# Homework Help: Flying disc moves up and down

1. Dec 1, 2016

### Vrbic

1. The problem statement, all variables and given/known data
The flying disc except rotation moves up and down. Show that such movement will be suppressed by frictional force of air, which has a force momentum relative to the center of disc $N=-K\omega$. What is the time profile of this process?

2. Relevant equations
$L= T - V$
$T=Tradial + Trot + Tupdown$, $V=0$

3. The attempt at a solution
I haven't seen such type of problem before. So I don't know if I should construct lagrangian and generally don't have idea how connect frictional force with this movement. Probably small kick should help and probably than I will able to solve it.
Please give me some kick :) Thank you.

2. Dec 1, 2016

### Vrbic

Hmm now I'm thinking that this movement seems to be similar to harmonic oscillator...so in that case there should be potential energy, but generally I don't know from what it should arise...in oscilator (if I take a moving point on a spring) this potential arise from the spring...in this disc I can't see any such mechanism...

3. Dec 1, 2016

### Nidum

We can only vaguely guess what this problem is all about . If it came from a book or course work then please show the complete text and diagrams .

4. Dec 1, 2016

### Vrbic

Understand, but unfortunately I have got only this text on paper. Nothing more. If we need something more maybe we can expect it and mention.

5. Dec 1, 2016

### Nidum

Perhaps we can synthesise a version of this problem that can be solved .

What do you think the physical system is ? Can you draw a picture ?

6. Dec 1, 2016

### Vrbic

I'm sorry I'm not good painter :) Now I don't know if you don't know what is happening there or you advise me this way.

File size:
31.7 KB
Views:
92
7. Dec 1, 2016

### Nidum

We'll have to leave this for now I think . Come back if you can find any more information about the problem .

8. Dec 1, 2016

### haruspex

I'm guessing this is a translation, and not a good one.
I assume it is essentially moving horizontally.
Does "except rotation" mean that it is not rotating about a vertical axis, or that, apart from such a rotation, its only movement is [up and down]?
What does "moving up and down" mean here? Translationally, or rotationally about a horizontal axis? What would sustain such a movement?
What is ω?
What is "force momentum", is that a force or a moment?

9. Dec 2, 2016

### Vrbic

It is frisbee. You throw it and it flies lets say in x-y plane (better only in x direction but it is not important in our case). When it flies it rotates around z axis with angular velocity $\omega$. But it also moves (vibrates - I dont know how call it), swing up and down in z direction. If you ever seen a frisbee you probably know what I mean.

"Force momentum" is typ sorry, there should be moment. Not momentum.

Is it a bit more understandable?

10. Dec 2, 2016

### haruspex

I think you mean its axis of rotation is not quite vertical, and precesses about the vertical. If so, there are two angular rates, the rate of spin about the instantaneous axis and the rate of precession of the axis.

Last edited: Dec 2, 2016
11. Dec 2, 2016

### Vrbic

Precession is the word which I was looking for :) Yes, it is as you said.

12. Dec 2, 2016

### haruspex

Ok, but I'm still not understanding something. Precession arises when there is a torque orthogonal to the angular momentum, such as in a leaning toy gyroscope. What is causing the precession here? I suppose the disc's mass centre might be a bit off to one side.
I used to play a lot of Frisbee, and do not recall much precession. There would be a lot of wobble initially, but I would say that was because the initial spin axis was not fully aligned with the axis of symmetry. Maybe that is what you mean, but I don't think it is precession.

13. Dec 2, 2016

### Vrbic

Ok. Wobble is better and as you said I believe it is caused by not aligned spin axis and axis of symmetry (for me because of wrong throw). And disc is homogeneous - mass center is in the point where rotation axis intersects the disc. I include a short video: . There you can see translation motion (forward and down), rotation motion and wobble with respect to x-y plane. And the last is it what I'm talking about.

14. Dec 2, 2016

### haruspex

Ok, I think I finally understand.
The wobble visible in the video is not precession. There's a blue dot on the side of the frisbee which rotates at the same rate as the wobble, so the plane of the frisbee is not normal to its axis of rotation.
The frictional forces act in the plane of the frisbee, so the frictional torque, as a vector, is normal to its plane.
So we have a torque set at an angle to the spin axis. This will result in precession, but quite a slow one I suggest. I think what you have to show is that this precession will have the effect of reducing the wobble.
See if you can write a vector equation encapsulating that.

15. Dec 5, 2016

### Vrbic

I suppose that the best way how to write such equations is start with Lagrangian and find equations without friction. Do you agree?

16. Dec 5, 2016

### haruspex

Different methods suit different people. I am more comfortable with forces, accelerations and torques.

17. Dec 5, 2016

### Vrbic

Ok, understand. I begin with list of forces which should be there:
1) The force which causes the translation movement. - I mean it is not important for us.
2) The centrifugal force due to rotation of disc.
3) The force which causes a wobble. - It should have a form of oscilator, I guess.
4) The friction force. - If it acts against a wobbling, it should have a form of a damping.
Is it all and do you agree with my comments to them?

18. Dec 5, 2016

### haruspex

There is no force causing the translational movement. It was thrown at some point. There will be a deceleration, but as you say, it is not relevant.
Centrifugal force is also irrelevant. There is nothing in danger of moving radially.
There is no force causing a wobble. The axis of rotation is not normal to the plane of the disc; that is the wobble, and it was there at the start.
The frictional force does not directly act against the wobble. Primarily it acts against the rotation of the disc, slowing it, but the key is that it takes the form of a torque which is normal to the plane of the disc.

If you have a torque which is not parallel to the angular momentum, what happens?
Hint: https://en.wikipedia.org/wiki/Rigid_body_dynamics#Rotation_in_three_dimensions

Suppose that at some point the disc is tilted at angle θ to the horizontal, and spinning at rate ω. You can take θ to be small.
What is the magnitude of the frictional torque? What is the consequence for the tilt angle?

Last edited: Dec 5, 2016
19. Dec 6, 2016

### Vrbic

Ok, I have read it and I thought about it...a long time but some things aren't much clear for me. They speak about precession and nutation. You said (and I agree) that it is not precession. Now I'm thinking if it is nutation and if a nutation can exists without precession?

20. Dec 6, 2016

### haruspex

The visible wobble is not precession, but there is precession, and this is what leads to the wobble reducing, but it will not be visibly obvious.

One thing is a bit confusing in this area, that the equation takes the form $\vec \tau=\vec \Omega_p\times \vec L$, yet the inputs to it are the angular momentum $\vec L$ and the applied torque $\tau$. The reason is that a general applied torque has two consequences. To the extent that it aligns with the angular momentum it merely accelerates that. The precession only comes from the orthogonal component. So in that equation, $\tau$ only stands for the component of the applied torque which is normal to $\vec L$.
In the present problem, we are given an expression for the magnitude of the frictional torque. If the disc is at angle $\theta$ to the horizontal, what is its component normal to the angular momentum?

21. Dec 7, 2016

### Vrbic

Ok, so now again me, for sorting things:
It wobbles, because of rotating axis and axis of disc are not aligned. If it is not aligned and there is some nonzero angle, in this case there arise a area for a friction which causes a precession and this force (torque) acts against wobbling. Is now my idea of situation right?

22. Dec 7, 2016

### haruspex

Nearly right. There will be a frictional torque whatever the angle to the horizontal. If the disc were horizontal the frictional torque would be aligned with the spin, so its only consequence would be to slow the spin down gradually. But because it is not fully aligned with the spin, the component of the frictional torque normal to the spin axis leads to precession. The nature of that precession is that it tends to rotate the axis of spin towards the vertical, reducing the wobble.
So, let's get that into equations.
At spin rate ω, we are told the frictional torque has magnitude Kω. If the disc is at angle θ to the horizontal, what is the component of the frictional torque normal to the spin axis?

23. Dec 7, 2016

### Vrbic

I believe that if there is given $N=-K\omega$ and it is parallel to the horizontal plane, normal component of $N$ to $L$ should be $N_n=N\sin{\theta}$. Then as you said, there is no normal torque if both axis are aligned. And the torque acts only in the horizontal plane it means it causes slowing of rotation.
Is it right?

24. Dec 7, 2016

### haruspex

Yes.
So what will the precession rate be?

25. Dec 7, 2016

### Vrbic

So it should be same $\omega$ as rotation. No?