# Law of Conservation of Energy on chain falling from table

As shown in the figure, a uniform chain of length 𝑳 is placed on a smooth horizontal table, and 𝑳/4 of it is hanging down on the edge of the table. After letting go, the chain starts to slide down the edge of the table from a standstill. What is the speed at which the chain slides to the edge of the table just after leaving the table (The chain is not on the ground)?

I used the law of conservation of energy to solve this problem. My attempt is:

I imagine the chain is similar to this situation:-

Therefore,

the gravitational potential energy of the L/4 chain = the kinetic energy of the whole chain

$$\frac{mgh}{4} = \frac{mv^{2}}{2} \\ v = \sqrt{\frac {gh}{2}}$$ However, the answer is $$𝒗 = \sqrt{\frac{15𝒈𝑳}{16}}$$

What's my mistake? How to solve this problem? Thank you in advance.

Your error is in ignoring the potential energy of the mass on the table. You can ignore it only if you take the point of reference as the table, which you haven't. Also, in your choice of reference, $$h$$ is not given in the question, so it is unclear where exactly your reference is.

You can solve this sum by consider the sections of the chain as particles located at their centres of mass. Considering table as reference, you will get initial potential energy as $$-\frac {mgL}{32}$$, and final potential energy as $$-\frac {mgL}{2}$$. You can then apply conservation of energy to get kinetic energy and the velocity of the chain.

• Oh, I see @dnaik . Could you explain more on the initial and final potential energy part? I don't understand it. Also, you explain that I could ignore the Ep of mass on the table if I take the table as the reference, so is it means that the potential energy that you calculated also ignores the mass on the table too? Perhaps could you give a more detailed answer, please? Thank you. Sep 6 '20 at 5:19

Let $$x$$ represent the downward axis with origin at the table edge. Also, let $$l_0$$ be the initial length over the edge of the table. We assume no friction here, and momentum is conserved.

$$F = \frac{x}{L}mg$$ and applying Newton's 2nd law, $$m\ddot{x} = \frac{x}{L}mg$$. This yields a simple, first order ODE, and upon applying the initial conditions $$x(0)=l_0$$ and $$\dot{x}(0) = 0$$, then letting $$k=\sqrt{g/L}$$,

$$x(t) = \frac{l_0}{2}\left(e^{kt} + e^{-kt} \right) = l_0 \cosh(kt)$$

The moment the last part of the chain leaves the table, time $$t^*$$, we have $$L = l_0 \cosh(kt^*)$$ or $$t^* = \frac{1}{k}\text{arcosh}(L/l_0)$$

Then, with $$x=L/l_0$$, and some identities

\begin{align} v^* &= l_0 k \sinh(\text{arcosh}(x)) \\ &= l_0 k \sinh(\ln (x + \sqrt{x^2 - 1})) \\ &= \frac{l_0 k }{2}\left(x + \sqrt{x^2 - 1} - \frac{1}{x + \sqrt{x^2 - 1}}\right)\\ &= \frac{l_0 k }{2}\left( 2\sqrt{x^2 - 1}\right) \\ \end{align}

Substituting in the values and simplifying yields the final answer,

$$v^* = \sqrt{\frac{15 L g}{16}}$$