Question

A rod of length $6\text{m}$ is placed along the $x-$ axis between $x=0$ and $x=6\text{m}$. The linear density (mass/length) $\lambda$ of the rod varies with the distance $x$ from the origin as $\lambda =k(10-x)$. Here $k$ is a positive constant. Find the position of the centre of mass of this rod.

• A. $(1.32\text{m},0\text{m},0\text{m})$
• B. $(3\text{m},0\text{m},0\text{m})$
• C. $(2.57\text{m},0\text{m},0\text{m})$
• D. $(3.9\text{m},0\text{m},0\text{m})$

Answer 1

The correct option is C $(2.57\text{m},0\text{m},0\text{m})$

Here, given length of rod $L=6\text{m}$


Take the element $dx$ of mass $dm$ length situated at $x$ distance from origin.
Then, $dm=\lambda dx$
$dm=k(10-x)dx$

The COM of the element has the co-ordinates $(x,0,0)$
Therefore, $x-$ co-ordinate of COM of the rod will be
$x_{COM}=\frac{\int xdm}{\int dm}=\frac{\underset{0}{\overset{6\text{m}}{\int }}x.k(10-x)dx}{\underset{0}{\overset{6\text{m}}{\int }}k(10-x)dx}$
$=\frac{k\underset{0}{\overset{6}{\int }}(10x-x^2)dx}{k\underset{0}{\overset{6}{\int }}(10-x)dx}=\frac{{[\frac{10x^2}{2}-\frac{x^3}{3}]}_0^6}{{[10x-\frac{x^2}{2}]}_0^6}$
$=\frac{\frac{10\times 36}{2}-\frac{216}{3}}{10\times 6-\frac{36}{2}}=\frac{180-72}{10\times 6-18}$
$\therefore x_{COM}=\frac{108}{42}=2.57\text{m}$

The $y-$ co-ordinate of COM of the rod is $y_{COM}=\frac{\int ydm}{\int dm}=0$
The $z-$ co-ordinate of COM of the rod is $z_{COM}=\frac{\int zdm}{\int dm}=0$
So, the center of mass of the rod lies at $(2.57,0,0)\text{m}$