the problem appears to be simple: "a flexible rope of length 1 meter slides from a frictionless table top. the rope is initially released from rest with 30cm hanging over the edge of the table. find the time at which the left end of the rope reaches the edge of the table." (thornton and marion, 5th edition, chapter 9, problem 21, page 381.) i have attempted to solve this problem in a variety of ways: method 1) breaking up the rope into two sections and applying newton's laws. i find the mass of each section as a function of time, find the momentum of each side, find the time derivative of the momentum of each side and use dP1/dt = M1 g - T, dP2/dt = T, where T is the tension in the rope, P1 and M1 are the momentum and mass, respectively, for the part of the rope that is hanging off of the side, and P2 is the momentum of the part of the rope that is still on the table. all said and done, i got ddx/dtdt = (x+0.3)*g. (that's the second derivative of x with respect to time on the left hand side.) solving the diff eq, i get two exponentials and a constant term, but this leads to a transcendental equation, so i figured it was time to try something else. 2) same precedure, but i replaced the second derivative of x with dx/dt * d(dx/dt)/dx, using the chain rule. but this set up another messy differential equation, since i have to take a square root, then divide by the right hand side. 3) conservation of energy. i found the mass and center of mass of the hanging portion in its intial position and then calculated the potential energy. then i found the mass and center of mass of the hanging portion after 'x' more meters of rope fell over the side. so U_before = U_after + K, where K is the kinetic energy. i got dx/dt = (gx(x+0.6)/(x+0.3))^(1/2). this also is a mess! what am i missing that makes this problem soluble?