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

Moderate

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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