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

Answer 1

Answer: (D)

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