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 $x=0$ ens is :-

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

Answer 1

The correct option is D $3L/4$

Let the width of disc at a distance $x$ is $dx$ and the radius of the cylinder is $r$.

The mass of the disc is given as,

$dm=\pi r^{2}dx\times \frac{{\rho }_0{x}^2}{L^2}$

The centre of mass is given as,

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

$=\frac{{\int }_0^L\pi r^{2}dx\times \frac{{\rho }_0{x}^3}{L^2}}{{\int }_0^L\pi r^{2}dx\times \frac{{\rho }_0{x}^2}{L^2}}$

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

$=\frac{\frac{L^4}{4}}{\frac{L^3}{3}}$

$=\frac{3L}{4}$

Thus, the position of centre of mass from $x=0$ end is $\frac{3L}{4}$.