Suppose there is a uniform circular disc of m | vedclass

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(B) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc is $M = \sigma \pi r^2$.Each cut-out disc has a radius $r' = r

Vedclass · Accepted Answer

$109$

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(B) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc is $M = \sigma \pi r^2$.Each cut-out disc has a radius $r' = r/4$. Its mass $m$ is given by $m = \sigma \pi (r/4)^2 = \sigma \pi r^2 / 16 = M/16$.The moment of inertia of the original disc about axis $A$ is $I_0 = \frac{1}{2} Mr^2$.The moment of inertia of one cut-out disc about its own central axis is $I_{cm} = \frac{1}{2} m (r/4)^2 = \frac{1}{2} m (r^2/16) = \frac{mr^2}{32}$.Using the parallel axis theorem, the moment of inertia of one cut-out disc about axis $A$ is $I_{cut} = I_{cm} + md^2$, where $d = 3r/4$.$I_{cut} = \frac{mr^2}{32} + m(3r/4)^2 = \frac{mr^2}{32} + \frac{9mr^2}{16} = \frac{mr^2 + 18mr^2}{32} = \frac{19mr^2}{32}$.The moment of inertia of the remaining part is $I_{rem} = I_0 - 2 	imes I_{cut}$.$I_{rem} = \frac{1}{2} Mr^2 - 2 	imes \left( \frac{19mr^2}{32} ight) = \frac{1}{2} Mr^2 - \frac{19mr^2}{16}$.Substituting $m = M/16$, we get $I_{rem} = \frac{1}{2} Mr^2 - \frac{19(M/16)r^2}{16} = \frac{1}{2} Mr^2 - \frac{19}{256} Mr^2$.$I_{rem} = \frac{128}{256} Mr^2 - \frac{19}{256} Mr^2 = \frac{109}{256} Mr^2$.Comparing this with $\frac{x}{256} Mr^2$, we find $x = 109$.

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