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:

Answer 1

Let the width of the distance $x$ is $dx$ and the radius of the cylinder is $r$. So, the mass of the disc is given as:

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

The centre of the mass is given as:
$CM=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}$

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

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

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

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

Hence, $\left(D\right)$ is the correct answer.