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

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(A) The moment of inertia of the original disc about the central axis is $I_1 = \frac{1}{2} M R^2$. The radius of the hole is $r = \frac{R}{2}$. The m

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

Vedclass · Answer

(A) The moment of inertia of the original disc about the central axis is $I_1 = \frac{1}{2} M R^2$. The radius of the hole is $r = \frac{R}{2}$. The mass of the hole is $m = \frac{M}{\pi R^2} 	imes \pi r^2 = \frac{M}{\pi R^2} 	imes \pi (\frac{R}{2})^2 = \frac{M}{4}$. The moment of inertia of the hole about its own central axis is $I_{hole, cm} = \frac{1}{2} m r^2 = \frac{1}{2} (\frac{M}{4}) (\frac{R}{2})^2 = \frac{M R^2}{32}$. Using the parallel axis theorem,the moment of inertia of the hole about the central axis of the original disc is $I_2 = I_{hole, cm} + m d^2$,where $d = \frac{R}{2}$ is the distance between the axes. $I_2 = \frac{M R^2}{32} + (\frac{M}{4}) (\frac{R}{2})^2 = \frac{M R^2}{32} + \frac{M R^2}{16} = \frac{3 M R^2}{32}$. The moment of inertia of the remaining part is $I = I_1 - I_2 = \frac{1}{2} M R^2 - \frac{3 M R^2}{32} = \frac{16 M R^2 - 3 M R^2}{32} = \frac{13 M R^2}{32}$.

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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