Masses m, (1/2)(m/2), (1/2)^2(m/3), dots, (1/ | vedclass

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(A) The position of the centre of mass $X_{CM}$ is given by the formula $X_{CM} = \frac{\sum m_i x_i}{M}$.Substituting the given values:$X_{CM} = \fra

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$(\frac{2m}{M}, 0, 0)$

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(A) The position of the centre of mass $X_{CM}$ is given by the formula $X_{CM} = \frac{\sum m_i x_i}{M}$.Substituting the given values:$X_{CM} = \frac{m(1) + (1/2)(m/2)(2) + (1/2)^2(m/3)(3) + \dots + (1/2)^{N-1}(m/N)(N) + \dots}{M}$$X_{CM} = \frac{m + (1/2)m + (1/2)^2m + \dots + (1/2)^{N-1}m + \dots}{M}$$X_{CM} = \frac{m}{M} [1 + 1/2 + (1/2)^2 + \dots + (1/2)^{N-1} + \dots]$The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/2$.The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.Therefore,$X_{CM} = \frac{m}{M} 	imes 2 = \frac{2m}{M}$.Since the masses are placed only along the $X$-axis,$Y_{CM} = 0$ and $Z_{CM} = 0$.Thus,the centre of mass is $(\frac{2m}{M}, 0, 0)$.

Masses $m, (1/2)(m/2), (1/2)^2(m/3), \dots, (1/2)^{N-1}(m/N), \dots \infty$ are placed at $x = 1, 2, 3, \dots, N, \dots \infty$ respectively. If the total mass is $M$,then the centre of mass of the system is:

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