A student sits on a freely rotating stool holding two dumbbells, each of mass 3.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 3.08 kg (see figure below). When his arms are extended horizontally (figure a), the dumbbells are 1.09 m from the axis of rotation and the student rotates with an angular speed of 0.753 rad/s. The moment of inertia of the student plus stool is 2.74 kg m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.310 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 3.08 kg (see figure below). When his arms are extended horizontally (figure a), the dumbbells are 1.09 m from the axis of rotation and the student rotates with an angular speed of 0.753 rad/s. The moment of inertia of the student plus stool is 2.74 kg ½ m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.310 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

Explanation / Answer

Initially, we add the moments of inertia of the two dumbells, U = mr^2 for each,

I1 = 2.74 kg m^2 + 2* 3.08 kg * (1.09 m)^2

I1 = 10.059 kg m^2

In the final case, only the given I is relevant,

I2 = 2.74 kg m^2

Now, by conservation of angular momentum,

I1 w1 = I2 w2 ---> w2 = I1 w1 /I2

Thus,

w2 = 2.76 rad/s   [ANSWER, PART A]

**************

Note that

KE = 1/2 I w^2

Thus,

KEbefore = 1/2 I1 w1^2 = 2.85 J

KEafter = 1/2 I2 w2^2 = 10.4 J

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