Question

Four identical particles each of mass $m$ are arranged at the corners of a square of side length $L$. If one of the masses is doubled, the shift in the centre of mass of the system w.r.t the diagonally opposite mass is

• A. $\frac{L}{\sqrt{2}}$
• B. $\frac{3\sqrt{2}L}{5}$
• C. $\frac{L}{4\sqrt{2}}$
• D. $\frac{L}{5\sqrt{2}}$

Answer 1

The correct option is D $\frac{L}{5\sqrt{2}}$

Initially we have



$(x_{cm},y_{cm})=(\frac{L}{2},\frac{L}{2})$

$O{C}_1=\sqrt{{\left(\frac{L}{2}\right)}^2+{\left(\frac{L}{2}\right)}^2}$

$O{C}_1=\sqrt{2}\frac{L}{2}.....\left(1\right)$

When we double the mass we have,



$x_{cm}^{\prime }=\frac{m\times 0+m\times 0+mL+2mL}{5m}=\frac{3L}{5}$

$y_{cm}^{\prime }=\frac{m\times 0+m\times 0+mL+2mL}{5m}=\frac{3L}{5}$

$O{C}_2=\sqrt{{\left(\frac{3L}{5}\right)}^2+{\left(\frac{3L}{5}\right)}^2}$

$O{C}_2=\frac{3\sqrt{2}L}{5}.....\left(2\right)$

Shift in COM $=O{C}_2-O{C}_1$

Shift in COM $=\sqrt{2}L[\frac{3}{5}-\frac{1}{2}]=\frac{\sqrt{2}L}{10}=\frac{L}{5\sqrt{2}}$