Physics Puzzle Solution: Pulling up the Rope

April 28th, 2005 | Categories: Physics Puzzles - Solutions

This is the solution to the “Pulling up the Rope” puzzle.

We start by introducing a bit of notation (see figure below):

  • Let y=y(t) be the height of the pulled up rope.  Since the rope is being pulled with a constant velocity, v, then y(t) = vt.
  • Let λ be the density of the rope (mass per unit length).
  • Let L be the length of the rope, so λL is the total mass of the rope.
  • Let F be the upward force that pulls the rope up.
  • Let N be the normal force exerted on the rope by the surface on which it lies.

ropepullingupsolution

There are three forces acting upon the rope: the force F that pulls it up; its weight, λLg ; and the normal force exerted upon it by the floor, which is equal to

N=λ(L-y)g.

The rope’s momentum, p(t), is a function of time. At times t and t+Δt, it is:

  • p(t) = λy(t)v = λv²t
  • p(t+Δt) = λy(t+Δt)v = λv²t + λv²Δt

The change in momentum, Δp(t) = p(t+Δt) - p(t) = λv²Δt, must equal the impulse according to Newton’s 2nd law (dp/dt = F), so:

  • λv² = Δp(t)/Δt = F(t) + N - λLg = F(t) + λ(L-y)g - λLg = F(t) - λyg

This allows us to solve for the force F(t), as a function of time:

  • F(t) = λv² + λyg = λv² + λvtg = λv(v + gt)

This holds as long as there’s rope left on the surface. Once the rope leaves the surface it moves at a constant velocity (zero acceleration) upward, so its equation of motion becomes 0 = F - λLg, so F = λLg. This happens at a time t1 such that y(t1) = L, or t1=L/v. This is all the information we need to plot F(t):

ropepullingupsolution2

This completes the first part of the question. Next we compute the work W done by the force F(t) in lifting up the entire length of the rope:

W = \int_0^{L/v} F(t) v_{cm}(t) dt

Remember, the work done on a system of particles must be calculated using the center of mass variables. Luckily, in our case, the velocity of the center of mass is v as well. Here is a short proof:

v_{cm}(t) = \frac{1}{\lambda L} \int_0^{y(t)} \lambda v(y) dy = \frac{1}{\lambda L} \int_0^{vt} \lambda v dy = v

Using this, we can evaluate W:

W = \int_0^{L/v} \lambda v(v + gt) v dt= \frac{\lambda g L^2}{2} + \lambda L v^2= MgY_{cm} + Mv^2

Here I’ve used M = λL, the total mass of the rope, and Y_{cm} = L/2 to denote the position of the center of mass right after the rope is completely lifted off the table. The final answer is comprised of two terms, potential and kinetic, only the kinetic term looks funny - it’s missing the ½ in front of it! Compare the work done to the mechanical energy of the rope right after it has been lifted off the table:

E = Potential + Kinetic = ½MgL + ½Mv² < W

That means more work has been done than strictly necessary. Where did the energy go to waste? Well, when the chain was being pulled up, we’ve assumed that each bit of chain goes instantly from rest to having velocity v upward. Think of a particle at rest suddently being given some velocity v. This would cause its kinetic energy to jump from 0 to ½Mv² (where M is the particle’s mass). Where did this kinetic energy come from? Such a discontinuous jump can be caused by a collision, but to conserve energy it would mean there would have to be some other particle that would lose energy. In our model, the “particle” that loses energy is the hand that’s maintaining the force F(t): it has to put in extra energy to impart this initial velocity to every bit of chain. We can verify this in another way. Note that

W - E = ½Mv²

The gain in kinetic energy that results from taking a piece of rope Δy = vΔt and giving it a velocity v is:

ΔEk = ½λ(Δy)v² = ½λv³Δt

Integrating over time from 0 to L/v yields the total gain in kinetic energy solely due to our implicit assumption that every bit of rope starts off with an upward velocity v:

\int_0^{L/v} \frac{1}{2}\lambda v^3 dt = \frac{1}{2} \lambda v^3 (L/v) = \frac{Mv^2}{2}

which is precisely the “missing extra energy” W-E.

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