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A ring of radius 2 m and mass 100 kg undergoes pure rolling motion on a horizontal floor. The speed of its centre of mass is 20 cm/s. How much work (in joules) must be done to bring the ring to rest?
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Correct answer: 4
Solution
For a ring rolling without slipping: moment of inertia about centre = mr². Angular velocity omega = v/r where v = speed of CM. Total KE = (1/2)mv² (translational) + (1/2)*mr²*(v/r)² (rotational) = (1/2)mv² + (1/2)mv² = mv². Given: m = 100 kg, v = 20 cm/s = 0.20 m/s. Total KE = 100 * (0.20)² = 100 * 0.04 = 4 J. Work done to stop = 4 J.
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