You stand on a frictional platform that is rotating at 1.0 rev/s. Your arms are

Q: You stand on a frictional platform that is rotating at 1.0 rev/s. Your arms are

apply fron the aw of cnservation of rotatiional energy ' I1W1^2 = I2W2^2 W2^2 = 7.7 * 1*1/(3.5*3.5) W2^2 = 0.628 W2 = 0.792 rev/s -------------------------------- Change in KE = 0.5 I W1^2 - 0.5 I2W2^2 KE = 7.7 * 1^2 - (3.5 * 0.792^2) KE = 5.504 J

ID: 1468752 • Letter: Y

Question

You stand on a frictional platform that is rotating at 1.0 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 7.7 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 3.5 kg · m2.

(a) What is the resulting angular speed of the platform?

(b) What is the change in kinetic energy of the system?

(c) Where did this increase in energy come from? (Select all that apply.)

I know c is the internal energy

Explanation / Answer

apply fron the aw of cnservation of rotatiional energy '

I1W1^2 = I2W2^2

W2^2 = 7.7 * 1*1/(3.5*3.5)

W2^2 = 0.628

W2 = 0.792 rev/s
--------------------------------

Change in KE = 0.5 I W1^2 - 0.5 I2W2^2

KE = 7.7 * 1^2 - (3.5 * 0.792^2)

KE = 5.504 J

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You stand on a frictional platform that is rotating at 1.0 rev/s. Your arms are

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