In the figure shown find out the distance of& | vedclass

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Question

$\bar{x}=\frac{m_{1} x_{1}+\left(-m_{2}\right) x_{2}}{m_{1}+\left(-m_{2}\right)}=\frac{A_{1} x_{1}+\left(-A_{2}\right) x_{2}}{A_{1}+\left(-A_{2}\right

Vedclass · Accepted Answer

$R/4$

Vedclass · Answer

$\bar{x}=\frac{m_{1} x_{1}+\left(-m_{2}ight) x_{2}}{m_{1}+\left(-m_{2}ight)}=\frac{A_{1} x_{1}+\left(-A_{2}ight) x_{2}}{A_{1}+\left(-A_{2}ight)}$

$A_{1}=\pi(3 R)^{2}, A_{2}=\pi R^{2}$

$x_{1}=0, x_{2}=2 R$

$	herefore \quad \bar{x}=-\mathrm{R} / 4$

In the figure shown find out the distance of centre of mass of a system of a uniform circular plate of radius $3R$ from $O$ in which a hole of radius $R$ is cut whose centre is at $2R$ distance from centre of large circular plate

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