Question

The linear mass density of a thin rod $\text{AB}$ of length $L$ varies from $\text{A}$ to $\text{B}$ as $\lambda \left(x\right)={\lambda }_0(1+\frac{x}{L}),$ where $x$ is the distance from $\text{A}$. If $M$ is the mass of the rod then its moment of inertia about an axis passing through $\text{A}$ and perpendicular to the rod is :

• A. $\frac{5}{12}M{L}^2$
• B. $\frac{7}{18}M{L}^2$
• C. $\frac{2}{5}M{L}^2$
• D. $\frac{3}{7}M{L}^2$

Answer 1

The correct option is B $\frac{7}{18}M{L}^2$


Mass of the small element of the rod
$dm=\lambda dx$

Moment of inertia of small element,
$dI=dm{x}^2={\lambda }_0(1+\frac{x}{L}).x^{2}dx$

Moment of inertia of the complete rod can be obtained by integration

$I={\lambda }_0{\int }_0^L(x^2+\frac{x^3}{L})dx$

$={\lambda }_0{|\frac{x^3}{3}+\frac{x^4}{4L}|}_0^L={\lambda }_0[\frac{L^3}{3}+\frac{L^3}{4}]$

$\Rightarrow I=\frac{7{\lambda }_0{L}^3}{12}.......\left(i\right)$

Mass of the thin rod,

$M={\int }_0^L\lambda dx={\int }_0^L{\lambda }_0(1+\frac{x}{L})dx=\frac{3{\lambda }_{0}L}{2}$

$\therefore {\lambda }_0=\frac{2M}{3L}$

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

Hence, option $\left(\text{B}\right)$ is correct.