Question

Four identical particles each of mass 'm' are kept at the four corners of a square of side 'a'. If one of the particles is removed, the shift in the position of the centre of mass is:

• A. \(\sqrt{2}a\)
• B. \(\frac{3a}{\sqrt{2}}\)
• C. \(\frac{a}{\sqrt{2}}\)
• D. \(\frac{a}{3\sqrt{2}}\)

Hint:

The center of mass of a system of particles is the average position of the particles, weighted by their masses.

Answer 1

The Correct Option is D

Let the vertices of the square be A(0,0), B(a,0), C(a,a), D(0,a). The coordinates of the center of mass when all four particles are present is: $$x_{cm} = \frac{m(0) + m(a) + m(a) + m(0)}{4m} = \frac{2ma}{4m} = \frac{a}{2}$$ $$y_{cm} = \frac{m(0) + m(0) + m(a) + m(a)}{4m} = \frac{2ma}{4m} = \frac{a}{2}$$ The center of mass is $(\frac{a}{2}, \frac{a}{2})$. Now, let's remove the particle at A(0,0). The new coordinates of the center of mass are: $$x_{cm}' = \frac{m(a) + m(a) + m(0)}{3m} = \frac{2ma}{3m} = \frac{2a}{3}$$ $$y_{cm}' = \frac{m(0) + m(a) + m(a)}{3m} = \frac{2ma}{3m} = \frac{2a}{3}$$ The new center of mass is $(\frac{2a}{3}, \frac{2a}{3})$. The shift in the position of the center of mass is the distance between the two points: $$d = \sqrt{\left(\frac{2a}{3} - \frac{a}{2}\right)^2 + \left(\frac{2a}{3} - \frac{a}{2}\right)^2}$$ $$d = \sqrt{2\left(\frac{4a-3a}{6}\right)^2} = \sqrt{2\left(\frac{a}{6}\right)^2} = \sqrt{2} \cdot \frac{a}{6} = \frac{a\sqrt{2}}{6} = \frac{a}{3\sqrt{2}}$$