A thin bar of length L has a mass per unit length λ , that increases linearly with distance from one end. If…

1. $\frac{\text{L}}{3}+\frac{{\lambda }_0{\text{L}}^2}{8\text{M}}$
2. $\frac{\text{L}}{3}+\frac{{\lambda }_0{\text{L}}^2}{4\text{M}}$
3. $\frac{\text{L}}{2}-\frac{{\lambda }_0{\text{L}}^2}{4\text{M}}$
4. $\frac{2\text{L}}{3}-\frac{{\lambda }_0{\text{L}}^2}{6\text{M}}$

Solution

$\lambda \propto \text{x}$

$\lambda =\text{Kx}+{\lambda }_0$

$\text{dm}=\lambda \text{dx}$

$\therefore {\text{X}}_{\text{cm}}=\frac{\underset{0}{\overset{\text{L}}{\int }}\text{dm}·\text{x}}{\underset{0}{\overset{\text{L}}{\int }}\text{dm}}=\frac{\underset{0}{\overset{\text{L}}{\int }}\left(\text{Kx}+{\lambda }_0\right)\text{dx.x} }{\underset{0}{\overset{\text{L}}{\int }}\left(\text{Kx}+{\lambda }_0\right)\text{dx}}$

$=\frac{\text{K}·{\left[\frac{{\text{x}}^3}{3}\right]}_0^{\text{L}}+{\lambda }_0{\left[\frac{{\text{x}}^2}{2}\right]}_0^{\text{L}}}{\text{K}{\left[\frac{{\text{x}}^2}{2}\right]}_0^{\text{L}}+{\lambda }_0 \text{L}}$

$=\frac{\frac{{\text{KL}}^3}{3}+\frac{{\lambda }_0{\text{L}}^2}{2}}{\frac{{\text{KL}}^2}{2}+{\lambda }_0\text{L}}$

$=\frac{\left(\frac{\text{KL}}{3}+\frac{{\lambda }_0}{2}\right){\text{L}}^2}{\frac{{\text{KL}}^2}{2}+{\lambda }_0\text{L}}$

Now, $\text{M}=\underset{0}{\overset{\text{L}}{\int }}\text{dm}$

$=\frac{{\text{KL}}^2}{2}+{\lambda }_0\text{L}$

$2\text{M}={\text{KL}}^2+2{\lambda }_0\text{L}$

$\text{K}=\frac{2\left(\text{M}-{\lambda }_0\text{L}\right)}{{\text{L}}^2}$

Putting the value of K.

${\text{X}}_{\text{cm}}=\frac{\left\{\frac{2\left(\text{M}-{\lambda }_0\text{L}\right)}{3\text{L}}+\frac{{\lambda }_0}{2}\right\}{\text{L}}^2}{\text{M}}$

$=\frac{2{\text{L}}^2\left(\text{M}-{\lambda }_0\text{L}\right)}{3\text{LM}}+\frac{{\lambda }_0{\text{L}}^2}{2\text{M}}$

$=\frac{2\text{ML}}{3\text{M}}-\frac{2{\lambda }_0{\text{L}}^2}{3\text{M}}+\frac{{\lambda }_0{\text{L}}^2}{2\text{M}}$

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

Hence, $\frac{2\text{L}}{3}-\frac{{\lambda }_0{\text{L}}^2}{6\text{M}}$

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