Question

Find the moment of inertia of a non-uniform rod having mass per unit length $\lambda ={\lambda }_0(1+\frac{x}{L})$ about an axis $\perp$ to the length of the rod and passing through an end point $L$. Length of the rod is $L$ and $m$ is mass of the rod.

• A. $\frac{m{L}^2}{3}$
• B. $\frac{7m{L}^2}{12}$
• C. $\frac{7m{L}^2}{18}$
• D. $\frac{m{L}^2}{18}$

Answer 1

The correct option is C $\frac{7m{L}^2}{18}$


Total mass of the rod,
$m=\int dm$
$m=\int \lambda dx$
$=\underset{0}{\overset{L}{\int }}{\lambda }_o(1+\frac{x}{L})dx$
$={\lambda }_o\left[\underset{0}{\overset{L}{\int }}dx+\frac{1}{L}\underset{0}{\overset{L}{\int }}xdx\right]$
$={\lambda }_o{[x+\frac{x^2}{2L}]}_0^L$
$={\lambda }_o[L+\frac{L}{2}]$
$\Rightarrow m=\frac{3{\lambda }_{o}L}{2}$
Now, moment of inertia of rod
$I=\int dm{x}^2$
$=\int {\lambda }_o(1+\frac{x}{L})dx{x}^2$
$={\lambda }_o\left[\underset{0}{\overset{L}{\int }}{x}^{2}dx+\frac{1}{L}\underset{0}{\overset{L}{\int }}{x}^{3}dx\right]$
$={\lambda }_o[\frac{L^3}{3}+\frac{L^4}{4L}]$
$I=\frac{7{\lambda }_o{L}^3}{12}$
$=\frac{7}{12}\left[\frac{3{\lambda }_{o}L}{2}\right]\times \frac{2}{3}{L}^2$
$=\frac{7}{18}m{L}^2$