A student sits on a freely rotating stool holding two dumbbells, each of mass 2.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 2.99 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.01 m from the axis of rotation and the student rotates with an angular speed of 0.743 rad/s. The moment of inertia of the student plus stool is 2.70 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.303 m from the rotation axis (Figure b).

(a) Find the new angular speed of the student.
rad/s

(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.

Kbefore =  J Kafter =  J A student sits on a freely rotating stool holding two dumbbells, each of mass 2.99 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.01 m from the axis of rotation and the student rotates with an angular speed of 0.743 rad/s. The moment of inertia of the student plus stool is 2.70 kg A?A? m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.303 m from the rotation axis (Figure b). A student sits on a freely rotating stool holding (a) Find the new angular speed of the student. rad/s (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. Kbefore = J Kafter = J

Explanation / Answer

conserve angular momentum

I1w1 = I2w2

I1 = 2.7 + 2 * 2.99 * 1.012    ( added the moment of inertia of two dumbells )

I2 = 2.7 + 2 * 2.99* 0.3432    (added the moment of inertia of two dumbells after pulled in)

since w1 is known we can find w2.

Initial energy is 0.5 I1 w1*w1

Final energy = 0.5 * I2 *w2*w2

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