Physics problem: rope sliding off the fence

Question

A rope of length l<h is thrown over a fence of height h. Initially, the rope is in equilibrium position. At some moment, the equilibrium is lost and the rope slides off the fence. Find how much time will pass from the moment when all of the rope to slides off the fence to the moment when it reaches the ground. Ignore the friction.

Answer 1

Let

m (unknown) be the mass of the rope,

g be gravitational acceleration,

v (unknown) be the rope's velocity at the moment the rope cleared the top of the fence,

t (to be computed) be the time from the moment one end of the rope clears the fence till the other end touches the ground.

I will assume that the rope's center of gravity is at its half-point.

This is the strategy:

compute v using conservation of energy,

compute t using Newton's Laws.

COMPUTE v

In the absence of friction, in the initial equilibrium position

the rope's center of gravity must have been on top of the fence.

Its potential energy was mgh.

Then the moment the rope clears the fence its potential energy was mg(h-l/2),

and its kinetic energy was mv²/2.

By conservation of energy

mg(h-l/2) + mv²/2 = mgh

From that

v = √(gl) (1)

COMPUTE t

The moment one end of the rope clears the fence,

the other end is distance h-l from the ground.

By free fall it covers that distance in the time t.

Without deriving it, the equation for free fall gives us

h-l = vt + gt²/2

That gives the quadratic equation

gt²/2 + vt - (h-l) = 0

The solution is

t = (-v ± √(v² + 2g(h-l)))/g

Substitute from (1)

t = (-√(gl) ± √(gl + 2gh - 2gl))/g = (-√(gl) ± √(2gh - gl))/g = (-√l ± √(2h - l))/√g

While the negative solution does have a physical significance,

only the positive solution corresponds to the given problem.

So

t = (-√l + √(2h - l)))/√g