How Does Lowering Half a Rope Affect Its Potential Energy?

[Shifty]

A uniform 3.60- kg rope 23.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top let's down half of it to help his partner climb up the cliff.

What was the change in potential energy of the rope during this maneuver?

-This in a moment of inertia assignment, so I'm guessing it uses it somewhere. Don't have a clue after that.

-Help would be appreciated :)

 

[ideasrule]

It may be a moment of inertia question, but this is a center of mass problem. I'll give you a hint: you can consider an object's mass to be concentrated at its center of mass when calculating gravitational potential energy.

 

[Shifty]

It may be a moment of inertia question, but this is a center of mass problem. I'll give you a hint: you can consider an object's mass to be concentrated at its center of mass when calculating gravitational potential energy.

Thanks for the reply :), but I already tried that and calculated the change to be -203 J, unfortunately it didn't yield a correct answer.

\Delta U=-mgy=203

y=23m/4

EDIT!: looking at this reply I realized the total mass would only be half since only half of the rope was lowered. That solved it, thanks :)!

m=3.6kg/2, implies \Delta U=102J