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\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.

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\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.

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\subsection{Torque and Lever Arm}
This subsection introduces torque as the rotational effect of a force about a chosen point or axis. In AP mechanics, the key computational idea is that only the part of the force with a nonzero lever arm contributes to the turning effect.
\dfn{Torque vector, lever arm, and fixed-axis sign convention}{Let $O$ denote the pivot or reference point. Let $\vec{r}$ denote the position vector from $O$ to the point where a force $\vec{F}$ is applied. The \emph{torque vector} of $\vec{F}$ about $O$ is the vector quantity that measures the tendency of the force to cause rotation about $O$.
Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$. The \emph{lever arm} $\ell$ is the perpendicular distance from the pivot to the line of action of the force, so
\[
\ell=r\sin\phi,
\]
where $r=|\vec{r}|$.
For a fixed-axis problem, choose an axis with unit vector $\hat{k}$ and declare a positive sense of rotation by the right-hand rule. The corresponding signed scalar torque is
\[
\tau=(\vec{r}\times \vec{F})\cdot \hat{k}.
\]
If the force tends to rotate the object in the chosen positive sense, then $\tau>0$; if it tends to rotate the object in the opposite sense, then $\tau<0$.}
\thm{Torque formulas}{Let $\vec{r}$ denote the position vector from the pivot to the point of application of a force $\vec{F}$. Let $\phi$ denote the angle between $\vec{r}$ and $\vec{F}$, let $r=|\vec{r}|$, let $F=|\vec{F}|$, and let $\ell$ denote the lever arm. Then the torque vector is
\[
\vec{\tau}=\vec{r}\times \vec{F},
\]
and its magnitude is
\[
|\vec{\tau}|=rF\sin\phi=F\ell.
\]
For rotation about a chosen fixed axis with unit vector $\hat{k}$,
\[
\tau=(\vec{r}\times \vec{F})\cdot \hat{k},
\]
so the scalar $\tau$ is positive or negative according to the declared sign convention.}
\pf{Why $rF\sin\phi$ equals $F\ell$}{From the cross-product magnitude formula,
\[
|\vec{\tau}|=|\vec{r}\times \vec{F}|=rF\sin\phi.
\]
But by geometry, the lever arm is the perpendicular distance from the pivot to the line of action of the force, so
\[
\ell=r\sin\phi.
\]
Substituting this into the previous expression gives
\[
|\vec{\tau}|=rF\sin\phi=F\ell.
\]
Thus torque can be found either from the perpendicular component of the force or from the full force multiplied by the lever arm.}
\cor{If the line of action passes through the pivot, the torque is zero}{If the line of action of $\vec{F}$ passes through the pivot, then the perpendicular distance from the pivot to that line is $\ell=0$. Therefore
\[
|\vec{\tau}|=F\ell=0.
\]
Equivalently, in this case $\phi=0$ or $\phi=\pi$, so $rF\sin\phi=0$. Thus a force directed exactly through the pivot can change the net force on an object without producing any torque about that pivot.}
\qs{Worked example}{A door rotates about a vertical hinge through its left edge. View the door from above and choose counterclockwise rotation as positive. Let $r=0.90\,\mathrm{m}$ be the distance from the hinge to the point where the force is applied at the outer edge. A student pushes with force magnitude $F=40\,\mathrm{N}$. The force lies in the horizontal plane and makes an angle $\phi=30^\circ$ with the door, so it tends to rotate the door counterclockwise.
Find:
\begin{enumerate}[label=(\alph*)]
\item the lever arm $\ell$,
\item the signed torque $\tau$ about the hinge, and
\item the magnitude of a force applied perpendicular to the door at the same point that would produce the same torque.
\end{enumerate}}
\sol Because the position vector from the hinge to the point of application lies along the door, the given angle $\phi=30^\circ$ is the angle between $\vec{r}$ and $\vec{F}$.
For part (a), the lever arm is
\[
\ell=r\sin\phi=(0.90\,\mathrm{m})\sin 30^\circ.
\]
Since $\sin 30^\circ=0.50$,
\[
\ell=(0.90)(0.50)=0.45\,\mathrm{m}.
\]
So the lever arm is
\[
\ell=0.45\,\mathrm{m}.
\]
For part (b), the torque magnitude is
\[
|\tau|=rF\sin\phi=F\ell.
\]
Using either form,
\[
|\tau|=(0.90\,\mathrm{m})(40\,\mathrm{N})\sin 30^\circ=(40\,\mathrm{N})(0.45\,\mathrm{m}).
\]
Thus
\[
|\tau|=18\,\mathrm{N\cdot m}.
\]
Because the force tends to rotate the door counterclockwise and counterclockwise was chosen as positive,
\[
\tau=+18\,\mathrm{N\cdot m}.
\]
For part (c), let $F_{\perp}$ denote the force magnitude that would act perpendicular to the door at the same point. A perpendicular force has lever arm equal to the full distance $r=0.90\,\mathrm{m}$, so its torque magnitude is
\[
|\tau|=rF_{\perp}.
\]
Set this equal to the required $18\,\mathrm{N\cdot m}$:
\[
18\,\mathrm{N\cdot m}=(0.90\,\mathrm{m})F_{\perp}.
\]
Therefore,
\[
F_{\perp}=\frac{18}{0.90}=20\,\mathrm{N}.
\]
The results are
\[
\ell=0.45\,\mathrm{m},
\qquad
\tau=+18\,\mathrm{N\cdot m},
\qquad
F_{\perp}=20\,\mathrm{N}.
\]
This example shows that the same torque can be produced either by a larger force with a shorter lever arm or by a smaller force applied more effectively.

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\subsection{Moment of Inertia and Mass Distribution}
This subsection introduces the rotational analog of mass for fixed-axis motion. In AP mechanics, the key idea is that both the amount of mass and how far that mass lies from the chosen axis determine how strongly an object resists angular acceleration.
\dfn{Moment of inertia for discrete and continuous mass distributions}{Consider a rigid body about a chosen fixed axis. For a discrete collection of particles, let particle $i$ have mass $m_i$, let $\vec{r}_i$ denote its position vector relative to a point on the axis, and let $r_{\perp,i}$ denote its perpendicular distance to the axis. The \emph{moment of inertia} of the body about that axis is
\[
I=\sum_i m_i r_{\perp,i}^2.
\]
For a continuous mass distribution, let $dm$ denote a small mass element and let $r_\perp$ denote that element's perpendicular distance to the same axis. Then
\[
I=\int r_\perp^2\,dm.
\]
For a single point mass,
\[
I=mr_\perp^2.
\]
The SI unit of moment of inertia is $\mathrm{kg\cdot m^2}$. Because the distance to the axis is squared, mass farther from the axis contributes much more strongly to $I$.}
\thm{Key fixed-axis relations and axis dependence}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $I$ denote the moment of inertia about that same axis. Then
\[
\tau_{\mathrm{net}}=I\alpha,
\qquad
\vec{\tau}_{\mathrm{net}}=I\vec{\alpha}.
\]
Thus, for the same net torque, a larger moment of inertia gives a smaller angular acceleration.
Now let $M$ denote the total mass of the body, let $I_{\mathrm{cm}}$ denote the moment of inertia about an axis through the center of mass, and let $d$ denote the perpendicular distance from that axis to a second axis parallel to it. Then the parallel-axis theorem states
\[
I=I_{\mathrm{cm}}+Md^2.
\]
Therefore moment of inertia depends on the chosen axis as well as on the mass distribution. Moving the axis farther from the center of mass increases $I$.}
\ex{Illustrative example}{A light turntable carries two small clay balls, each of mass $m=0.40\,\mathrm{kg}$. In arrangement A, each ball is at distance $r_A=0.10\,\mathrm{m}$ from the axis. In arrangement B, each ball is at distance $r_B=0.20\,\mathrm{m}$ from the axis.
For arrangement A,
\[
I_A=2mr_A^2=2(0.40)(0.10)^2=8.0\times 10^{-3}\,\mathrm{kg\cdot m^2}.
\]
For arrangement B,
\[
I_B=2mr_B^2=2(0.40)(0.20)^2=3.2\times 10^{-2}\,\mathrm{kg\cdot m^2}.
\]
So doubling the distance from the axis makes the moment of inertia four times as large, even though the total mass is unchanged.}
\nt{Equal mass does not guarantee equal rotational response. Two objects can have the same total mass but different moments of inertia if their mass is distributed differently or if the axis is changed. In fixed-axis rotation, the object with larger $I$ has smaller angular acceleration for the same net torque.}
\qs{Worked example}{A light rigid rod of length $L=0.80\,\mathrm{m}$ lies on a horizontal frictionless table and can rotate about a vertical axle. A small mass $m_1=0.50\,\mathrm{kg}$ is attached at the left end, and a small mass $m_2=1.50\,\mathrm{kg}$ is attached at the right end. First, the axle passes through the center of the rod, so each mass is at distance $r=0.40\,\mathrm{m}$ from the axis. A string pulls perpendicularly on the right end with force magnitude $F=3.0\,\mathrm{N}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item the moment of inertia $I$ about the center axle,
\item the torque magnitude $\tau$ due to the pull,
\item the angular acceleration magnitude $\alpha$, and
\item the new moment of inertia and angular acceleration if the axle is moved to the left end of the rod while the same perpendicular force is still applied at the right end.
\end{enumerate}}
\sol For parts (a)--(c), the axis passes through the center of the rod, so each mass is $r=0.40\,\mathrm{m}$ from the axis. Because the rod is light, treat its mass as negligible and include only the two point masses.
For part (a), the moment of inertia is
\[
I=\sum m_i r_i^2=m_1r^2+m_2r^2.
\]
Substitute the given values:
\[
I=(0.50)(0.40)^2+(1.50)(0.40)^2.
\]
Since $(0.40)^2=0.16$,
\[
I=(0.50)(0.16)+(1.50)(0.16)=0.08+0.24=0.32\,\mathrm{kg\cdot m^2}.
\]
So the moment of inertia about the center axle is
\[
I=0.32\,\mathrm{kg\cdot m^2}.
\]
For part (b), the pull is perpendicular to the rod, so the torque magnitude is
\[
\tau=rF.
\]
Here the force is applied at the right end, which is $0.40\,\mathrm{m}$ from the center axle. Therefore,
\[
\tau=(0.40\,\mathrm{m})(3.0\,\mathrm{N})=1.2\,\mathrm{N\cdot m}.
\]
For part (c), use the rotational form of Newton's second law:
\[
\tau=I\alpha.
\]
Hence,
\[
\alpha=\frac{\tau}{I}=\frac{1.2}{0.32}=3.75\,\mathrm{rad/s^2}.
\]
So for the center axle,
\[
\alpha=3.75\,\mathrm{rad/s^2}.
\]
For part (d), move the axle to the left end. Then the left mass is on the axis, so its distance is $r_1=0$, and the right mass is $r_2=L=0.80\,\mathrm{m}$ from the axis. The new moment of inertia is
\[
I' = m_1r_1^2+m_2r_2^2.
\]
Thus,
\[
I'=(0.50)(0)^2+(1.50)(0.80)^2.
\]
Since $(0.80)^2=0.64$,
\[
I'=0+(1.50)(0.64)=0.96\,\mathrm{kg\cdot m^2}.
\]
Now the same perpendicular force is applied at the right end, which is $0.80\,\mathrm{m}$ from the new axis, so the torque magnitude is
\[
\tau'=(0.80\,\mathrm{m})(3.0\,\mathrm{N})=2.4\,\mathrm{N\cdot m}.
\]
Then
\[
\alpha'=\frac{\tau'}{I'}=\frac{2.4}{0.96}=2.50\,\mathrm{rad/s^2}.
\]
Therefore, with the axle at the left end,
\[
I'=0.96\,\mathrm{kg\cdot m^2},
\qquad
\alpha'=2.50\,\mathrm{rad/s^2}.
\]
Even though the torque becomes larger, the angular acceleration becomes smaller because much more of the mass is farther from the axis, so the moment of inertia increases substantially.

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\subsection{Rotational Equilibrium}
This subsection treats the equilibrium case of fixed-axis rotational dynamics. In AP statics, a rigid body remains at rest only when both the translational and rotational balances are satisfied.
\dfn{Rotational equilibrium and static equilibrium}{Consider a rigid body about a chosen fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $\vec{F}_{\mathrm{net}}$ denote the net external force on the body.
The body is in \emph{rotational equilibrium} if
\[
\vec{\alpha}=\vec{0},
\]
so its angular velocity is constant. In the common AP statics case, the body is at rest and is also in \emph{static equilibrium}, meaning that both its translational and rotational motion remain unchanged:
\[
\vec{F}_{\mathrm{net}}=\vec{0},
\qquad
\vec{\alpha}=\vec{0}.
\]
Thus rotational equilibrium is the $\alpha=0$ special case of rotational dynamics, and static equilibrium is the at-rest case in which there is also no translational acceleration.}
\thm{Equilibrium conditions for a rigid body}{Let $\vec{F}_{\mathrm{net}}$ denote the net external force on a rigid body, and let $\vec{\tau}_{\mathrm{net},O}$ denote the net external torque about a chosen point $O$. For static equilibrium in an inertial frame,
\[
\sum \vec{F}=\vec{0},
\qquad
\sum \vec{\tau}_O=\vec{0}.
\]
For a planar fixed-axis problem, choose a positive sense of rotation about the axis. Then the equivalent scalar conditions are
\[
\sum F_x=0,
\qquad
\sum F_y=0,
\qquad
\sum \tau_O=0,
\]
where positive and negative torques are assigned by the declared sign convention. Force balance enforces translational equilibrium, and torque balance enforces rotational equilibrium.}
\nt{In equilibrium problems, the pivot can be chosen wherever it is most convenient. A smart choice is often a point through which one or more unknown forces act, because those forces then contribute zero torque about that point. After using $\sum \tau_O=0$ to solve for a remaining unknown such as a tension, use $\sum \vec{F}=\vec{0}$ to find the support-force components. If $\sum \vec{F}=\vec{0}$, then torque balance about one point is equivalent to torque balance about any other point, so changing the pivot changes the algebra, not the physics. For fixed-axis AP problems, signed scalar torques are fully acceptable once a sign convention such as counterclockwise positive has been declared.}
\pf{Short explanation from $\tau_{\mathrm{net}}=I\alpha$}{Let $I$ denote the moment of inertia of the rigid body about the chosen fixed axis. Fixed-axis rotational dynamics gives
\[
\tau_{\mathrm{net}}=I\alpha.
\]
If the body is in rotational equilibrium, then $\alpha=0$, so
\[
\tau_{\mathrm{net}}=0.
\]
Now let $m$ denote the total mass of the body, and let $\vec{a}_{\mathrm{cm}}$ denote the acceleration of its center of mass. For statics the body also has no translational acceleration, so Newton's second law gives
\[
\sum \vec{F}=m\vec{a}_{\mathrm{cm}}=\vec{0}.
\]
Therefore a rigid body at rest must satisfy both torque balance and force balance. Conversely, if both balances hold, the body has neither angular acceleration nor translational acceleration, so an initially resting body remains in static equilibrium.}
\qs{Worked example}{A uniform horizontal beam has length $L=4.0\,\mathrm{m}$ and weight $W_b=200\,\mathrm{N}$. It is hinged to a wall at its left end. A light cable is attached to the right end of the beam and makes an angle $\theta=30^\circ$ above the beam. A sign of weight $W_s=300\,\mathrm{N}$ hangs from the beam at a point $x_s=3.0\,\mathrm{m}$ from the hinge. Let $T$ denote the cable tension. Let the hinge force on the beam be $\vec{H}=H_x\hat{i}+H_y\hat{j}$, where $\hat{i}$ points horizontally to the right and $\hat{j}$ points vertically upward. Choose counterclockwise torque as positive.
Find:
\begin{enumerate}[label=(\alph*)]
\item the cable tension $T$,
\item the horizontal component $H_x$, and
\item the vertical component $H_y$ of the hinge force.
\end{enumerate}}
\sol Draw the beam's free-body diagram. The external forces on the beam are the hinge force $\vec{H}$ at the left end, the cable force $\vec{T}$ at the right end, the beam's weight $\vec{W}_b$ downward at its center, and the sign's weight $\vec{W}_s$ downward at $x_s=3.0\,\mathrm{m}$.
Choose the hinge as the pivot. Then the unknown hinge force produces zero torque about that point, which is why this pivot choice is efficient.
The beam's center is at
\[
\frac{L}{2}=2.0\,\mathrm{m}
\]
from the hinge. The horizontal component of the cable force acts along the beam, so its line of action passes through the hinge and it produces zero torque about the hinge. Thus only the cable's vertical component contributes to the torque balance:
\[
\sum \tau_{\mathrm{hinge}}=0.
\]
Using counterclockwise as positive,
\[
(T\sin\theta)L-W_b\left(\frac{L}{2}\right)-W_sx_s=0.
\]
Substitute the given values:
\[
(T\sin 30^\circ)(4.0\,\mathrm{m})-(200\,\mathrm{N})(2.0\,\mathrm{m})-(300\,\mathrm{N})(3.0\,\mathrm{m})=0.
\]
Since $\sin 30^\circ=0.50$,
\[
(0.50T)(4.0)-400-900=0.
\]
So
\[
2.0T-1300=0,
\]
which gives
\[
T=650\,\mathrm{N}.
\]
Now apply force balance in the horizontal direction:
\[
\sum F_x=0.
\]
The cable pulls the beam toward the wall, so its horizontal component is to the left. Therefore,
\[
H_x-T\cos 30^\circ=0.
\]
Hence
\[
H_x=T\cos 30^\circ=(650\,\mathrm{N})(0.866)\approx 5.63\times 10^2\,\mathrm{N}.
\]
So the horizontal hinge-force component is
\[
H_x\approx 563\,\mathrm{N}
\]
to the right.
Now apply force balance in the vertical direction:
\[
\sum F_y=0.
\]
Thus
\[
H_y+T\sin 30^\circ-W_b-W_s=0.
\]
Substitute the values:
\[
H_y+(650\,\mathrm{N})(0.50)-200\,\mathrm{N}-300\,\mathrm{N}=0.
\]
So
\[
H_y+325-500=0,
\]
which gives
\[
H_y=175\,\mathrm{N}.
\]
Thus the vertical hinge-force component is upward.
The final answers are
\[
T=650\,\mathrm{N},
\qquad
H_x\approx 563\,\mathrm{N}\text{ to the right},
\qquad
H_y=175\,\mathrm{N}\text{ upward}.
\]
This is the standard statics strategy: pair $\sum \vec{F}=\vec{0}$ with $\sum \tau=0$, choose a pivot that eliminates unknown torque contributions, solve the torque equation first, and then use force balance to determine the support forces.

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\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.