marsultor/physics-handbook
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

checkpoint 1

Browse Source
This commit is contained in:
Krishna Ayyalasomayajula
2026-04-29 21:21:22 -05:00
commit eb6959b896
113 changed files with 14240 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

0
concepts/mechanics/u2/.gitkeep Normal file
View File

133
concepts/mechanics/u2/m2-1-newton-laws.tex Normal file
View 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
View 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
View 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
View 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
View 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
View 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
View 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
View 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
\[
a_r\approx 8.69\,\mathrm{m/s^2}.
\]