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

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(A) Mass per unit area of the disc is $\sigma = \frac{M}{\pi R^2}$.The radius of the removed circular hole is $r = \frac{R}{2}$.The mass of the remove

Vedclass · Accepted Answer

$\frac{13M{R^2}}{32}$

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(A) Mass per unit area of the disc is $\sigma = \frac{M}{\pi R^2}$.The radius of the removed circular hole is $r = \frac{R}{2}$.The mass of the removed portion is $M' = \sigma 	imes \pi r^2 = \frac{M}{\pi R^2} 	imes \pi \left(\frac{R}{2}ight)^2 = \frac{M}{4}$.The moment of inertia of the removed portion about its own central axis (perpendicular to the plane) is $I_{cm} = \frac{1}{2} M' r^2 = \frac{1}{2} \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2 = \frac{MR^2}{32}$.Using the parallel axis theorem,the moment of inertia of the removed portion about the axis passing through the center $O$ of the original disc is $I'_{0} = I_{cm} + M' d^2$,where $d = \frac{R}{2}$ is the distance between the center of the disc and the center of the hole.$I'_{0} = \frac{MR^2}{32} + \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$.The moment of inertia of the complete disc about the center $O$ is $I_{0} = \frac{1}{2} MR^2$.The moment of inertia of the remaining part is $I = I_{0} - I'_{0} = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

From a disc of radius $R$ and mass $M,$ a circular hole of diameter $R,$ whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis,passing through the centre?

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