A disk of radius R is removed from a larger d | vedclass

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(A) Let the surface mass density of the disk be $\sigma$.Mass of the removed disk $(m_1)$ = $\sigma \pi R^2$.The center of mass of the removed disk is

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$1/3$

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(A) Let the surface mass density of the disk be $\sigma$.Mass of the removed disk $(m_1)$ = $\sigma \pi R^2$.The center of mass of the removed disk is at a distance $x_1 = R$ from the center of the larger disk.Mass of the remaining part $(m_2)$ = $\sigma \pi (2R)^2 - \sigma \pi R^2 = 3 \sigma \pi R^2$.Let the center of mass of the remaining part be at a distance $x_2$ from the center of the larger disk.Using the principle of center of mass for the system:$m_1 x_1 = m_2 x_2$$(\sigma \pi R^2) R = (3 \sigma \pi R^2) x_2$$x_2 = R/3$Therefore,the ratio $x/R = 1/3$.

$A$ disk of radius $R$ is removed from a larger disk of radius $2R$. The circumferences of both disks are tangent to each other. The center of mass of the new disk is at a distance $x$ from the center of the larger disk. Find the value of $x/R$.

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