\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.
