A circular hole of radius left(frac{a}{2}righ | vedclass

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(C) Let $\sigma$ be the uniform surface mass density of the disc.Let $M_1$ be the mass of the full disc of radius $a$ and $M_2$ be the mass of the rem

Vedclass · Accepted Answer

$\frac{5}{6} a$

Vedclass · Answer

(C) Let $\sigma$ be the uniform surface mass density of the disc.Let $M_1$ be the mass of the full disc of radius $a$ and $M_2$ be the mass of the removed circular hole of radius $\frac{a}{2}$.$M_1 = \sigma \pi a^2$,with its center of mass at $x_1 = a$.$M_2 = \sigma \pi \left(\frac{a}{2}ight)^2 = \frac{\sigma \pi a^2}{4}$,with its center of mass at $x_2 = \frac{3a}{2}$.The center of mass of the remaining portion is given by:$x_{COM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$$x_{COM} = \frac{(\sigma \pi a^2)(a) - (\frac{\sigma \pi a^2}{4})(\frac{3a}{2})}{\sigma \pi a^2 - \frac{\sigma \pi a^2}{4}}$$x_{COM} = \frac{a - \frac{3a}{8}}{1 - \frac{1}{4}} = \frac{\frac{5a}{8}}{\frac{3}{4}} = \frac{5a}{8} 	imes \frac{4}{3} = \frac{5a}{6}$

$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be:

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