Question

Linear mass density of a bar is $\rho =\frac{{\rho }_o}{L_o}L+{\rho }_o$ where ${\rho }_o$ is constant and $L_o$ is length of bar. Its centre of mass from one of the end is
(Assume the other dimensions of the bar to be unity)

• A. $\frac{3L_o}{5}$
• B. $\frac{4L_o}{7}$
• C. $\frac{5L_o}{9}$
• D. $\frac{2L_o}{5}$

Answer 1

The correct option is C $\frac{5L_o}{9}$

Let us take a small mass $dm$ at distance $L$ from one of the end of width $dl$


We know, C.O.M of continuous body,
$r_{\text{C.O.M}}=\frac{\int Ldm}{\int dm}$
$=\frac{\underset{0}{\overset{L_o}{\int }}L(\frac{{\rho }_o}{L_o}L+{\rho }_o)dL}{\underset{0}{\overset{L_o}{\int }}(\frac{{\rho }_o}{L_o}L+{\rho }_o)dL}$ $[\because \rho =\frac{dm}{dL}=\frac{{\rho }_o}{L_o}L+{\rho }_o]$
$r_{\text{C.O.M}}=\frac{{[\frac{{\rho }_o}{L_o}\times \frac{L^3}{3}+\frac{{\rho }_o{L}^2}{2}]}_0^{L_o}}{{[\frac{{\rho }_o}{L_o}\times \frac{L^2}{2}+{\rho }_{o}L]}_0^{L_o}}$
$=\frac{\frac{{\rho }_o{L}_o^2}{3}+\frac{{\rho }_o{L}_o^2}{2}}{\frac{{\rho }_o{L}_o}{2}+{\rho }_o{L}_o}$
$r_{\text{C.O.M}}=\frac{5}{9}{L}_o$
Centre of mass from one of the end is $\frac{5L_o}{9}$.