Question

A disk-shaped merry-go-round of radius 2.13 m and mass 185 kg rotates freely with an angular speed of 0.631 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.31 m/s jumps onto ts rim and holds on. Before jumping on the merry-go- round, the person was moving in the same direction as the merry-go-round's rim

Answer 1

Radius of the disk is r = 2.13m

mass of the disk is m = 185 kg

angular speed is ω_0 = 0.631 rev /s

                                  = 3.96 rad /s

mass of the person is M = 59.4 kg

speed v = 3.31 m/s

Moment of inertia is I = 1/2 mr^2

                                    = 1/2 * 185 kg * ( 2.13m)^2

                                    = 419.7 kg m^2

Initial angular momentum is L = I ω + M r v

                      = 419.7 kg m^2 * 3.96 rad /s + 59.4 kg * 2.13m * 3.31 m/s

                      = 2080.79 kg m^2 /s

Final angular momentum is L = Iω + mr^2

= 419.7 + 59.4* ( 2.13m)^2

L = 689.19

   final angular speed is ω_f = 2080.79 kg m^2 /s / 689.19

= 3.02 rad /s (Answer)

I hope help you !!