Pulling a Puck Inward: Angular Momentum Versus Energy
A 0.50 kg puck on a level air table is tied to a string that passes through a small central hole. The puck moves in a circle of radius 0.40 m at 2.0 m/s. A student pulls the string slowly downward until the radius is 0.20 m. Student 1 predicts the final speed is 2.0 m/s by conserving kinetic energy. Student 2 predicts 4.0 m/s by conserving angular momentum. Student 3 predicts 4.0 m/s because the tension does work on the puck. The measured final speed is 3.9 m/s. A force sensor records the initial tension as 5.0 N and the final tension as 38 N. The pull takes 3.0 s, and the puck makes 12 revolutions during the pull. The air table is level but not perfectly frictionless.
1. Which student's prediction is correct, and what is the final speed? Identify the conservation law that applies and explain why the other students' reasoning is incomplete.
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Student 2's angular-momentum approach is correct. The tension is directed toward the hole, so its torque about the hole is zero; angular momentum L = mvr is conserved. Therefore v_f = v_i(r_i/r_f) = 2.0(0.40/0.20) = 4.0 m/s. Student 1 is wrong because kinetic energy is not conserved: as the radius decreases, the tension has a component along the radial displacement and does positive work. Student 3 has the right final speed but incomplete reasoning: the work done by tension is not known in advance, so energy conservation alone cannot predict v_f unless that work is calculated. Angular momentum gives v_f directly. The measured 3.9 m/s is 2.5% below the ideal value, consistent with a small amount of friction or air drag.
2. Calculate the change in kinetic energy and the work done by the tension. Compare the ideal final tension with the measured 38 N and give a plausible reason for the difference.
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Initial kinetic energy K_i = ½(0.50)(2.0)² = 1.0 J. Final ideal kinetic energy K_f = ½(0.50)(4.0)² = 4.0 J. Thus ΔK = 3.0 J, so the ideal work done by the tension is 3.0 J. The ideal final tension is T = mv²/r = (0.50)(4.0)²/0.20 = 40 N; the initial tension is (0.50)(2.0)²/0.40 = 5.0 N, matching the sensor. The measured 38 N is 5% low. Plausible reasons include friction or air drag doing negative work, the final radius being slightly larger than 0.20 m, the speed being 3.9 m/s rather than 4.0 m/s, a force-sensor calibration offset, or a non-quasi-static pull that gives the puck some radial velocity and makes the tension differ from mv_t²/r.
3. The student claims the pull is 'slow' because it takes 3.0 s. Is that claim justified? What would change in the analysis if the string were yanked quickly?
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The claim is reasonable but should be checked quantitatively. Twelve revolutions in 3.0 s is about 4 rev/s, or an angular speed of roughly 25 rad/s. The radial speed is only about 0.20 m/3.0 s = 0.067 m/s, much smaller than the tangential speed of 2–4 m/s, so the path is nearly circular and the quasi-static approximation is fair. If the string were yanked quickly, the puck would follow a visible spiral, radial kinetic energy would become significant, and the final total speed would include a radial component. Angular momentum would still be conserved because the force remains central, but L = mrv_t would give only the tangential part; the total kinetic energy and the work done by tension would be larger than in the slow-pull case.