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  4. A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity ω. Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

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Q. A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $\omega$. Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

KEAMKEAM 2008System of Particles and Rotational Motion

Solution:

According to conservation of angular momentum,
$ I\omega =constant $ ie, we can write $ {{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}} $
or $ M{{R}^{2}}\omega =(M+4m){{R}^{2}}{{\omega }_{2}} $
or $ {{\omega }_{2}}=\left( \frac{M}{M+4m} \right)\omega $