Combine translation and rotation in rolling

Race two bodies with the same mass and radius

A solid cylinder and a thin hoop start from rest on the same incline. They have equal mass and outer radius. Predict whether they arrive together when both roll without slipping. Compare how their mass is distributed, not only their total mass.

The experiment uses a 2 kg symmetric rigid body with radius R = 0.5 m. It travels 4 m along a stationary incline under teaching g = 10 m/s². The control β specifies its inertia about the center: I = βmR². β is dimensionless. A uniform solid sphere has β = 0.4, a uniform solid cylinder β = 0.5 and a thin hoop β = 1. Other settings represent ideal intermediate mass distributions. The surface’s static-friction coefficient is fixed at 0.4.

Before using any speed prediction, check the friction comparison. The model displays a rolling endpoint only when the required static friction is available. There is no drag, deformation, rolling resistance or driving motor.

Link rotation to travel

Rolling without slipping on a stationary surface imposes v = Rω, using speed magnitudes here. v is the center’s speed in m/s and ω its angular speed in rad/s. During an angle change of one radian, the center travels one radius along the surface. For a constant R, center acceleration and angular acceleration magnitudes similarly satisfy a = Rα.

This condition is not true for a spinning wheel that slides. Nor does it say every point of the body has speed v. The material point instantaneously touching the stationary ramp has zero velocity relative to it. Its rotational velocity relative to the center cancels the center’s velocity. At the opposite rim point the velocities add, giving speed 2v along the ramp. A different material point becomes the contact point as the body turns; zero instantaneous velocity does not mean one point remains fixed there.

Divide the available energy

The body’s kinetic energy is the sum of ½mv² for center-of-mass translation and ½Iω² for rotation about the center. This decomposition describes two distinct contributions. It does not add particle kinetic energy a second time to an equivalent rotational expression about a fixed axis.

At 30 degrees, the vertical drop after 4 m along the incline is h = 4 sin(30°) = 2 m. The gravitational-energy decrease is mgh = 2 × 10 × 2 = 40 J. Without losses, these 40 J become translational plus rotational kinetic energy.

For β = 0.5, substituting I = βmR² and ω = v/R gives rotational energy equal to β times translational energy. Therefore 1.5 times the translational energy is 40 J. Translation receives 26.667 J and rotation 13.333 J. The center speed is √(2 × 26.667/2) = 5.164 m/s and angular speed is 5.164/0.5 = 10.328 rad/s.

The OpenStax rolling-motion discussion develops the no-slip constraint and the combined translational and rotational energy account. In this lab, the energy bars use a shared 0–60 J scale so equal totals and differing partitions remain visible.

Compare the inertia factors

Keep the incline at 30 degrees and compare β = 0.4, 0.5 and 1. All three pass the supplied friction condition. The sphere model receives 28.571 J of translational energy and has center speed 5.345 m/s. The cylinder has 26.667 J and 5.164 m/s. The hoop receives 20 J and reaches 4.472 m/s. Its other 20 J are rotational energy.

Capture the cylinder and hoop results. Explain why the hoop’s larger rotational-energy share leaves less energy for translation at the same drop. A larger inertia factor produces a smaller center acceleration and a longer travel time in this release, provided no slipping occurs.

The radius cancels from the speed formula v² = 2gh/(1+β) for this ideal comparison. It does not cancel from ω = v/R. Equal center speed would mean different angular speed for different radii. In real apparatus, bearing losses, deformation and different contact conditions can prevent this cancellation from predicting an observed race.

Recognize the level-surface boundary

Set the incline to zero. The downslope component of gravity is zero and the required static friction is zero. The body starts at rest and remains there; there is no finite travel time to the point 4 m away. Reporting zero seconds would confuse a journey that never occurs with an instantaneous arrival.

A body already rolling on an ideal level surface could continue at constant speed without friction. That is a different initial condition. This lab always starts from rest and does not inject a push or initial spin.

Transfer to another release

A solid cylinder with β = 0.5 and radius 0.25 m starts from rest and drops vertically by 1.5 m while rolling without slipping. Use g = 10 m/s² and assume the friction condition is satisfied. Predict center speed, angular speed and the fraction of final kinetic energy in rotation. Does changing only the mass affect this ideal speed?

Check the rolling energy transfer

v² = 2 × 10 × 1.5/1.5 = 20 m²/s², so v ≈ 4.472 m/s. Angular speed is v/0.25 ≈ 17.889 rad/s. The rotational share is β/(1+β) = 1/3. Mass cancels from the energy equation for this ideal release; it changes the energy amounts and forces, but not the predicted speed when the same assumptions hold.