\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.
