Questions
A girl of mass M stands on the rim of a frictionless merry go-round of radius R and rotational inertia I that is not moving. She throws a rock of mass m horizontally in a direction that is tangent to the outer edge of the merry go-round. The speed of the rock, relative to the ground, is v. Afterward, the linear speed of the girl is
detailed solution
Correct option is A
The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merry-go-round is I+MR2ω,which we will take to be positive. The final angular momentum we associate with the thrown rock is negative: -mRv, where v is the speed (positive, by definition) of the rock relative to the ground.Angular momentum conservation leads to0=I+MR2ω−mRv⇒ω=mRvI+MR2The girl's linear speed is Rω=mvR2I+MR2Talk to our academic expert!
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A boy stands over the centre of a horizontal platform which is rotating freely with a speed of 2 revolutions/s about a vertical axis through the centre of the platform and straight up through the boy. He holds 2 kg masses in each of his hands close to his body. The. combined moment of inertia of the system is 1 kgmetre2. The boy now stretches his arms so as to hold the masses far from his body. In this situation, the moment of inertia of the system increases to 2kgmetre2. The kinetic energy of the system in the latter case as compared with that in the previous case will
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