Question

A uniform chain of mass $M$ and length $L$ has a part $\frac{L}{3}$ hanging over the edge of a table. If the friction between the table and the chain is neglected, the kinetic energy of the chain as it completely falls off the edge of the table is

• A. $\frac{2\text{MgL}}{3}$
• B. $\frac{4\text{MgL}}{9}$
• C. $\frac{3}{2}\text{MgL}$
• D. $\text{MgL}$

Answer 1

The correct option is B $\frac{4\text{MgL}}{9}$

Mass per unit length of the chain is $\frac{M}{L}$. Consider a small element of length dx at a distance x below the edge of the table. The mass of this element is $m=\left(\frac{M}{L}\right)dx$. The potential energy of the hanging part of the chain is
$U=-{\int }_0^{L/3}\frac{Mg}{L}xdx$
$=-\frac{Mg}{L}{\left|\frac{x^2}{2}\right|}_0^{L/3}=-\frac{MgL}{18}$
Let us take the potential energy to be zero when the chain is on the table. Therefore, the total initial P.E. of the chain is
$U_t=0+U=-\frac{MgL}{18}$
If the full chain (of length L) were to fall, the P.E. would be
$U_f={\int }_0^L\frac{Mg}{L}xdx=-\frac{MgL}{2}$
$\therefore$ Loss in P.E = $U_i-U_f=\frac{MgL}{2}-\frac{MgL}{18}=\frac{4}{9}\text{MgL}$
Loss in P.E. = Gain in K.E.
$K_f-K_i=\frac{4}{9}\text{MgL}$
But initially the chain was at rest. So $K_i=0$
$K_f=\frac{4}{9}\text{MgL}$