Q.
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.m2 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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a
1.5 rad/s
b
2.5 rad/s
c
2.0 rad/s
d
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 byItotal = Iweights + Istudent = 2(mr2)+5.0 kg.m2Before: r = 1.0 mThus, Ii = 2(5.0 kg)(1.0 m)2+5.0 kg.m2 = 15 kg.m2 After: r = 0.50 mThus, If = 2(5.0 kg)(0.50 m)2+5.0 kg.m2 = 7.50 kg.m2We now use conservation of angular momentum.Ifωf = Iiωior ωf = (IiIf)ωi = (15.07.5)(1.0 rad/s) = 2 rad/s
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