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(B) The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4}MR^2$. Given that this

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$6$

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(B) The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4}MR^2$. Given that this value is $I$,we have $MR^2 = 4I$. To find the moment of inertia about an axis perpendicular to the plane and passing through a point on the rim,we use the parallel axis theorem. First,the moment of inertia about an axis perpendicular to the plane and passing through the center of mass is $I_{cm} = \frac{1}{2}MR^2$. By the parallel axis theorem,$I_{rim} = I_{cm} + Md^2$,where $d = R$ is the distance between the axes. $I_{rim} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$. Substituting $MR^2 = 4I$ into the expression: $I_{rim} = \frac{3}{2}(4I) = 6I$.

The moment of inertia of a uniform circular disc about a diameter is $I$. What is its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim (in $I$)?

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