# Lagrange's Equations for a Tetherball

• hanyas
In summary, you are attempting to solve the equations of motion for a tetherball moving in 3D around a pole while the string is getting shorter. You've started with lagrange equations and found the potential energy when the ball has not wrapped around the pole to be zero. You then applied the Lagrange derivative and found a linear velocity that is somehow increasing. You say you would appreciate some feedback, so let's move on to that.
hanyas
I'm trying to write down the equations of motion for a tetherball moving in 3D around a pole while the string is getting shorter.I've started with lagrange equations:

$x(t)=l(t) \sin (\theta) \cos (\phi)\\ y(t)=l(t) \sin (\theta) \sin (\phi)\\ z(t)=h(t)+l(t) \cos(\theta)\\ \\ T = \frac{1}{2}m(\dot x^2 +\dot y^2+\dot z^2)\\ U=m g l(t)(1-\cos(\theta)) + mg(S-(h(t)+l(t)))$

where $l(t)$ is the length of the string. Here only I am assuming the radius of the pole is really small compared to $l$. The polar angle is $\theta(t)$. $h(t)$ is the change in height due to the string wrapping on the pole. $S$ is the length of string when unwrapped.

The change in length along is given by:
$\dot l(t) = -\frac{r\dot\phi}{ \sin(\theta)}$

where $r$ is the radius of the pole. And the sliding pivot point is given by:
$\dot h(t) = \frac{r\dot\phi}{ \tan(\theta)}$

After plugging those in $T$ I apply the Lagrange derivative to $L = T-U$ and solve for $\ddot \theta$ and $\ddot \phi$

Now when I simulate the results I get a linear velocity that is some how increasing which is not supposed to happen for a tetherball because no new energy has been introduced to the system and the angular momentum is not conserved.
I would appreciate some feedback

Last edited:
Hello hanyas, and welcome to PF. The template is there for good reasons (see guidelines), so don't edit it away.

Can you explain the U equation ?

Oh I'm sorry I thought the template was just there for guidance.

Well the first part of $U$ is the normal pendulum like potential energy. I might have chosen a weird zero point though. I thought I want the minimum of potential energy when the ball is hanging down. The maximum would be then when you move the ball the way all the way up to $\theta=π/2$.

I'm not sure about the second part. Here too I am saying the potential energy when the ball has not wrapped around the pole at all is equal to zero. And when you take the extreme case of the $l(t)$ being equal to zero (completely wrapped) you have to have some potential energy in order for the ball to swing in the opposite direction.

Wouldn't it be easier to take U = mgz ?

You say you assume r << l. Where do you need that ?

Also, h is not "the change in ..." but simply the position of the point where the wire separates from the pole.

yes I guess U = mgz would be easier. But isn't it just a convention issue? I've actually already tried that with similar results. I know only the difference in energy is actually what matters. it is that just at some point i wanted to have it in a form I could imagine. And maximum potential energy when the ball is up seemed good.

About h(t): I agree and this is what I wrote or what I meant to communicate. The differential equation for h(t) only describes how the "pivot" is moving down the pole.

Sorry, I meant U = - mgz in case you want to have g = 9.81 m/s[sup2[/sup]. You have z and g pointing down. Difference is indeed what matters, but the sign simply has to be correct.

So now there is a differential eqn. Apparently you have some simulator that comes with uneexpected results? Can you show or describe some in a bit more detail ? (initial conditions, numerical values, etc.)

I have a small Mathematica script that solves the equations numerically. Another thing that I'm not sure about is how to use the differential equations of l(t) and h(t). Do they come in the Lagrangian as constraints?

But you can see what I did clearly in the script: http://pastebin.com/wUMj0Zyr

If you can run the script you will see the that angle theta keeps growing. Depending on the initial value it could even go over pi/2. Which does not really make sense. Also the energy is increasing over time.

Don't have Mathematica here. Do I see line 12 active ? And what is the l(t) doing in line 13 ?

If you write your Lagrangian in terms of ##\theta## and ##\phi## there are no l and h any more !

So what does L look like when fully written out ?
And the equations of motion?
Having such a tool is nice, but you learn a lot more when you do it by hand (*). Whatever, a lot of checking is essential!

I'm not sure what to think of the coordinates. You have x, y, z and l(t) (**) as a constraint, so that leaves you with two generalized coordinates ##\theta## and ##\phi##.

h and l are found once ##\theta(t)## and ##\phi(t)## are known from solving the equations of motion.

Only genuine constraint I see would be S but it's quite a difficult one ( l + some integral of h = S)! A good second is then l, which you have already processed.

(*) This is the place where I would make use of l >> r to keep it reasonable. Later on, if it all works, you can let Mathematica do the work of seeing the effect of bigger r.

(**) Only genuine constraint I see would be S but it's quite a difficult one ( l + some integral of h = S)! A good second is then l, which you have already processed.

line 12 is active. That's where U is being calculated. I've just checked and it does not matter if I use line 12 or 13 for U. The l(t) in line 13 is an error.

In lines 14,15 I am practically plugging the differential equations of l(t) and h(t) into T because it contains the derivatives. Is that the right way to do it or do I have to introduce the equations of l(t) and h(t) somehow as extra constraints with lagrangian multipliers maybe?

If you write your Lagrangian in terms of θ and ϕ there are no l and h any more !
I'm not sure what you mean here. I can only substitute the derivatives of l(t) and h(t). But l(t) and h(t) still come up in the Lagrangian. Here is T and L while taking into consideration that l>>r.

$T = 0.5 m \left(l(t)^2 \left(\phi '(t)^2 (0.5\, -0.5 \cos (2 \theta (t)))+ \theta '(t)^2\right)+ \\ h'(t) \left(2l'(t) \cos (\theta (t))-2 l(t) \theta '(t) \sin (\theta (t))\right)+ h'(t)^2+ l'(t)^2\right) \\ L = m \left(l(t)^2 \left(\phi '(t)^2 (0.25\, -0.25 \cos (2 \theta (t)))+0.5 \theta '(t)^2\right)+g h(t)+ \\ l(t) \cos (\theta (t)) \left(g- r \theta '(t) \phi '(t)\right)+0.5 r^2 \phi '(t)^2\right)$

I am uploading plots from Mathematica for a result of simulation. The plots are for [theta, phi, l , T, U, L] respectively. You can see how theta keeps growing and how the energy is also increasing.

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You're right. So you end up with four equations of motion and two Lagrange multipliers. Awkward...

Nice plots. I'll have to sleep over it and might chew on it some more tomorrow.
Perhaps others are listening in and have something constructive to say ? (I hope so..")

1 person
I made some progress if you are interested. Instead of plugging the differential equations of l(t) and h(t) I built the extended Lagrangian with the constraints and the multipliers. Now the energy is conserved and the angle theta does not increase randomly.

Here is link to the new mathematica code: http://pastebin.com/MZ2zEpZ7

There still seems to be a a problem that l(t) does not converge to 0 (not wrapped completely). Which is something new.

## 1. What are Lagrange's Equations for a Tetherball?

Lagrange's Equations for a Tetherball are a set of equations developed by mathematician Joseph-Louis Lagrange to describe the motion of a tetherball. They take into account the forces acting on the ball, such as gravity and tension in the rope, to determine its position and velocity at any given time.

## 2. How do Lagrange's Equations apply to a tetherball?

Lagrange's Equations apply to a tetherball because they are based on the principle of least action, which states that the path a system takes between two points is the one that minimizes the action or energy required. In the case of a tetherball, this means that the ball will follow the path that minimizes the energy it expends to move along the rope.

## 3. Can Lagrange's Equations be used for all types of tetherball games?

Yes, Lagrange's Equations can be used for all types of tetherball games, as they are general equations that can be applied to any system with constraints, such as a ball attached to a rope. The specific values for the variables in the equations may vary depending on the specific game or setup, but the equations themselves can be used for any tetherball game.

## 4. What are the advantages of using Lagrange's Equations for a tetherball?

One of the main advantages of using Lagrange's Equations for a tetherball is that they take into account all of the forces acting on the ball, allowing for a more accurate prediction of its motion. They also provide a systematic and elegant approach to solving the equations of motion, making it easier to analyze the system and make predictions about its behavior.

## 5. Are there any limitations to using Lagrange's Equations for a tetherball?

One limitation of using Lagrange's Equations for a tetherball is that they assume the rope to be inextensible, meaning it does not stretch or bend. In reality, most ropes used for tetherball games will have some degree of stretch, which can affect the accuracy of the equations. Additionally, the equations may become more complex when applied to more complicated tetherball setups, such as multiple balls or different types of movement.

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