\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.}

\wc{Static friction does not always oppose motion}{Static friction opposes \emph{impending relative motion at the contact point}, not the motion of the object as a whole. In rolling without slipping down an incline, the object moves downward but static friction points \emph{up} the incline, because without friction the contact point would slide downward. The friction prevents that slip by pulling the contact backward.}

\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.

1. For pure rolling, $s_{\mathrm{cm}}=R\theta$, $v_{\mathrm{cm}}=R\omega$, $a_{t,\mathrm{cm}}=R\alpha$.

2. 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 $K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2$.

3. 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, $Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$.

4. 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 $Mg\sin\beta-f_s=Ma_{\mathrm{cm}}$, $f_sR=I_{\mathrm{cm}}\alpha$, $a_{\mathrm{cm}}=R\alpha$. Therefore, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$.
For an object accelerating down the incline, the static friction force on the object points up the incline.}

\nt{When several objects roll without slipping down the same incline from the same height, the one with the smallest $I_{\mathrm{cm}}/(MR^2)$ ratio reaches the bottom first. Race order: solid sphere ($\tfrac25=0.40$) $>$ solid cylinder ($\tfrac12=0.50$) $>$ hollow cylinder ($\simeq1$) $>$ thin hoop ($1.0$). Mass and radius do not appear in the comparison --- only the shape-dependent coefficient matters.}

\wc{Rolling race order depends on moment of inertia}{Objects with a \emph{smaller} $I/(MR^2)$ ratio accelerate faster and reach the bottom first. For example: solid sphere ($\tfrac25$) beats solid cylinder ($\tfrac12$) beats thin hoop ($1$). The mass and radius cancel out --- only the \emph{shape distribution} matters. More mass near the rim means a larger fraction of energy goes into rotation, leaving less for translation.}

\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 $\mu_s N$.
