Question

A nonuniform rod $OM$ 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 $AB$ , 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 $AB\left({\left(\text{I}\right)}_{AB}\right)$ is minimum?

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

Answer 1

Answer: (D)

$I_{AB}=\int y^{2}dm=\frac{{\lambda }_0}{e^3}{\int }_{-x}^{l-x}(y+x{)}^3{y}^{2}dy$
$\int (y+x{)}^3{y}^{2}dy=y^2\frac{(y+x{)}^4}{4}-\int \frac{1}{2}y(y+x{)}^{4}dy$
$=\frac{y^2(y+x{)}^2}{4}-\frac{y}{10}(y+x{)}^5+\frac{1}{10}(y+x{)}^{5}dy$
$=\frac{y^2(y+x{)}^2}{4}-\frac{y(y+x{)}^5}{10}+\frac{(y+x{)}^2}{60}$
${\int }_{-x_0}^{(-x}{y}^2(y+x{)}^{3}dx=\frac{(\ell -x{)}^2(\ell {)}^4}{4}-\frac{(\ell -x)t^5}{10}+\frac{{\ell }^6}{60}$
$I_{AB}={\lambda }_0\left[\frac{(\ell -x{)}^2}{4}\ell -\frac{1}{10}(\ell -x){\ell }^2+\frac{1}{60}{\ell }^3\right]$
$={\lambda }_0\ell \left[\frac{(\ell -x{)}^2}{4}\ell -\frac{1}{10}(\ell -x)+\frac{1}{60}{\ell }^2\right]$
$\frac{d{l}_{AB}}{dx}=0\Rightarrow -\frac{(\ell -x)}{2}+\frac{\ell }{10}=0$
$\Rightarrow 5x=4\ell \Rightarrow x=\frac{4l}{5}$