The Loop-the-Loop Marble That Leaves the Track
A physics club builds a loop-the-loop from flexible plastic channel. The loop's inner radius is R = 0.180 m, measured to the inside of the channel, and the bottom of the loop rests on the table. A steel marble of radius r = 0.012 m is released from rest at height h = 0.460 m above the table, measured to the marble's center, and it rolls without slipping down the straight ramp into the loop. Two groups make predictions. One compares 0.460 m with 2R = 0.360 m and concludes the marble must clear the loop. The other says the marble loses energy to rotation and will not. In the trial run the marble leaves the track near the top of the loop. The advisor asks for a calculation that settles the question, plus an explanation of why a frictionless sliding block released from the same height would behave differently.
1. Calculate the minimum release height (to the marble's center) for the rolling marble to maintain contact at the top of the loop, and compare it with the actual 0.460 m release height.
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Two conditions must be met at the top of the loop.
Contact condition. The marble's center travels a circle of radius R − r = 0.168 m. At the limiting case the normal force is zero, so gravity alone supplies the centripetal force: mg = mv²/(R − r), giving v² ≥ g(R − r) = 9.81 × 0.168 = 1.65 m²/s².
Energy condition. For a solid sphere rolling without slipping, I = (2/5)mr² and v = ωr, so the kinetic energy is ½mv² + ½Iω² = (7/10)mv². Energy conservation from release to the top of the loop (height 2R = 0.360 m at the center's path, approximately) gives (7/10)v² = g(h − 2R).
At h = 0.460 m: v² = (10/7)(9.81)(0.100) = 1.40 m²/s², which is below the required 1.65. The marble leaves the track - consistent with the trial run.
Setting v² = g(R − r) gives h_min = 2R + (7/10)(R − r) = 0.360 + 0.118 = 0.478 m. The release point is about 1.8 cm too low.
2. Explain why a frictionless sliding block released from the same 0.460 m would stay on the track, and identify precisely which energy term is responsible.
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For a block sliding without friction, all the gravitational energy becomes translational kinetic energy: ½v² = g(h − 2R), so v² = 2 × 9.81 × 0.100 = 1.96 m²/s², above the 1.65 m²/s² threshold. The block clears.
The responsible term is rotational kinetic energy, ½Iω² = (1/5)mv² for the solid sphere, which is 2/7 of the marble's total kinetic energy. That energy is stored in the spin, not in forward motion, so for the same drop the marble's translational speed at the top is smaller by the factor sqrt(5/7) ≈ 0.85 in v, or 5/7 = 0.71 in v².
The counterintuitive point is that rolling friction does no net work here; the marble is not 'losing energy' in the sense of dissipation. The energy is still in the marble, simply partitioned into a form that does not help it round the loop. On the loop itself, static friction transfers energy back and forth between translation and rotation, and near the top, where the normal force approaches zero, the available friction also approaches zero - so the rolling-without-slipping assumption is itself strained right where the contact condition matters.
3. The club replaces the steel marble with a hollow ball of the same radius, and separately considers whether the rolling assumption can be trusted at the top. Address both.
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Hollow ball. A thin spherical shell has I = (2/3)mr², so the kinetic energy is ½mv²(1 + 2/3) = (5/6)mv² and v² = (6/5)g(h − 2R) = 1.2 × 0.981 = 1.18 m²/s², further below 1.65 than the solid marble's 1.40. Same release height, same radius, worse outcome: the shell puts a larger fraction (2/5) of its kinetic energy into rotation than the solid sphere (2/7), because its mass sits farther from the center. Required release height: h_min = 2R + (5/6)(R − r) = 0.360 + 0.140 = 0.500 m.
Rolling assumption. At the top the normal force approaches zero, so the maximum static friction (μN) also approaches zero, while the marble still needs a torque to keep its angular speed matched to its translational speed. In practice the marble may slip near the top, which removes the rotational constraint: the marble then behaves more like the sliding block, and the strict v² ≥ g(R − r) condition with the rolling energy split no longer describes it. The calculation should therefore be presented as an idealization whose margin (0.478 m vs 0.460 m) is small enough that slipping could change the outcome; an empirical trial is the honest test.