Question

A 38.2 kg child stands on a spinning platform that has been well maintained (neglect friction in the axle). Initially the platform is spinning at 2.8 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 in rev/s 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 1.3 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.

please draw a picture and show your work with reasoning

Answer 1

By conservation of angular momentum,

i1w1 = i2w2 where w1,w2 are initial and final angular velocities, i1,i2 are moment of inertia.

(0.5*38.2*(0.8/2)^2 +2*2*1.3^2)*2.8 = (0.5*38.2*(0.8/2)^2)*w2

w2 = (0.5*38.2*(0.8/2)^2 +2*2*1.3^2)*2.8/ (0.5*38.2*(0.8/2)^2 + 2*2*(0.8/2)^2)

= 7.436 rev/s answer

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