Rankers Physics

A circular ring of mass $M$ and radius $R$ is rotating about its axis with constant angular velocity $\omega$. Two particles each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The angular velocity of the ring will now become: (1998)

$\frac{M - 2m}{m\omega}$
$\frac{m\omega}{M - 2m}$
$\frac{M\omega}{M + 2m}$
$\frac{M - 2m}{m}$

Solution:

Since the particles are attached gently, external torque is zero, meaning angular momentum is conserved.
$I_1\omega_1 = I_2\omega_2 \implies (MR^2)\omega = (MR^2 + 2mR^2)\omega'$.
Solving for the new angular velocity yields $\omega' = \frac{M\omega}{M + 2m}$.

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