From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of

Question

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° 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

Accepted Answer

Given that, Mass of Ring = M; Radius of Ring = R. Now 90° arc is removed from circular ring, then mass removed = M/4. Mass of remaining portion = 3M/4. Moment of inertia of remaining part = ∫dmr^2. Since r=R, I = R^2 ∫dm = 3MR^2/4. So the value of K is 3/4.

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