A child with mass m is standing at the edge of a mery-go-round having moment of inertia I, radius R and initial angular velocity ω as shown in the figure. The child jumps off the edge of the merry-go-round with tangential velocity v with respect to the ground. The new angular velocity of the merry-go-round is :

Engineering

Physics

Moment of Inertia

Angular Momentum and Conservation of Angular Momentum

Circular Motion in Vertical Plane

Question

A child with mass m is standing at the edge of a mery-go-round having moment of inertia I, radius R and initial angular velocity ω as shown in the figure. The child jumps off the edge of the merry-go-round with tangential velocity v with respect to the ground. The new angular velocity of the merry-go-round is :

$\sqrt{\frac{{Iω}^{2} - {mv}^{2}}{I}}$

$\sqrt{\frac{(I + {mR}^{2}) ω^{2} - {mv}^{2}}{I}}$

$\frac{Iω - mvR}{I}$

$\frac{(I + {mR}^{2}) ω - mvR}{I}$

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This problem involves conservation of angular momentum. The total angular momentum before and after the child jumps must be equal. Initially, the child is part of the rotating system, so the total moment of inertia is I + mR². The initial angular momentum is (I + mR²)ω.

After the jump, the child has linear momentum, contributing an angular momentum of mvR about the center. The merry-go-round alone has angular momentum Iω'. Setting initial and final angular momenta equal:

$(I + m R^{2}) ω = I ω^{'} + m v R$

Solving for the new angular velocity ω':

$ω^{'} = \frac{(I + m R^{2}) ω - m v R}{I}$

Final Answer: $\frac{(I + {mR}^{2}) ω - mvR}{I}$