Question

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

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

Answer 1

Answer: (D)

$I=\int r^{2}dm=\int x^2\lambda dx$
$I=\underset{0}{\overset{L}{\int }}{x}^2{\lambda }_0\left(1+\frac{x}{L}\right)dx$
$I={\lambda }_0{\int }_0^L\left(x^2+\frac{x^3}{L}\right)dx$
$I=\lambda \left[\frac{L^3}{3}+\frac{L^3}{4}\right]$
$I=\frac{7L^3{\lambda }_0}{12}$ ___(i)
$M=\underset{0}{\overset{L}{\int }}\lambda dx=\underset{0}{\overset{L}{\int }}{\lambda }_0\left(1+\frac{x}{L}\right)dx$
$M={\lambda }_0\left(L+\frac{L}{2}\right)={\lambda }_0\frac{3L}{2}$
$\frac{2}{3}M=\left({\lambda }_{0}L\right)$ ___(ii)
From (i) & (ii)
$I=\frac{7}{12}\left(\frac{2}{3}M\right)L^2$
$=\frac{7M{L}^2}{18}$