In a playground, there is a small merry-go-round of radius 1.20 m and mass 170 k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 170 kg. Its radius of gyration is 91.0 cm. A child of mass 44.0 kg runs at a speed of 5.00 m/s along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round.
(a) Calculate the rotational inertia of the merry-go-round about its axis of rotation.
kg·m2

(b) Calculate the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round.
kg·m2/s

(c) Calculate the angular speed of the merry-go-round and child after the child has jumped on.
rad/s

Explanation / Answer

radius R = 1.2 m mass M = 170 kg radius of gyration Rg = 91 cm = 0.91 m mass of child m = 44 kg speed v = 5 m/s a) the rotational inertia of the merry-go-round about its axis of rotation is        I = MRg2           = (170)(0.91)2           = 140.77 kg.m2 b) the magnitude of the angular momentum of the child is       L = mvR           = (44)(5)(1.2)           = 264 kg.m2/s c) from conservation of angular momentum ,      mvR = (I + mR2)        L = (I + mR2) the angular speed = L/(I + mR2)                                 = 264 / (140.77 + 44*1.2*1.2)                                 = 1.293 rad/s                                ˜ 1.3 rad/s b) the magnitude of the angular momentum of the child is       L = mvR           = (44)(5)(1.2)           = 264 kg.m2/s c) from conservation of angular momentum ,      mvR = (I + mR2)        L = (I + mR2) the angular speed = L/(I + mR2)                                 = 264 / (140.77 + 44*1.2*1.2)                                 = 1.293 rad/s                                ˜ 1.3 rad/s

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