You stand on a frictional platform that is rotating at 2.0 rev/s. Your arms are

Question

You stand on a frictional platform that is rotating at 2.0 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 7.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 3.6 kg · m2.
(a) What is the resulting angular speed of the platform?
? rev/s

(b) What is the change in kinetic energy of the system?
? J
(c) Where did this increase in energy come from? (Select all that apply.)
your internal energy
gravity
air resistance
mass of the weights
kinetic energy of the platform

Explanation / Answer

(a) The total angular momentum of the system of man, weights, &platform is conserved Li = Lf Iiw i =Ifw f w f = (Ii /If )w i = [(7.8kg· m2 ) / (3.6kg· m2 )](2.0rev/s) w f = 4.333 rev/s (b)
Remember that the Kinetic energy of a rotating body is½Iw2. Ki = ½Iiwi2 = 1/2*7.8*(22)^2       = 616.02 J Kf = ½If wf2        = 1334.51 J the change in energy, KE = Kf - Ki             = 1334.51 - 616.02            = 718.49 J (c) The added kinetic energy came from the man doing work indecreasing the rotational inertia by moving the weights closer to his body.

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