Moment of Inertia of Cut Ring – Rankers Physics
Topic: Rotational Motion
Subtopic: Moment of Inertia

Moment of Inertia of Cut Ring

From a circular ring of mass \(M\) and radius \(R\) an arc corresponding to a \(90^\circ\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \(K\) times \(MR^2\). Then the value of \(K\) is
1/8
3/4
7/8
1/4

Solution:

The remaining mass of the ring after removing a quarter (\(90^\circ\) out of \(360^\circ\)) is \(M' = \frac{3}{4}M\). Since all parts of the remaining ring are still at a perpendicular distance \(R\) from the center axis, the moment of inertia is \(I = M' R^2 = \frac{3}{4} M R^2\). Thus, \(K = \frac{3}{4}\).

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