# Tetherball - Classical Mechanics

## Homework Statement

A small ball is attached to a massless string of lenght L, the other end of which is attached to a very thin pole. The ball is thrown so that it initially travels in a horizontal circle, with the string making an angle $$\theta_0$$ with the vertical. As time goes on, the string wraps itself around the pole. Assume that the pole is thin enough so that the length of string in the air decreases very slowly, so that the ball's motion may always be approximated as a cricle. Assume that the pole has enough friction so that the string does not slide on the pole, once it touches it. Show that the ratio of the ball's final speed (right before it hits the pole) to initial speed is $$v_f / v_i = \sin(\theta_0)$$

## Homework Equations

Conservation of energy applies. Conservation of angular momentum doesn't (the friction provides torque).

## The Attempt at a Solution

I couldn't really do much on this problem. The $$\Delta L$$ after the mass has traveled once around the pole is:

$$\Delta L = - \frac{2 \pi d}{\sin(\theta)}$$

I had forgot about the slow descending of the lowest contact point of the string with the pole. Without considering it, finding the velocity of the mass as a function of $$\theta$$ I could find (using the tension in the string and balancing weight):

$$v^2 = g L \frac{(\sin(\theta))^2}{\cos(\theta)}$$

Differentiating this relation, and using energy after having expressed $$\Delta h$$ as a function of $$\Delta \theta$$, after a lot of calculation I found:

$$(v_{n+1}^2 - v_n^2) = 8 \pi g d \frac{\cos{\theta_n}}{(\sin{\theta_n})^3}$$

Anyway this relation is probably false for the mistake I spoke about before, and even if it was true I didn't know how to use it to find the final velocity. I'd like to find some sort of relation with $$d(v)$$ and $$d(\theta)$$ in order to integrate that.

The problem comes from "Introduction to classical mechanics" by David Morin. It has 4 stars, which means it is quite hard. Thank you for any help, I have no idea on how to solve it .

Shooting Star
Homework Helper
Conservation of energy applies. Conservation of angular momentum doesn't (the friction provides torque).

The angular momentum IS conserved, the rotational KE is NOT. It's the same as a spinning skater when he pulls his hand inward. (But where does the KE go or come from?) The phrase "thin pole" has been mentioned to prevent any torque from acting on the ball by the pole.

The answer is one step from here.

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The angular momentum IS conserved, the rotational KE is NOT. It's the same as a spinning skater when he pulls his hand inward. (But where does the KE go or come from?) The phrase "thin pole" has been mentioned to prevent any torque from acting on the ball by the pole.

The answer is one step from here.

If angular momentum is conserved, shouldn't the velocity of the ball be infinite when the radius becomes 0? This would not agree with the thesis...

## Homework Equations

Conservation of energy applies. Conservation of angular momentum doesn't (the friction provides torque).

## The Attempt at a Solution

I couldn't really do much on this problem. The $$\Delta L$$ after the mass has traveled once around the pole is:

$$\Delta L = - \frac{2 \pi d}{\sin(\theta)}$$

I had forgot about the slow descending of the lowest contact point of the string with the pole. Without considering it, finding the velocity of the mass as a function of $$\theta$$ I could find (using the tension in the string and balancing weight):

$$v^2 = g L \frac{(\sin(\theta))^2}{\cos(\theta)}$$

......

The problem comes from "Introduction to classical mechanics" by David Morin. It has 4 stars, which means it is quite hard. Thank you for any help, I have no idea on how to solve it .

This is indeed a **** question, and you have made an excellent attempt at it.

Energy of the mass is conserved because the string does no work
(velocity perpendicular to the string), and angular momentum is not
because the tension has a moment about the axis of the pole.
(This was NOT the reason for specifying that the pole was thin)

"$$\Delta L = - \frac{2 \pi d}{\sin(\theta)}$$"
Well done, except d should mean radius of pole.

"I had forgot about the slow descending of the lowest contact point of the string with the pole".
This is crucial, because it is needed to find the change in height and thus
potential energy of the mass. (It's $$\Delta$$L cos$$\theta$$)

"$$v^2 = g L \frac{(\sin(\theta))^2}{\cos(\theta)}$$"
Well done again.

If you find the change of height of the mass, you can find the change in kinetic
energy.
Keep going:)
David

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Thank you all for the help. I can't work on the problem today cause I have a test this afternoon, I will try again in the next days. Hopefully this problem won't be in the test (the teacher sometime chooses some of the problems from that book).

Of course $$d$$ was the radius of the pole, I just forgot to say that.

I think is he may be right, actually. Please correct me if I say something false.

1) It is obvious that the kinetic energy is not conserved. Anyway, the total energy (gravitational + kinetic) of the system is conserved. This is a well-known fact about tetherball.
2) If the angular momentum is conserved, when the ball is very near the pole, it will have very high speed, because radius times speed would be a constant. This means that it can't have just $$1/ \sin{\theta_0}$$ times the speed it had at the beginning. There is no need to search for singularities (of course I didn't mean to put radius = 0 in some equation, I just meant to say that the velocity grows too much when the radius becomes very little).
3) The problem is in the chapter 5 of the book, which is the chapter about the conservation of energy. Angular momentum will be defined in chapter 8. Of course this is not a Physics argument, but I think that means that we shouldn't use angular momentum to solve the problem, while energy might be useful.

If you are not familiar with the rudiments of mechanics, then you should not post on that topic and unnecessarily confuse the learners.

Why the angular momentum is concerved and why the kinetic energy is not has been explained in my previous post. The string does do work but not in the transverse direction and the tension along the string has no moment about the pole since it passes through the pole, which was precisely the reason for approximating it as a thin pole.

I can only request you to refrain from posting in this thread.

I was perhaps too tactful in not pointing out more explicitly why you were
wrong.

To make it simple, consider a frictionless puck on ice rotating on a string
which is winding itself round a pole of radius r.

If x is the unwrapped length of the string is x, the tension T=mv^2/x

The angular momentum L = mvx

dx/dt = -vr/x

-Tr=dL/dt

so -rmv^2/x = mx(dv/dt) + mv(-vr/x)
giving dv/dt =0

dL/dx = dL/dt* dt/dx = (-Tr)(-x/vr) = mv
L=mvx
Now ain't that a cute result?
I can't see r in it!

I think you owe me an apology.

David

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I think is he may be right, actually. Please correct me if I say something false.

1) It is obvious that the kinetic energy is not conserved. Anyway, the total energy (gravitational + kinetic) of the system is conserved. This is a well-known fact about tetherball.
2) If the angular momentum is conserved, when the ball is very near the pole, it will have very high speed, because radius times speed would be a constant. This means that it can't have just $$1/ \sin{\theta_0}$$ times the speed it had at the beginning. There is no need to search for singularities (of course I didn't mean to put radius = 0 in some equation, I just meant to say that the velocity grows too much when the radius becomes very little).
3) The problem is in the chapter 5 of the book, which is the chapter about the conservation of energy. Angular momentum will be defined in chapter 8. Of course this is not a Physics argument, but I think that means that we shouldn't use angular momentum to solve the problem, while energy might be useful.

Spot on again.
I hope the simpler example of an ice puck on a string winding itself round
a pole will convince Shooting Star.
The "skater" analogy applies if the strnig was being pulled through a small hole
in the ice. Then it is doing work as the radius decreases, and the torque is zero.
Someone/thing is pulling the string through the hole, and this is where the increase
in KE comes from.
If the energy was increasing while the string was wrapping itself round a pole,
where (pray) is it coming from???

David

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Shooting Star
Homework Helper
I seem to have made quite a blunder and solved a very similar but not quite this problem! (Thanks to Dick.) My solution pertains to the case where the string is being shortened manually or otherwise by pulling at it. In the problem as given here, the non-zero radius of the pole contributes to a torque on the ball which changes the angular momentum. In this case the energy is conserved because the ball is tethered to an infinitely massive object and there's no place for the energy to go.

David, my profuse apologies. But I would have liked it better had you been less tactful and pointed out the mistake, instead of high-handedly writing an indirect post, which had disturbed me quite a lot at that time, and hence my impatience toward you.

I have deleted a couple of posts with my wrong explanation.

One good thing to come out of all this is the http://www.youtube.com/watch?v=cSVSnw3s1wo&feature=related" which David sent me a link to.

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David, my profuse apologies. But I would have liked it better had you been less tactful and pointed out the mistake, instead of high-handedly writing an indirect post, which had disturbed me quite a lot at that time, and hence my impatience toward you.

One good thing to come out of all this is the http://www.youtube.com/watch?v=cSVSnw3s1wo&feature=related" which David sent me a link to.

No Problems THX for the apology, and glad you appreciated my favourite Dylan track.

David

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David, my profuse apologies. But I would have liked it better had you been less tactful and pointed out the mistake, instead of high-handedly writing an indirect post, which had disturbed me quite a lot at that time, and hence my impatience toward you.

Read the first five lines of my first post

PS That is assuming you haven't edited it with or without help from your Dick:)