Infinite rods of uniform mass density and len | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$L/3$

Vedclass · Answer

(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.

Similar Questions

An isolated particle of mass $m$ is moving in a horizontal plane $(x-y)$ along the $x$-axis,at a certain height above the ground. It suddenly explodes into two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = +15 \ cm$. The larger fragment at this instant is at

Two bodies of mass $1 \ kg$ and $3 \ kg$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$,respectively. The centre of mass of this system has a position vector:

For a system consisting of two objects,when can they be treated as point particles?

$A$ uniform thin rod of length $L$ and mass $M$ is placed along the $x$-axis with one end at the origin. Find the position of the centre of mass of the rod.

Vedclass Test Series

Exam Paper Generator

Online Exam Module