(A) The initial moment of inertia of the ring is $I = MR^2$. The initial angular momentum is $L = I\omega = MR^2\omega$.When $4$ point masses of mass

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$\left( {\frac{M}{{M + 4m}}} \right)\omega $

(A) The initial moment of inertia of the ring is $I = MR^2$. The initial angular momentum is $L = I\omega = MR^2\omega$.When $4$ point masses of mass $m$ are placed at the ends of two mutually perpendicular diameters,they are all at a distance $R$ from the axis of rotation.The new moment of inertia of the system becomes $I' = MR^2 + 4(mR^2) = (M + 4m)R^2$.Since there is no external torque acting on the system,the angular momentum remains conserved $(L_{initial} = L_{final})$.$MR^2\omega = (M + 4m)R^2\omega'$Dividing both sides by $R^2$,we get $M\omega = (M + 4m)\omega'$.Therefore,the new angular velocity is $\omega' = \left( \frac{M}{M + 4m} \right)\omega$.

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

Difficult

$A$ thin ring of mass $M$ and radius $R$ is rotating with an angular velocity $\omega$ about an axis passing through its center and perpendicular to its plane. If $4$ point masses each of mass $m$ are placed gently at the opposite ends of two mutually perpendicular diameters,what will be the new angular velocity of the system?

A
$\left( {\frac{M}{{M + 4m}}} \right)\omega $
B
$\left( {\frac{{M + 4m}}{M}} \right)\omega $
C
$\left( {\frac{{M - 4m}}{{M + 4m}}} \right)\omega $
D
$\left( {\frac{M}{{4m}}} \right)\omega $

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

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MediumAIPMT 1998

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If the earth suddenly shrinks to $\frac{1}{64}$th of its original volume with its mass remaining the same,the period of rotation of earth becomes $\frac{24}{x} \text{ h}$. The value of $x$ is $.......$

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