Discussion
A mouse of mass m jumps on a freely rotating disc of moment of inertia I and radius R. If \(\omega_{0}\) and \(\omega\) are the angular velocities of the disc before and after mouse jumps, then the ratio \(\omega/\omega_{0}\) is :
Answer: C
The ratio is I/(I + mR\(^{2}\)), printed as 'I/I + mR\(^{2}\)'.
No external torque acts about the axis, so angular momentum is conserved. The mouse lands at the rim, adding a moment of inertia \(mR^{2}\):
\[ \begin{aligned} I\omega_{0} &= (I + mR^{2})\omega \\ \frac{\omega}{\omega_{0}} &= \frac{I}{I + mR^{2}} \end{aligned} \]
The disc slows down, so the ratio is less than 1. \(I/mR^{2}\) and \((I - mR^{2})/I\) do not follow from conservation, and \(I/(I - mR^{2})\) would mean the disc speeds up when mass is added.
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