A circular portion of radius R_2 has been rem | vedclass

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(A) Let the original disc have radius $R_1$ and mass $M_1 = \sigma \pi R_1^2$,where $\sigma$ is the surface mass density. Its center of mass is at the

Vedclass · Accepted Answer

$-\frac{R_2^2}{R_1+R_2}$

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(A) Let the original disc have radius $R_1$ and mass $M_1 = \sigma \pi R_1^2$,where $\sigma$ is the surface mass density. Its center of mass is at the origin $(0,0)$.Let the removed circular portion have radius $R_2$ and mass $M_2 = \sigma \pi R_2^2$. Its center is at $(R_1 - R_2, 0)$ as shown in the figure.The center of mass of the remaining portion is given by:$x_{CM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$Substituting the values:$x_{CM} = \frac{(\sigma \pi R_1^2)(0) - (\sigma \pi R_2^2)(R_1 - R_2)}{\sigma \pi R_1^2 - \sigma \pi R_2^2}$$x_{CM} = \frac{-R_2^2(R_1 - R_2)}{R_1^2 - R_2^2}$$x_{CM} = \frac{-R_2^2(R_1 - R_2)}{(R_1 - R_2)(R_1 + R_2)}$$x_{CM} = -\frac{R_2^2}{R_1 + R_2}$

$A$ circular portion of radius $R_2$ has been removed from one edge of a circular disc of radius $R_1$. The correct expression for the centre of mass for the remaining portion of the disc is

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