Rolling without slipping

A body that rolls without slipping does two things at once: its center moves with speed $v$, and it spins with angular speed $\omega$. The rolling condition $v=R\omega$ ties them together. Because the body both moves and turns, its kinetic energy is the sum of a translational part and a rotational part.

The everyday picture is a ball rolling down a ramp. Gravity sets how much energy the body gains, but its shape sets how that energy splits between moving and spinning, and so sets the speed at the bottom. A solid sphere, a disk, and a hoop released together do not tie.

Four habits carry the topic: keep both kinetic-energy terms, link the spin to the speed with $v=R\omega$, rank rolling bodies by the shape factor $I/mR^2$ rather than mass or radius, and remember that the static friction does no work.

Rolling without slipping means the contact point is not sliding: at each instant the bottom of the wheel is momentarily at rest on the ground. Combine the center's forward motion with the spin, and the result is $v=R\omega$, so $\omega=\dfrac{v}{R}$. Dropping the radius and writing $\omega=v$ is the classic slip.

Differentiate once and the same link holds for the accelerations: $a=R\alpha$. Two consequences follow: the contact point has zero velocity, while the top of the wheel moves at $2v$, twice the center's speed.

• $v=R\omega$, so $\omega=\dfrac{v}{R}$ (not $\omega=v$).
• $a=R\alpha$ for the matching accelerations.

Rolling without slipping ties the center speed to the spin: $v=R\omega$. The contact point is instantaneously at rest; the top moves at $2v$.

A rolling body's kinetic energy is the sum of a translational term and a rotational term:

• $K=\tfrac{1}{2}mv^2+\tfrac{1}{2}I\omega^2$.

Substitute the constraint $\omega=v/R$ and write the moment of inertia as $I=\beta mR^2$, where $\beta=I/mR^2$ is the dimensionless shape factor ($\tfrac{2}{5}$ for a solid sphere, $\tfrac{1}{2}$ for a disk, $1$ for a hoop). The rotational term becomes $\tfrac{1}{2}\beta mv^2$, so

• $K=\tfrac{1}{2}mv^2\,(1+\beta)$.

For a solid sphere ($\beta=\tfrac{2}{5}$) rolling at $v$, $K=\tfrac{1}{2}mv^2+\tfrac{1}{5}mv^2=\tfrac{7}{10}mv^2$. Reporting only $\tfrac{1}{2}mv^2$, or only the rotational part, is the most common error.

Rolling kinetic energy is $\tfrac{1}{2}mv^2+\tfrac{1}{2}I\omega^2$. With $\omega=v/R$ that is $\tfrac{1}{2}mv^2(1+\beta)$; both terms are always there.

Release a body from rest and let it roll without slipping down a height $h$. The static friction does no work, so mechanical energy is conserved: $mgh=\tfrac{1}{2}mv^2(1+\beta)$. Solving for the speed at the bottom,

• $v=\sqrt{\dfrac{2gh}{1+\beta}}$.

The mass cancels and the radius cancels: the bottom speed depends only on the height and the shape factor $\beta$. A smaller $\beta$ leaves more energy for translation, so the body is faster. Every shape arrives with the same total kinetic energy $mgh$; only the split between moving and spinning differs.

Down a ramp, $v=\sqrt{2gh/(1+\beta)}$: independent of mass and radius, set only by the shape factor. A smaller $\beta$ wins.

For a body rolling without slipping on a stationary surface, the friction at the contact is static, and it does no work. The contact point is instantaneously at rest, so the friction force acts through zero displacement; with no sliding there is no heating, and mechanical energy is conserved.

On an incline the static friction points up the slope. It supplies the torque that spins the body up to keep $v=R\omega$. Without friction there is no such torque, and the body would slide instead of roll. Friction is essential to rolling, not a drain on the energy.

• Rolling without slipping: the friction is static and does zero work.
• On a ramp it points up the slope and provides the rolling torque.

Static friction in rolling without slipping does no work, because the contact point does not move. It points up the incline and provides the torque that makes the body roll.

Rolling motion invites a familiar cluster of errors: keeping only one kinetic-energy term, breaking the $v=R\omega$ link, ranking a race by mass, or blaming static friction for lost energy.

• One KE termUsing only $\tfrac{1}{2}mv^2$ or only $\tfrac{1}{2}I\omega^2$ for a rolling body. Fix: $K=\tfrac{1}{2}mv^2+\tfrac{1}{2}I\omega^2$.
• Constraint slipBreaking the link $v=R\omega$: dropping the $R$, inverting it, or reading a stationary contact point as no spin. Fix: $\omega=v/R$.
• Ranked by massRanking an incline race by mass or radius. Fix: rank by the shape factor $\beta=I/mR^2$; mass and radius cancel.
• Friction drains energyTreating the static friction as if it removed energy. Fix: it does no work in rolling without slipping, so energy is conserved.

Keep both energy terms, link the spin to the speed, rank by shape, and let the friction do no work. The whole topic comes down to those four habits.