Question

A rod of length $3m$ and its mass per unit length is directly proportional to the distance $x$ from its one end. The centre of gravity of the rod from that end will be at

• A. $3m$
• B. $2.5m$
• C. $2m$
• D. $1.5m$

Answer 1

The correct option is C $2m$

Step 1: Draw a rough diagram of non-uniform rod.
Given: $l=3m,\lambda \propto x$
Let the origin be at left end
Assume a small element dm at distance $x$ from left end,

Step 2: Find centre of mass of rod.
Formula used: $x_{cm}=\frac{\int xdm}{\int dm}$
Mass per unit length, i.e., linear mass density
$\lambda \propto x\lambda ={\lambda }_{0}x,{\lambda }_0$ is costant
$dm=\lambda dx={\lambda }_{o}xdx$

$x_{cm}=\frac{\int xdm}{\int dm}=\frac{{\int }_0^{3}x{\lambda }_{0}xdx}{{\int }_0^3{\lambda }_{0}xdx}$

$x_{cm}=\frac{\int x^{2}dx}{\int xdx}=\frac{{\left(\frac{x^3}{3}\right)}_0^3}{{\left(\frac{x^2}{2}\right)}_0^3}=2m$