Question

A 36 kg child stands on a spinning platform that has been well maintained (neglect friction in the axle). Initially the platform is spinning at 2.4 rev/s while the child's arms are extended outward (holding some weights). The child then pulls the weights back to close to their body. What is the final angular speed of the platform? Assume that the child can be considered a cylinder with a diameter of 0.8 m, spinning along its center axis and the weights are point masses (2 kg each) that are initially 1.4 m away from the center of the child. Assume that the weights are at the outer edge of the "cylinder" (i.e., the child) in the final moment. Enter your answer to the tenth's place

Answer 1

a)

M = mass of child = 36 kg

R = radius of cylinder = diameter/2 = 0.8/2 = 0.4 m

I_child = moment of inertia of child = (0.5) MR² = (0.5) (36) (0.4)² = 2.88 kgm²

m = mass of each weight = 2 kg

r = distance of each weight from axis = 1.4 m

I_weights = moment of inertia of weights = 2 mr² = (2) (2) (1.4)² = 7.84 kgm²

w_i = initial angular speed = 2.4 rev/s

w_f = final angular speed = ?

using conservation of angular momentum

(I_child + I_weights) w_i = I_child w_f

(2.88 + 7.84) (2.4) = (2.88) w_f

w_f = 8.93 rev/s