5.8 Rolling Motion

Rolling Without Slipping

For an object that rolls without slipping, the contact point has zero velocity relative to the surface. This gives the constraint v_cm = Rω and a_cm = Rα, linking translational and rotational motion. Static friction enables rolling but does no work and does not dissipate energy in this case.

Energy of Rolling Objects

A rolling object has both translational and rotational kinetic energy: KE = ½mv_cm² + ½Iω². Using v_cm = Rω, this becomes KE = ½mv_cm²(1 + I/mR²). Objects with larger moment of inertia (relative to mR²) convert less energy into translational speed, so they roll down inclines more slowly.

Dynamics on an Incline

Applying Newton's second law and torque equations to a rolling object on an incline (using friction f and the constraint a = Rα) gives a_cm = g sinθ/(1 + I/mR²). Objects with smaller I/mR² (like solid spheres) accelerate faster than hollow or larger-I objects (like hoops) when released from rest.

Key terms

Rolling without slipping condition

v_cm = Rω and a_cm = Rα; the contact point has zero velocity relative to the surface.

Total KE of a rolling object

KE = ½mv_cm² + ½Iω² = ½mv_cm²(1 + I/mR²) using the rolling constraint.

Role of static friction in rolling

Provides torque to maintain the no-slip condition; does zero work on the object.

Acceleration of a rolling object on an incline

a_cm = g sinθ / (1 + I/mR²); smaller I/mR² means faster acceleration down the incline.

Why a solid sphere beats a hoop in a race down an incline

The solid sphere has a smaller moment of inertia relative to mR², so more energy goes into translational motion, giving greater acceleration.

Try an AP-style question

A solid sphere rolls without slipping down an incline of angle θ. What is its acceleration?

1. g sin θ
2. (1/2) g sin θ
3. (5/7) g sin θ
4. (2/3) g sin θ