Question

If linear density of a rod of length $3m$ varies as $\lambda =2+x$, then the position of the centre of mass of the rod is $\frac{P}{7}m$. Find the value of $P$

Answer 1

Answer: 1.714

Linear density of the rod
varies with distance $\frac{dm}{dx}=\lambda$
$\therefore dm=\lambda dx$
Position of centre of mass
$x_{cm}=\frac{\underset{0}{\overset{3}{\int }}(\lambda dx)\times x}{\underset{0}{\overset{3}{\int }}\lambda dx}=\frac{\underset{0}{\overset{3}{\int }}(2+x)\times xdx}{\underset{0}{\overset{3}{\int }}(2+x)dx}$
$=\frac{[x^2+\frac{x^3}{3}{]}_0^3}{[2x+\frac{x^2}{2}{]}_0^3}=\frac{12}{7}$m