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

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

Vedclass · Answer

(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 disc) is $I_{cm}' = \frac{1}{2} M' r^2 = \frac{1}{2} \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2 = \frac{M R^2}{32}$.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 centers.$I_0' = \frac{M R^2}{32} + \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2 = \frac{M R^2}{32} + \frac{M R^2}{16} = \frac{3 M R^2}{32}$.The moment of inertia of the complete disc about the center $O$ is $I_0 = \frac{1}{2} M R^2$.The moment of inertia of the remaining part is $I_{rem} = I_0 - I_0' = \frac{1}{2} M R^2 - \frac{3 M R^2}{32} = \frac{16 M R^2 - 3 M R^2}{32} = \frac{13 M R^2}{32}$.

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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