Four identical particles each of mass m are k | vedclass

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(D) Let the origin be at the centre of the square. The initial centre of mass $(CM_1)$ of the four particles is at the origin $(0, 0)$.When one partic

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$\frac{a}{3 \sqrt{2}}$

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(D) Let the origin be at the centre of the square. The initial centre of mass $(CM_1)$ of the four particles is at the origin $(0, 0)$.When one particle of mass $m$ at corner $C$ is removed,the remaining system consists of three particles of mass $m$ at corners $A, B,$ and $D$.The new centre of mass $(CM_2)$ will shift towards the centroid of the triangle formed by the remaining three particles.The distance of the centre of the square from any corner is $r = \frac{a}{\sqrt{2}}$.Using the formula for the centre of mass of the remaining system: $R_{CM} = \frac{\sum m_i r_i}{\sum m_i}$.Since the mass at $C$ is removed,we can treat this as a system of four particles with one mass being $-m$ at $C$ and the original system having total mass $4m$ at the origin.The shift in the centre of mass is given by $\Delta R = \frac{|m_C \cdot r_C|}{M_{remaining}} = \frac{m \cdot (a/\sqrt{2})}{3m} = \frac{a}{3\sqrt{2}}$.

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

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