Question

A nonuniform rod $O M$ of length $l$ has linear mass density that varies with the distance $x$ from left end of the rod according to $\lambda = \left(\lambda \right)_{0} \left(\frac{x^{3}}{L^{3}}\right)$ ; Where $\lambda _{0}$ is constant, is kept along $x$ -axis and is rotating about an axis $A B$ , which is perpendicular to the rod as shown in the figure. What is the value of $x$ so that moment of inertia of rod about axis $A B \, \left(\left(\text{I}\right)_{A B}\right)$ is minimum?

• A. $\frac{7 \ell }{15}$
• B. $\frac{2 \ell }{5}$
• C. $\frac{8 l}{15}$
• D. $\frac{4 \ell }{5}$

Answer 1

Answer: (D)

$I_{A B}=\int y^{2} d m=\frac{\lambda_{0}}{e^{3}} \int_{-x}^{l-x}(y+x)^{3} y^{2} d y$
$\int(y+x)^{3} y^{2} d y=y^{2} \frac{(y+x)^{4}}{4}-\int \frac{1}{2} y(y+x)^{4} d y$
$=\frac{y^{2}(y+x)^{2}}{4}-\frac{y}{10}(y+x)^{5}+\frac{1}{10}(y+x)^{5} d y$
$=\frac{y^{2}(y+x)^{2}}{4}-\frac{y(y+x)^{5}}{10}+\frac{(y+x)^{2}}{60}$
$\int_{-x_{0}}^{(-x} y^{2}(y+x)^{3} d x=\frac{(\ell-x)^{2}(\ell)^{4}}{4}-\frac{(\ell-x) t^{5}}{10}+\frac{\ell^{6}}{60}$
$I_{A B}=\lambda_{0}\left[\frac{(\ell-x)^{2}}{4} \ell-\frac{1}{10}(\ell-x) \ell^{2}+\frac{1}{60} \ell^{3}\right]$
$=\lambda_{0} \ell\left[\frac{(\ell-x)^{2}}{4} \ell-\frac{1}{10}(\ell-x)+\frac{1}{60} \ell^{2}\right]$
$\frac{d l_{A B}}{d x}=0 \Rightarrow -\frac{(\ell-x)}{2}+\frac{\ell}{10}=0$
$\Rightarrow 5 x=4 \ell \Rightarrow x=\frac{4 l}{5}$