Question 70 – Rankers Physics

Uncategorized: Practice Problem & Solution

A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity: (2009)
$\frac{\omega M}{M + 2m}$
$\frac{\omega (M + 2m)}{M}$
$\frac{\omega M}{M + m}$
$\frac{\omega (M - 2m)}{M + 2m}$

Solution Explained:

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of angular momentum, initial angular momentum equals final angular momentum ($I_1\omega_1 = I_2\omega_2$).
Initially, $I_1 = MR^2$ and $\omega_1 = \omega$. After attaching masses, $I_2 = MR^2 + 2mR^2 = (M+2m)R^2$.
Equating them: $MR^2\omega = (M+2m)R^2\omega' \implies \omega' = \frac{M\omega}{M + 2m}$.

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