A merry-go-round has a mass of 1440 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.00 s? Assume it is a solid cylinder.
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This is Giancoli Answers with Mr. Dychko. The net work done on this merry-go-round is the change in its kinetic energy. So that's the final kinetic energy minus the initial and there is no initial kinetic energy so this whole term here disappears and we have the change then is just the final kinetic energy of one-half moment of inertia times final angular velocity squared. The moment of inertia of a uniform disk or a uniform cylinder is one-half mass times radius squared. So we substitute that in for I and then one-half times a half makes a quarter. So we have mr squaredω f squared over 4 and we substitute in numbers being careful to convert this revolutions per second into radians per second by multiplying 1 revolution for every 7 seconds by 2π radians per second or 2π radians per revolution; revolution's cancel and you square that result and then square the 7.5 meters times by 1440 kilograms giving us 1.63 times 10 to the 4 joules.
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