The moment of inertia of a ring about one of | vedclass

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(D) The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{2}MR^2$.According to the parallel axis the

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$3I$

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(D) The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{2}MR^2$.According to the parallel axis theorem,the moment of inertia $I_{tangent}$ about a tangent parallel to the diameter is $I_{tangent} = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the center of mass (diameter) and $d = R$ is the distance between the parallel axes.Substituting the values,we get $I_{tangent} = I + MR^2$.Since $I = \frac{1}{2}MR^2$,we have $MR^2 = 2I$.Therefore,$I_{tangent} = I + 2I = 3I$.

The moment of inertia of a ring about one of its diameters is $I$. What will be its moment of inertia about a tangent parallel to the diameter?

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