Question

A uniform rod of mass $M$ and length $L$ is hanging from point $P$ as shown in the figure. Find the elongation in its length due to the self weight of the rod. The Young's modulus of elasticity of the rod is $Y$ and cross-sectional area of the rod is $A$.

• A. $\frac{MgL}{3AY}$
• B. $\frac{2MgL}{3AY}$
• C. $\frac{2MgL}{AY}$
• D. $\frac{MgL}{2AY}$

Answer 1

The correct option is D $\frac{MgL}{2AY}$

Since mass $M$ is uniformly distributed along its length.
So, Linear mass density ($\lambda$) of the rod $=\frac{M}{L}$

Consider a small section $dx$ of the rod at a distance $x$ from $Q$. The force acting at the section $dx$ is,
$W=\frac{Mg}{L}x=\lambda gx$
From Hooke's law, elongation in section $dx$ will be
$dL=\frac{W}{AY}dx=\frac{\lambda gx}{AY}dx$
So, Total elongation in the rod can be obtained by integrating the above expression from $x=0$ to $x=L$

$\Delta L={\int }_0^{L}dL={\int }_0^L\frac{\lambda g}{AY}xdx$
$=\frac{\lambda g}{AY}$ ${\int }_0^{L}xdx=\frac{\lambda g{L}^2}{2AY}$
$\therefore \Delta L=\frac{MgL}{2AY}$ $[\because \lambda =\frac{M}{L}]$