A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley (?Fig. E9.43?), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Solution 47E Since the system is released from rest, therefore there will be loss in potential energy of the system and gain in its kinetic energy. Therefore, Loss in potential energy = gain in (kinetic energy of the pulley + kinetic energy of the stone)…..(1) Let the stone falls by a distance of h. Given, mass of the pulley disk m = 2.p0 kg Its radius r = 20.0 cm = 0.2 m 1 2 Its moment of inertia I = m r2 p 2 2 I = 2 × 2.50 kg × 0.2 m 2 I = 0.05 kg.m Given, kinetic energy of the pulley disk = 4.50 J If is the angular velocity of the pulley, 1 2 2I = 4.50 J 1 × 0.05 kg.m × = 4.50 J 2 2 9 2 2 = 0.05rad /s = 180 rad /s 2 2 2 2 2 2 v = 180 × (0.2) m /s 2 2 2 v = 7.2 m /s Given, mass of the stone m = 1.s0 kg 1 2 2 Its kinetic energy = 2 × 1.50 kg × 7.2 m /s = 5.4 J 2 Potential energy loss of the stone = 1,50 kg × 9.80 m/s × h (a) Therefore, from equation (1), w get 1.50 kg × 9.8 m/s × h = 4.5 J + 5.4 J h = 9.9 J 1.50 kg×9.8 m/s h = 0.673 m So, the stone must fall by a distance of 0.673 m. (b) Total kinetic energy of the system = 9.9 J Kinetic energy of the pulley = 4.50 J 4.50 J The pulley’s fraction of kinetic energy = 9.9 J = 0.455 = 45.5% Therefore, the pulley has 45.5% of the total kinetic energy.