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\subsection{Rotational Kinetic Energy and Work by Torque}
This subsection introduces the energy description of fixed-axis rotation. In AP mechanics, the key idea is that a net torque acting through an angular displacement changes a rigid body's rotational kinetic energy in the same way that a net force acting through a displacement changes translational kinetic energy.
\dfn{Rotational kinetic energy and incremental work by torque}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the body's moment of inertia about that axis, let $\vec{\omega}=\omega\hat{k}$ denote its angular velocity, and let $\omega=|\vec{\omega}|$ denote the angular speed. The \emph{rotational kinetic energy} of the rigid body is
\[
K_{\mathrm{rot}}=\tfrac12 I\omega^2.
\]
Now let the body undergo an infinitesimal angular displacement $d\theta$ about the same axis, with the positive direction chosen consistently with $\hat{k}$. Let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis. Then the incremental work done by the net torque is
\[
dW=\tau_{\mathrm{net}}\,d\theta.
\]
If $\tau_{\mathrm{net}}$ is constant over a finite angular displacement $\Delta\theta$, then
\[
W=\tau_{\mathrm{net}}\Delta\theta.
\]
The SI unit of both rotational kinetic energy and work is the joule, where $1\,\mathrm{J}=1\,\mathrm{N\cdot m}$.}
\thm{Rotational work-energy relation for fixed-axis motion}{Consider a rigid body rotating about a fixed axis with moment of inertia $I$. Let $\omega_i$ and $\omega_f$ denote the initial and final angular speeds, let $\theta_i$ and $\theta_f$ denote the corresponding angular positions, and let $\tau_{\mathrm{net}}$ denote the signed net external torque about the axis. Then the net rotational work from the initial state to the final state is
\[
W_{\mathrm{net}}=\int_{\theta_i}^{\theta_f} \tau_{\mathrm{net}}\,d\theta,
\]
and the rotational work-energy theorem is
\[
W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}=K_{\mathrm{rot},f}-K_{\mathrm{rot},i}
=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
\]
Thus a positive net torque doing positive work increases rotational kinetic energy, while negative net work decreases it.}
\nt{This result is the rotational analog of the translational work-energy theorem $W_{\mathrm{net}}=\Delta K$ with $K=\tfrac12 mv^2$. The correspondence is
\[
\vec{F}_{\mathrm{net}}\leftrightarrow \vec{\tau}_{\mathrm{net}},
\qquad
x\leftrightarrow \theta,
\qquad
v\leftrightarrow \omega,
\qquad
\tfrac12 mv^2\leftrightarrow \tfrac12 I\omega^2.
\]
In this subsection, assume a rigid body rotating about one fixed axis so that every part of the body shares the same angular displacement $\theta$ and angular speed $\omega$, and so that $I$ stays constant about that axis. The sign convention for $\tau_{\mathrm{net}}$ must match the sign convention for $\theta$.}
\pf{Why $K_{\mathrm{rot}}=\tfrac12 I\omega^2$ and $W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}$}{Model the rigid body as particles labeled by an index $i$. Let particle $i$ have mass $m_i$ and perpendicular distance $r_{\perp,i}$ from the fixed axis. Because the body is rigid, each particle has speed
\[
v_i=r_{\perp,i}\omega.
\]
Therefore the total kinetic energy is
\[
K_{\mathrm{rot}}=\sum_i \tfrac12 m_i v_i^2
=\sum_i \tfrac12 m_i(r_{\perp,i}\omega)^2
=\tfrac12 \omega^2\sum_i m_i r_{\perp,i}^2.
\]
Since
\[
I=\sum_i m_i r_{\perp,i}^2,
\]
it follows that
\[
K_{\mathrm{rot}}=\tfrac12 I\omega^2.
\]
For the work-energy relation, start with the fixed-axis rotational form of Newton's second law,
\[
\tau_{\mathrm{net}}=I\alpha,
\]
where $\alpha=d\omega/dt$. Multiply both sides by $d\theta$:
\[
\tau_{\mathrm{net}}\,d\theta=I\alpha\,d\theta.
\]
Using $\omega=d\theta/dt$, we have $d\theta=\omega\,dt$, so
\[
I\alpha\,d\theta=I\frac{d\omega}{dt}(\omega\,dt)=I\omega\,d\omega.
\]
Thus
\[
dW=\tau_{\mathrm{net}}\,d\theta=I\omega\,d\omega=d\!\left(\tfrac12 I\omega^2\right)=dK_{\mathrm{rot}}.
\]
Integrating from the initial state to the final state gives
\[
W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}.
\]}
\qs{Worked example}{A wheel of radius $R=0.25\,\mathrm{m}$ rotates about a frictionless fixed axle. Its moment of inertia about the axle is $I=1.0\,\mathrm{kg\cdot m^2}$. A light string is wrapped around the rim, and a student pulls tangentially on the string with constant force magnitude $F=12\,\mathrm{N}$. The string does not slip, and a length $s=2.0\,\mathrm{m}$ of string unwinds. Initially the wheel is already spinning in the same direction as the applied torque with angular speed $\omega_i=4.0\,\mathrm{rad/s}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item the torque magnitude $\tau$ applied by the string,
\item the angular displacement $\Delta\theta$ during the pull,
\item the work done on the wheel by the torque, and
\item the final angular speed $\omega_f$.
\end{enumerate}}
\sol Because the pull is tangential to the rim, the lever arm equals the radius $R$. Therefore the applied torque magnitude is
\[
\tau=RF=(0.25\,\mathrm{m})(12\,\mathrm{N})=3.0\,\mathrm{N\cdot m}.
\]
Since the string does not slip, the unwound length equals the arc length at the rim:
\[
s=R\Delta\theta.
\]
Hence
\[
\Delta\theta=\frac{s}{R}=\frac{2.0\,\mathrm{m}}{0.25\,\mathrm{m}}=8.0\,\mathrm{rad}.
\]
The torque is constant and acts in the direction of rotation, so the work done on the wheel is
\[
W=\tau\Delta\theta=(3.0\,\mathrm{N\cdot m})(8.0\,\mathrm{rad})=24\,\mathrm{J}.
\]
This also agrees with $W=Fs=(12\,\mathrm{N})(2.0\,\mathrm{m})=24\,\mathrm{J}$.
Now apply the rotational work-energy theorem:
\[
W=\Delta K_{\mathrm{rot}}=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
\]
Substitute the known values:
\[
24=\tfrac12 (1.0)\omega_f^2-\tfrac12 (1.0)(4.0)^2.
\]
Since
\[
\tfrac12 (1.0)(4.0)^2=8,
\]
we get
\[
24=\tfrac12 \omega_f^2-8.
\]
Add $8$ to both sides:
\[
32=\tfrac12 \omega_f^2.
\]
Multiply by $2$:
\[
\omega_f^2=64.
\]
Therefore,
\[
\omega_f=8.0\,\mathrm{rad/s}.
\]
So the results are
\[
\tau=3.0\,\mathrm{N\cdot m},
\qquad
\Delta\theta=8.0\,\mathrm{rad},
\qquad
W=24\,\mathrm{J},
\qquad
\omega_f=8.0\,\mathrm{rad/s}.
\]
The key idea is that the applied torque does positive work, so the wheel's rotational kinetic energy increases.

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\subsection{Angular Momentum and Angular Impulse}
This subsection connects torque to rotational motion in the same way that linear impulse connects force to linear momentum. The key AP idea is that a net external torque acting for a time interval changes angular momentum about a chosen point.
\dfn{Angular momentum and angular impulse}{Let $O$ denote a chosen reference point. Let a particle of mass $m$ have position vector $\vec{r}$ measured from $O$, velocity $\vec{v}$, and linear momentum $\vec{p}=m\vec{v}$. The angular momentum of the particle about $O$ is
\[
\vec{L}_O=\vec{r}\times \vec{p}.
\]
For a system of particles, the total angular momentum about $O$ is
\[
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
\]
Let $\vec{\tau}_{\mathrm{ext},O}(t)$ denote the net external torque about the same point $O$ over a time interval from $t_i$ to $t_f$. The angular impulse about $O$ over that interval is
\[
\vec{J}_{\tau,O}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
Its SI unit is $\mathrm{N\cdot m\cdot s}$, which is equivalent to $\mathrm{kg\cdot m^2/s}$. For a rigid body rotating about a fixed axis with moment of inertia $I$ about that axis and angular velocity $\vec{\omega}$, the angular momentum simplifies to
\[
\vec{L}=I\vec{\omega}.
\]
}
\thm{Torque-angular momentum relation and angular impulse theorem}{Let $O$ denote a chosen reference point. Let $\vec{L}_O(t)$ be the total angular momentum of a particle or system about $O$, and let $\vec{\tau}_{\mathrm{ext},O}(t)$ be the net external torque about $O$. Then
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrating from $t_i$ to $t_f$ gives
\[
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{J}_{\tau,O}.
\]
For fixed-axis rotation along a chosen axis with unit vector $\hat{k}$, let $\vec{L}=L\hat{k}$ and let $\vec{\tau}_{\mathrm{ext}}=\tau_{\mathrm{ext}}\hat{k}$. Then
\[
\frac{dL}{dt}=\tau_{\mathrm{ext}},
\qquad
\Delta L=\int_{t_i}^{t_f} \tau_{\mathrm{ext}}\,dt,
\]
and if $\tau_{\mathrm{ext}}$ is constant,
\[
\Delta L=\tau_{\mathrm{ext}}\Delta t.
\]
}
\nt{Angular momentum and torque depend on the chosen point $O$. The same object can have different $\vec{L}_O$ and $\vec{\tau}_{\mathrm{ext},O}$ when a different origin is chosen, so use the same reference point consistently throughout a problem. The direction of $\vec{L}$ and $\vec{\tau}$ is set by the right-hand rule and is perpendicular to the plane of the relevant cross product. In many AP fixed-axis problems, this vector bookkeeping reduces to signed scalars along one axis: counterclockwise may be chosen positive and clockwise negative. The simplification $\vec{L}=I\vec{\omega}$ is valid for rigid rotation about that fixed axis with the stated moment of inertia about the same axis.}
\pf{Short derivation from $\vec{L}=\vec{r}\times \vec{p}$}{For one particle about point $O$,
\[
\vec{L}_O=\vec{r}\times \vec{p}.
\]
Differentiate with respect to time:
\[
\frac{d\vec{L}_O}{dt}=\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r}\times \frac{d\vec{p}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times \vec{F}_{\mathrm{net}}.
\]
Because $\vec{v}\times \vec{v}=\vec{0}$,
\[
\frac{d\vec{L}_O}{dt}=\vec{r}\times \vec{F}_{\mathrm{net}}=\vec{\tau}_{\mathrm{net},O}.
\]
For a system, summing over all particles gives the same form with the net external torque:
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrating from $t_i$ to $t_f$ yields
\[
\Delta \vec{L}_O=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
For a rigid body rotating about a fixed axis with constant moment of inertia $I$,
\[
\vec{L}=I\vec{\omega},
\]
so along that axis the relation becomes
\[
\vec{\tau}_{\mathrm{ext}}=I\vec{\alpha}.
\]
}
\qs{Worked example}{A flywheel rotates about a frictionless fixed axle through its center. Choose counterclockwise as positive, so the axis direction is $\hat{k}$ by the right-hand rule. The flywheel has moment of inertia $I=0.80\,\mathrm{kg\cdot m^2}$ about the axle. Initially its angular velocity is
\[
\vec{\omega}_i=12\,\hat{k}\,\mathrm{rad/s}.
\]
From $t=0$ to $t=1.5\,\mathrm{s}$, a brake pad exerts a constant external torque
\[
\vec{\tau}_{\mathrm{ext}}=-3.2\,\hat{k}\,\mathrm{N\cdot m}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the angular impulse delivered by the brake,
\item the flywheel's final angular momentum,
\item the flywheel's final angular velocity, and
\item whether the flywheel reverses direction during the interval.
\end{enumerate}}
\sol First compute the initial angular momentum using $\vec{L}=I\vec{\omega}$:
\[
\vec{L}_i=I\vec{\omega}_i=(0.80\,\mathrm{kg\cdot m^2})(12\,\hat{k}\,\mathrm{rad/s})=9.6\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
For part (a), the torque is constant, so the angular impulse is
\[
\vec{J}_\tau=\vec{\tau}_{\mathrm{ext}}\Delta t.
\]
Thus
\[
\vec{J}_\tau=(-3.2\,\hat{k}\,\mathrm{N\cdot m})(1.5\,\mathrm{s})=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s}.
\]
Using $1\,\mathrm{N\cdot m\cdot s}=1\,\mathrm{kg\cdot m^2/s}$,
\[
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
For part (b), apply the angular impulse theorem:
\[
\Delta \vec{L}=\vec{J}_\tau=\vec{L}_f-\vec{L}_i.
\]
So
\[
\vec{L}_f=\vec{L}_i+\vec{J}_\tau=(9.6-4.8)\,\hat{k}\,\mathrm{kg\cdot m^2/s}=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
For part (c), again use $\vec{L}=I\vec{\omega}$:
\[
\vec{\omega}_f=\frac{\vec{L}_f}{I}=\frac{4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}}{0.80\,\mathrm{kg\cdot m^2}}=6.0\,\hat{k}\,\mathrm{rad/s}.
\]
For part (d), the final angular velocity still points in the $+\hat{k}$ direction, so the flywheel is still rotating counterclockwise. It slows down, but it does not reverse direction during the $1.5\,\mathrm{s}$ interval.
The results are
\[
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s},
\qquad
\vec{L}_f=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s},
\qquad
\vec{\omega}_f=6.0\,\hat{k}\,\mathrm{rad/s}.
\]
This example shows the key AP idea: a torque acting over time changes angular momentum directly, and the sign of the torque determines whether the wheel speeds up or slows down.

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\subsection{Conservation of Angular Momentum}
This subsection gives the rotational conservation law that parallels conservation of linear momentum. In AP mechanics, the key question is whether the net \emph{external} torque about a chosen origin or axis is zero over the interval of interest.
\dfn{Angular-momentum-isolated system}{Consider a system of particles labeled by an index $i=1,2,\dots,N$. Choose an origin $O$. Let $\vec{r}_i$ denote the position vector of particle $i$ relative to $O$, let $m_i$ denote its mass, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its linear momentum. The total angular momentum of the system about $O$ is
\[
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
\]
Let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about the same origin. The system is said to be \emph{isolated for angular momentum about $O$} over a time interval if the external torque impulse about $O$ is zero:
\[
\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{0}.
\]
In particular, if $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ at every instant in the interval, then the system is angular-momentum-isolated about $O$. For a rigid body rotating about a fixed axis with unit vector $\hat{k}$, one often writes $\vec{L}=L\hat{k}$ and $\vec{\omega}=\omega\hat{k}$, so that in the fixed-axis case $L=I\omega$.}
\thm{Conservation law for angular momentum}{Let $\vec{L}_O$ denote the total angular momentum of a system about a chosen origin $O$, and let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about that same origin. Then
\[
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
Therefore, if the net external torque is zero throughout the interval, or more generally if the external torque impulse is zero, then
\[
\vec{L}_{O,f}=\vec{L}_{O,i},
\]
so total angular momentum about $O$ is conserved. In the common fixed-axis AP case,
\[
L_i=L_f
\qquad\Rightarrow\qquad
I_i\omega_i=I_f\omega_f
\]
when the net external torque about that axis is zero.}
\nt{Angular momentum is conserved only about the origin or axis for which the net external torque is zero, so the choice of origin matters. Internal forces and internal torques can redistribute angular momentum among parts of the system and can change the moment of inertia $I$, but they do not change the system's \emph{total} angular momentum about the chosen origin when $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$. Also, conservation of angular momentum does \emph{not} imply conservation of kinetic energy: for example, a skater can pull in her arms, decrease $I$, increase $\omega$, and increase rotational kinetic energy by doing internal work.}
\pf{Short derivation from $d\vec{L}/dt=\vec{\tau}_{\mathrm{ext}}$}{Start with the angular-momentum form of Newton's second law about the chosen origin $O$:
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrate from $t_i$ to $t_f$:
\[
\int_{t_i}^{t_f} \frac{d\vec{L}_O}{dt}\,dt=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
This gives
\[
\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
If $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ throughout the interval, or if the integral on the right is zero, then $\vec{L}_{O,f}-\vec{L}_{O,i}=\vec{0}$. Hence
\[
\vec{L}_{O,f}=\vec{L}_{O,i},
\]
which is the conservation of angular momentum. In a fixed-axis problem, this reduces to $I_i\omega_i=I_f\omega_f$.}
\qs{Worked example}{An ice skater spins about a vertical axis with unit vector $\hat{k}$. Assume the net external torque about that axis is negligible. With her arms extended, her moment of inertia is $I_i=3.0\,\mathrm{kg\cdot m^2}$ and her angular velocity is $\vec{\omega}_i=(2.0\,\mathrm{rad/s})\hat{k}$. She then pulls her arms inward so that her final moment of inertia is $I_f=1.2\,\mathrm{kg\cdot m^2}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item the final angular velocity vector $\vec{\omega}_f$,
\item the initial and final angular momentum vectors, and
\item the initial and final rotational kinetic energies.
\end{enumerate}
Explain briefly why the energy result does not contradict conservation of angular momentum.}
\sol Because the net external torque about the vertical axis is negligible, angular momentum about that axis is conserved:
\[
\vec{L}_i=\vec{L}_f.
\]
For fixed-axis rotation, $\vec{L}=I\vec{\omega}$, so
\[
I_i\vec{\omega}_i=I_f\vec{\omega}_f.
\]
For part (a), solve for the final angular velocity:
\[
\vec{\omega}_f=\frac{I_i}{I_f}\vec{\omega}_i
=\frac{3.0}{1.2}(2.0\hat{k})
=(2.5)(2.0\hat{k})
=5.0\hat{k}\,\mathrm{rad/s}.
\]
So,
\[
\boxed{\vec{\omega}_f=(5.0\,\mathrm{rad/s})\hat{k}}.
\]
For part (b), compute the angular momentum before and after:
\[
\vec{L}_i=I_i\vec{\omega}_i=(3.0)(2.0\hat{k})=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
Because angular momentum is conserved,
\[
\vec{L}_f=\vec{L}_i=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
Thus,
\[
\boxed{\vec{L}_i=\vec{L}_f=(6.0\,\mathrm{kg\cdot m^2/s})\hat{k}}.
\]
For part (c), use $K_{\mathrm{rot}}=\tfrac12 I\omega^2$.
Initially,
\[
K_{\mathrm{rot},i}=\tfrac12 I_i\omega_i^2
=\tfrac12 (3.0)(2.0)^2
=\tfrac12 (3.0)(4.0)
=6.0\,\mathrm{J}.
\]
Finally,
\[
K_{\mathrm{rot},f}=\tfrac12 I_f\omega_f^2
=\tfrac12 (1.2)(5.0)^2
=\tfrac12 (1.2)(25)
=15\,\mathrm{J}.
\]
So,
\[
\boxed{K_{\mathrm{rot},i}=6.0\,\mathrm{J}},
\qquad
\boxed{K_{\mathrm{rot},f}=15\,\mathrm{J}}.
\]
The rotational kinetic energy increases even though angular momentum stays constant. This does not contradict conservation of angular momentum because the skater does internal work while pulling in her arms. That internal work increases $K_{\mathrm{rot}}$ while the external torque remains negligible, so $\vec{L}$ is still conserved.

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\subsection{Rolling Without Slipping}
This subsection introduces pure rolling, where a rigid body both translates and rotates while the contact point does not slide relative to the surface.
\dfn{Pure rolling and the rolling constraint}{Consider a rigid body of mass $M$ and radius $R$ moving on a fixed surface. Let $s_{\mathrm{cm}}$ denote the distance traveled by its center of mass along the surface, let $v_{\mathrm{cm}}=|\vec{v}_{\mathrm{cm}}|$ denote the speed of the center of mass, and let $a_{t,\mathrm{cm}}$ denote the component of $\vec{a}_{\mathrm{cm}}$ tangent to the surface. Let $\theta$ denote the angular displacement of the body, let $\omega$ denote its angular speed, and let $\alpha$ denote its angular acceleration.
A body \emph{rolls without slipping} if the point in contact with the surface is instantaneously at rest relative to the surface. For pure rolling,
\[
s_{\mathrm{cm}}=R\theta,
\qquad
v_{\mathrm{cm}}=R\omega,
\qquad
a_{t,\mathrm{cm}}=R\alpha.
\]
These relations are called the \emph{rolling constraint}. They apply only when there is no slipping at the contact.}
\nt{Let $P$ denote the point on the rim that touches the ground at some instant. Its velocity relative to the ground is the vector sum of the center-of-mass velocity $\vec{v}_{\mathrm{cm}}$ and the velocity of $P$ relative to the center due to rotation. In pure rolling, these cancel exactly at the contact point, so $P$ is instantaneously at rest even though the body is moving.
Also, the friction in rolling without slipping is \emph{static} friction. Let $\vec{N}$ denote the normal force, let $N=|\vec{N}|$ denote its magnitude, and let $f_s$ denote the magnitude of the static friction force. Static friction is not automatically equal to $\mu_s N$. Instead, its magnitude is whatever value is required to prevent slipping, provided that value satisfies $f_s\le \mu_s N$. On level ground at constant speed, the needed static friction can even be zero.}
\ex{Illustrative example}{A wheel of radius $R=0.30\,\mathrm{m}$ rolls without slipping on level ground with angular speed $\omega=8.0\,\mathrm{rad/s}$. Find the speed of its center of mass and the speed of the top point of the wheel relative to the ground.
From the rolling constraint,
\[
v_{\mathrm{cm}}=R\omega=(0.30)(8.0)=2.4\,\mathrm{m/s}.
\]
At the top of the wheel, the translational velocity and the rotational velocity point in the same direction, so the top-point speed is
\[
v_{\mathrm{top}}=v_{\mathrm{cm}}+R\omega=2v_{\mathrm{cm}}=4.8\,\mathrm{m/s}.
\]
Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves at $4.8\,\mathrm{m/s}$ relative to the ground.}
\mprop{Rolling kinematics, energy, and incline dynamics}{Consider a rigid body of mass $M$, radius $R$, and moment of inertia $I_{\mathrm{cm}}$ about its center of mass. Let $s_{\mathrm{cm}}$, $v_{\mathrm{cm}}$, and $a_{t,\mathrm{cm}}$ denote the center-of-mass distance, speed, and tangential acceleration, and let $\theta$, $\omega$, and $\alpha$ denote the corresponding angular variables. Let $K_i$ and $U_i$ denote the initial kinetic and potential energies, and let $K_f$ and $U_f$ denote the final kinetic and potential energies.
\begin{enumerate}[label=\textbf{\arabic*.}]
\item For pure rolling,
\[
s_{\mathrm{cm}}=R\theta,
\qquad
v_{\mathrm{cm}}=R\omega,
\qquad
a_{t,\mathrm{cm}}=R\alpha.
\]
\item The total kinetic energy is the sum of translational and rotational parts:
\[
K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
\]
Using $v_{\mathrm{cm}}=R\omega$, this may also be written as
\[
K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2.
\]
\item If no nonconservative force removes mechanical energy from the system, then for rolling without slipping,
\[
K_i+U_i=K_f+U_f.
\]
For a vertical drop of magnitude $h$ from rest,
\[
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
\]
\item For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then
\[
Mg\sin\beta-f_s=Ma_{\mathrm{cm}},
\qquad
f_sR=I_{\mathrm{cm}}\alpha,
\qquad
a_{\mathrm{cm}}=R\alpha.
\]
Therefore,
\[
a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)},
\qquad
f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}.
\]
For an object accelerating down the incline, the static friction force on the object points up the incline.}
\qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item the acceleration magnitude of the center of mass,
\item the magnitude and direction of the static friction force,
\item the speed of the center of mass after the drop, and
\item the minimum coefficient of static friction required for rolling without slipping.
\end{enumerate}}
\sol Let $a_{\mathrm{cm}}$ denote the acceleration magnitude of the center of mass down the incline, and let $f_s$ denote the magnitude of the static friction force. For a solid cylinder,
\[
I_{\mathrm{cm}}=\tfrac12 MR^2.
\]
Choose the positive axis down the incline. The forces parallel to the incline are the downslope component of the weight and the upslope static friction force, so Newton's second law for translation gives
\[
Mg\sin\beta-f_s=Ma_{\mathrm{cm}}.
\]
The only torque about the center of mass is due to static friction, so
\[
f_sR=I_{\mathrm{cm}}\alpha.
\]
Because the cylinder rolls without slipping,
\[
a_{\mathrm{cm}}=R\alpha.
\]
Substitute $\alpha=a_{\mathrm{cm}}/R$ and $I_{\mathrm{cm}}=\tfrac12 MR^2$ into the torque equation:
\[
f_sR=\left(\tfrac12 MR^2\right)\frac{a_{\mathrm{cm}}}{R}.
\]
Thus,
\[
f_s=\tfrac12 Ma_{\mathrm{cm}}.
\]
Now substitute this into the translational equation:
\[
Mg\sin\beta-\tfrac12 Ma_{\mathrm{cm}}=Ma_{\mathrm{cm}}.
\]
So,
\[
Mg\sin\beta=\tfrac32 Ma_{\mathrm{cm}},
\]
and therefore
\[
a_{\mathrm{cm}}=\frac{2}{3}g\sin\beta.
\]
With $g=9.8\,\mathrm{m/s^2}$ and $\sin 30^\circ=0.50$,
\[
a_{\mathrm{cm}}=\frac{2}{3}(9.8)(0.50)=3.27\,\mathrm{m/s^2}.
\]
This is the acceleration magnitude of the center of mass, directed down the incline.
For the static friction force,
\[
f_s=\tfrac12 Ma_{\mathrm{cm}}=\tfrac12 (2.0)(3.27)=3.27\,\mathrm{N}.
\]
Its direction is up the incline, because it must provide the clockwise torque that increases the cylinder's rotation as the cylinder moves downward.
Now find the speed after the cylinder drops through height $h=0.75\,\mathrm{m}$. Since the cylinder rolls without slipping, static friction does no work at the contact point, so mechanical energy is conserved:
\[
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
\]
Using $I_{\mathrm{cm}}=\tfrac12 MR^2$ and $v_{\mathrm{cm}}=R\omega$,
\[
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12\left(\tfrac12 MR^2\right)\left(\frac{v_{\mathrm{cm}}}{R}\right)^2.
\]
So,
\[
Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac14 Mv_{\mathrm{cm}}^2
=\tfrac34 Mv_{\mathrm{cm}}^2.
\]
Cancel $M$ and solve for $v_{\mathrm{cm}}$:
\[
gh=\tfrac34 v_{\mathrm{cm}}^2,
\qquad
v_{\mathrm{cm}}^2=\frac{4}{3}gh.
\]
Substitute the numbers:
\[
v_{\mathrm{cm}}^2=\frac{4}{3}(9.8)(0.75)=9.8.
\]
Hence,
\[
v_{\mathrm{cm}}=\sqrt{9.8}=3.13\,\mathrm{m/s}.
\]
Finally, find the minimum coefficient of static friction. The normal force is
\[
N=Mg\cos\beta=(2.0)(9.8)\cos 30^\circ=17.0\,\mathrm{N}
\]
to three significant figures. For rolling without slipping, the needed static friction must satisfy
\[
f_s\le \mu_s N.
\]
Thus the minimum value is
\[
\mu_{s,\min}=\frac{f_s}{N}=\frac{3.27}{17.0}=0.192.
\]
Therefore,
\[
a_{\mathrm{cm}}=3.27\,\mathrm{m/s^2}\text{ down the incline},
\qquad
f_s=3.27\,\mathrm{N}\text{ up the incline},
\]
\[
v_{\mathrm{cm}}=3.13\,\mathrm{m/s},
\qquad
\mu_{s,\min}=0.192.
\]
The key ideas are the rolling constraint $v_{\mathrm{cm}}=R\omega$, the split of kinetic energy into translational and rotational parts, and the fact that static friction adjusts to the amount needed for no slipping rather than automatically equaling $\mu_s N$.

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\subsection{Circular Orbits, Satellite Speed, and Orbital Energy}
This subsection focuses on Newtonian circular orbits around a much more massive central body. The main AP results are the orbital-speed formula and the linked kinetic, potential, and total-energy relations for a satellite in a circular orbit.
\dfn{Circular-orbit setup and gravitational potential-energy reference}{Let a satellite of mass $m$ move in a circular orbit around a spherically symmetric body of mass $M$. Let $O$ denote the center of the central body. Let $\vec{r}$ denote the satellite's position vector from $O$, let $r=|\vec{r}|$ denote the constant orbital radius, let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector, let $\vec{v}$ denote the satellite's velocity, and let $v=|\vec{v}|$ denote its speed. Let $\vec{L}_O=\vec{r}\times m\vec{v}$ denote the satellite's angular momentum about $O$. Let $K$ denote kinetic energy, let $U$ denote gravitational potential energy, and let $E=K+U$ denote total mechanical energy.
Choose the gravitational potential-energy reference so that $U=0$ when the separation is infinite. Then for separation $r$,
\[
U(r)=-\frac{GMm}{r}.
\]
The gravitational force on the satellite is
\[
\vec{F}_g=-\frac{GMm}{r^2}\hat{r}.
\]
}
\thm{Circular-orbit speed and energy relations}{Let a satellite of mass $m$ move in a circular orbit of radius $r$ around a spherically symmetric body of mass $M$. Because $\vec{F}_g$ is parallel to $\vec{r}$,
\[
\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0},
\]
so the orbital angular momentum about the center is conserved.
For a circular orbit, gravity supplies the centripetal force:
\[
\frac{GMm}{r^2}=\frac{mv^2}{r}.
\]
Therefore,
\[
v=\sqrt{\frac{GM}{r}}.
\]
The kinetic energy is then
\[
K=\frac12 mv^2=\frac{GMm}{2r}.
\]
Using $U=-GMm/r$, the total mechanical energy is
\[
E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}.
\]
Thus, for a circular Newtonian orbit,
\[
v=\sqrt{\frac{GM}{r}},
\qquad
K=\frac{GMm}{2r},
\qquad
U=-\frac{GMm}{r},
\qquad
E=-\frac{GMm}{2r}.
\]
}
\ex{Illustrative example}{Two satellites of the same mass orbit the same planet in circular orbits. Satellite A has orbital radius $r$, and satellite B has orbital radius $4r$.
Since $v=\sqrt{GM/r}$,
\[
v_B=\sqrt{\frac{GM}{4r}}=\frac12\sqrt{\frac{GM}{r}}=\frac12 v_A.
\]
Since $K=GMm/(2r)$,
\[
K_B=\frac{GMm}{2(4r)}=\frac14 K_A.
\]
Also,
\[
U_B=-\frac{GMm}{4r}=\frac14 U_A,
\qquad
E_B=-\frac{GMm}{8r}=\frac14 E_A.
\]
So a larger circular orbit has a lower speed and a less negative total energy.}
\nt{With the reference choice $U(\infty)=0$, any bound gravitational orbit has negative total mechanical energy. For a circular orbit specifically, $E=-K=U/2<0$. Also, the formulas $v=\sqrt{GM/r}$, $K=GMm/(2r)$, and $E=-GMm/(2r)$ are for \emph{circular} Newtonian orbits only. In a noncircular orbit, the speed is not constant, so these same expressions do not apply at every point of the motion.}
\qs{Worked example}{An Earth satellite of mass $m=850\,\mathrm{kg}$ moves in a circular orbit at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let Earth's mass be $M_E=5.97\times 10^{24}\,\mathrm{kg}$, Earth's radius be $R_E=6.37\times 10^6\,\mathrm{m}$, and the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\cdot m^2/kg^2}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item why the satellite's angular momentum about Earth's center is conserved,
\item the orbital speed $v$,
\item the kinetic energy $K$,
\item the gravitational potential energy $U$, and
\item the total mechanical energy $E$.
\end{enumerate}}
\sol Let $r$ denote the orbital radius measured from Earth's center. First compute it from the given altitude:
\[
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
\]
For part (a), the only significant force on the satellite is Earth's gravitational force, which points along the line from Earth to the satellite. Therefore $\vec{F}_g$ is parallel to $\vec{r}$, so the torque about Earth's center is
\[
\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0}.
\]
Since
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_O,
\]
it follows that
\[
\frac{d\vec{L}_O}{dt}=\vec{0},
\]
so the satellite's angular momentum about Earth's center is conserved.
For part (b), use the circular-orbit speed formula:
\[
v=\sqrt{\frac{GM_E}{r}}.
\]
Substitute the values:
\[
v=\sqrt{\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)}{6.77\times 10^6}}.
\]
This gives
\[
v\approx 7.67\times 10^3\,\mathrm{m/s}.
\]
For part (c), the kinetic energy is
\[
K=\frac12 mv^2.
\]
Using $m=850\,\mathrm{kg}$ and the value of $v^2=GM_E/r$,
\[
K=\frac12(850)\left(7.67\times 10^3\right)^2\,\mathrm{J}
\approx 2.50\times 10^{10}\,\mathrm{J}.
\]
For part (d), the gravitational potential energy is
\[
U=-\frac{GM_E m}{r}.
\]
Substitute the numbers:
\[
U=-\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)(850)}{6.77\times 10^6}\,\mathrm{J}
\approx -5.00\times 10^{10}\,\mathrm{J}.
\]
For part (e), the total mechanical energy is
\[
E=K+U.
\]
So,
\[
E=(2.50\times 10^{10})+(-5.00\times 10^{10})\,\mathrm{J}
\approx -2.50\times 10^{10}\,\mathrm{J}.
\]
This agrees with the circular-orbit relation
\[
E=-\frac{GM_E m}{2r}.
\]
Therefore,
\[
v\approx 7.67\times 10^3\,\mathrm{m/s},
\qquad
K\approx 2.50\times 10^{10}\,\mathrm{J},
\]
\[
U\approx -5.00\times 10^{10}\,\mathrm{J},
\qquad
E\approx -2.50\times 10^{10}\,\mathrm{J}.
\]
The negative total energy shows that the satellite is in a bound orbit.