Find the moment of inertia of the ring shown | vedclass

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Question

(A) The moment of inertia of a ring about its diameter is given by $I_{dia} = \frac{1}{2} MR^2$.The axis $AB$ is parallel to the diameter of the ring

Vedclass · Accepted Answer

$\frac{3}{2} MR^2$

Vedclass · Answer

(A) The moment of inertia of a ring about its diameter is given by $I_{dia} = \frac{1}{2} MR^2$.The axis $AB$ is parallel to the diameter of the ring and is at a distance $R$ from the center of the ring.According to the parallel axis theorem,the moment of inertia about an axis parallel to the diameter is $I_{AB} = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the center of mass (which is the diameter in this case) and $d = R$ is the distance between the axes.Substituting the values,we get:$I_{AB} = \frac{1}{2} MR^2 + MR^2$$I_{AB} = \frac{3}{2} MR^2$

Find the moment of inertia of the ring shown in the figure about the axis $AB$.

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