Question

13. 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 :-
(1) $\frac{2L}{3}$
(2) $\frac{L}{2}$
(3) $\frac{L}{3}$
(4) $\frac{3L}{4}$

Answer 1

Dear Student,

$$\begin{gathered} lettakeadiscofdxwidthatdis\tan cexfromthex=0. \\ lettheradiusberofthecylinder. \\ massofthisdisc \\ dm={\mathrm{\pi r}}^2\mathrm{dx}\times \frac{{\mathrm{\rho }}_0{\mathrm{x}}^2}{{\mathrm{L}}^2} \\ \mathrm{CM}=\frac{{\int }_0^{\mathrm{L}}\mathrm{xdm}}{{\int }_0^{\mathrm{L}}\mathrm{dm}}=\frac{{\int }_0^{\mathrm{L}}{\mathrm{\pi r}}^2\mathrm{dx}\times \frac{{\mathrm{\rho }}_0{\mathrm{x}}^3}{{\mathrm{L}}^2}}{{\int }_0^{\mathrm{L}}{\mathrm{\pi r}}^2\mathrm{dx}\times \frac{{\mathrm{\rho }}_0{\mathrm{x}}^2}{{\mathrm{L}}^2}}=\frac{{\int }_0^{\mathrm{L}}{\mathrm{x}}^3\mathrm{dx}}{{\int }_0^{\mathrm{L}}{\mathrm{x}}^2\mathrm{dx}}=\frac{{\displaystyle \frac{{\mathrm{L}}^4}{4}}}{{\displaystyle \frac{{\mathrm{L}}^3}{3}}}=\frac{3\mathrm{L}}{4} \\ \mathrm{option}4 \\ \mathrm{regards} \end{gathered}$$