A circular plate A of radius 1.5 r is removed | vedclass

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(B) Let the surface mass density of the plate be $\sigma$. The mass of the original plate $B$ of radius $R = 2r$ is $M_B = \sigma \pi (2r)^2 = 4 \sigm

Vedclass · Accepted Answer

$\frac{9 r}{14}$

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(B) Let the surface mass density of the plate be $\sigma$. The mass of the original plate $B$ of radius $R = 2r$ is $M_B = \sigma \pi (2r)^2 = 4 \sigma \pi r^2$.Let the mass of the removed plate $A$ of radius $r_A = 1.5r$ be $M_A = \sigma \pi (1.5r)^2 = 2.25 \sigma \pi r^2$.The centre of mass of the original plate $B$ is at its centre (origin,$x_B = 0$).The centre of the removed plate $A$ is at a distance $d = R - r_A = 2r - 1.5r = 0.5r$ from the centre of plate $B$.The centre of mass of the remaining portion $X_{cm}$ is given by:$X_{cm} = \frac{M_B x_B - M_A x_A}{M_B - M_A}$$X_{cm} = \frac{(4 \sigma \pi r^2)(0) - (2.25 \sigma \pi r^2)(0.5r)}{4 \sigma \pi r^2 - 2.25 \sigma \pi r^2}$$X_{cm} = \frac{-1.125 \sigma \pi r^3}{1.75 \sigma \pi r^2} = -\frac{1.125}{1.75} r = -\frac{1125}{1750} r = -\frac{9}{14} r$.The magnitude of the distance is $\frac{9r}{14}$.

$A$ circular plate $A$ of radius $1.5 r$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of the centre of mass of the remaining portion from the centre of the plate $B$ is

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