Question

A uniform rope of mass m and lenght l is placed on a smooth table such that one fourth of its lenght hangs down. The rope begins to slide down. The speed of the rope when it complitly slips of the tabel is

Answer 1

Given :

Mass of the uniform rope placed on a smooth table  = m

Length of the rope = l

Part of the rope length hanging down the table = $\frac{l}{4}$

To Find :

Speed of the rope when it completely slips of the table = ?

Solution :

Let the height of the table be :

= h

Since the table is smooth , i.e.  there is no friction .

So, we can use energy conservation here :

⇒ Initial energy = final  energy

Now , since the $\frac{3}{4}$ th of the rope is on the table , so its potential energy = $\frac{3}{4} mgh$

Also $\frac{l}{4}$ length of the rope is hanging and its centre of mass will be at $\frac{l}{8}$ .

Hence its potential energy will be = $\frac{mg}{4}\times (h - \frac{l}{8})$

And the final energy will be as :

=$\frac{1}{2}m v^2 + mg(h - \frac{l}{2})$

Here v is the velocity with which the rope falls .

So, now using the energy conservation we can write :

$\frac{3}{4}mgh + \frac{mg}{4}(h-\frac{l}{8}) = \frac{1}{2}mv^2+ mg(h-\frac{l}{2})$

Or, $\frac{3}{4}gh + \frac{g}{4}(h-\frac{l}{8})= \frac{1}{2}v^2+ gh - \frac{gl}{2}$

Or,$gh - \frac{gl}{32} = \frac{1}{2} v^2 + gh - \frac{gl}{2}$

Or, $\frac{16gl}{16\times2} -\frac{gl}{32}= \frac{v^2}{2}$

Or ,$\frac{v^2}{2} = \frac{15gl}{32}$

Or ,$v = \frac{\sqrt{15gl} }{4}\frac{m}{s}$

So, finally the speed of the rope when it completely slips of the table is $v = \frac{\sqrt{15gl} }{4}\frac{m}{s}$ .