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

Question

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

Infinity Learn · Accepted Answer

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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