A student sits on a freely rotating stool holding two dumbbells, each of mass 5.0 kg (Fig. a). When his arms are extended horizontally (Fig. a), the dumbbells are 1.0 m from the axis of rotation and the student rotates with an angular speed of 1.0 rad/s. The moment of inertia of the student plus stool is 5.0 $kg . m^{2}$ and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.50 m from the rotation axis (Fig. b). The new angular speed of the student is

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1.5 rad/s

2.5 rad/s

2.0 rad/s

1.25 rad/s

answer is C.

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Detailed Solution

The total angular momentum of the system of the student, the stool, and the weights about the axis of rotation is given by

$I_{total} = I_{weights} + I_{student} = 2 ({mr}^{2}) + 5.0 kg . m^{2}$

Before: r = 1.0 m

Thus, $I_{i} = 2 (5.0 kg) {(1.0 m)}^{2} + 5.0 kg . m^{2} = 15 kg . m^{2}$

After: $r = 0.50 m$

Thus, $I_{f} = 2 (5.0 kg) {(0.50 m)}^{2} + 5.0 kg . m^{2} = 7.50 kg . m^{2}$

We now use conservation of angular momentum.

$I_{f} ω_{f} = I_{i} ω_{i}$

or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{15.0}{7.5}) (1.0 rad / s) = 2 rad / s$

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