Question

Calculate the centre of mass of a non-uniform rod whose linear mass density $\left(\lambda \right)$ varies as $\lambda =\frac{{\lambda }_o}{L}{x}^2$, where ${\lambda }_0$ is a constant,$L$ is the length of the rod and $x$ distance is measured from one end of the rod..

Answer 1

Let OX, rod be placed along x - axis

consider a small element $AB=x△\theta$

$OA=x$ & $OB=x+dx$

mass of element = $\lambda =\frac{\lambda }{L}{x}^2$

$\therefore x=$ distance of centre of mass from O, the

$X=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}$

$=\frac{{\int }_0^{L}x\frac{{\lambda }_0}{L}{x}^{2}dx}{{\int }_0^L\frac{\lambda 0}{L}{x}^{2}dx}$

$\frac{\frac{\lambda 0}{L}}{\frac{\lambda 0}{L}}\frac{{\int }_0^L{x}^{3}dx}{{\int }_0^L{x}^{2}dx}$

$=\frac{\frac{x^4}{4}{|}_0^L}{\frac{x^3}{3}{|}_0^L}\Rightarrow \frac{L^4}{4}.\frac{3}{L^3}=\frac{3L}{4}$

centre of mass $\Rightarrow \frac{3L}{4}$