A uniform circular disc of radius a is taken. | vedclass

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(B) Let the original circular disc of radius $a$ be considered as the combination of the removed circular section of radius $b$ and the remaining part

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$\frac{{ - c{b^2}}}{{\left( {{a^2} - {b^2}} \right)}}$

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(B) Let the original circular disc of radius $a$ be considered as the combination of the removed circular section of radius $b$ and the remaining part.Let the line of symmetry joining the centres $O$ and $O_1$ be the $x$-axis with $O$ as the origin.The centre of mass of the original disc of radius $a$ is at the origin $O$,so $X_{CM} = 0$.The formula for the centre of mass is:$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \dots (i)$Let $\sigma$ be the surface mass density of the disc.The mass of the removed circular portion is $m_1 = \pi b^2 \sigma$ and its centre is at $x_1 = c$.The mass of the remaining part is $m_2 = \pi (a^2 - b^2) \sigma$ and its centre of mass is at $x_2$.The total mass is $M = m_1 + m_2 = \pi a^2 \sigma$.Substituting these values into equation $(i)$:$0 = \frac{(\pi b^2 \sigma)(c) + (\pi (a^2 - b^2) \sigma)(x_2)}{\pi a^2 \sigma}$$0 = \pi b^2 \sigma c + \pi (a^2 - b^2) \sigma x_2$$x_2 = \frac{-c b^2}{a^2 - b^2}$Thus,the distance $x_2$ of the centre of mass of the remaining part from the initial centre of mass $O$ is $\frac{-c b^2}{a^2 - b^2}$.

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