}=\vec{0}$,
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{r}\times \vec{F}_{\mathrm{net}}=\vec{\tau}_{\mathrm{net},O}.
|
||||
\]
|
||||
For a system, summing over all particles gives the same form with the net external torque:
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
|
||||
\]
|
||||
Integrating from $t_i$ to $t_f$ yields
|
||||
\[
|
||||
\Delta \vec{L}_O=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
|
||||
\]
|
||||
For a rigid body rotating about a fixed axis with constant moment of inertia $I$,
|
||||
\[
|
||||
\vec{L}=I\vec{\omega},
|
||||
\]
|
||||
so along that axis the relation becomes
|
||||
\[
|
||||
\vec{\tau}_{\mathrm{ext}}=I\vec{\alpha}.
|
||||
\]
|
||||
}
|
||||
|
||||
\qs{Worked example}{A flywheel rotates about a frictionless fixed axle through its center. Choose counterclockwise as positive, so the axis direction is $\hat{k}$ by the right-hand rule. The flywheel has moment of inertia $I=0.80\,\mathrm{kg\cdot m^2}$ about the axle. Initially its angular velocity is
|
||||
\[
|
||||
\vec{\omega}_i=12\,\hat{k}\,\mathrm{rad/s}.
|
||||
\]
|
||||
From $t=0$ to $t=1.5\,\mathrm{s}$, a brake pad exerts a constant external torque
|
||||
\[
|
||||
\vec{\tau}_{\mathrm{ext}}=-3.2\,\hat{k}\,\mathrm{N\cdot m}.
|
||||
\]
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the angular impulse delivered by the brake,
|
||||
\item the flywheel's final angular momentum,
|
||||
\item the flywheel's final angular velocity, and
|
||||
\item whether the flywheel reverses direction during the interval.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol First compute the initial angular momentum using $\vec{L}=I\vec{\omega}$:
|
||||
\[
|
||||
\vec{L}_i=I\vec{\omega}_i=(0.80\,\mathrm{kg\cdot m^2})(12\,\hat{k}\,\mathrm{rad/s})=9.6\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
|
||||
\]
|
||||
|
||||
For part (a), the torque is constant, so the angular impulse is
|
||||
\[
|
||||
\vec{J}_\tau=\vec{\tau}_{\mathrm{ext}}\Delta t.
|
||||
\]
|
||||
Thus
|
||||
\[
|
||||
\vec{J}_\tau=(-3.2\,\hat{k}\,\mathrm{N\cdot m})(1.5\,\mathrm{s})=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s}.
|
||||
\]
|
||||
Using $1\,\mathrm{N\cdot m\cdot s}=1\,\mathrm{kg\cdot m^2/s}$,
|
||||
\[
|
||||
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
|
||||
\]
|
||||
|
||||
For part (b), apply the angular impulse theorem:
|
||||
\[
|
||||
\Delta \vec{L}=\vec{J}_\tau=\vec{L}_f-\vec{L}_i.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
\vec{L}_f=\vec{L}_i+\vec{J}_\tau=(9.6-4.8)\,\hat{k}\,\mathrm{kg\cdot m^2/s}=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
|
||||
\]
|
||||
|
||||
For part (c), again use $\vec{L}=I\vec{\omega}$:
|
||||
\[
|
||||
\vec{\omega}_f=\frac{\vec{L}_f}{I}=\frac{4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}}{0.80\,\mathrm{kg\cdot m^2}}=6.0\,\hat{k}\,\mathrm{rad/s}.
|
||||
\]
|
||||
|
||||
For part (d), the final angular velocity still points in the $+\hat{k}$ direction, so the flywheel is still rotating counterclockwise. It slows down, but it does not reverse direction during the $1.5\,\mathrm{s}$ interval.
|
||||
|
||||
The results are
|
||||
\[
|
||||
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s},
|
||||
\qquad
|
||||
\vec{L}_f=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s},
|
||||
\qquad
|
||||
\vec{\omega}_f=6.0\,\hat{k}\,\mathrm{rad/s}.
|
||||
\]
|
||||
This example shows the key AP idea: a torque acting over time changes angular momentum directly, and the sign of the torque determines whether the wheel speeds up or slows down.
|
||||
121
concepts/mechanics/u6/m6-3-angular-momentum-conservation.tex
Normal file
121
concepts/mechanics/u6/m6-3-angular-momentum-conservation.tex
Normal file
@@ -0,0 +1,121 @@
|
||||
\subsection{Conservation of Angular Momentum}
|
||||
|
||||
This subsection gives the rotational conservation law that parallels conservation of linear momentum. In AP mechanics, the key question is whether the net \emph{external} torque about a chosen origin or axis is zero over the interval of interest.
|
||||
|
||||
\dfn{Angular-momentum-isolated system}{Consider a system of particles labeled by an index $i=1,2,\dots,N$. Choose an origin $O$. Let $\vec{r}_i$ denote the position vector of particle $i$ relative to $O$, let $m_i$ denote its mass, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its linear momentum. The total angular momentum of the system about $O$ is
|
||||
\[
|
||||
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
|
||||
\]
|
||||
Let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about the same origin. The system is said to be \emph{isolated for angular momentum about $O$} over a time interval if the external torque impulse about $O$ is zero:
|
||||
\[
|
||||
\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{0}.
|
||||
\]
|
||||
In particular, if $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ at every instant in the interval, then the system is angular-momentum-isolated about $O$. For a rigid body rotating about a fixed axis with unit vector $\hat{k}$, one often writes $\vec{L}=L\hat{k}$ and $\vec{\omega}=\omega\hat{k}$, so that in the fixed-axis case $L=I\omega$.}
|
||||
|
||||
\thm{Conservation law for angular momentum}{Let $\vec{L}_O$ denote the total angular momentum of a system about a chosen origin $O$, and let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about that same origin. Then
|
||||
\[
|
||||
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
|
||||
\]
|
||||
Therefore, if the net external torque is zero throughout the interval, or more generally if the external torque impulse is zero, then
|
||||
\[
|
||||
\vec{L}_{O,f}=\vec{L}_{O,i},
|
||||
\]
|
||||
so total angular momentum about $O$ is conserved. In the common fixed-axis AP case,
|
||||
\[
|
||||
L_i=L_f
|
||||
\qquad\Rightarrow\qquad
|
||||
I_i\omega_i=I_f\omega_f
|
||||
\]
|
||||
when the net external torque about that axis is zero.}
|
||||
|
||||
\nt{Angular momentum is conserved only about the origin or axis for which the net external torque is zero, so the choice of origin matters. Internal forces and internal torques can redistribute angular momentum among parts of the system and can change the moment of inertia $I$, but they do not change the system's \emph{total} angular momentum about the chosen origin when $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$. Also, conservation of angular momentum does \emph{not} imply conservation of kinetic energy: for example, a skater can pull in her arms, decrease $I$, increase $\omega$, and increase rotational kinetic energy by doing internal work.}
|
||||
|
||||
\pf{Short derivation from $d\vec{L}/dt=\vec{\tau}_{\mathrm{ext}}$}{Start with the angular-momentum form of Newton's second law about the chosen origin $O$:
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
|
||||
\]
|
||||
Integrate from $t_i$ to $t_f$:
|
||||
\[
|
||||
\int_{t_i}^{t_f} \frac{d\vec{L}_O}{dt}\,dt=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
|
||||
\]
|
||||
This gives
|
||||
\[
|
||||
\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
|
||||
\]
|
||||
If $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ throughout the interval, or if the integral on the right is zero, then $\vec{L}_{O,f}-\vec{L}_{O,i}=\vec{0}$. Hence
|
||||
\[
|
||||
\vec{L}_{O,f}=\vec{L}_{O,i},
|
||||
\]
|
||||
which is the conservation of angular momentum. In a fixed-axis problem, this reduces to $I_i\omega_i=I_f\omega_f$.}
|
||||
|
||||
\qs{Worked example}{An ice skater spins about a vertical axis with unit vector $\hat{k}$. Assume the net external torque about that axis is negligible. With her arms extended, her moment of inertia is $I_i=3.0\,\mathrm{kg\cdot m^2}$ and her angular velocity is $\vec{\omega}_i=(2.0\,\mathrm{rad/s})\hat{k}$. She then pulls her arms inward so that her final moment of inertia is $I_f=1.2\,\mathrm{kg\cdot m^2}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the final angular velocity vector $\vec{\omega}_f$,
|
||||
\item the initial and final angular momentum vectors, and
|
||||
\item the initial and final rotational kinetic energies.
|
||||
\end{enumerate}
|
||||
|
||||
Explain briefly why the energy result does not contradict conservation of angular momentum.}
|
||||
|
||||
\sol Because the net external torque about the vertical axis is negligible, angular momentum about that axis is conserved:
|
||||
\[
|
||||
\vec{L}_i=\vec{L}_f.
|
||||
\]
|
||||
For fixed-axis rotation, $\vec{L}=I\vec{\omega}$, so
|
||||
\[
|
||||
I_i\vec{\omega}_i=I_f\vec{\omega}_f.
|
||||
\]
|
||||
|
||||
For part (a), solve for the final angular velocity:
|
||||
\[
|
||||
\vec{\omega}_f=\frac{I_i}{I_f}\vec{\omega}_i
|
||||
=\frac{3.0}{1.2}(2.0\hat{k})
|
||||
=(2.5)(2.0\hat{k})
|
||||
=5.0\hat{k}\,\mathrm{rad/s}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\boxed{\vec{\omega}_f=(5.0\,\mathrm{rad/s})\hat{k}}.
|
||||
\]
|
||||
|
||||
For part (b), compute the angular momentum before and after:
|
||||
\[
|
||||
\vec{L}_i=I_i\vec{\omega}_i=(3.0)(2.0\hat{k})=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
|
||||
\]
|
||||
Because angular momentum is conserved,
|
||||
\[
|
||||
\vec{L}_f=\vec{L}_i=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\boxed{\vec{L}_i=\vec{L}_f=(6.0\,\mathrm{kg\cdot m^2/s})\hat{k}}.
|
||||
\]
|
||||
|
||||
For part (c), use $K_{\mathrm{rot}}=\tfrac12 I\omega^2$.
|
||||
|
||||
Initially,
|
||||
\[
|
||||
K_{\mathrm{rot},i}=\tfrac12 I_i\omega_i^2
|
||||
=\tfrac12 (3.0)(2.0)^2
|
||||
=\tfrac12 (3.0)(4.0)
|
||||
=6.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Finally,
|
||||
\[
|
||||
K_{\mathrm{rot},f}=\tfrac12 I_f\omega_f^2
|
||||
=\tfrac12 (1.2)(5.0)^2
|
||||
=\tfrac12 (1.2)(25)
|
||||
=15\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
So,
|
||||
\[
|
||||
\boxed{K_{\mathrm{rot},i}=6.0\,\mathrm{J}},
|
||||
\qquad
|
||||
\boxed{K_{\mathrm{rot},f}=15\,\mathrm{J}}.
|
||||
\]
|
||||
|
||||
The rotational kinetic energy increases even though angular momentum stays constant. This does not contradict conservation of angular momentum because the skater does internal work while pulling in her arms. That internal work increases $K_{\mathrm{rot}}$ while the external torque remains negligible, so $\vec{L}$ is still conserved.
|
||||
191
concepts/mechanics/u6/m6-4-rolling.tex
Normal file
191
concepts/mechanics/u6/m6-4-rolling.tex
Normal file
@@ -0,0 +1,191 @@
|
||||
\subsection{Rolling Without Slipping}
|
||||
|
||||
This subsection introduces pure rolling, where a rigid body both translates and rotates while the contact point does not slide relative to the surface.
|
||||
|
||||
\dfn{Pure rolling and the rolling constraint}{Consider a rigid body of mass $M$ and radius $R$ moving on a fixed surface. Let $s_{\mathrm{cm}}$ denote the distance traveled by its center of mass along the surface, let $v_{\mathrm{cm}}=|\vec{v}_{\mathrm{cm}}|$ denote the speed of the center of mass, and let $a_{t,\mathrm{cm}}$ denote the component of $\vec{a}_{\mathrm{cm}}$ tangent to the surface. Let $\theta$ denote the angular displacement of the body, let $\omega$ denote its angular speed, and let $\alpha$ denote its angular acceleration.
|
||||
|
||||
A body \emph{rolls without slipping} if the point in contact with the surface is instantaneously at rest relative to the surface. For pure rolling,
|
||||
\[
|
||||
s_{\mathrm{cm}}=R\theta,
|
||||
\qquad
|
||||
v_{\mathrm{cm}}=R\omega,
|
||||
\qquad
|
||||
a_{t,\mathrm{cm}}=R\alpha.
|
||||
\]
|
||||
These relations are called the \emph{rolling constraint}. They apply only when there is no slipping at the contact.}
|
||||
|
||||
\nt{Let $P$ denote the point on the rim that touches the ground at some instant. Its velocity relative to the ground is the vector sum of the center-of-mass velocity $\vec{v}_{\mathrm{cm}}$ and the velocity of $P$ relative to the center due to rotation. In pure rolling, these cancel exactly at the contact point, so $P$ is instantaneously at rest even though the body is moving.
|
||||
|
||||
Also, the friction in rolling without slipping is \emph{static} friction. Let $\vec{N}$ denote the normal force, let $N=|\vec{N}|$ denote its magnitude, and let $f_s$ denote the magnitude of the static friction force. Static friction is not automatically equal to $\mu_s N$. Instead, its magnitude is whatever value is required to prevent slipping, provided that value satisfies $f_s\le \mu_s N$. On level ground at constant speed, the needed static friction can even be zero.}
|
||||
|
||||
\ex{Illustrative example}{A wheel of radius $R=0.30\,\mathrm{m}$ rolls without slipping on level ground with angular speed $\omega=8.0\,\mathrm{rad/s}$. Find the speed of its center of mass and the speed of the top point of the wheel relative to the ground.
|
||||
|
||||
From the rolling constraint,
|
||||
\[
|
||||
v_{\mathrm{cm}}=R\omega=(0.30)(8.0)=2.4\,\mathrm{m/s}.
|
||||
\]
|
||||
At the top of the wheel, the translational velocity and the rotational velocity point in the same direction, so the top-point speed is
|
||||
\[
|
||||
v_{\mathrm{top}}=v_{\mathrm{cm}}+R\omega=2v_{\mathrm{cm}}=4.8\,\mathrm{m/s}.
|
||||
\]
|
||||
Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves at $4.8\,\mathrm{m/s}$ relative to the ground.}
|
||||
|
||||
\mprop{Rolling kinematics, energy, and incline dynamics}{Consider a rigid body of mass $M$, radius $R$, and moment of inertia $I_{\mathrm{cm}}$ about its center of mass. Let $s_{\mathrm{cm}}$, $v_{\mathrm{cm}}$, and $a_{t,\mathrm{cm}}$ denote the center-of-mass distance, speed, and tangential acceleration, and let $\theta$, $\omega$, and $\alpha$ denote the corresponding angular variables. Let $K_i$ and $U_i$ denote the initial kinetic and potential energies, and let $K_f$ and $U_f$ denote the final kinetic and potential energies.
|
||||
|
||||
\begin{enumerate}[label=\textbf{\arabic*.}]
|
||||
\item For pure rolling,
|
||||
\[
|
||||
s_{\mathrm{cm}}=R\theta,
|
||||
\qquad
|
||||
v_{\mathrm{cm}}=R\omega,
|
||||
\qquad
|
||||
a_{t,\mathrm{cm}}=R\alpha.
|
||||
\]
|
||||
|
||||
\item The total kinetic energy is the sum of translational and rotational parts:
|
||||
\[
|
||||
K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
|
||||
\]
|
||||
Using $v_{\mathrm{cm}}=R\omega$, this may also be written as
|
||||
\[
|
||||
K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2.
|
||||
\]
|
||||
|
||||
\item If no nonconservative force removes mechanical energy from the system, then for rolling without slipping,
|
||||
\[
|
||||
K_i+U_i=K_f+U_f.
|
||||
\]
|
||||
For a vertical drop of magnitude $h$ from rest,
|
||||
\[
|
||||
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
|
||||
\]
|
||||
|
||||
\item For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then
|
||||
\[
|
||||
Mg\sin\beta-f_s=Ma_{\mathrm{cm}},
|
||||
\qquad
|
||||
f_sR=I_{\mathrm{cm}}\alpha,
|
||||
\qquad
|
||||
a_{\mathrm{cm}}=R\alpha.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)},
|
||||
\qquad
|
||||
f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}.
|
||||
\]
|
||||
For an object accelerating down the incline, the static friction force on the object points up the incline.}
|
||||
|
||||
\qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the acceleration magnitude of the center of mass,
|
||||
\item the magnitude and direction of the static friction force,
|
||||
\item the speed of the center of mass after the drop, and
|
||||
\item the minimum coefficient of static friction required for rolling without slipping.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Let $a_{\mathrm{cm}}$ denote the acceleration magnitude of the center of mass down the incline, and let $f_s$ denote the magnitude of the static friction force. For a solid cylinder,
|
||||
\[
|
||||
I_{\mathrm{cm}}=\tfrac12 MR^2.
|
||||
\]
|
||||
|
||||
Choose the positive axis down the incline. The forces parallel to the incline are the downslope component of the weight and the upslope static friction force, so Newton's second law for translation gives
|
||||
\[
|
||||
Mg\sin\beta-f_s=Ma_{\mathrm{cm}}.
|
||||
\]
|
||||
The only torque about the center of mass is due to static friction, so
|
||||
\[
|
||||
f_sR=I_{\mathrm{cm}}\alpha.
|
||||
\]
|
||||
Because the cylinder rolls without slipping,
|
||||
\[
|
||||
a_{\mathrm{cm}}=R\alpha.
|
||||
\]
|
||||
Substitute $\alpha=a_{\mathrm{cm}}/R$ and $I_{\mathrm{cm}}=\tfrac12 MR^2$ into the torque equation:
|
||||
\[
|
||||
f_sR=\left(\tfrac12 MR^2\right)\frac{a_{\mathrm{cm}}}{R}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
f_s=\tfrac12 Ma_{\mathrm{cm}}.
|
||||
\]
|
||||
|
||||
Now substitute this into the translational equation:
|
||||
\[
|
||||
Mg\sin\beta-\tfrac12 Ma_{\mathrm{cm}}=Ma_{\mathrm{cm}}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
Mg\sin\beta=\tfrac32 Ma_{\mathrm{cm}},
|
||||
\]
|
||||
and therefore
|
||||
\[
|
||||
a_{\mathrm{cm}}=\frac{2}{3}g\sin\beta.
|
||||
\]
|
||||
With $g=9.8\,\mathrm{m/s^2}$ and $\sin 30^\circ=0.50$,
|
||||
\[
|
||||
a_{\mathrm{cm}}=\frac{2}{3}(9.8)(0.50)=3.27\,\mathrm{m/s^2}.
|
||||
\]
|
||||
This is the acceleration magnitude of the center of mass, directed down the incline.
|
||||
|
||||
For the static friction force,
|
||||
\[
|
||||
f_s=\tfrac12 Ma_{\mathrm{cm}}=\tfrac12 (2.0)(3.27)=3.27\,\mathrm{N}.
|
||||
\]
|
||||
Its direction is up the incline, because it must provide the clockwise torque that increases the cylinder's rotation as the cylinder moves downward.
|
||||
|
||||
Now find the speed after the cylinder drops through height $h=0.75\,\mathrm{m}$. Since the cylinder rolls without slipping, static friction does no work at the contact point, so mechanical energy is conserved:
|
||||
\[
|
||||
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
|
||||
\]
|
||||
Using $I_{\mathrm{cm}}=\tfrac12 MR^2$ and $v_{\mathrm{cm}}=R\omega$,
|
||||
\[
|
||||
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12\left(\tfrac12 MR^2\right)\left(\frac{v_{\mathrm{cm}}}{R}\right)^2.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac14 Mv_{\mathrm{cm}}^2
|
||||
=\tfrac34 Mv_{\mathrm{cm}}^2.
|
||||
\]
|
||||
Cancel $M$ and solve for $v_{\mathrm{cm}}$:
|
||||
\[
|
||||
gh=\tfrac34 v_{\mathrm{cm}}^2,
|
||||
\qquad
|
||||
v_{\mathrm{cm}}^2=\frac{4}{3}gh.
|
||||
\]
|
||||
Substitute the numbers:
|
||||
\[
|
||||
v_{\mathrm{cm}}^2=\frac{4}{3}(9.8)(0.75)=9.8.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
v_{\mathrm{cm}}=\sqrt{9.8}=3.13\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Finally, find the minimum coefficient of static friction. The normal force is
|
||||
\[
|
||||
N=Mg\cos\beta=(2.0)(9.8)\cos 30^\circ=17.0\,\mathrm{N}
|
||||
\]
|
||||
to three significant figures. For rolling without slipping, the needed static friction must satisfy
|
||||
\[
|
||||
f_s\le \mu_s N.
|
||||
\]
|
||||
Thus the minimum value is
|
||||
\[
|
||||
\mu_{s,\min}=\frac{f_s}{N}=\frac{3.27}{17.0}=0.192.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
a_{\mathrm{cm}}=3.27\,\mathrm{m/s^2}\text{ down the incline},
|
||||
\qquad
|
||||
f_s=3.27\,\mathrm{N}\text{ up the incline},
|
||||
\]
|
||||
\[
|
||||
v_{\mathrm{cm}}=3.13\,\mathrm{m/s},
|
||||
\qquad
|
||||
\mu_{s,\min}=0.192.
|
||||
\]
|
||||
The key ideas are the rolling constraint $v_{\mathrm{cm}}=R\omega$, the split of kinetic energy into translational and rotational parts, and the fact that static friction adjusts to the amount needed for no slipping rather than automatically equaling $\mu_s N$.
|
||||
159
concepts/mechanics/u6/m6-5-orbits.tex
Normal file
159
concepts/mechanics/u6/m6-5-orbits.tex
Normal file
@@ -0,0 +1,159 @@
|
||||
\subsection{Circular Orbits, Satellite Speed, and Orbital Energy}
|
||||
|
||||
This subsection focuses on Newtonian circular orbits around a much more massive central body. The main AP results are the orbital-speed formula and the linked kinetic, potential, and total-energy relations for a satellite in a circular orbit.
|
||||
|
||||
\dfn{Circular-orbit setup and gravitational potential-energy reference}{Let a satellite of mass $m$ move in a circular orbit around a spherically symmetric body of mass $M$. Let $O$ denote the center of the central body. Let $\vec{r}$ denote the satellite's position vector from $O$, let $r=|\vec{r}|$ denote the constant orbital radius, let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector, let $\vec{v}$ denote the satellite's velocity, and let $v=|\vec{v}|$ denote its speed. Let $\vec{L}_O=\vec{r}\times m\vec{v}$ denote the satellite's angular momentum about $O$. Let $K$ denote kinetic energy, let $U$ denote gravitational potential energy, and let $E=K+U$ denote total mechanical energy.
|
||||
|
||||
Choose the gravitational potential-energy reference so that $U=0$ when the separation is infinite. Then for separation $r$,
|
||||
\[
|
||||
U(r)=-\frac{GMm}{r}.
|
||||
\]
|
||||
The gravitational force on the satellite is
|
||||
\[
|
||||
\vec{F}_g=-\frac{GMm}{r^2}\hat{r}.
|
||||
\]
|
||||
}
|
||||
|
||||
\thm{Circular-orbit speed and energy relations}{Let a satellite of mass $m$ move in a circular orbit of radius $r$ around a spherically symmetric body of mass $M$. Because $\vec{F}_g$ is parallel to $\vec{r}$,
|
||||
\[
|
||||
\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0},
|
||||
\]
|
||||
so the orbital angular momentum about the center is conserved.
|
||||
|
||||
For a circular orbit, gravity supplies the centripetal force:
|
||||
\[
|
||||
\frac{GMm}{r^2}=\frac{mv^2}{r}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
v=\sqrt{\frac{GM}{r}}.
|
||||
\]
|
||||
The kinetic energy is then
|
||||
\[
|
||||
K=\frac12 mv^2=\frac{GMm}{2r}.
|
||||
\]
|
||||
Using $U=-GMm/r$, the total mechanical energy is
|
||||
\[
|
||||
E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}.
|
||||
\]
|
||||
Thus, for a circular Newtonian orbit,
|
||||
\[
|
||||
v=\sqrt{\frac{GM}{r}},
|
||||
\qquad
|
||||
K=\frac{GMm}{2r},
|
||||
\qquad
|
||||
U=-\frac{GMm}{r},
|
||||
\qquad
|
||||
E=-\frac{GMm}{2r}.
|
||||
\]
|
||||
}
|
||||
|
||||
\ex{Illustrative example}{Two satellites of the same mass orbit the same planet in circular orbits. Satellite A has orbital radius $r$, and satellite B has orbital radius $4r$.
|
||||
|
||||
Since $v=\sqrt{GM/r}$,
|
||||
\[
|
||||
v_B=\sqrt{\frac{GM}{4r}}=\frac12\sqrt{\frac{GM}{r}}=\frac12 v_A.
|
||||
\]
|
||||
Since $K=GMm/(2r)$,
|
||||
\[
|
||||
K_B=\frac{GMm}{2(4r)}=\frac14 K_A.
|
||||
\]
|
||||
Also,
|
||||
\[
|
||||
U_B=-\frac{GMm}{4r}=\frac14 U_A,
|
||||
\qquad
|
||||
E_B=-\frac{GMm}{8r}=\frac14 E_A.
|
||||
\]
|
||||
So a larger circular orbit has a lower speed and a less negative total energy.}
|
||||
|
||||
\nt{With the reference choice $U(\infty)=0$, any bound gravitational orbit has negative total mechanical energy. For a circular orbit specifically, $E=-K=U/2<0$. Also, the formulas $v=\sqrt{GM/r}$, $K=GMm/(2r)$, and $E=-GMm/(2r)$ are for \emph{circular} Newtonian orbits only. In a noncircular orbit, the speed is not constant, so these same expressions do not apply at every point of the motion.}
|
||||
|
||||
\qs{Worked example}{An Earth satellite of mass $m=850\,\mathrm{kg}$ moves in a circular orbit at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let Earth's mass be $M_E=5.97\times 10^{24}\,\mathrm{kg}$, Earth's radius be $R_E=6.37\times 10^6\,\mathrm{m}$, and the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\cdot m^2/kg^2}$.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item why the satellite's angular momentum about Earth's center is conserved,
|
||||
\item the orbital speed $v$,
|
||||
\item the kinetic energy $K$,
|
||||
\item the gravitational potential energy $U$, and
|
||||
\item the total mechanical energy $E$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Let $r$ denote the orbital radius measured from Earth's center. First compute it from the given altitude:
|
||||
\[
|
||||
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For part (a), the only significant force on the satellite is Earth's gravitational force, which points along the line from Earth to the satellite. Therefore $\vec{F}_g$ is parallel to $\vec{r}$, so the torque about Earth's center is
|
||||
\[
|
||||
\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{\tau}_O,
|
||||
\]
|
||||
it follows that
|
||||
\[
|
||||
\frac{d\vec{L}_O}{dt}=\vec{0},
|
||||
\]
|
||||
so the satellite's angular momentum about Earth's center is conserved.
|
||||
|
||||
For part (b), use the circular-orbit speed formula:
|
||||
\[
|
||||
v=\sqrt{\frac{GM_E}{r}}.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
v=\sqrt{\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)}{6.77\times 10^6}}.
|
||||
\]
|
||||
This gives
|
||||
\[
|
||||
v\approx 7.67\times 10^3\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
For part (c), the kinetic energy is
|
||||
\[
|
||||
K=\frac12 mv^2.
|
||||
\]
|
||||
Using $m=850\,\mathrm{kg}$ and the value of $v^2=GM_E/r$,
|
||||
\[
|
||||
K=\frac12(850)\left(7.67\times 10^3\right)^2\,\mathrm{J}
|
||||
\approx 2.50\times 10^{10}\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
For part (d), the gravitational potential energy is
|
||||
\[
|
||||
U=-\frac{GM_E m}{r}.
|
||||
\]
|
||||
Substitute the numbers:
|
||||
\[
|
||||
U=-\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)(850)}{6.77\times 10^6}\,\mathrm{J}
|
||||
\approx -5.00\times 10^{10}\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
For part (e), the total mechanical energy is
|
||||
\[
|
||||
E=K+U.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
E=(2.50\times 10^{10})+(-5.00\times 10^{10})\,\mathrm{J}
|
||||
\approx -2.50\times 10^{10}\,\mathrm{J}.
|
||||
\]
|
||||
This agrees with the circular-orbit relation
|
||||
\[
|
||||
E=-\frac{GM_E m}{2r}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
v\approx 7.67\times 10^3\,\mathrm{m/s},
|
||||
\qquad
|
||||
K\approx 2.50\times 10^{10}\,\mathrm{J},
|
||||
\]
|
||||
\[
|
||||
U\approx -5.00\times 10^{10}\,\mathrm{J},
|
||||
\qquad
|
||||
E\approx -2.50\times 10^{10}\,\mathrm{J}.
|
||||
\]
|
||||
The negative total energy shows that the satellite is in a bound orbit.
|
||||
0
concepts/mechanics/u7/.gitkeep
Normal file
0
concepts/mechanics/u7/.gitkeep
Normal file
176
concepts/mechanics/u7/m7-1-shm.tex
Normal file
176
concepts/mechanics/u7/m7-1-shm.tex
Normal file
@@ -0,0 +1,176 @@
|
||||
\subsection{Simple Harmonic Motion and Its Governing ODE}
|
||||
|
||||
This subsection introduces simple harmonic motion as one-dimensional motion about a stable equilibrium under a linear restoring law.
|
||||
|
||||
\dfn{Simple harmonic motion and the equilibrium coordinate}{Let an object move along a line with fixed unit vector $\hat{u}$. Let
|
||||
\[
|
||||
\vec{r}(t)=q(t)\hat{u}
|
||||
\]
|
||||
denote the object's displacement from a stable equilibrium position, where $q(t)$ is the signed equilibrium coordinate. The motion is called \emph{simple harmonic motion} (SHM) if the net restoring force is proportional to the displacement and points toward equilibrium:
|
||||
\[
|
||||
\vec{F}_{\mathrm{net}}=-kq\hat{u}
|
||||
\]
|
||||
for some constant $k>0$.
|
||||
|
||||
Equivalently, in scalar form along the chosen axis,
|
||||
\[
|
||||
F_{\mathrm{net}}=-kq.
|
||||
\]
|
||||
The negative sign shows that when $q>0$ the force is negative, and when $q<0$ the force is positive, so the force always points back toward $q=0$.}
|
||||
|
||||
\thm{SHM ODE, standard solution, and period relations}{Let an object of mass $m$ move in SHM with equilibrium coordinate $q(t)$ and restoring constant $k>0$. Define
|
||||
\[
|
||||
\omega=\sqrt{\frac{k}{m}}.
|
||||
\]
|
||||
Then the governing differential equation is
|
||||
\[
|
||||
q''+\omega^2 q=0.
|
||||
\]
|
||||
Its standard solution may be written as
|
||||
\[
|
||||
q(t)=C\cos(\omega t)+D\sin(\omega t),
|
||||
\]
|
||||
where $C$ and $D$ are constants set by the initial conditions, or equivalently as
|
||||
\[
|
||||
q(t)=A\cos(\omega t+\phi)
|
||||
\]
|
||||
for amplitude $A\ge 0$ and phase constant $\phi$.
|
||||
|
||||
The period $T$ and frequency $f$ are
|
||||
\[
|
||||
T=\frac{2\pi}{\omega},
|
||||
\qquad
|
||||
f=\frac{1}{T}=\frac{\omega}{2\pi}.
|
||||
\]}
|
||||
|
||||
\pf{Short derivation from the linear restoring law}{For SHM, the net force along the line of motion is
|
||||
\[
|
||||
F_{\mathrm{net}}=-kq.
|
||||
\]
|
||||
Newton's second law gives
|
||||
\[
|
||||
m\frac{d^2q}{dt^2}=-kq.
|
||||
\]
|
||||
Divide by $m$:
|
||||
\[
|
||||
\frac{d^2q}{dt^2}+\frac{k}{m}q=0.
|
||||
\]
|
||||
If we define
|
||||
\[
|
||||
\omega^2=\frac{k}{m},
|
||||
\]
|
||||
then the equation becomes
|
||||
\[
|
||||
q''+\omega^2 q=0.
|
||||
\]
|
||||
The solutions of this constant-coefficient ODE are sinusoidal, so one may write
|
||||
\[
|
||||
q(t)=C\cos(\omega t)+D\sin(\omega t).
|
||||
\]
|
||||
Because sine and cosine repeat after an angle change of $2\pi$, one full cycle takes time
|
||||
\[
|
||||
T=\frac{2\pi}{\omega},
|
||||
\]
|
||||
and therefore $f=1/T=\omega/(2\pi)$.}
|
||||
|
||||
\ex{Illustrative example}{A particle's equilibrium coordinate satisfies
|
||||
\[
|
||||
q''+25q=0.
|
||||
\]
|
||||
Identify $\omega$, the period, and the frequency.
|
||||
|
||||
Compare this with the SHM form $q''+\omega^2 q=0$. Then
|
||||
\[
|
||||
\omega^2=25
|
||||
\qquad \Rightarrow \qquad
|
||||
\omega=5.0\,\mathrm{rad/s}.
|
||||
\]
|
||||
So the period is
|
||||
\[
|
||||
T=\frac{2\pi}{\omega}=\frac{2\pi}{5.0}\,\mathrm{s}=1.26\,\mathrm{s},
|
||||
\]
|
||||
and the frequency is
|
||||
\[
|
||||
f=\frac{1}{T}=\frac{5.0}{2\pi}\,\mathrm{Hz}=0.796\,\mathrm{Hz}.
|
||||
\]
|
||||
Thus this motion is SHM with angular frequency $5.0\,\mathrm{rad/s}$, period $1.26\,\mathrm{s}$, and frequency $0.796\,\mathrm{Hz}$.}
|
||||
|
||||
\qs{Worked example}{For one-dimensional SHM about equilibrium, let the equilibrium coordinate be
|
||||
\[
|
||||
q(t)=(0.080\,\mathrm{m})\cos\!\left(4\pi t-\tfrac{\pi}{3}\right)
|
||||
\]
|
||||
with $t$ in seconds.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the amplitude,
|
||||
\item the angular frequency,
|
||||
\item the period and frequency,
|
||||
\item the displacement at $t=0$, and
|
||||
\item the governing differential equation in the form $q''+\omega^2 q=0$.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol From
|
||||
\[
|
||||
q(t)=A\cos(\omega t+\phi),
|
||||
\]
|
||||
we identify the amplitude as the coefficient of the cosine and the angular frequency as the coefficient of $t$ inside the cosine.
|
||||
|
||||
For part (a),
|
||||
\[
|
||||
A=0.080\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For part (b),
|
||||
\[
|
||||
\omega=4\pi\,\mathrm{rad/s}.
|
||||
\]
|
||||
|
||||
For part (c), the period is
|
||||
\[
|
||||
T=\frac{2\pi}{\omega}=\frac{2\pi}{4\pi}=0.50\,\mathrm{s}.
|
||||
\]
|
||||
Therefore the frequency is
|
||||
\[
|
||||
f=\frac{1}{T}=\frac{1}{0.50\,\mathrm{s}}=2.0\,\mathrm{Hz}.
|
||||
\]
|
||||
|
||||
For part (d), substitute $t=0$ into the position function:
|
||||
\[
|
||||
q(0)=(0.080)\cos\!\left(-\frac{\pi}{3}\right)\,\mathrm{m}.
|
||||
\]
|
||||
Since $\cos(-\pi/3)=\cos(\pi/3)=1/2$,
|
||||
\[
|
||||
q(0)=(0.080)\left(\frac12\right)=0.040\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
For part (e), SHM always satisfies
|
||||
\[
|
||||
q''+\omega^2 q=0.
|
||||
\]
|
||||
Here $\omega=4\pi\,\mathrm{rad/s}$, so
|
||||
\[
|
||||
\omega^2=(4\pi)^2=16\pi^2.
|
||||
\]
|
||||
Thus the governing ODE is
|
||||
\[
|
||||
q''+16\pi^2 q=0.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
A=0.080\,\mathrm{m},
|
||||
\qquad
|
||||
\omega=4\pi\,\mathrm{rad/s},
|
||||
\]
|
||||
\[
|
||||
T=0.50\,\mathrm{s},
|
||||
\qquad
|
||||
f=2.0\,\mathrm{Hz},
|
||||
\qquad
|
||||
q(0)=0.040\,\mathrm{m},
|
||||
\]
|
||||
and the motion is governed by
|
||||
\[
|
||||
q''+16\pi^2 q=0.
|
||||
\]
|
||||
174
concepts/mechanics/u7/m7-2-spring-oscillator.tex
Normal file
174
concepts/mechanics/u7/m7-2-spring-oscillator.tex
Normal file
@@ -0,0 +1,174 @@
|
||||
\subsection{The Spring-Mass Oscillator}
|
||||
|
||||
This subsection models a mass attached to an ideal spring, using displacement measured from equilibrium so that both horizontal motion and vertical motion about equilibrium take the same mathematical form.
|
||||
|
||||
\dfn{Spring-mass oscillator and equilibrium coordinate}{Let $m$ denote the mass of an object and let $k_{\mathrm{eff}}>0$ denote the effective spring constant of the spring system attached to it. Let the motion occur along a line with positive direction given by the unit vector $\hat{u}$.
|
||||
|
||||
Define $x(t)$ to be the signed displacement of the mass from its equilibrium position, measured along that line. Then a \emph{spring-mass oscillator} is a system for which the net restoring force is proportional to $x$ and opposite its sign.
|
||||
|
||||
For a horizontal spring, $x$ is measured directly from the equilibrium position on the track. For a vertical spring, let $y(t)$ denote the displacement measured from the spring's unstretched length and let $y_{\mathrm{eq}}$ denote the static equilibrium displacement. The equilibrium coordinate is then
|
||||
\[
|
||||
x=y-y_{\mathrm{eq}}.
|
||||
\]
|
||||
Using $x$ rather than $y$ makes the vertical oscillator look exactly like the horizontal one.}
|
||||
|
||||
\thm{Governing equation and period of a spring-mass oscillator}{Let $m$ denote the mass, let $k_{\mathrm{eff}}$ denote the effective spring constant, and let $x(t)$ denote displacement from equilibrium. Then the motion satisfies
|
||||
\[
|
||||
m\ddot{x}+k_{\mathrm{eff}}x=0.
|
||||
\]
|
||||
Equivalently,
|
||||
\[
|
||||
\ddot{x}+\omega^2 x=0,
|
||||
\qquad
|
||||
\omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}}.
|
||||
\]
|
||||
Thus the motion is simple harmonic. If $T$ denotes the period and $f$ denotes the frequency, then
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{m}{k_{\mathrm{eff}}}},
|
||||
\qquad
|
||||
f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k_{\mathrm{eff}}}{m}}.
|
||||
\]
|
||||
One convenient position model is
|
||||
\[
|
||||
x(t)=A\cos(\omega t+\phi),
|
||||
\]
|
||||
where $A$ is the amplitude and $\phi$ is a phase constant. Consequently,
|
||||
\[
|
||||
v_{\max}=A\omega,
|
||||
\qquad
|
||||
a_{\max}=A\omega^2.
|
||||
\]}
|
||||
|
||||
\pf{Why the equation is the same horizontally and vertically}{For horizontal motion, let $x$ denote displacement from equilibrium along $\hat{u}$. The spring force is
|
||||
\[
|
||||
\vec{F}_s=-k_{\mathrm{eff}}x\hat{u}.
|
||||
\]
|
||||
By Newton's second law,
|
||||
\[
|
||||
m\ddot{x}=-k_{\mathrm{eff}}x,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
m\ddot{x}+k_{\mathrm{eff}}x=0.
|
||||
\]
|
||||
|
||||
For vertical motion, choose downward as positive. Let $y$ denote the downward displacement from the unstretched length, and let $y_{\mathrm{eq}}$ denote the equilibrium value. At equilibrium,
|
||||
\[
|
||||
mg-k_{\mathrm{eff}}y_{\mathrm{eq}}=0.
|
||||
\]
|
||||
Now write the actual position as $y=y_{\mathrm{eq}}+x$, where $x$ is displacement from equilibrium. Then
|
||||
\[
|
||||
m\ddot{y}=mg-k_{\mathrm{eff}}y=mg-k_{\mathrm{eff}}(y_{\mathrm{eq}}+x).
|
||||
\]
|
||||
Since $mg-k_{\mathrm{eff}}y_{\mathrm{eq}}=0$ and $\ddot{y}=\ddot{x}$, this becomes
|
||||
\[
|
||||
m\ddot{x}=-k_{\mathrm{eff}}x.
|
||||
\]
|
||||
So in either viewpoint,
|
||||
\[
|
||||
m\ddot{x}+k_{\mathrm{eff}}x=0.
|
||||
\]
|
||||
Comparing with $\ddot{x}+\omega^2x=0$ gives $\omega=\sqrt{k_{\mathrm{eff}}/m}$, and then $T=2\pi/\omega$ and $f=1/T$.}
|
||||
|
||||
\ex{Illustrative example}{A block of mass $m=0.40\,\mathrm{kg}$ oscillates on a frictionless horizontal surface attached to a spring system with effective spring constant $k_{\mathrm{eff}}=100\,\mathrm{N/m}$. Find the angular frequency, period, and frequency.
|
||||
|
||||
Use
|
||||
\[
|
||||
\omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}}=\sqrt{\frac{100}{0.40}}=\sqrt{250}=15.8\,\mathrm{rad/s}.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{m}{k_{\mathrm{eff}}}}=2\pi\sqrt{\frac{0.40}{100}}=0.397\,\mathrm{s},
|
||||
\]
|
||||
and
|
||||
\[
|
||||
f=\frac{1}{T}=\frac{1}{0.397}=2.52\,\mathrm{Hz}.
|
||||
\]
|
||||
So the oscillator has angular frequency $15.8\,\mathrm{rad/s}$, period $0.397\,\mathrm{s}$, and frequency $2.52\,\mathrm{Hz}$.}
|
||||
|
||||
\qs{Worked AP-style problem}{A mass $m=0.60\,\mathrm{kg}$ hangs from a vertical spring with spring constant $k=150\,\mathrm{N/m}$. Choose downward as positive. Let $y$ denote the mass's downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium displacement, and let
|
||||
\[
|
||||
x=y-y_{\mathrm{eq}}
|
||||
\]
|
||||
denote displacement from equilibrium.
|
||||
|
||||
The mass is pulled downward $0.080\,\mathrm{m}$ from equilibrium and released from rest.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the equilibrium displacement $y_{\mathrm{eq}}$,
|
||||
\item the differential equation for $x(t)$ together with the angular frequency and period, and
|
||||
\item the maximum speed of the mass.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol At static equilibrium, the acceleration is zero, so the net force is zero:
|
||||
\[
|
||||
mg-ky_{\mathrm{eq}}=0.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
y_{\mathrm{eq}}=\frac{mg}{k}=\frac{(0.60\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{150\,\mathrm{N/m}}=0.0392\,\mathrm{m}.
|
||||
\]
|
||||
So the equilibrium position is $3.92\times 10^{-2}\,\mathrm{m}$ below the unstretched length.
|
||||
|
||||
Now write the motion in terms of displacement from equilibrium:
|
||||
\[
|
||||
x=y-y_{\mathrm{eq}}.
|
||||
\]
|
||||
The net force is
|
||||
\[
|
||||
F_{\mathrm{net}}=mg-ky=mg-k(y_{\mathrm{eq}}+x).
|
||||
\]
|
||||
Using $mg-ky_{\mathrm{eq}}=0$, this becomes
|
||||
\[
|
||||
F_{\mathrm{net}}=-kx.
|
||||
\]
|
||||
Apply Newton's second law:
|
||||
\[
|
||||
m\ddot{x}=-kx.
|
||||
\]
|
||||
Hence the differential equation is
|
||||
\[
|
||||
0.60\,\ddot{x}+150x=0,
|
||||
\]
|
||||
or equivalently,
|
||||
\[
|
||||
\ddot{x}+250x=0.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{150}{0.60}}=\sqrt{250}=15.8\,\mathrm{rad/s}.
|
||||
\]
|
||||
The period is
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.60}{150}}=0.397\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
Because the mass is released from rest $0.080\,\mathrm{m}$ from equilibrium, the amplitude is
|
||||
\[
|
||||
A=0.080\,\mathrm{m}.
|
||||
\]
|
||||
For simple harmonic motion,
|
||||
\[
|
||||
v_{\max}=A\omega.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
v_{\max}=(0.080)(15.8)=1.26\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Thus,
|
||||
\[
|
||||
y_{\mathrm{eq}}=0.0392\,\mathrm{m},
|
||||
\qquad
|
||||
\ddot{x}+250x=0,
|
||||
\qquad
|
||||
\omega=15.8\,\mathrm{rad/s},
|
||||
\qquad
|
||||
T=0.397\,\mathrm{s},
|
||||
\]
|
||||
and the maximum speed is
|
||||
\[
|
||||
v_{\max}=1.26\,\mathrm{m/s}.
|
||||
\]
|
||||
153
concepts/mechanics/u7/m7-3-shm-energy.tex
Normal file
153
concepts/mechanics/u7/m7-3-shm-energy.tex
Normal file
@@ -0,0 +1,153 @@
|
||||
\subsection{Energy in Simple Harmonic Motion}
|
||||
|
||||
This subsection uses energy to describe how a frictionless spring-mass oscillator trades energy between motion and spring deformation.
|
||||
|
||||
\dfn{Kinetic, potential, and total energy in spring SHM}{Consider a block of mass $m$ attached to an ideal spring of spring constant $k$ and moving frictionlessly along the $x$-axis. Let $x$ denote the signed displacement from equilibrium, let $\vec{v}=\dot{x}\hat{\imath}$ denote the block's velocity, let $v=|\vec{v}|=|\dot{x}|$ denote its speed, and let $A>0$ denote the amplitude of the motion.
|
||||
|
||||
The kinetic energy is
|
||||
\[
|
||||
K=\tfrac12 mv^2=\tfrac12 m\dot{x}^2.
|
||||
\]
|
||||
The spring potential energy is
|
||||
\[
|
||||
U_s=\tfrac12 kx^2.
|
||||
\]
|
||||
The total mechanical energy is
|
||||
\[
|
||||
E=K+U_s=\tfrac12 m\dot{x}^2+\tfrac12 kx^2.
|
||||
\]}
|
||||
|
||||
\thm{Conserved-energy relation for SHM}{For the frictionless spring-mass oscillator above, the total mechanical energy is constant:
|
||||
\[
|
||||
E=\tfrac12 m\dot{x}^2+\tfrac12 kx^2=\tfrac12 kA^2.
|
||||
\]
|
||||
Therefore, at any displacement $x$,
|
||||
\[
|
||||
\dot{x}^2=\frac{k}{m}\left(A^2-x^2\right),
|
||||
\qquad
|
||||
v=\sqrt{\frac{k}{m}\left(A^2-x^2\right)}.
|
||||
\]
|
||||
In particular, the maximum speed occurs at equilibrium $x=0$:
|
||||
\[
|
||||
v_{\max}=\sqrt{\frac{k}{m}}\,A.
|
||||
\]}
|
||||
|
||||
\nt{At the turning points $x=\pm A$, the block reverses direction, so $v=0$, $K=0$, and all the mechanical energy is spring potential energy:
|
||||
\[
|
||||
U_s=E=\tfrac12 kA^2.
|
||||
\]
|
||||
At equilibrium $x=0$, the spring is neither stretched nor compressed, so $U_s=0$ and all the energy is kinetic:
|
||||
\[
|
||||
K=E=\tfrac12 kA^2.
|
||||
\]
|
||||
Thus SHM continually swaps energy between kinetic and potential forms. If $x(t)=A\cos(\omega t+\phi)$, then $U_s\propto \cos^2(\omega t+\phi)$ and $K\propto \sin^2(\omega t+\phi)$, so the two energy curves are out of phase and each repeats twice during one full oscillation.}
|
||||
|
||||
\pf{Short derivation from conservation of mechanical energy}{For a frictionless spring-mass system, the only horizontal interaction is the spring force
|
||||
\[
|
||||
\vec{F}_s=-kx\hat{\imath},
|
||||
\]
|
||||
which is conservative. Therefore the mechanical energy $E=K+U_s$ is constant. Using the spring potential-energy function,
|
||||
\[
|
||||
U_s=\tfrac12 kx^2,
|
||||
\]
|
||||
the total energy at any instant is
|
||||
\[
|
||||
E=\tfrac12 m\dot{x}^2+\tfrac12 kx^2.
|
||||
\]
|
||||
At a turning point, $x=\pm A$ and $\dot{x}=0$, so
|
||||
\[
|
||||
E=\tfrac12 kA^2.
|
||||
\]
|
||||
Equating the two expressions for $E$ gives
|
||||
\[
|
||||
\tfrac12 m\dot{x}^2+\tfrac12 kx^2=\tfrac12 kA^2.
|
||||
\]
|
||||
Solving for $\dot{x}^2$ yields
|
||||
\[
|
||||
\dot{x}^2=\frac{k}{m}\left(A^2-x^2\right),
|
||||
\]
|
||||
and taking the positive square root gives the speed formula for the magnitude $v=|\dot{x}|$.}
|
||||
|
||||
\qs{Worked example}{A block of mass $m=0.40\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $k=160\,\mathrm{N/m}$ and oscillates frictionlessly with amplitude $A=0.10\,\mathrm{m}$. At one instant, the block is at displacement $x=+0.060\,\mathrm{m}$ from equilibrium.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the total mechanical energy of the oscillator,
|
||||
\item the spring potential energy and kinetic energy at $x=+0.060\,\mathrm{m}$,
|
||||
\item the speed of the block at that displacement, and
|
||||
\item the maximum speed and where it occurs.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Let $E$ denote the total mechanical energy. Because the motion is frictionless,
|
||||
\[
|
||||
E=\tfrac12 kA^2.
|
||||
\]
|
||||
Substitute $k=160\,\mathrm{N/m}$ and $A=0.10\,\mathrm{m}$:
|
||||
\[
|
||||
E=\tfrac12 (160)(0.10)^2=80(0.010)=0.80\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
So the oscillator's total mechanical energy is
|
||||
\[
|
||||
0.80\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
At $x=+0.060\,\mathrm{m}$, the spring potential energy is
|
||||
\[
|
||||
U_s=\tfrac12 kx^2=\tfrac12 (160)(0.060)^2.
|
||||
\]
|
||||
Since $(0.060)^2=0.0036$,
|
||||
\[
|
||||
U_s=80(0.0036)=0.288\,\mathrm{J}.
|
||||
\]
|
||||
Then the kinetic energy is
|
||||
\[
|
||||
K=E-U_s=0.80-0.288=0.512\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Now use kinetic energy to find the speed:
|
||||
\[
|
||||
K=\tfrac12 mv^2.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
0.512=\tfrac12 (0.40)v^2=0.20v^2.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
v^2=\frac{0.512}{0.20}=2.56,
|
||||
\qquad
|
||||
v=1.60\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
For the maximum speed, use the equilibrium position $x=0$, where all the energy is kinetic:
|
||||
\[
|
||||
\tfrac12 mv_{\max}^2=E=0.80\,\mathrm{J}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
0.20v_{\max}^2=0.80,
|
||||
\qquad
|
||||
v_{\max}^2=4.0,
|
||||
\qquad
|
||||
v_{\max}=2.0\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Equivalently,
|
||||
\[
|
||||
v_{\max}=\sqrt{\frac{k}{m}}\,A=\sqrt{\frac{160}{0.40}}(0.10)=20(0.10)=2.0\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
E=0.80\,\mathrm{J},
|
||||
\qquad
|
||||
U_s=0.288\,\mathrm{J},
|
||||
\qquad
|
||||
K=0.512\,\mathrm{J},
|
||||
\]
|
||||
\[
|
||||
v=1.60\,\mathrm{m/s},
|
||||
\qquad
|
||||
v_{\max}=2.0\,\mathrm{m/s}\text{ at }x=0.
|
||||
\]
|
||||
164
concepts/mechanics/u7/m7-4-simple-pendulum.tex
Normal file
164
concepts/mechanics/u7/m7-4-simple-pendulum.tex
Normal file
@@ -0,0 +1,164 @@
|
||||
\subsection{The Simple Pendulum}
|
||||
|
||||
This subsection models a bob of mass on a light string, using angular displacement from the vertical as the natural coordinate.
|
||||
|
||||
\dfn{Simple pendulum and angular coordinate}{Let $m$ denote the bob's mass, let $\ell>0$ denote the string length, let $g$ denote the magnitude of the gravitational field, and let $\theta(t)$ denote the angular displacement from the downward vertical, measured in radians and taken positive in the counterclockwise direction.
|
||||
|
||||
A \emph{simple pendulum} is an idealized system consisting of a point mass $m$ attached to a massless string of fixed length $\ell$, swinging without friction in a uniform gravitational field. The bob moves along a circular arc of radius $\ell$. If $s(t)$ denotes the arc displacement from equilibrium, then
|
||||
\[
|
||||
s=\ell\theta.
|
||||
\]
|
||||
The equilibrium position is $\theta=0$.}
|
||||
|
||||
\thm{Exact pendulum equation and small-angle SHM model}{For the simple pendulum above, the exact rotational equation of motion is
|
||||
\[
|
||||
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
|
||||
\]
|
||||
This equation is nonlinear, so the motion is not exactly simple harmonic for arbitrary amplitude.
|
||||
|
||||
If the oscillation remains at small angles so that $|\theta|\ll 1$ radian and $\sin\theta\approx\theta$, then the motion is approximated by
|
||||
\[
|
||||
\ddot{\theta}+\frac{g}{\ell}\theta=0.
|
||||
\]
|
||||
Thus the pendulum behaves approximately like SHM with angular frequency
|
||||
\[
|
||||
\omega=\sqrt{\frac{g}{\ell}},
|
||||
\]
|
||||
small-angle period
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{\ell}{g}},
|
||||
\]
|
||||
and small-angle frequency
|
||||
\[
|
||||
f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.
|
||||
\]}
|
||||
|
||||
\pf{Short derivation from torque and linearization}{About the pivot, the gravitational torque on the bob is restoring, so
|
||||
\[
|
||||
\tau=-mg\ell\sin\theta.
|
||||
\]
|
||||
The bob acts like a point mass at distance $\ell$, so its moment of inertia about the pivot is
|
||||
\[
|
||||
I=m\ell^2.
|
||||
\]
|
||||
Using rotational Newton's second law, $\sum\tau=I\ddot{\theta}$, gives
|
||||
\[
|
||||
m\ell^2\ddot{\theta}=-mg\ell\sin\theta.
|
||||
\]
|
||||
Divide by $m\ell^2$:
|
||||
\[
|
||||
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
|
||||
\]
|
||||
For small oscillations with $|\theta|\ll 1$ radian, use the small-angle approximation $\sin\theta\approx\theta$. Then
|
||||
\[
|
||||
\ddot{\theta}+\frac{g}{\ell}\theta=0,
|
||||
\]
|
||||
which is the standard SHM equation with $\omega^2=g/\ell$.}
|
||||
|
||||
\ex{Illustrative example}{A pendulum oscillates through small angles. Its length is changed from $\ell_1=0.50\,\mathrm{m}$ to $\ell_2=2.00\,\mathrm{m}$. How do the period and frequency change?
|
||||
|
||||
For small-angle motion,
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{\ell}{g}}.
|
||||
\]
|
||||
Therefore $T\propto\sqrt{\ell}$. Since
|
||||
\[
|
||||
\frac{\ell_2}{\ell_1}=\frac{2.00}{0.50}=4,
|
||||
\]
|
||||
the new period is multiplied by
|
||||
\[
|
||||
\sqrt{4}=2.
|
||||
\]
|
||||
So the period doubles. Because $f=1/T$, the frequency is cut in half.}
|
||||
|
||||
\qs{Worked AP-style problem}{A simple pendulum has length $\ell=0.90\,\mathrm{m}$. It is pulled aside to a maximum angle $\theta_{\max}=0.10\,\mathrm{rad}$ and released from rest. Take $g=9.8\,\mathrm{m/s^2}$.
|
||||
|
||||
Assume the small-angle model is valid.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the exact equation of motion and the small-angle approximate equation,
|
||||
\item the angular frequency, period, and frequency,
|
||||
\item the time required to move from maximum displacement to equilibrium, and
|
||||
\item the maximum linear speed of the bob.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol Let $\theta(t)$ denote the angular displacement from the downward vertical.
|
||||
|
||||
For part (a), the exact pendulum equation is
|
||||
\[
|
||||
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
|
||||
\]
|
||||
Substitute $g=9.8\,\mathrm{m/s^2}$ and $\ell=0.90\,\mathrm{m}$:
|
||||
\[
|
||||
\ddot{\theta}+\frac{9.8}{0.90}\sin\theta=0.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\ddot{\theta}+10.9\sin\theta=0
|
||||
\]
|
||||
to three significant figures.
|
||||
|
||||
Under the small-angle approximation $\sin\theta\approx\theta$, the motion is modeled by
|
||||
\[
|
||||
\ddot{\theta}+\frac{9.8}{0.90}\theta=0,
|
||||
\]
|
||||
or
|
||||
\[
|
||||
\ddot{\theta}+10.9\theta=0.
|
||||
\]
|
||||
|
||||
For part (b), compare the small-angle equation with
|
||||
\[
|
||||
\ddot{\theta}+\omega^2\theta=0.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
\omega=\sqrt{\frac{g}{\ell}}=\sqrt{\frac{9.8}{0.90}}=3.30\,\mathrm{rad/s}.
|
||||
\]
|
||||
Then the period is
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{\ell}{g}}=\frac{2\pi}{\omega}=\frac{2\pi}{3.30}=1.90\,\mathrm{s}.
|
||||
\]
|
||||
The frequency is
|
||||
\[
|
||||
f=\frac{1}{T}=\frac{1}{1.90}=0.526\,\mathrm{Hz}.
|
||||
\]
|
||||
|
||||
For part (c), a pendulum in SHM takes one-quarter of a cycle to move from an endpoint to equilibrium. Therefore,
|
||||
\[
|
||||
t=\frac{T}{4}=\frac{1.90}{4}=0.475\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
For part (d), the maximum angular speed in SHM is
|
||||
\[
|
||||
\dot{\theta}_{\max}=\omega\theta_{\max}.
|
||||
\]
|
||||
So
|
||||
\[
|
||||
\dot{\theta}_{\max}=(3.30)(0.10)=0.330\,\mathrm{rad/s}.
|
||||
\]
|
||||
The bob's linear speed is related by $v=\ell\dot{\theta}$, so the maximum linear speed is
|
||||
\[
|
||||
v_{\max}=\ell\dot{\theta}_{\max}=(0.90)(0.330)=0.297\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
\ddot{\theta}+10.9\sin\theta=0,
|
||||
\qquad
|
||||
\ddot{\theta}+10.9\theta=0,
|
||||
\]
|
||||
\[
|
||||
\omega=3.30\,\mathrm{rad/s},
|
||||
\qquad
|
||||
T=1.90\,\mathrm{s},
|
||||
\qquad
|
||||
f=0.526\,\mathrm{Hz},
|
||||
\]
|
||||
and
|
||||
\[
|
||||
t=0.475\,\mathrm{s},
|
||||
\qquad
|
||||
v_{\max}=0.297\,\mathrm{m/s}.
|
||||
\]
|
||||
155
concepts/mechanics/u7/m7-5-physical-pendulum.tex
Normal file
155
concepts/mechanics/u7/m7-5-physical-pendulum.tex
Normal file
@@ -0,0 +1,155 @@
|
||||
\subsection{Physical Pendulum and Small-Angle Linearization}
|
||||
|
||||
This subsection models the small oscillations of a rigid body that swings about a fixed pivot under gravity.
|
||||
|
||||
\dfn{Physical pendulum, pivot-to-CM distance, and angular coordinate}{Let a rigid body of mass $m$ swing in a vertical plane about a fixed pivot point $O$. Let $C$ denote the center of mass of the body, let
|
||||
\[
|
||||
d=OC
|
||||
\]
|
||||
denote the distance from the pivot to the center of mass, and let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position.
|
||||
|
||||
Such a system is called a \emph{physical pendulum}. Unlike a simple pendulum, the body's mass is distributed throughout the rigid object, so its rotational inertia must be included in the dynamics.}
|
||||
|
||||
\thm{Exact torque equation and small-angle SHM model}{Let $m$ denote the mass of the rigid body, let $d$ denote the distance from the pivot to the center of mass, let $I$ denote the moment of inertia of the body about the pivot, let $g$ denote the magnitude of the gravitational field, and let $\theta(t)$ denote the angular displacement from stable equilibrium.
|
||||
|
||||
Then the exact rotational equation of motion is
|
||||
\[
|
||||
I\ddot{\theta}=-mgd\sin\theta,
|
||||
\]
|
||||
or equivalently,
|
||||
\[
|
||||
I\ddot{\theta}+mgd\sin\theta=0.
|
||||
\]
|
||||
|
||||
For small angular displacements, use the linearization $\sin\theta\approx\theta$ to obtain
|
||||
\[
|
||||
I\ddot{\theta}+mgd\,\theta=0.
|
||||
\]
|
||||
Therefore the motion is approximately simple harmonic with angular frequency
|
||||
\[
|
||||
\omega=\sqrt{\frac{mgd}{I}}
|
||||
\]
|
||||
and period
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{I}{mgd}}.
|
||||
\]
|
||||
|
||||
A simple pendulum is the special case in which all the mass is concentrated a distance $L$ from the pivot, so $I=mL^2$ and $d=L$.}
|
||||
|
||||
\pf{Short derivation from torque and linearization}{The weight $m\vec{g}$ acts at the center of mass. When the body is displaced by angle $\theta$, the gravitational torque about the pivot is restoring, so
|
||||
\[
|
||||
\tau=-mgd\sin\theta.
|
||||
\]
|
||||
For rotation about a fixed axis, Newton's second law for rotation gives
|
||||
\[
|
||||
\sum \tau=I\ddot{\theta}.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
I\ddot{\theta}=-mgd\sin\theta,
|
||||
\]
|
||||
which is the exact equation.
|
||||
|
||||
If the oscillations are small, then $\sin\theta\approx\theta$, so the equation becomes
|
||||
\[
|
||||
I\ddot{\theta}+mgd\,\theta=0.
|
||||
\]
|
||||
Divide by $I$ to get
|
||||
\[
|
||||
\ddot{\theta}+\frac{mgd}{I}\theta=0.
|
||||
\]
|
||||
Comparing with the SHM form $q''+\omega^2 q=0$ shows that
|
||||
\[
|
||||
\omega^2=\frac{mgd}{I},
|
||||
\qquad
|
||||
\omega=\sqrt{\frac{mgd}{I}}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{mgd}}.
|
||||
\]}
|
||||
|
||||
\ex{Illustrative example}{Show that the simple pendulum is a special case of the physical pendulum formula.
|
||||
|
||||
For a point mass $m$ at distance $L$ from the pivot,
|
||||
\[
|
||||
I=mL^2,
|
||||
\qquad
|
||||
d=L.
|
||||
\]
|
||||
Substitute into the physical-pendulum period formula:
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{mL^2}{mgL}}=2\pi\sqrt{\frac{L}{g}}.
|
||||
\]
|
||||
This is exactly the small-angle period of a simple pendulum.}
|
||||
|
||||
\qs{Worked AP-style problem}{A uniform rod of mass $m=1.50\,\mathrm{kg}$ and length $L=0.90\,\mathrm{m}$ is pivoted about one end and allowed to swing in a vertical plane. Let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position. Assume the oscillations are small.
|
||||
|
||||
Find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item the pivot-to-center-of-mass distance $d$ and the rod's moment of inertia $I$ about the pivot,
|
||||
\item the small-angle differential equation for $\theta(t)$, and
|
||||
\item the period of oscillation.
|
||||
\end{enumerate}}
|
||||
|
||||
\sol For a uniform rod pivoted about one end, the center of mass is at the midpoint, so
|
||||
\[
|
||||
d=\frac{L}{2}=\frac{0.90\,\mathrm{m}}{2}=0.45\,\mathrm{m}.
|
||||
\]
|
||||
|
||||
The moment of inertia of a uniform rod about one end is
|
||||
\[
|
||||
I=\frac{1}{3}mL^2.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
I=\frac{1}{3}(1.50)(0.90)^2\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
Since $(0.90)^2=0.81$,
|
||||
\[
|
||||
I=\frac{1}{3}(1.50)(0.81)=0.405\,\mathrm{kg\cdot m^2}.
|
||||
\]
|
||||
|
||||
For small oscillations, a physical pendulum satisfies
|
||||
\[
|
||||
I\ddot{\theta}+mgd\,\theta=0.
|
||||
\]
|
||||
Now compute $mgd$:
|
||||
\[
|
||||
mgd=(1.50)(9.8)(0.45)=6.615.
|
||||
\]
|
||||
So the differential equation is
|
||||
\[
|
||||
0.405\,\ddot{\theta}+6.615\,\theta=0.
|
||||
\]
|
||||
Divide by $0.405$:
|
||||
\[
|
||||
\ddot{\theta}+16.3\,\theta=0.
|
||||
\]
|
||||
|
||||
Thus,
|
||||
\[
|
||||
\omega=\sqrt{16.3}=4.04\,\mathrm{rad/s}.
|
||||
\]
|
||||
The period is
|
||||
\[
|
||||
T=\frac{2\pi}{\omega}=\frac{2\pi}{4.04}=1.56\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
Equivalently, using the period formula directly,
|
||||
\[
|
||||
T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{0.405}{6.615}}=1.56\,\mathrm{s}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
d=0.45\,\mathrm{m},
|
||||
\qquad
|
||||
I=0.405\,\mathrm{kg\cdot m^2},
|
||||
\]
|
||||
and the small-angle motion is governed by
|
||||
\[
|
||||
\ddot{\theta}+16.3\,\theta=0,
|
||||
\qquad
|
||||
T=1.56\,\mathrm{s}.
|
||||
\]
|
||||
Reference in New Issue
Block a user