A 5.4 m diameter merry-go-round is rotatingfreely with an angular velocity of 0.

Question

   A 5.4 m diameter merry-go-round is rotatingfreely with an angular velocity of 0.95 rad/s. Its totalmoment of inertia is 1850 kg m^2. Four people standing on theground, each of mass 85 kg, suddenly step onto the edge of themerry-go-round. What is the angular velocity of teh merry-go-round now? What if the people were on it initially and then jumped off ina radial direction (relative to the merry-go-round)? Please show all steps    A 5.4 m diameter merry-go-round is rotatingfreely with an angular velocity of 0.95 rad/s. Its totalmoment of inertia is 1850 kg m^2. Four people standing on theground, each of mass 85 kg, suddenly step onto the edge of themerry-go-round. What is the angular velocity of teh merry-go-round now? What if the people were on it initially and then jumped off ina radial direction (relative to the merry-go-round)? Please show all steps

Explanation / Answer

           Given that the mass of the each person is m = 85 kg             The diameter of the merry-go-round is d = 5.4 m             The radius of the platform is r = d/2 = 2.7 m             The moment of inertia of the merry-go-round is I1 = 1850kg.m2             The anglular velocity of the platform is 1 = 0.95rad/s     --------------------------------------------------------------------------------------            since there is no external torque on the system then the angularmomentum is conserved                                   initial angular momentum = final angular momentum                                                       I1*1 = I2*2                                                         I1*1= (I1+ 4 mr2 ) *2                                                              2 =  I1*1 / (I1+ 4 mr2)                                                                     =----------- rad/s         The intial kineticenergy before walking is K1 = (1/2)I1*12                                                                              =---------- J                    The final kinetic energy after walking is K2 =(1/2)I2*22                                                                           =---------- J                                   The diameter of the merry-go-round is d = 5.4 m             The radius of the platform is r = d/2 = 2.7 m             The moment of inertia of the merry-go-round is I1 = 1850kg.m2             The anglular velocity of the platform is 1 = 0.95rad/s     --------------------------------------------------------------------------------------            since there is no external torque on the system then the angularmomentum is conserved                                   initial angular momentum = final angular momentum                                                       I1*1 = I2*2                                                         I1*1= (I1+ 4 mr2 ) *2                                                              2 =  I1*1 / (I1+ 4 mr2)                                                                     =----------- rad/s         The intial kineticenergy before walking is K1 = (1/2)I1*12                                                                              =---------- J                    The final kinetic energy after walking is K2 =(1/2)I2*22                                                                           =---------- J                                                                                                    =---------- J                                                                                      =---------- J           

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