Final angular velocity of the ring – Rankers Physics

Final angular velocity of the ring

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. The objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by: (2010 Mains)

$\frac{(M+2m)\omega}{2m}$
$\frac{2M\omega}{M+2m}$
$\frac{(M+2m)\omega}{M}$
$\frac{M\omega}{M+2m}$

Solution:

By conservation of angular momentum, $I_{1}\omega_{1} = I_{2}\omega_{2}$. Initially, $I_{1} = Mr^{2}$. Finally, the moment of inertia is $I_{2} = Mr^{2} + 2mr^{2} = (M+2m)r^{2}$. Equating the two yields $Mr^{2}\omega = (M+2m)r^{2}\omega_{2}$, so $\omega_{2} = \frac{M\omega}{M+2m}$.

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