A merry-go-round rotates at the rate of 0.20 rev/s with an 75-kg manstanding at

Question

A merry-go-round rotates at the rate of 0.20 rev/s with an 75-kg manstanding at a point 2.0 m from the axis of rotation. (a) What isthe new angular speed when the man walks to a point 0.8 m from the center? Assume that the merry-go-roundis a solid 25-kg cylinder of radius 2.0 m.
1 rad/sec

(b) Calculate the change in kinetic energy due to thismovement.
2 J

How do you account for this change in kineticenergy? 3 Wind resistance varies as the square ofthe speed.
Work is done by the merry-go-round on theman as he moves towards the center.
Work is done by the man on the system ashe moves towards the center.
The moment arm of the man about the axisof rotation decreases. 3

Explanation / Answer

          Given that the initial angular velocity is 1 = 0.20rev/s           mass of man is m = 75 kg           Distance between the axis of rotation and man is d1 =2.0m           Final distance between the axis of rotation and man is d2 =0.8 m           mass of solid cylinder is M = 25 kg and R = 2.0m     ------------------------------------------------------------------------------------------         The moment ofinertia of the merry-go-round is I = M R2/2                                                                                   = ----------- kg.m2                Sincethere is no external force on the system then the angular momentumis conserved                             initial angular momentum = final angular momentum                                       (I + 2md12 )1 =   ( I + 2md22)2                            Then                      2 = ( I + 2md12 )1/ ( I +2md22 )                                                                = ---------- rad/s      The kinetic energy before collsion isK1 = (1/2)( I + 2md12)12                                                                   =------------ J      The kinetic energy beforecollsion is K2 = (1/2)( I + 2md22)22                                                                   =------------ J            Change in kinetic energy is K = K2-K1                                                     =------- J    Work isdone by the man on the system as he moves towards thecenter.                             initial angular momentum = final angular momentum                                       (I + 2md12 )1 =   ( I + 2md22)2                            Then                      2 = ( I + 2md12 )1/ ( I +2md22 )                                                                = ---------- rad/s      The kinetic energy before collsion isK1 = (1/2)( I + 2md12)12                                                                   =------------ J      The kinetic energy beforecollsion is K2 = (1/2)( I + 2md22)22                                                                   =------------ J            Change in kinetic energy is K = K2-K1                                                     =------- J    Work isdone by the man on the system as he moves towards thecenter.                                                                   =------------ J      The kinetic energy beforecollsion is K2 = (1/2)( I + 2md22)22                                                                   =------------ J            Change in kinetic energy is K = K2-K1                                                                   =------------ J                                                           =------- J    Work isdone by the man on the system as he moves towards thecenter.

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