Question

The variation of density of a cylindrical thick and long rod is $\rho ={\rho }_0\frac{x^2}{L^2}$, then position of its centre of mass from x = 0 end is:-

• A. $2L/3$
• B. $L/2$
• C. $L/3$
• D. $3L/4$

Answer 1

The correct option is D $3L/4$

$COM=\frac{{\int }^{}{\rho }_0.x.dx}{{\int }^{}\rho .dx}=\frac{{\int }_0^L{\rho }_0.\frac{x^3}{L^2}.dx}{{\int }_0^L{\rho }_0\frac{x^2}{L^2}.dx}$
$=\frac{\frac{{\rho }_0}{L^2}\times \frac{1}{4}{L}^4}{\frac{{\rho }_0}{L^2}\times \frac{1}{3}{L}^3}=\frac{3}{4}L$