What is the Vertical Velocity of a Rope Sliding off a Peg?

[NewbieofPhysic]

Homework Statement

A limp rope with a mass of 2.5 kg and a length of 1.2 m is hung, initially at rest, on a frictionless peg that has a negligible radius. The rope is hung such that 0.8m hangs off the longer end, and 0.4m off the lower end. What is the vertical velocity of the rope just as the end slides off the peg?

Homework Equations

PE = mgh
KE = 1/2mv^2

The Attempt at a Solution

Because the kinematics of this system seemed incredibly complicated, I figured it best to use conservation of energy in the system. Knowing that the center of mass will fall 0.4 meters, I assumed:

mgΔH = 1/2mv^2
2.5*9.81*0.4 = 1/2*2.5*v^2
v = 2.8 m/s

Unfortunately, this seems to be incorrect. I assume this has something to do with the counterweight applied by the short end of the rope, but I'm not sure how to account for this.

 

[NewbieofPhysic]

Aha, got it! I knew the initial problem had to do with my initial potential energy, so I realized I could set it equal to:

PE = mh*g(0.8/1.2)

As the initial acceleration is dependent upon the ratio of mass at any given end and gravity. Using this, I could go straight to kinetic energy and solve.

 

[voko]

The second attempt has the correct initial C. M.