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(A) According to the principle of conservation of angular momentum, since no external torque acts on the system, the initial angular momentum equals t

Vedclass · Accepted Answer

$\frac{m R^2}{I+m R^2}$

Vedclass · Answer

(A) According to the principle of conservation of angular momentum, since no external torque acts on the system, the initial angular momentum equals the final angular momentum.Initial angular momentum $L_i = I \omega_0$, where $\omega_0$ is the initial angular velocity.When the mouse of mass $m$ lands on the edge at radius $R$, the new moment of inertia of the system becomes $I' = I + m R^2$.Let the new angular velocity be $\omega$. Then, $L_f = (I + m R^2) \omega$.Equating $L_i = L_f$, we get $I \omega_0 = (I + m R^2) \omega$.Thus, $\omega = \frac{I \omega_0}{I + m R^2}$.The fractional loss of angular velocity is given by $\frac{\omega_0 - \omega}{\omega_0} = 1 - \frac{\omega}{\omega_0}$.Substituting the value of $\omega$, we get $1 - \frac{I}{I + m R^2} = \frac{I + m R^2 - I}{I + m R^2} = \frac{m R^2}{I + m R^2}$.

$A$ mouse of mass $m$ jumps on the outside edge of a rotating ceiling fan of moment of inertia $I$ and radius $R$. The fractional loss of angular velocity of the fan as a result is

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