\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.