Question

A thin bar of length L has a mass per unit length $\lambda$, that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is ${\lambda }_0$, then the distance of the centre of mass from the lighter end is :

• A. $\frac{L}{2}-\frac{{\lambda }_0{L}^2}{4M}$
• B. $\frac{L}{3}+\frac{{\lambda }_0{L}^2}{4M}$
• C. $\frac{L}{3}+\frac{{\lambda }_0{L}^2}{8M}$
• D. $\frac{2L}{3}-\frac{{\lambda }_0{L}^2}{6M}$

Answer 1

The correct option is D $\frac{2L}{3}-\frac{{\lambda }_0{L}^2}{6M}$

Let the bar be oriented along the x axis with its lighter end at the origin.
We have the mass per unit length as $\lambda \left(x\right)={\lambda }_o+ax$
Thus we have the mass M as

$M={\int }_o^L\lambda \left(x\right)dx={\lambda }_{o}x+\frac{a{x}^2}{2}{|}_o^L$

$={\lambda }_{o}L+\frac{a{L}^2}{2}$

or
$a=\frac{2(M-{\lambda }_{o}L)}{L^2}$

Thus we get the center of mass $\overline{x}$ as

$\overline{x}=\frac{\int x\lambda \left(x\right)dx}{M}$

$=\frac{({\lambda }_o{x}^2/2+a{x}^3/3){|}_o^L}{M}$

$=\frac{{\lambda }_o{L}^2/2+a{L}^3/3}{M}$

Substituting for $a$ we get

$x=\frac{\frac{{\lambda }_o{L}^2}{2}+\frac{2(M-{\lambda }_{o}L)L}{3}}{M}$

$=\frac{2L}{3}-\frac{{\lambda }_o{L}^2}{6M}$