Question

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

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

Answer 1

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

Let the bar be oriented along the $x-$ axis with its lighter end at the origin.
We have the mass per unit length as

$\lambda \left(x\right)={\lambda }_0+ax$

So mass,

$M={\int }_0^L\lambda \left(x\right)dx={\int }_0^L({\lambda }_0+ax)dx$
$={[{\lambda }_{0}x+\frac{a{x}^2}{2}]}_0^L={\lambda }_{0}L+\frac{a{L}^2}{2}$

or $a=\frac{2(M-{\lambda }_{0}L)}{L^2}$

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

$\overline{x}=\frac{{\int }_0^{L}x\lambda \left(x\right)dx}{M}=\frac{{\int }_0^{L}x({\lambda }_0+ax)dx}{M}$
$=\frac{{[\frac{{\lambda }_0{x}^2}{2}+\frac{a{x}^3}{3}]}_0^L}{M}$
$=\frac{(\frac{{\lambda }_0{L}^2}{2}+\frac{a{L}^3}{3})}{M}$

Substitutig for $a$ we get

$\overline{x}=\frac{\frac{{\lambda }_0{L}^2}{2}+\frac{2(M-{\lambda }_{0}L)}{L^2}\times \frac{L^3}{3}}{M}$

$\overline{x}=\frac{2L}{3}-\frac{{\lambda }_0{L}^2}{6M}$

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