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A uniform disc radius R is put over another uniform disc of radius 2R of same thickness and density . The peripheres of the two discs touch each other . The position of their center of mass is

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at $\frac{R}{5}$ from the center of the bigger towards the center of the smaller disc

none of the above

at $\frac{R}{3}$ from the center of the bigger disc towards the center of the smaller disc

at $\frac{2 R}{5}$ from the center of the bigger disc towards the center of the smaller disc

answer is B.

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Detailed Solution

First method

$d = \frac{M}{π (4 R^{2})}$

$M_{1} = (0, 0)$

$M_{2} = \frac{M}{4 {πR}^{2}} \times {πR}^{2} = \frac{M}{4}$

$x_{cm} = \frac{0 + \frac{M}{4} (R)}{M + \frac{M}{4}} = \frac{\frac{RM}{4}}{\frac{5 M}{4}}$

$= \frac{RM}{4} \times \frac{4}{5 M} = \frac{R}{5}$

$x_{cm} = \frac{R}{5}$

Second method:

When we add the mass

$m_{initial} = M, m_{add} = \frac{M}{4}$

$X_{shift} = [\frac{m_{add} . d}{m_{initial} + m_{add}}]$

$= \frac{[\frac{M}{4} . R]}{[M + \frac{M}{4}]} = \frac{\frac{MR}{4}}{\frac{5 M}{4}} = \frac{R}{5}$

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