Question

The flexible biycle type chain of length $\frac{\pi r}{2}$ and mass per unit length $p$ is released from rest with $\theta =0^{\circ }$ in the smooth circulur channel and falls through the hole in the supporting surface. Determine the velocity $v$ of the chain as the last link leaves the slot.

• A.
  $\sqrt{rg(\frac{\pi }{2}+\frac{4}{\pi })}$
• B.
  $\sqrt{rg(\frac{\pi }{4}+\frac{3}{\pi })}$
• C.
  $\sqrt{rg(1+\frac{2}{\pi })}$
• D.
  
  $\sqrt{rg(\frac{3}{\pi }+\frac{\pi }{2})}$

Answer 1

The correct option is A

$\sqrt{rg(\frac{\pi }{2}+\frac{4}{\pi })}$
Given,
mass per unit length of the chain$=p$

To calculate the initial potential energy of chain, let us consider a small element of mass $dm$ substended at an angle $d\theta$ from the centre as shown in figure,


mass of the element, $dm=p\left(rd\theta \right)$
height of the element, $h_e=r\cos \theta$

Taking the horizontal plane as the reference, gravitational potential energy of the element,
$d{U}_i=\left(dm\right)g\left(h_e\right)=p\left(rd\theta \right)g\left(r\cos \theta \right)$

Total potential energy of chain,
$U_i=\int d{U}_i={\int }_{\theta =0^o}^{\theta =\pi /2}\left(rd\theta \right)\left(\rho \right)\left(g\right)\left(r\cos \theta \right)$

$\Rightarrow U_i=pg{r}^2{\left[\sin \theta \right]}_0^{\pi /2}=pg{r}^2$

Now, when the chain leaves the slot, the center of mass of the chain is at a distance of $\frac{\pi r}{4}$ from the reference (horizontal plane) .

Thus the gravitational potential energy of the chain is
$U_f=(\frac{\pi r}{2}\times p)\left(g\right)(-\frac{\pi r}{4})=-\frac{{\pi }^2{r}^{2}pg}{8}$

So, loss in potential energy of chain
$\Delta U=U_i-U_f=r^{2}pg(1+\frac{{\pi }^2}{8})$

Gain in kinetic energy of the chain, $K.E=\frac{1}{2}m{v}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> According to work energy theorem,

$\Delta U=K.E$

$\Rightarrow r^{2}pg(1+\frac{{\pi }^2}{8})=\frac{1}{2}\left(\frac{\pi r}{2}\right)\left(p\right)v^2$

$\Rightarrow v=\sqrt{4rg(\frac{1}{\pi }+\frac{\pi }{8})}$

$\therefore v=\sqrt{rg(\frac{\pi }{2}+\frac{4}{\pi })}$

Hence, option (a) is correct answer.

Why this question? This question checks the application part of work-energy theorem or energy conservation principle. It also gives an idea of calculating potential energy if body of distributed mass is given.