Questions
A portion of a ring of radius R has been removed as shown in figure. Mass of the remaining portion is m. Centre of the ring is at origin O. Let IA and IO be the moments of inertia passing through points A and O are perpendicular to the plane of the ring. Then,
detailed solution
Correct option is D
Whole mass has equal distance from the centre O. Hence, IO = mR2 . Further, centre of mass of the remaining portion will be to the left of point O. More the distance of axis from centre of mass, more is the moment of inertia. Hence, IA > IO.Talk to our academic expert!
Similar Questions
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its mid-point and perpendicular to its length is I0 , Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is
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