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Rolling
Without Slipping

Explore the physics of rolling motion, where translational and rotational kinematics unite through a single elegant constraint.

AP Physics 1

What Is Rolling Motion?

Rolling is a combination of translational motion (the center of mass moves forward) and rotational motion (the object spins about its own axis). A wheel on a road, a bowling ball down a lane, or a cylinder on a ramp: all exhibit rolling.

Two Motions in One

Translational
The center of mass moves with velocity \( v_{\text{cm}} \). Governed by \( F_{\text{net}} = Ma_{\text{cm}} \).
Rotational
The object spins with angular velocity \( \omega \). Governed by \( \tau_{\text{net}} = I\alpha \).

The Rolling Constraint

\( R \) contact point \( v_{\text{cm}} \) \( \omega \) \( s = R\theta \)

Why the Contact Point Is at Rest

Consider any point on the rim. Its velocity is the sum of two contributions:

1
Translation: Every point shares \( v_{\text{cm}} \) (forward).
2
Rotation about the center: The bottom point moves backward at speed \( R\omega \).
3
Result at bottom: \( v_{\text{cm}} - R\omega = 0 \). The contact point has zero velocity.

Velocity at three points

For ideal rolling in the ground frame, translation and rotation add at the top and cancel at the contact point. Both panels are needed for the vector construction; zero contact speed does not mean zero acceleration.

The photograph shows mounted apparatus. The modeled arrows describe ideal rightward rolling on a stationary surface: top speed 2v, center speed v, contact speed zero. The required vector-sum companion explains translation plus rotation; zero speed does not imply zero acceleration.
The photograph shows mounted apparatus. The modeled arrows describe ideal rightward rolling on a stationary surface: top speed 2v, center speed v, contact speed zero. The required vector-sum companion explains translation plus rotation; zero speed does not imply zero acceleration.
In the ground frame, v_point = v_CM + ω × r. Top: v + v = 2v; center: v + 0 = v; contact: v − v = 0. Zero contact velocity does not imply zero contact acceleration.
In the ground frame, v_point = v_CM + ω × r. Top: v + v = 2v; center: v + 0 = v; contact: v − v = 0. Zero contact velocity does not imply zero contact acceleration.

Kinetic Energy of a Rolling Object

The total kinetic energy is the sum of translational KE and rotational KE about the center of mass:

Using \( v_{\text{cm}} = R\omega \), substitute \( \omega = v_{\text{cm}} / R \):

Rolling vs Sliding

Rolling Without Slipping Sliding (No Rolling)
Constraint \( v_{\text{cm}} = R\omega \) \( \omega = 0 \) or uncoupled
Contact point At rest (zero velocity) Moving (kinetic friction acts)
Friction type Static (no slip) Kinetic (sliding)
Energy lost to friction None (static friction does no work) Yes (heat generated)
KE expression \( \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \) \( \frac{1}{2}Mv^2 \) only

Common Moments of Inertia

Solid Sphere
\( I = \frac{2}{5}MR^2 \)
Hollow Sphere
\( I = \frac{2}{3}MR^2 \)
Solid Cylinder
\( I = \frac{1}{2}MR^2 \)
Hollow Cylinder
\( I = MR^2 \)

Rolling Down an Incline

\( \theta \) \( h \) \( Mg \) \( f_s \)

The Role of Static Friction

On an incline, static friction provides the torque needed to spin the object. Without it, the object would simply slide.

1
Gravity pulls the CM downhill: component \( Mg\sin\theta \) along the ramp.
2
Static friction acts uphill at the contact point, creating a net torque about the CM.
3
No energy is lost: static friction does zero work because the contact point has zero displacement.

How to Solve Rolling Problems

1
Draw a free-body diagram. Include gravity, normal force, and friction at the contact point.
2
Apply Newton's second law for translation: \( F_{\text{net}} = Ma_{\text{cm}} \).
3
Apply the torque equation about the CM: \( \tau_{\text{net}} = I_{\text{cm}}\alpha \).
4
Use the rolling constraint: \( a_{\text{cm}} = R\alpha \) to link the two equations.
5
Solve the system for unknowns (acceleration, friction, velocity).

Example 1: Solid Sphere on a Ramp

A solid sphere of mass \( 2 \text{ kg} \) and radius \( 0.1 \text{ m} \) rolls without slipping from rest down a ramp of height \( 3 \text{ m} \). Find its speed at the bottom.

Example 2: Race Down an Incline

A solid cylinder, a solid sphere, and a hollow cylinder are released simultaneously from the same height. Which reaches the bottom first?

Object \( I_{\text{cm}} / MR^2 \) \( a_{\text{cm}} \) Finish Order
Solid Sphere \( \frac{2}{5} \) \( \frac{5}{7}g\sin\theta \) 1st
Solid Cylinder \( \frac{1}{2} \) \( \frac{2}{3}g\sin\theta \) 2nd
Hollow Cylinder \( 1 \) \( \frac{1}{2}g\sin\theta \) 3rd

A tactile rolling race with visible mass distribution

A tabletop ramp with a hoop, disk and sphere, the latter two showing conceptual exposed interiors.For bodies rolling without slipping from the same initial center-of-mass height, mass distribution determines acceleration and the partition of kinetic energy. Equal elapsed time and equal height loss are separate comparisons.

Rolling race at equal elapsed time

Rolling race at equal elapsed time. Equal-mass, equal-radius bodies start together on a 30° ramp. At 0.6 s, the hoop travels 0.441 m, the disk 0.588 m and the solid sphere 0.630 m, with no slipping.Equal-mass, equal-radius bodies start together on a 30° ramp. At 0.6 s, the hoop travels 0.441 m, the disk 0.588 m and the solid sphere 0.630 m, with no slipping.

Velocities around a rolling wheel

Velocities around a rolling wheel. In the stationary ground frame, translation and rotation combine to give zero velocity at the contact point, v at the center and 2v at the top. Zero instantaneous velocity does not imply zero acceleration.In the stationary ground frame, translation and rotation combine to give zero velocity at the contact point, v at the center and 2v at the top. Zero instantaneous velocity does not imply zero acceleration.

Example 3: Finding Friction Force

A solid cylinder (\( M = 4 \text{ kg} \), \( R = 0.2 \text{ m} \)) rolls without slipping down a \( 30^\circ \) incline. Find the friction force and the acceleration.

1
Translation: \( Mg\sin\theta - f_s = Ma_{\text{cm}} \)
2
Rotation about CM: \( f_s R = I_{\text{cm}}\alpha = \frac{1}{2}MR^2 \cdot \frac{a_{\text{cm}}}{R} \) → \( f_s = \frac{1}{2}Ma_{\text{cm}} \)
3
Substitute: \( Mg\sin\theta = Ma_{\text{cm}} + \frac{1}{2}Ma_{\text{cm}} = \frac{3}{2}Ma_{\text{cm}} \)

Common Mistakes in Rolling

✕ Forgetting Rotational KE
Using only \( \frac{1}{2}Mv^2 \) ignores the energy stored in rotation, giving a speed that is too high.
✕ Using Kinetic Friction
Rolling without slipping involves static friction, not kinetic. Static friction does no work and dissipates no energy.
✕ Thinking Mass Matters in a Race
On an incline, \( M \) and \( R \) cancel. Only the ratio \( I_{\text{cm}} / MR^2 \) (shape) determines the winner.

Summary of Rolling Formulas

Quantity Formula Notes
Rolling constraint \( v_{\text{cm}} = R\omega \) Links translation and rotation
Total KE \( \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \) Both terms always needed
Speed from height \( h \) \( \sqrt{\frac{2gh}{1 + I/(MR^2)}} \) Energy conservation shortcut
Acceleration on incline \( \frac{g\sin\theta}{1 + I/(MR^2)} \) Independent of \( M \) and \( R \)
Contact point velocity \( 0 \) Defines "without slipping"
Top point velocity \( 2v_{\text{cm}} \) Translation + rotation add

Question 1 of 5

A wheel rolls without slipping with \( v_{\text{cm}} = 4 \text{ m/s} \). What is the velocity of the contact point with the ground?

Question 2 of 5

A solid sphere and a hollow sphere of equal mass and radius roll down the same incline from rest. Which reaches the bottom first?

Question 3 of 5

A solid cylinder rolls without slipping down a ramp of height \( h = 5 \text{ m} \). What is \( v_{\text{cm}} \) at the bottom? (Use \( g = 10 \text{ m/s}^2 \))

Question 4 of 5

During rolling without slipping on a level surface, what type of friction acts on the object and how much work does it do?

Question 5 of 5

A rolling wheel has \( v_{\text{cm}} = 3 \text{ m/s} \). What is the speed of the topmost point of the wheel relative to the ground?