Question

If linear mass density of a rod of length $2$m varies as $\lambda =3x+2$, then the position of the centre of gravity of the rod is?

• A. $1$m
• B. $1.5$m
• C. $1.2$m
• D. $2$m

Answer 1

The correct option is C $1.2$m

Given length l=2 m

Considering an elemental section dx at a distance x from any one free end

Mass of the section dx=$\lambda dx$=(3x+2) dx

Now the position of center of gravity of rod

$X_{C.G}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}=\frac{{\int }_0^{2}x(3x+2)dx}{{\int }_0^2(3x+2)dx}=\frac{2^3+2^2}{3\left(\frac{2^2}{2}\right)+2^2}=1.2m$