Moment of inertia of a disc about an axis whi | vedclass

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(A) The moment of inertia of a disc about its diameter is $I_d = \frac{1}{4}MR^2$.Using the parallel axis theorem,the moment of inertia about a tangen

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$ \frac{6}{5}I $

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(A) The moment of inertia of a disc about its diameter is $I_d = \frac{1}{4}MR^2$.Using the parallel axis theorem,the moment of inertia about a tangent parallel to its plane is $I = I_d + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.From this,we find $MR^2 = \frac{4}{5}I$.Now,the moment of inertia of a disc about its central axis perpendicular to its plane is $I_c = \frac{1}{2}MR^2$.Using the parallel axis theorem,the moment of inertia about a tangent perpendicular to its plane is $I' = I_c + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Substituting $MR^2 = \frac{4}{5}I$ into the expression for $I'$,we get $I' = \frac{3}{2} 	imes (\frac{4}{5}I) = \frac{6}{5}I$.

Moment of inertia of a disc about an axis which is tangent and parallel to its plane is $I$. Then the moment of inertia of disc about a tangent,but perpendicular to its plane will be

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