Question

)A 3.5-m diameter merry-go-round with a mass of 275 kg is spinning at 22 rpm. John runs around the merry-go-round at 5.0 m/s in the opposite direction that the merry-go-round is turning, and jumps onto the outer edge. John’s mass is 30 kgAfter John jumps on, a torque is applied that eventually slows down the merry-go-round. If the merry-go-round comes to a stop 15.6 seconds after the torque is applied, what is the magnitude of the torque?

Answer 1

w = 22 rpm = 22*2pi/60 = 2.3 rad/s

angular momentum of the merry go L1 = (1/2)*M*R^2*w

angular momentum of John L2 = -m*v*R

Initial angular momentum of the system Li = L1 + L2

after john jumping on to the merrygo

final angular momentum Lf = ((1/2)*M*R^2+ m*R^2)*w'

total angular momentum is conserved

Lf = Li

((1/2)*M*R^2+ m*R^2)*w' = (1/2)*M*R^2*w - m*v*R

((1/2)*275*1.75^2 + 30*1.75^2)*wf = (1/2)*275*1.75^2*2.3 + 30*5*1.75

angular speed = w' = 1.4 rad/s

merry go comes to rest wf = 0

angular acceleration alpha = (wf - w')/t = (0-1.4)/15.6 = -0.089 rad/s^2

torque = I*alpha

I = ((1/2)*M*R^2+ m*R^2)

torque = ((1/2)*M*R^2+ m*R^2)*alpha

torque = ((1/2)*275*1.75^2 + 30*1.75^2)*0.089

torque = 45.6 Nm <<<<<---------------ANSWER