From a circular disc of radius R and mass 9M, | vedclass

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(B) The moment of inertia of a circular disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I

Vedclass · Accepted Answer

$\frac{40}{9} M R^2$

Vedclass · Answer

(B) The moment of inertia of a circular disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.For the original disc: $M_1 = 9M$,$R_1 = R$. Thus,$I_1 = \frac{1}{2} (9M) R^2 = \frac{9}{2} M R^2$.For the removed disc: $M_2 = M$,$R_2 = R/3$. Thus,$I_2 = \frac{1}{2} (M) (R/3)^2 = \frac{1}{2} M (R^2/9) = \frac{1}{18} M R^2$.The moment of inertia of the remaining disc is $I = I_1 - I_2$.$I = \frac{9}{2} M R^2 - \frac{1}{18} M R^2$.$I = \frac{81 M R^2 - M R^2}{18} = \frac{80 M R^2}{18} = \frac{40}{9} M R^2$.

From a circular disc of radius $R$ and mass $9M$,a small disc of mass $M$ and radius $R/3$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

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