Question

The density of a rod having length $l$ varies as $\rho =c+dx$, where $x$ is the distance from the left end. The centre of mass is:

• A. $\frac{3cl+2D{l}^2}{3(2c+Dl)}$
• B. $\frac{2cl+3D{l}^2}{2(4c+8l)}$
• C. $\frac{2cl+3D{l}^2}{3(2c+l)}$
• D. $\frac{cl+D{l}^2}{3(2c+Dl)}$

Answer 1

The correct option is A $\frac{3cl+2D{l}^2}{3(2c+Dl)}$

Centre of mass of the rod $x_{cm}=\frac{{\int }_o^{l}x\left(\rho Adx\right)}{{\int }_o^l\rho Adx}$ where $A$ is the cross-section area of rod

$⟹$ $x_{cm}=\frac{{\int }_o^{l}x\left(\rho dx\right)}{{\int }_o^l\rho dx}$

We calculate : $I_1={\int }_o^{l}x\left(\rho dx\right)$

$\therefore$ $I_1={\int }_o^{l}x(c+xD)dx={\int }_o^l(cx+x^{2}D)dx$

OR $I_1=c\frac{x^2}{2}+\frac{x^3}{3}D{|}_o^l$

$⟹$ $I_1=c\frac{l^2}{2}+\frac{l^3}{3}D=(\frac{c}{2}+\frac{lD}{3})l^2=\frac{(3c+2lD)l^2}{6}$

We calculate : $I_2={\int }_o^l\rho dx$

$\therefore$ $I_2={\int }_o^l(c+xD)dx$

OR $I_2=cx+\frac{x^2}{2}D{|}_o^l$

$⟹$ $I_2=cl+\frac{l^2}{2}D=(c+\frac{lD}{2})l=\frac{(2c+lD)l}{2}$

$⟹$ $x_{cm}=\frac{(3c+2lD)l^2/6}{(2c+lD)l/2}=\frac{3cl+2D{l}^2}{3(2c+Dl)}$