Question

The density of a thin rod of length $L$ varies with the distance $x$ form one end as $\rho ={\rho }_0\frac{x^2}{L^2}$. Find the position of centre of mass of rod.

Answer 1

Let A = Area of cross section of rod

$x_{cm}=\frac{{\int }_0^{1}xdm}{{\int }_0^{1}dm}$

$=\frac{{\int }_0^1\left(x\right)\left(Adx\right)po\frac{x^2}{L^2}}{{\int }_0^1\left(Adx\right)po\frac{x^2}{L^2}}$

$=\frac{3l}{4}$