Question

A rod of mass $m$, uniform cross-sectional area $A$ and length $L$ is accelerated by applying force $F$ as shown in the figure on a smooth surface. The Young's modulus of elasticity of the material of rod is $Y$. Consider $x$ as measured from the right end, strain in the rod is $\frac{\alpha F}{\beta AY}$. Find $(\alpha +\beta )$.

Answer 1

At any distance $x$ from edge, let there be a cross-section for which $FBD$ is given as:

For complete rod: $ma=F\Rightarrow a=\frac{F}{m}\dots \left(i\right)$

For portion $BC:m_{BC}a=T\Rightarrow T=\left(\frac{mx}{L}\right)a$

$\Rightarrow T=\left(\frac{mx}{L}\right)\frac{F}{m}=\frac{F}{L}x$

$\Rightarrow \text{Stress}=\frac{Fx}{AL}$[using $\left(i\right)$]

$\frac{dl}{dx}=\frac{\text{Stress}}{Y}=\frac{\frac{Fx}{LA}}{Y}=\frac{Fx}{LAY}$

$\Rightarrow \Delta l={\int }_0^L\frac{Fx}{LAY}dx\Rightarrow \frac{\Delta l}{L}=\frac{F}{2AY}$

Final Answer: $3$