Question

A straight rod of length $L$ is made of a material having mass per unit length $m(x)=\lambda |x|$, where $x$ is measured from the centre of rod. The moment of inertia about an axis perpendicular to the rod and passing through one end of the rod will be, $L=1m$ and $\lambda =16kg/m^2$

• A. $\frac{32}{3}kg-m^2$
• B. $40kg-m^2$
• C. $\frac{36}{5}kg-m^2$
• D. $246kg-m^2$

Answer 1

Answer: (A)

Mass per unit length,
$\frac{dM}{dx}=\lambda |x|$
$\Rightarrow dM=\lambda |x|dx$

Moment of inertia about axis $AA$
$I_{AA}=\int x^{2}dM$
Put the value of $dM$
$I_{AA}=\underset{-L/2}{\overset{+L/2}{\int }}{x}^2\times \lambda |x|dx=\underset{-L/2}{\overset{+L/2}{\int }}{x}^2\times 16|x|dx$
$\left[\because \lambda =16kg/m^2\text{ (given) }\right]$
$=16\underset{-L/2}{\overset{+L/2}{\int }}{x}^2|x|dx$
$=16\left[\underset{-L/2}{\overset{0}{\int }}{x}^2(-x)dx+\underset{0}{\overset{+L/2}{\int }}{x}^2(+x)dx\right]$

So, $I_{AA}=16\left[\underset{-L/2}{\overset{0}{\int }}\left(-x^3\right)dx+\underset{0}{\overset{+L/2}{\int }}\left(x^3\right)dx\right]$
$=16\left[-{\left[\frac{x^4}{4}\right]}_{-L/2}^0+{\left[\frac{x^4}{4}\right]}_0^{+L/2}\right]$
$=4\left[-{\left[x^4\right]}_{-L/2}^0+{\left[x^4\right]}_0^{+L/2}\right]$
$=4\left[-\left\{(0{)}^4-{\left(-\frac{L}{2}\right)}^4\right\}+\left\{{\left(+\frac{L}{2}\right)}^4-(0{)}^4\right\}\right]$
$=4\left[\left\{0-\frac{L^4}{16}\right\}+\left\{\frac{L^4}{16}-0\right\}\right]$
$=4\left[\frac{L^4}{16}+\frac{L^4}{16}\right]=4\left[\frac{L^4}{8}\right]=\frac{L^4}{2}$
Now by the theorem of parallel axes, the moment of inertia about axis $BB$, $I_{BB}=I_{COM}+\int {\left(\frac{L}{2}\right)}^{2}dM=I_{COM}+\frac{L^2}{4}\int dM$
Put the value of $dM$ and $I_{\text{com }}$
$I_{BB}=I_M+\frac{L^2}{4}\underset{0}{\overset{L/2}{\int }}\lambda |x|dx=\frac{L^4}{2}+\frac{L^2}{4}\lambda \underset{0}{\overset{L/2}{\int }}|x|dx$
$=\frac{L^4}{2}+\frac{L^2}{4}\times 16\underset{0}{\overset{L/2}{\int }}(+x)dx$
$[\because \lambda =16kg/m^2$ (given) $]$
$=\frac{L^4}{2}+4L^2{\left[\frac{x^2}{2}\right]}_0^{L/2}=\frac{L^4}{2}+2L^2{\left[x^2\right]}_0^{l/2}$
$=\frac{L^4}{2}+2L^2\left[{\left(\frac{L}{2}\right)}^2-(0{)}^2\right]$
$=\frac{L^4}{2}+2L^2\left[\frac{L^2}{4}-0\right]$
$=\frac{L^4}{2}+\frac{L^4}{2}=L^4=1^4=1kg-m^2$