Loop-the-loop: is the ball a point particle or a rolling body?
A solid steel ball of radius 8.0 mm and mass 5.2 g is released from rest and rolls without slipping down a straight ramp that joins a vertical circular track of radius R = 0.35 m tangentially. Students bracket the minimum release height, measured from the track's lowest point to the ball's centre, by trial: the ball stays in contact through the loop when released from 0.98 m, but leaves the track before reaching the top when released from 0.90 m. High-speed video shows the ball's centre tracing a circle of radius 0.342 m while it is inside the loop. One student argues that because the ball is only 8 mm across, its rotation can be neglected and it can be treated as a sliding particle; she predicts contact down to 0.875 m and says the 0.90 m trial must have been a bad release. The ramp height is measured to plus or minus 5 mm, and the group has no way to release the ball from intermediate heights without adding a launch tube.
1. Evaluate the student's claim that the ball may be treated as a sliding particle. Calculate the minimum release height for (i) a frictionless sliding particle and (ii) a ball rolling without slipping, and state which model the observations support.
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Sliding particle. At the top the track can only push, so N ≥ 0 with N + mg = mv²/R gives v² ≥ gR. From mg(h − 2R) = ½mv², h ≥ 2R + R/2 = 2.5R = 0.875 m.
Rolling ball. The centre travels on radius R − r = 0.342 m, so N + mg = mv²/(R − r), giving v² ≥ 9.81 × 0.342 = 3.36 m² s⁻². Rolling without slipping gives KE = ½mv² + ½(2/5 mr²)(v/r)² = 0.70 mv². The centre rises from h to 2R − r = 0.692 m, so 9.81(h − 0.692) = 0.70v² ≥ 0.70 × 3.36 = 2.35 J kg⁻¹, i.e. h ≥ 0.692 + 0.239 = 0.93 m.
Verdict. The particle model forbids failure at 0.90 m, which is exactly what was observed, so it is refuted. The rolling model predicts failure at 0.90 m and success at 0.98 m, matching the bracket. The 'small ball' argument fails because the rotational share, (2/5)/(1 + 2/5) = 29 %, is independent of radius - only the shape matters.
2. Using the forces on the ball at the top of the loop, explain why contact is lost. State what the friction force is doing at that instant and why it does not appear in your energy equation.
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At the top, the normal force and gravity both point towards the centre, so the radial equation is N + mg = mv²/(R − r). As the release height falls, v falls and N shrinks; it reaches zero when v² = g(R − r) = 3.36 m² s⁻². Below that the track would have to pull the ball inwards, which it cannot, so the ball separates - in practice slightly before the apex, where the inward component of gravity is smaller and the available centripetal force is less.
At the apex the tangential component of gravity is zero, and N is purely radial, so the tangential acceleration is zero and hence the angular acceleration is zero. Rolling without slipping therefore requires zero friction at that instant. Elsewhere on the loop friction is non-zero: it enforces v = ωr, converting part of the gravitational drop into rotational energy. It does no work in the energy equation because the contact point of a rolling body is instantaneously at rest, so static friction has zero power.
3. The group wants to report h_min = 0.93 m. Comment on the precision that is justified and describe one modification that would sharpen the result.
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The bracket 0.90–0.98 m is ±0.04 m, about 4 %, which is far worse than the ±5 mm ruler uncertainty - the limiting factor is the coarse set of release heights, not the length measurement. Reporting 0.93 ± 0.05 m is honest; reporting 0.930 m is not.
There is also a one-sided bias to discuss: the ideal value assumes no losses, whereas rolling resistance, the ramp-to-loop junction and air drag all remove energy, so a measured threshold is an upper bound on the ideal 0.93 m. Any precise experiment will therefore land slightly above the theory value, and that offset is a measurement of the losses rather than a failure of the model.
Improvement: replace the stepped ramp with a continuously adjustable launcher (a spring-loaded plunger with a fine height screw), repeat each height several times, and detect loss of contact with a force sensor on the track surface or by tracking the centre of mass in the video, so that h_min is found by interpolation rather than by bracketing.