A chain of mass $M$ and length $L$ is held on a frictionless table with a fraction $\frac{1}{n}$ of its length hanging over the edge. If the chain is released, what is the velocity of the chain as it completely leaves the table?

Solution

1. The initial potential energy of the hanging portion of length $L/n$ is $U_i=-(\frac{M}{n})g(\frac{L}{2n})$, where the center of mass is at $L/2n$ below the table.
2. When the chain completely leaves the table, its center of mass is at $L/2$ below the table, so the final potential energy is $U_f=-Mg(\frac{L}{2})$.
3. By conservation of energy, the change in potential energy equals the final kinetic energy: $$\begin{array}{rl}\frac{1}{2}M{v}^2& =U_i-U_f\\ & =MgL(\frac{1}{2}-\frac{1}{2n^2})\end{array}$$.
4. Solving for velocity, we get $v^2=gL(1-\frac{1}{n^2})$, which simplifies to $v=\sqrt{gL(1-\frac{1}{n^2})}$.
5. Hence the answer is (A).