A merry-go-round rotates at the rate of 0.20 rev/s with an 85-kg man standing at

Question

A merry-go-round rotates at the rate of 0.20 rev/s with an 85-kg man standing at a point 2.0 m from the axis of rotation. (a) What is the new angular speed when the man walks to a point 1.0 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m. _______rad/sec (b) Calculate the change in kinetic energy due to this movement. _______ J How do you account for this change in kinetic energy? Choose the correct answer: a.)Work is done by the man on the system as he moves towards the center. b.)Wind resistance varies as the square of the speed. c.)The moment arm of the man about the axis of rotation decreases. d.)Work is done by the merry-go-round on the man as he moves towards the center.

Explanation / Answer

Im = ½mR² = ½*25*2² = 50 kg·m² Iman1 = m*R² = 85*2² = 340 kg·m² Iman2 = m*r² = 85*.6² = 30.6 kg·m² w1 = .15*2p = .9425 rad/sec w2 = w1*(50+340)/(50+30.6) = 4.839*w1 = 4.561 rad/sec (conservation of momentum) dKE = ½[I2*w2² - I1*w1²] = ½[(50+30.6)*4.561² - (50+340)*.9425²] = 665.13 J

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