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concepts/mechanics/u5/m5-2-linear-rotational-link.tex
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concepts/mechanics/u5/m5-2-linear-rotational-link.tex
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\subsection{Linear and Rotational Kinematic Correspondence}
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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.
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\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.
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The signed arc-length coordinate of $P$ along its circular path is
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\[
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s=r\theta.
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\]
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The tangential velocity component and tangential acceleration component of $P$ are
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\[
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v_t=\frac{ds}{dt},
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\qquad
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a_t=\frac{dv_t}{dt}.
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\]
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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
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\[
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\vec{a}=-a_r\hat{r}+a_t\hat{t},
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\]
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because the radial part points toward the axis.}
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\thm{Linear/rotational correspondence for fixed-axis rotation}{For the point $P$ above at radius $r$,
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\[
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s=r\theta,
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\qquad
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v_t=r\omega,
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\qquad
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a_t=r\alpha,
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\qquad
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a_r=\frac{v_t^2}{r}=r\omega^2.
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\]
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Thus
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\[
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\vec{v}=v_t\hat{t},
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\qquad
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\vec{a}=-r\omega^2\hat{r}+r\alpha\hat{t}.
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\]
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If $\vec{\omega}=\omega\hat{k}$, then the tangential-velocity relation may also be written as
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\[
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\vec{v}=\vec{\omega}\times \vec{r}.
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\]
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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.}
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\pf{Short derivation from $s=r\theta$}{Because the point stays a fixed distance $r$ from the axis, its arc-length coordinate satisfies
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\[
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s=r\theta.
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\]
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Differentiate with respect to time:
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\[
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v_t=\frac{ds}{dt}=r\frac{d\theta}{dt}=r\omega.
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\]
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Differentiate again:
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\[
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a_t=\frac{dv_t}{dt}=r\frac{d\omega}{dt}=r\alpha.
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\]
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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
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\[
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a_r=\frac{v_t^2}{r}.
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\]
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Substitute $v_t=r\omega$:
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\[
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a_r=\frac{(r\omega)^2}{r}=r\omega^2.
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\]
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Therefore
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\[
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s=r\theta,
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\qquad
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v_t=r\omega,
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\qquad
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a_t=r\alpha,
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\qquad
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a_r=r\omega^2.
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\]}
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\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:
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\[
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\frac{s_2}{s_1}=\frac{r_2}{r_1},
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\qquad
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\frac{v_{t,2}}{v_{t,1}}=\frac{r_2}{r_1},
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\qquad
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\frac{a_{t,2}}{a_{t,1}}=\frac{r_2}{r_1},
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\qquad
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\frac{a_{r,2}}{a_{r,1}}=\frac{r_2}{r_1}.
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\]
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So points farther from the axis move faster, have larger tangential acceleration for the same $\alpha$, and require larger inward acceleration for the same $\omega$.}
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\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 $r_B=0.150\,\mathrm{m}$.
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For each point, find:
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\begin{enumerate}[label=(\alph*)]
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\item the signed arc-length coordinate $s$,
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\item the tangential velocity component $v_t$,
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\item the tangential acceleration component $a_t$, and
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\item the radial acceleration magnitude $a_r$.
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\end{enumerate}}
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\sol Choose the positive tangential direction to be counterclockwise, since the wheel's rotation is positive in that sense. Use
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\[
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s=r\theta,
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\qquad
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v_t=r\omega,
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\qquad
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a_t=r\alpha,
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\qquad
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a_r=r\omega^2.
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\]
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For point $A$, $r_A=0.050\,\mathrm{m}$.
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Its arc-length coordinate is
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\[
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s_A=r_A\theta=(0.050\,\mathrm{m})(1.20)=0.060\,\mathrm{m}.
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\]
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Its tangential velocity component is
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\[
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v_{t,A}=r_A\omega=(0.050\,\mathrm{m})(6.0\,\mathrm{rad/s})=0.30\,\mathrm{m/s}.
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\]
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Its tangential acceleration component is
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\[
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a_{t,A}=r_A\alpha=(0.050\,\mathrm{m})(-2.0\,\mathrm{rad/s^2})=-0.10\,\mathrm{m/s^2}.
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\]
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The negative sign means the tangential acceleration is opposite the positive tangential direction at that instant.
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Its radial acceleration magnitude is
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\[
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a_{r,A}=r_A\omega^2=(0.050\,\mathrm{m})(6.0\,\mathrm{rad/s})^2
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=(0.050)(36)=1.8\,\mathrm{m/s^2}.
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\]
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This radial part points inward, toward the axle.
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For point $B$, $r_B=0.150\,\mathrm{m}$.
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Its arc-length coordinate is
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\[
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s_B=r_B\theta=(0.150\,\mathrm{m})(1.20)=0.180\,\mathrm{m}.
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\]
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Its tangential velocity component is
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\[
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v_{t,B}=r_B\omega=(0.150\,\mathrm{m})(6.0\,\mathrm{rad/s})=0.90\,\mathrm{m/s}.
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\]
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Its tangential acceleration component is
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\[
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a_{t,B}=r_B\alpha=(0.150\,\mathrm{m})(-2.0\,\mathrm{rad/s^2})=-0.30\,\mathrm{m/s^2}.
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\]
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Its radial acceleration magnitude is
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\[
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a_{r,B}=r_B\omega^2=(0.150\,\mathrm{m})(6.0\,\mathrm{rad/s})^2
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=(0.150)(36)=5.4\,\mathrm{m/s^2}.
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\]
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Therefore,
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\[
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s_A=0.060\,\mathrm{m},
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\qquad
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v_{t,A}=0.30\,\mathrm{m/s},
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\qquad
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a_{t,A}=-0.10\,\mathrm{m/s^2},
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\qquad
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a_{r,A}=1.8\,\mathrm{m/s^2},
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\]
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\[
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s_B=0.180\,\mathrm{m},
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\qquad
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v_{t,B}=0.90\,\mathrm{m/s},
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\qquad
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a_{t,B}=-0.30\,\mathrm{m/s^2},
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\qquad
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a_{r,B}=5.4\,\mathrm{m/s^2}.
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\]
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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.
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