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

Linear density of the rod
varies with distance $\frac{dm}{dx} = \lambda$
$\therefore dm = \lambda dx$
Position of centre of mass
$x_{cm} = \frac{\int\limits_0^3 (\lambda dx ) \times x}{\int\limits_0^3 \lambda dx} = \frac{\int\limits_0^3 ( 2 + x)\times xdx}{\int\limits_0^3 ( 2 + x)dx}$
$ = \frac{[x^2 + \frac{x^3}{3}]_{0}^{3}}{[2x + \frac{x^2}{2}]_{0}^{3}} = \frac{12}{7}$m