# The linear mass density of a thin rod

Question:

The linear mass density of a thin rod $\mathrm{AB}$ of length $\mathrm{L}$ varies from $\mathrm{A}$ to $\mathrm{B}$ as $\lambda(\mathrm{x})=\lambda_{0}\left(1+\frac{\mathrm{x}}{\mathrm{L}}\right)$, where $\mathrm{x}$ is the distance from A. If $\mathrm{M}$ is the mass of the rod then its moment of inertia about an axis passing through $\mathrm{A}$ and perpendicular to the rod is:

1. $\frac{5}{12} \mathrm{ML}^{2}$

2. $\frac{3}{7} \mathrm{ML}^{2}$

3. $\frac{2}{5} \mathrm{ML}^{2}$

4. $\frac{7}{18} \mathrm{ML}^{2}$

Correct Option: , 4

Solution:

$I=\int r^{2} d m=\int x^{2} \lambda d x$

$I=\int_{0}^{L} x^{2} \lambda_{0}\left(1+\frac{x}{L}\right) d x$

$\mathrm{I}=\lambda_{0} \int_{0}^{\mathrm{L}}\left(\mathrm{x}^{2}+\frac{\mathrm{x}^{3}}{\mathrm{~L}}\right) \mathrm{dx}$

$I=\lambda\left[\frac{L^{3}}{3}+\frac{L^{3}}{4}\right]$

$\mathrm{I}=\frac{7 \mathrm{~L}^{3} \lambda_{0}}{12}$..(1)

$\mathrm{M}=\int_{0}^{\mathrm{L}} \lambda \mathrm{dx}=\int_{0}^{\mathrm{L}} \lambda_{0}\left(1+\frac{\mathrm{x}}{\mathrm{L}}\right) \mathrm{dx}$

$\mathrm{M}=\lambda_{0}\left(\mathrm{~L}+\frac{\mathrm{L}}{2}\right)=\lambda_{0} \frac{3 \mathrm{~L}}{2}$

$\frac{2}{3} \mathrm{M}=\left(\lambda_{0} \mathrm{~L}\right)$..(2)

From (i) \& (ii)

$\mathrm{I}=\frac{7}{12}\left(\frac{2}{3} \mathrm{M}\right) \mathrm{L}^{2}=\frac{7 \mathrm{ML}^{2}}{18}$

Ans. (4)