Question

The mass per unit length of a non-uniform rod of length $L$ varies as $m=\lambda x$ where $\lambda$ is constant. The centre of mass of the rod will be at :

• A. $\frac{2}{3}L$
• B. $\frac{3}{2}L$
• C. $\frac{1}{2}L$
• D. $\frac{4}{3}L$

Answer 1

Answer: (A)

$\frac{2}{3}L$

$m=\lambda x$

taking differentiable on both side

$dm=\lambda xdx$

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

$=\frac{{\int }_0^{L}xx\lambda dx}{{\int }_0^L\lambda xdx}$

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

$={\left[\frac{x^3/3}{x^2/2}\right]}_0^L$

$=\frac{L^3}{3}\times \frac{2}{L^2}$

$x_{cm}=\frac{2}{3}L$.