Moment of inertia of a disc about a diameter | vedclass

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(A) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4} M R^2$. From this,we get $M R^2 = 4 I$.

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$6$

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(A) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4} M R^2$. From this,we get $M R^2 = 4 I$. To find the moment of inertia about an axis perpendicular to the plane and passing through 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 (center of mass) is $I_{cm} = \frac{1}{2} M R^2$. By the parallel axis theorem,$I_{rim} = I_{cm} + M R^2$. Substituting $I_{cm} = \frac{1}{2} M R^2$,we get $I_{rim} = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$. Substituting $M R^2 = 4 I$ into the expression,we get $I_{rim} = \frac{3}{2} (4 I) = 6 I$.

Moment of inertia of a disc about a diameter is $I$. Find the moment of inertia of the disc about an axis perpendicular to its plane and passing through its rim. (in $I$)

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