Rolling Cylinder Versus Hoop on a Ramp
A solid cylinder and a thin hoop have the same mass, 0.250 kg, and the same outer radius, 2.5 cm. Each is released from rest at the top of a 0.80 m long ramp inclined at 8.0° above horizontal. Both roll down the ramp without any applied torque; a photogate at the bottom records transit times. The solid cylinder takes 1.36 s and the hoop takes 1.52 s. The student predicts that the solid cylinder will arrive first because it has less rotational inertia, and concludes that both objects rolled without slipping because the measured times are close to the calculated values. The ramp surface is clean but not perfectly rigid; the release mechanism may impart a small initial speed. The cylinder and hoop have slightly different masses in reality: 0.252 kg and 0.248 kg.
1. Calculate the acceleration and ramp time predicted for each object if it rolls without slipping. Compare with the measured times.
Read the worked answer
For rolling without slipping, the acceleration down the incline is
a = g sin θ / (1 + I/(mR²)).
Here θ = 8.0°, so sin θ = 0.1392, and g sin θ = (9.8)(0.1392) = 1.364 m/s².
Solid cylinder: I = ½mR², so
a = 1.364/(1 + 0.5) = 0.909 m/s².
From rest over 0.80 m,
t = sqrt(2s/a) = sqrt(1.60/0.909) = 1.33 s.
Hoop: I = mR², so
a = 1.364/(1 + 1) = 0.682 m/s²,
t = sqrt(1.60/0.682) = 1.53 s.
The measured values are 1.36 s for the cylinder and 1.52 s for the hoop. The cylinder is about 2% slower than predicted, while the hoop is about 1% faster. The mass difference is only about 1% and does not affect the rolling prediction, because mass cancels in I/(mR²). The times are close to the rolling predictions, but they are not exact.
2. Does the data prove that both objects rolled without slipping? What additional measurement would test the claim?
Read the worked answer
No. Matching the predicted times supports rolling without slipping, but it does not prove it. If an object slips, friction does less to spin it up, so its translational acceleration approaches the frictionless value
a = g sin θ = 1.36 m/s²,
which would give a time of about sqrt(1.60/1.36) = 1.08 s. The measured times are much closer to the rolling predictions than to the sliding prediction, so slipping is not dominant. The hoop’s slightly short time could hint at a small amount of slipping, but it is also consistent with timing error or a small initial speed from the release.
To test rolling directly, measure the angular speed ω at the bottom and compare it with the translational speed v. Rolling without slipping requires v = ωR. That can be done with two photogates, a video with a marked radius, or a rolling sensor. Repeating the releases and measuring the actual initial speed would also separate release effects from slipping.
3. Predict what would happen if the ramp were frictionless and both objects were released from rest. Would that result support the student’s reasoning?
Read the worked answer
Without friction, neither object can acquire angular acceleration, so both would slide without rolling. Each would have the same translational acceleration
a = g sin θ = 1.36 m/s²,
and both would take
t = sqrt(1.60/1.36) = 1.08 s.
They would arrive together, not with the solid cylinder first. Therefore the observed time difference, 1.36 s versus 1.52 s, is evidence that friction is causing rolling and that rotational inertia matters.
The student’s conclusion that the solid cylinder arrives first is correct for rolling, but the reason should be stated in terms of I/(mR²): the solid cylinder has the smaller value, so more gravitational energy goes into translational motion and its acceleration is larger. The frictionless case would not support that reasoning; it would eliminate the difference.