# Rotational motion: air puck revolving Is it possible?

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1. May 3, 2015

### **Mariam**

question here

Hello, this isn't really a homework question as I understand how to solve it. But just out of curiosity, is it possible for this to actually be set up in real life and for the 1 kg mass to be in equilibrium?

because when I imagine such a situation I feel that the revolving puck will slowly decrease in radius as the 1 kg mass pulls it down.
what am I missing in the picture??

2. May 3, 2015

### Orodruin

Staff Emeritus
The wording "frictionless" and "ideal" should already give it away. You may be able to obtain a good approximation with very little dissipation of energy out of the system, in which case the situation will be quite well described. However, if there is dissipation of energy out of the system due to friction etc, then it will not go on forever.

Edit: That is, it is about as possible as getting an air puck to travel indefinitely with constant momentum.

3. May 3, 2015

### Merlin3189

Now that someone has made the first answer, I can come in.
First,
as Orodruin says, no! This is an "ïdeal" situation, and like all "frictionless", "light", "inextensible", decriptions, is intended to simplify the situation, either to bring out a principle or to make a calculation easier.

But
I wondered why you feel this? If it were just the friction, you'd hardly be puzzled: the problem specifies a frictionless table, you know that is practically impossible, so you would not be puzzled. So do you think that even with a frictionless table, the 1kg should pull the puck inwards?

The 1kg mass is pulling the puck towards the centre. This force is causing the puck to accelerate towards the centre. The puck is accelerating towards the centre and that is why it moves in a circle - of constant radius.
The puck equally is pulling on the string and the string on the 1kg mass. So the mass has a gravitational force down and the string force up, giving no net force and no acceleration - as the question says, it is in equilibrium.

Now, if you just set this up and gave the puck a good push to set it in orbit, you might not push it hard enough, then indeed, the mass would fall and the puck move in towards the centre. Or you might push too hard and the puck moves outwards pulling the mass up. But the question says you did it just right to get equilibrium, so you can calculate how fast that is.

4. May 3, 2015

### Orodruin

Staff Emeritus
Just to add to this: In the ideal situation, the problem also exhibits rotational symmetry, resulting in conservation of angular momentum. Thus, if the system has non zero angular momentum to start with, the puck will never be able to fall through the hole, even if the trajectory is not a perfect circle.

5. May 3, 2015

### CWatters

In the real world...

You could set the puck rotating fast enough to actually raise the 1kg mass (mv2/r > 1kg * g). As radius increases the tension on the string reduces. That continues until they are in equilibrium (might happen rapidly). Then as friction slows the puck the tension would reduce and the 1kg mass would descend.

6. May 3, 2015

### Orodruin

Staff Emeritus
What do you mean by "equilibrium". In the idealised situation, the puck will not tend to a fixed radius, but oscillate around the radius appropriate for the angular momentum in the system. In the "real world" with friction, the same thing will happen apart from the fact that the system will gradually lose angular momentum as well as energy.

7. May 3, 2015

8. May 3, 2015

### Orodruin

Staff Emeritus
It is basically nothing else than a motion in a central potential with potential $m_2 g r$. The full effective potential is $L^2/(2m_1 r^2) + m_2 gr$, where $L$ is the angular momentum and the mass for the kinetic energy term in the $r$ direction is $m_1 + m_2$. Of course assuming I did not go awry with the algebra done in my head ...

9. May 3, 2015

### A.T.

This applet shows orbits for different types of central force. The one relevant here is the constant one (green). So that's what you get without friction, if you don't set it up for a perfectly circular orbit:

http://www.fastswf.com/kRR_YZ0 (hit reload on that page to get a different random initialization)

10. May 3, 2015

### Orodruin

Staff Emeritus
As luck would have it, my first random initialization made the green orbit almost perfectly circular - imagine my confusion before I read the last part
I also suspect I know the form of the force for the red orbit (although numerical implementation is not perfect, I still see orbital precession for very elongated orbits)

Edit: Oh, look at that, it actually says in the top left corner - I was right

11. May 3, 2015

### A.T.

Yes, that's an artifact. The red and blue orbits should always be closed, consistent with Bertrand's theorem:

http://en.wikipedia.org/wiki/Bertrand's_theorem

12. May 3, 2015

### Orodruin

Staff Emeritus
As a matter of curiosity, in this year's class I did the red orbits in lectures and let the students attack the blue orbits on the exam ...

13. May 3, 2015

### A.T.

The F ~ r orbits are also on the one pound note, at least suggestively because of the way the ellipse diagram overlays with the sun in the background: