[bettylin]

a flexible chain of length L slides off the edge of a frictionless table, as in Fig. // the diagram is better than this one. The chain of course is consistent.



Initially a length y hangs over the edge. (a) Find the acceleration of the chain as a function of y. (b) show that the velocity as the chain becomes completely vertical is v = √(g(L- (y(initial))^2/L))

[ebaines]

One way to approach this is using work and energy principles. Consider the chain in two parts - one initially hanging over the dge and the other initially on the table. You need to determine the amount of work done by gravity on each part, and then set that equal to the increase in kinetic energy of the chain.

The part that is initially hanging over the table falls a distance (L-y_1). Multiply that by the mass of this part and you have the work done on part 1.

For the part that is initially on the table you need to consider each segment dy, because the distance that each segment falls under gravity is dependent on its initial position on the table. The distance each infinitesssimal segment of tHE chain falls under gravity is L-y, where y is measured from the leading edge of the chain, and the mass of each segment dy times its density. So you can set up an integral to calculate the work doe on this part.

The total work for parts 1 and 2 must equal the change in KE of the chain. You will see that the density of the chain per unit length cancels out.