# Arc of rope flowing over drum

1. Jul 7, 2014

### haruspex

This is a problem I thought up. If I'm correct as far as I've gone, the answer is rather surprising.

A thin but 'massive' rope lies coiled on a floor. Treat the coil as a point.
One end passes vertically up and over a smooth cylindrical drum radius r.
The centre of the drum is height h above the floor.
The other side of the drum, the rope descends to a second coil height d below the first coil.
In steady state, the rope runs at constant speed from the first coil to the second.
1. What is the shape of the descending arc of rope?
2. How far horizontally from the centre of the drum is the second coil?

Using (s, ψ) co-ordinates, s is the distance along the rope from the point where it loses contact with the drum, and ψ is the angle to the vertical made by the arc.
T = T(s) is the tension.
ρ = mass per unit length.
v = speed of rope

Along the rope there is no acceleration: $(1): \frac{dT}{ds} = - \rho g \cos(\psi)$
Perpendicular to the rope there is centripetal acceleration: $(2): v^2\rho\frac{d\psi}{ds} = T\frac{d\psi}{ds} + \rho g \sin(\psi)$
Differentiating (2): $v^2\rho\frac{d^2\psi}{ds^2} = \frac{dT}{ds}\frac{d\psi}{ds} +T\frac{d^2\psi}{ds^2} + \rho g \cos(\psi)\frac{d\psi}{ds}$
Applying (1): $v^2\rho\frac{d^2\psi}{ds^2} = T\frac{d^2\psi}{ds^2}$
Since T is clearly not constant: $\frac{d^2\psi}{ds^2} = 0$
Whence the trajectory is a circular arc(!).
Do I have something wrong?

2. Jul 7, 2014

### MrAnchovy

Sorry I misread the bit behind the spoiler - now I am on a proper screen so I can see it all at once I realise you meant T is not constant wrt s, which of course is true.

Last edited: Jul 7, 2014
3. Jul 7, 2014

### AlephZero

Are you assuming there is sufficient friction between the drum and the rope, and the drum is turning at constant angular velocity to give your "steady state" solution?

I don't understand the objection to
Obviously the variation of T along the rope depends on the speed and shape of the rope.

I think one mistake is assuming $\frac{d}{ds}v^2 = 0$. The $\mathbf{v}$ vector has constant magnitude, but not constant direction. That seems like the same mistake as arguing the $v$ is constant for uniform circular motion, therefore centripetal acceleration doesn't exist.

The "circular arc" answer is clearly wrong. Your analysis doesn't include the vertical distance between the drum and the floor. If that distance was big enough, it is "obvious" the rope isn't going to form a circular coil in mid air and never touch the ground.

4. Jul 7, 2014

### MrAnchovy

Oops, now I have edited my previous post instead of posting a new one correcting it, sorry.

5. Jul 7, 2014

### MrAnchovy

Have you considered the solution $\psi = 0$?

6. Jul 7, 2014

### haruspex

I don't believe it requires any friction. There is is an energy loss in the pick-up of the rope from the higher floor. But if it makes you happier to assume friction then that's fine, as it won't change the shape of the trajectory.
I should have been clearer that I meant "not constant along the rope".
My v is a speed, not a vector, and equn 2 is purely scalar. I don't see any error there.
I thought that at first. The arc could not become vertical, let alone miss the floor. Then I realised that if you lower the floor, maybe the arc's radius will increase. There will be more energy available to throw the rope further.
Can you come up with a DE that gives a different answer?

Edit: Assuming the arc of circle solution is correct, the centre of curvature will always be below the height of the lower floor. From equn 2 applied to where the rope lands, the distance below will be v2/g.
If the difference in the floor heights is small, the centre of curvature will be a long way horizontally from the drum. The rope will leave the drum at almost vertical, and become only a little more vertical before hitting the floor. In the limit, the descent is a vertical straight line.

Last edited: Jul 7, 2014
7. Jul 8, 2014

### MrAnchovy

i.e. $\psi = 0$. I think you will find that that is the only stable solution.

What makes you think that the rope will leave contact with the pulley before it reaches vertical? What force keeps the rope against the pulley?

8. Jul 8, 2014

### haruspex

It will always leave the pulley before the vertical. It will have some horizontal momentum from passing over the pulley. A vertical descent would not apply any force to counter that momentum.

9. Jul 9, 2014

### MrAnchovy

When the rope is vertical it has no horizontal momentum. Immediately before the rope is vertical it has a small amount of horizontal momentum, but there is also a horizontal component of tension (from both sides).

Can you draw these forces and derive the angle at which the rope leaves the pulley?

10. Jul 9, 2014

### MrAnchovy

Hint: the centripetal force is provided by tension in the rope, but the tension force is perpendicular to the centre of rotation - how can this be?

Also note that I am not saying that there is no condition in which a rope will leave a pulley before reaching the vertical - above a certain speed this will always happen, and when it does the descending rope goes slack. Clearly this speed cannot be achieved through constant tension on the descending rope.

11. Jul 9, 2014

### haruspex

It's provided by a combination of tension and gravitational force. Because the rope forms an arc, the tension provides a force perpendicular to the rope: Tdψ/ds.
Quite so, above zero speed in fact.
It can't be slack. It has to exert an upward force equal to the weight of hanging rope below it.
The tension is constant over time, but not over distance along the rope.

12. Jul 10, 2014

### MrAnchovy

Can you show that? I think the angular velocity $\omega$ of the pulley radius $r$ must satisfy $r \omega^2 > g$ (and then the angle $\psi$ is given by $r \omega^2 \sin \psi = g$).

Why does it have to? Your statement is a tautology - the rope is slack (has zero tension) exactly when it does not exert an upward force equal to the weight of hanging rope below it. And when the rope in contact with the pulley is accelerating downwards at $g$ (at the point described above) this is exactly what happens.

There is a term for this condition which is critical to avoid when designing a winch (or casting a fishing line) - it is called overrun.

Last edited: Jul 10, 2014
13. Jul 10, 2014

### TSny

haruspex, I believe there is a sign error in equation (1) of the spoiler.

This problem appears to be related to the "chain fountain" phenomenon. See these videos:

A paper on the chain fountain is http://arxiv.org/pdf/1310.4056v2.pdf

See equations (0.11) and (0.12) in this paper.

Last edited by a moderator: Sep 25, 2014
14. Jul 10, 2014

### haruspex

Sorry, you're right. I need to modify my set-up to get to the case I'm interested in. I can eliminate a parameter at the same time:

A pile of rope lies on a smooth floor. The rope has uniform mass. It is arbitrarily thin, so the pile can be considered a point.
One end of the rope trails off the edge of the floor, down to a lower level. The edge of the floor is in the form of a smooth quadrant of a circle, radius r. Having reached the vertical, the surface descends a further distance h to the lower floor.
(Diagram attached.)

From considering the rate of change of momentum as the rope is drawn from the pile, and the rate of work done, v2 = g(r+h).
If the rope reaches the vertical without losing contact then consideration of centripetal acceleration at (or infinitesimally before) that point gives v2 ≤ gh, a contradiction.

I don't see that any part of the rope will be accelerating down at g. It will always be somewhat less. The rope will only be slack at a point if it is exerting no force on the section below it. Near the bottom, where the descent is closest to vertical, there will be hardly any downward acceleration. Remember, the speed of the rope along its length is constant. The only downward acceleration is the downward component of the centripetal acceleration.

15. Jul 10, 2014

### haruspex

You're right! I wasn't being consistent in my definition of ψ. If I take it to be the angle from vertically up, around to the direction of motion, then cos(ψ) is negative. If I take it from vertically down, measuring it in the other sense, then the signs are wrong in the second equation.

Thanks!

16. Jul 11, 2014

### MrAnchovy

This assumes all the work done by the system is in picking up the rope. What about the $\frac t2 \rho v^3$ of kinetic energy that is dumped by the rope hitting the floor?

17. Jul 11, 2014

### MrAnchovy

Hmmm, ignore my last comment - that energy is already counted.