question_answer

A non-uniform rod of length I has a linear mass density \[\lambda ={{\lambda }_{0}}+kx,\] is placed over X-axis with its one end at origin. The x-coordinate of centre of mass of rod is

A) \[\frac{{{\lambda }_{0}}l+2k{{l}^{2}}}{3(2{{\lambda }_{0}}+kl)}\]

B) \[\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{(2{{\lambda }_{0}}+kl)}\]

C) \[\frac{4{{\lambda }_{0}}l+k{{l}^{2}}}{3\,(2{{\lambda }_{0}}+kl)}\] 

D) \[\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{3\,(2{{\lambda }_{0}}+kl)}\]

Correct Answer: D

Solution :

We have, \[dm=\lambda dx=({{\lambda }_{0}}+kx)dx\]

\[\therefore \]\[{{X}_{CM}}=\frac{\int{xdm}}{\int{dm}}=\frac{\int\limits_{0}^{l}{x({{\lambda }_{0}}+kx)dx}}{\int\limits_{0}^{l}{({{\lambda }_{0}}+kx)dx}}\]

\[=\frac{{{\lambda }_{0}}\frac{{{l}^{2}}}{2}+\frac{k{{l}^{3}}}{3}}{{{\lambda }_{0}}l+\frac{k{{l}^{2}}}{2}}=\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{3(2{{\lambda }_{0}}+kl)}\]