A student sits on a freely rotating stool holding two dumbbells, each of mass 5.0 kg (Fig.). 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 $5.0 k g m^{2}$ and is assumed to be constant. The student pulls the dumbbells inwards horizontally to a position 0.50 m from the rotation axis (Fig.). The ne angular speed of the student is

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1.5rad/s

1.25rad/s

2.5rad/s

2.0rad/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 rotation is given by
$I_{t o t a l} = I_{w e i g h t} + I_{s t u d e n t} = 2 (m r^{2}) + 5.0 k g . m^{2}$
Before : r = 1.0 m
Thus, $I_{i} = 2 (5.0 k g) {(1.0 m)}^{2} + 5.0 k g . m^{2} = 15 k g . m^{2}$
After : r = 0.50 m
Thus, $I_{f} = 2 (5.0 k g) {(0.50 m)}^{2} + 5.0 k g . m^{2} = 7.50 k g . m^{2}$
We now use conservation of angular momentum.
$I_{f} ω_{f} = I_{i} ω_{i}$
or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{15.0}{7.5}) (1.0 r a d / s) = 2 r a d / s$