Solve the following :


A uniform disc of radius $\mathrm{R}$ is put over another uniform disc of radius $2 \mathrm{R}$ of the same thickness and density. The peripheries of the two discs touch each other. Locate the center of mass of the system.


C.O.M of $2 \mathrm{R}$ disc $=(2 \mathrm{R}, 0)$

C.O.M of $R$ disc $=(R, 0)$

$=\frac{m_{2 R^{x}}{ }^{x}+m_{R^{x} R}}{m_{n R}+m_{R}}$

$\left\{m_{2 R}=\rho .4 \pi(2 R)^{2}\right\}$

$\left\{m_{R}=\rho .4 \pi R^{2} l\right\}$

$=\frac{\left(\rho 4 \pi l R^{2}\right) \cdot(2 R)+\left(\rho 4 \pi l R^{2}\right) \cdot(R)}{\rho 4 \pi R^{2} l(4+1)}$

$=\frac{8 R+R}{5}=\frac{9 R}{5}$

Distance of C.O.M (system) from C.O.M of $2 R$ disc

$=2 R-\frac{9 R}{5}=\frac{R}{5}$

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