A Man Stands on a Platform A man stands on aplatform that is rotating (without f

Question

A Man Stands on a Platform A man stands on aplatform that is rotating (without friction) with a rotationalspeed of 0.8 rev/s. His arms areoutstretched and he holds a brick in each hand. The rotationalinertia of the system consisting of the man, bricks, and platformabout the central axis is 8.0kg·m2. Assume that by moving thebricks the man decreases the rotational inertia of the system to2.0kg·m2. (a) What is the resultingangular speed of the platform?
1 revs/s
(b) What is the ratio of the new kinetic energy of the system tothe original kinetic energy?
2
(c) What provided the added kineticenergy? 3 the man pulling the weights closer to hisbody
the momentum of the platform
None of these is correct.
the man moving the weights further awayfrom his body
None of these is correct. 3

Explanation / Answer

              Initialangular velocity is 1 = 0.8 rad/s               momentof inertia of stool + student + brick is I1 = 8.0kg.m2               Final moment of inertia of the system is I2 = 2.0kg.m2                  --------------------------------------------------------------------------           Since there is no external force on the system then the angularmomentum is conserved                             initial angular momentum = final angular momentum                                       I1* 1 =   I2*2                 Then                      2 = I1*1/ I2                                                      = ---------- rad/s      The ratio of new kinetic energy tooriginal kinetic enegy Kf / Ki =(1/2)*I2*22/(1/2)*I1*12                                                                                                         = ------------                         momentof inertia of stool + student + brick is I1 = 8.0kg.m2               Final moment of inertia of the system is I2 = 2.0kg.m2                  --------------------------------------------------------------------------           Since there is no external force on the system then the angularmomentum is conserved                             initial angular momentum = final angular momentum                                       I1* 1 =   I2*2                 Then                      2 = I1*1/ I2                                                      = ---------- rad/s      The ratio of new kinetic energy tooriginal kinetic enegy Kf / Ki =(1/2)*I2*22/(1/2)*I1*12                                                                                                         = ------------                                                                                                                   = ------------                

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