Question

A thin prismatic rod of length $1\text{m}$ is placed along $z$ axis with one of its ends located at the origin. If the density of rod varies from one end to the other as $\rho =5(1-z^2)\text{kg /m}$ , the position of centre of mass of rod is

• A. $\frac{3}{8}\text{m}$
• B. $\frac{3}{4}\text{m}$
• C. $\frac{1}{4}\text{m}$
• D. $\frac{1}{8}\text{m}$

Answer 1

The correct option is A $\frac{3}{8}\text{m}$

Given
Density of rod, $\rho =5(1-z^2)$
Let $dm$ be elemental mass of length $dz$.
$\therefore \frac{dm}{dz}=5(1-z^2)$
$dm=5(1-z^2)dz$
Total mass of the rod, $M=\int dm={\int }_0^{1}5(1-z^2)dz$
$=5{\int }_0^1(1-z^2)dz$
$=5{(z-\frac{z^3}{3})}_0^1$
$=5(1-\frac{1}{3})=\frac{10}{3}\text{kg}$
Centre of mass of Rod, $z_{cm}=\frac{\int zdm}{M}=\frac{{\int }_0^{1}z\left[5\right(1-z^2\left)\right]dz}{\frac{10}{3}}$
$=\frac{3}{10}{\int }_0^1(5z-5z^3)dz$
$=\frac{3}{10}[5{\left(\frac{z^2}{2}\right)}_0^1-5{\left(\frac{z^4}{4}\right)}_0^1]$
$=\frac{3}{10}(\frac{5}{2}-\frac{5}{4})=\frac{3}{8}\text{m}$