Rolling Without Slipping and Coupled Motion

Driving question

How can one contact condition lock a body's translation and rotation together?

Before: predict the race

A solid cylinder and a hoop have equal mass and radius and start from rest at the same ramp height. Predict which arrives first and name the evidence your model needs.

Hook: the contact point is instantaneously at rest

For pure rolling on a stationary surface, the wheel's bottom point has zero instantaneous velocity relative to the ground. The center translates at $v_{cm}$ while rotation contributes an equal backward speed $R\omega$ at contact:

$v_{cm}=R\omega.$

This is a kinematic constraint, not an extra force.

Coupled kinematics

When $R$ is constant and rolling remains no-slip,

$a_{cm}=R\alpha.$

The top point moves at $2v_{cm}$ relative to the ground, the center at $v_{cm}$, and the contact point at zero. These are instantaneous velocities; the material point at contact still has centripetal acceleration.

Worked example: wheel speed

A wheel of radius $0.25\text{ m}$ rolls without slipping at $v_{cm}=3.0\text{ m/s}$. Then

$\omega=\frac{v_{cm}}{R}=\frac{3.0}{0.25}=12\text{ rad/s}.$

If $a_{cm}=1.5\text{ m/s}^2$, then $\alpha=6.0\text{ rad/s}^2$.

Energy partitions in rolling

For a rigid object,

$K=\frac12Mv_{cm}^2+\frac12I_{cm}\omega^2.$

Writing $I_{cm}=kMR^2$ and using $v_{cm}=R\omega$ gives

$K=\frac12Mv_{cm}^2(1+k).$

Rolling from rest down through height $h$ with negligible losses:

$Mgh=\frac12Mv^2(1+k),\qquad v=\sqrt{\frac{2gh}{1+k}}.$

A solid cylinder has $k=1/2$ and a hoop has $k=1$, so the cylinder reaches a larger speed from the same height.

Numerical comparison

At $h=0.60\text{ m}$ with $g=9.8\text{ m/s}^2$:

$v_{cyl}=\sqrt{\frac{2(9.8)(0.60)}{1.5}}=2.80\text{ m/s},$

$v_{hoop}=\sqrt{\frac{2(9.8)(0.60)}{2}}=2.42\text{ m/s}.$

Static friction: constraint enforcer

Static friction prevents relative sliding at contact when the needed force does not exceed $\mu_sN$. On a passive object rolling down a ramp, friction points uphill and supplies the torque that increases $\omega$. In other driven situations it can point downhill or be zero. Determine its direction from the required no-slip dynamics—not from the center's motion alone.

For an object rolling down angle $\theta$,

$Mg\sin\theta-f_s=Ma,$

$f_sR=I\alpha,$

and $a=R\alpha$. Therefore,

$a=\frac{g\sin\theta}{1+I/(MR^2)}.$

During: retrieval pause

Without calculating, rank the acceleration of a hoop, solid cylinder, and solid sphere of equal radius on the same ramp. Explain using $k=I/(MR^2)$.

Coupled bodies

A cord that does not slip on a pulley couples linear and angular variables: $v=R\omega$ and $a=R\alpha$. The pulley needs a net torque, so tensions on its two sides generally differ. Treat each body and the pulley as separate systems, then connect their equations using the constraint.

Misconception checks

• Static friction need not dissipate mechanical energy. For ideal rolling on a stationary rigid surface, the contact point has zero instantaneous displacement.
• Zero contact-point velocity does not mean zero acceleration.
• Equal mass and radius do not imply equal rolling acceleration; mass distribution sets $I$.
• If slipping occurs, do not use $v=R\omega$ automatically.

After: explain and transfer

A wheel begins sliding and spinning, then transitions to rolling. State which constraint becomes valid only after slipping stops, and describe how you would test that claim from video data.

AP-style synthesis

For every rolling problem: define the system, decide whether no-slip is justified, draw forces, write translation and rotation equations or an energy model, impose the constraint only when valid, and check units plus limiting cases.

Further learning

See sources.yaml for the College Board alignment and OpenStax verification references. This lesson and all problems are original and do not imply College Board endorsement.