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

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(A) The mass per unit area of the disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed small disc of radius $r = \frac{R}{3}$ is $m = \sigma

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$4 MR^2$

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(A) The mass per unit area of the disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed small disc of radius $r = \frac{R}{3}$ is $m = \sigma \cdot \pi r^2 = \frac{9M}{\pi R^2} \cdot \pi \left(\frac{R}{3}\right)^2 = M$.The moment of inertia $(I)$ of the complete disc about the axis passing through $O$ is $I_{total} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$.The moment of inertia of the removed disc about its own center is $I_{cm} = \frac{1}{2}m r^2 = \frac{1}{2}M(\frac{R}{3})^2 = \frac{1}{18}MR^2$.Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through $O$ is $I_{removed} = I_{cm} + md^2$,where $d$ is the distance between the centers,$d = R - \frac{R}{3} = \frac{2R}{3}$.$I_{removed} = \frac{1}{18}MR^2 + M(\frac{2R}{3})^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \frac{1+8}{18}MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.The moment of inertia of the remaining disc is $I_{remaining} = I_{total} - I_{removed} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = 4MR^2$.

From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is:

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