A student sits on a freely rotating stool holding two weights, eachof mass 2.96

Question

A student sits on a freely rotating stool holding two weights, eachof mass 2.96 kg. When his armsare extended horizontally, the weights are 0.96 m from the axis of rotation and he rotateswith an angular speed of 0.746 rad/s. The moment of inertia of the studentplus stool is 2.96 kg·m2 and is assumedto be constant. The student pulls the weights inward horizontallyto a position 0.295 m from therotation axis.
(a) Find the new angular speedof the student.

(b) Find the kinetic energy ofthe rotating system before and after he pulls the weightsinward.
Before: After:
(a) Find the new angular speedof the student.

(b) Find the kinetic energy ofthe rotating system before and after he pulls the weightsinward.
Before: After: (a) Find the new angular speedof the student.

(b) Find the kinetic energy ofthe rotating system before and after he pulls the weightsinward.
Before: After:

Explanation / Answer

I1w = 2 * 2.96 *.962    = 5.456 kgm2 initial moment of inertia of weights I2w = 2 * 2.96 * .2952 = .515 kgm2    final moment of inertia ofweights I1 = I1w + 2.96 = 8.42 kgm2   initial total inertia I2 = I2w + 2.96 = 3.475 kgm2 final total moment of inertia 2 = I1 1 /I2 = .746 * 8.42 / 3.475 = 1.81 rad / sec KE2 = 1/2 I112 = (8.42 / 2) * .7462 =2.34 J KE2 = 1/2 I222 = (3.475 / 2) *1.812 = 5.69 J

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