Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown i

Question

Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

The original moment of inertia of the full disc is $I_{orig} = \frac{1}{2}Mr^2$. Based on the provided diagram, two circular holes of radius $r/4$ are cut out, with their centers located at a distance of $3r/4$ from the main axis. The mass of each cutout is $m = M 	imes \frac{\pi (r/4)^2}{\pi r^2} = \frac{M}{16}$. By the parallel axis theorem, the moment of inertia of one cutout about the main axis is $I_{hole} = \frac{1}{2}m(r/4)^2 + m(3r/4)^2 = \frac{1}{2}(\frac{M}{16})(\frac{r^2}{16}) + (\frac{M}{16})(\frac{9r^2}{16}) = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512}$. For two holes, $I_{cutouts} = 2 	imes \frac{19Mr^2}{512} = \frac{19Mr^2}{256}$. The remainder's moment of inertia is $I_{rem} = I_{orig} - I_{cutouts} = \frac{1}{2}Mr^2 - \frac{19}{256}Mr^2 = \frac{128 - 19}{256}Mr^2 = \frac{109}{256}Mr^2$. Thus, $x = 109$.