First slide

Rotational motion

Question

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

Difficult

A

$\frac{m v R^{2}}{I + M R^{2}}$
B

$\frac{(m + M) v R^{2}}{I + M R^{2}}$
C

$\frac{m v R^{2}}{I + (M + m) R^{2}}$
D

$\frac{m v R^{2}}{I + (M - m) R^{2}}$

Solution

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merry-go-round is $(I + M R^{2}) ω$ ,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 to
$0 = (I + M R^{2}) ω - m R v \Rightarrow ω = \frac{m R v}{I + M R^{2}}$
The girl's linear speed is $R ω = \frac{m v R^{2}}{I + M R^{2}}$

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