Three identical spheres each of radius 'R'are placed on a horizontal surface touching one another. If one of the spheres is removed, the shift in the center of mass of the system is

First method:

$x_{cm} = 0$ , $y_{cm} = \frac{R}{\sqrt{3}}$

$m (- R, 0),$ $m (R, 0),$ $m (0, \sqrt{3} R)$

Here centre of mass shift is $\frac{R}{\sqrt{3}}$

Second method:

$y_{cm} = \frac{m (0) + (m (0) + m (\sqrt{3} R))}{3 m}$

$= \frac{R}{\sqrt{3}}$

If 3^rd one is removed

$(X_{m}, Y_{m}) = (0, 0)$

So shift is $\frac{R}{\sqrt{3}}$

Three identical spheres each of radius $' R'$ are placed on a horizontal surface touching one another. If one of the spheres is removed, the shift in the center of mass of the system is