The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is: [2025]

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Solution

Q.
The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is: [2025]

1 $\frac{1}{2} M r^{2}$

2 $\frac{3}{8} M r^{2}$

3 $\frac{3}{2} M r^{2}$

4 $2 M r^{2}$

Ans.
(2)

$I_{1} = \frac{M R^{2}}{2}$

$I' = I_{1} + M R^{2} = \frac{3}{2} M R^{2}$

For r = 2R

$I' = \frac{3}{8} M r^{2}$

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