A man stands on a rotating platform with his arms stretched holding a $$5$$ kg weight in each hand. The angular speed of the platform is $$1.2$$ rev $$s-1$$. The moment of inertia of the man together with the platform may be taken to be constant and equal to $$6$$ kg $$m2$$. If the man brings his arms close to his chest with the distance of each weight from the axis changing from $$100$$ cm to $$20$$ cm. The new angular speed of the platform is

Question

A man stands on a rotating platform with his arms stretched holding a $$5$$ kg weight in each hand. The angular speed of the platform is $$1.2$$ rev $$s-1$$. The moment of inertia of the man together with the platform may be taken to be constant and equal to $$6$$ kg $$m2$$. If the man brings his arms close to his chest with the distance of each weight from the axis changing from $$100$$ cm to $$20$$ cm. The new angular speed of the platform is

Infinity Learn · Accepted Answer

Initial moment of inertia,              $$I1$$ $$=$$ $$6$$ $$+$$[$$2$$×$$5$$×(1)2$$] $$=$$ $$16$$ kg $$m2Initial$$ angular velocity, ω$$1$$ $$=$$ $$1.2$$ rev $$s-1Initial$$ angular momentum is    $$L1$$ $$=$$ $$I1$$ω$$1Final$$ moment of inertia,   $$I2$$ $$=$$ $$6+$$[$$2$$×$$5$$×$$(0.2)2$$ ]$$=$$ $$6.4$$ kg $$m2Final$$ angular speed $$=$$ ω$$2Final$$ angular momentum is  $$L2$$ $$=$$ $$I2$$ω$$2According$$ to law of conservation of angular momentum,   $$L1$$ $$=$$ $$L2$$ or $$I1$$ω$$1$$ $$=$$ $$I2$$ω$$2$$ω$$2$$ $$=$$ $$I1$$ω$$1I2$$ $$=$$ $$(16$$ kg $$m2)(1.2$$ rev $$s-1)(6.4$$ kg $$m2) $$=$$ $$3$$ rev $$s-1$$

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