Masses m left(frac{1}{3}right)^N frac{1}{N} a | vedclass

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(A) The mass at position $x=N$ is given by $m_N = m \left(\frac{1}{3}\right)^N \frac{1}{N}$.The centre of mass $X_{cm}$ along the $X$-axis is given by

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$\frac{1}{6} \frac{m}{M}$

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(A) The mass at position $x=N$ is given by $m_N = m \left(\frac{1}{3}ight)^N \frac{1}{N}$.The centre of mass $X_{cm}$ along the $X$-axis is given by the formula:$X_{cm} = \frac{\sum m_N x_N}{\sum m_N} = \frac{\sum_{N=2}^{\infty} \left[ m \left(\frac{1}{3}ight)^N \frac{1}{N} ight] 	imes N}{M}$$X_{cm} = \frac{m}{M} \sum_{N=2}^{\infty} \left(\frac{1}{3}ight)^N$This is an infinite geometric progression with the first term $a = \left(\frac{1}{3}ight)^2 = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$X_{cm} = \frac{m}{M} \left[ \frac{1/9}{1 - 1/3} ight] = \frac{m}{M} \left[ \frac{1/9}{2/3} ight] = \frac{m}{M} \left[ \frac{1}{9} 	imes \frac{3}{2} ight] = \frac{m}{6M}$.

Masses $m \left(\frac{1}{3}\right)^N \frac{1}{N}$ are placed at $x=N$,where $N=2, 3, 4, \ldots \infty$. If the total mass of the system is $M$,then the centre of mass is

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