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(B) Let the linear mass density of the rods be $\lambda$.Each rod has its centre of mass at its midpoint. For a rod of length $l_i$ starting from the

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$L/3$

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(B) Let the linear mass density of the rods be $\lambda$.Each rod has its centre of mass at its midpoint. For a rod of length $l_i$ starting from the $y-$axis,the $x-$coordinate of its centre of mass is $x_i = l_i / 2$.The mass of each rod is $m_i = \lambda l_i$.The lengths are $L, L/2, L/4, \dots$,so the masses are $M_1 = \lambda L, M_2 = \lambda L/2, M_3 = \lambda L/4, \dots$.The $x-$coordinates of the centres of mass are $x_1 = L/2, x_2 = L/4, x_3 = L/8, \dots$.The $x-$coordinate of the centre of mass of the system is given by:$X_{CM} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(\lambda L)(L/2) + (\lambda L/2)(L/4) + (\lambda L/4)(L/8) + \dots}{\lambda L + \lambda L/2 + \lambda L/4 + \dots}$$X_{CM} = \frac{\lambda L^2 (1/2 + 1/8 + 1/32 + \dots)}{\lambda L (1 + 1/2 + 1/4 + \dots)}$$X_{CM} = \frac{L}{2} \cdot \frac{(1 + 1/4 + 1/16 + \dots)}{(1 + 1/2 + 1/4 + \dots)}$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$:Numerator sum $= \frac{1}{1 - 1/4} = \frac{1}{3/4} = 4/3$.Denominator sum $= \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.$X_{CM} = \frac{L}{2} \cdot \frac{4/3}{2} = \frac{L}{2} \cdot \frac{2}{3} = L/3$.

Infinite rods of uniform mass density and lengths $L, L/2, L/4, \dots$ are placed one upon another up to infinity as shown in the figure. Find the $x-$ coordinate of the centre of mass.

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