If the moment of inertia of a thin circular r | vedclass

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Question

(A) Let $M$ be the mass and $R$ be the radius of the thin circular ring.The moment of inertia of the ring about an axis passing through its center and

Vedclass · Accepted Answer

$\frac{I}{4}$

Vedclass · Answer

(A) Let $M$ be the mass and $R$ be the radius of the thin circular ring.The moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = MR^2$.According to the parallel axis theorem, the moment of inertia about an axis passing through its edge and perpendicular to its plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$.Given that this value is $I$, we have $2MR^2 = I$, which implies $MR^2 = \frac{I}{2}$.The moment of inertia of the ring about its diameter is $I_{diameter} = \frac{1}{2}MR^2$.Substituting the value of $MR^2$, we get $I_{diameter} = \frac{1}{2} 	imes \frac{I}{2} = \frac{I}{4}$.

If the moment of inertia of a thin circular ring about an axis passing through its edge and perpendicular to its plane is $I$, then the moment of inertia of the ring about its diameter is

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