Can Friction Suppress the Up and Down Movement of a Flying Disc?

[Vrbic]

Homework Statement

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?

Homework Equations

$L= T - V$
$T=Tradial + Trot + Tupdown$, $V=0$

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.

 

[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...

 

[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 .

 

[Vrbic]

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

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.

 

[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 ?

 

[Vrbic]

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 ?

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.

 

[Nidum]

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

 

[haruspex]

The flying disc except rotation moves up and down.

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?

 

[Vrbic]

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?

It is frisbee. You throw it and it flies let's 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 don't 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?

 

[haruspex]

It is frisbee. You throw it and it flies let's 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 don't 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?

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.

 

[Vrbic]

I think you mean its axis is 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.

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

 

[haruspex]

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

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.

 

[Vrbic]

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.

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.

 

[haruspex]

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.

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.

 

[Vrbic]

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.

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

 

[haruspex]

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

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

 

[Vrbic]

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

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?

 

[haruspex]

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?

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?

 

[Vrbic]

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

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?

 

[haruspex]

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?

No, reread my post #14.
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?

 

[Vrbic]

No, reread my post #14.
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.

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?

 

[haruspex]

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?

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?

 

[Vrbic]

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?

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?

 

[haruspex]

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 $N_n=N\sin{\theta}$.

Yes.
So what will the precession rate be?

 

[Vrbic]

Yes.
So what will the precession rate be?

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

 

[haruspex]

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

No.
The equation, in scalars, is $\tau=\Omega_p L$. You know $\tau$ and want $\Omega_p$.
What is the angular momentum, L?

 

[Vrbic]

No.
The equation, in scalars, is $\tau=\Omega_p L$. You know $\tau$ and want $\Omega_p$.
What is the angular momentum, L?

The angular momentum of the disc? $L=J\omega=\frac{1}{2}mR^2\omega$.
So $\tau=N_n$
$\Omega_p \frac{1}{2}mR^2 \sin{\theta}=-K\omega \sin{\theta}$
$\Omega_p=-\frac{2K}{mR^2}\omega$
But I'm bit confused now ...

 

[Vrbic]

No.
The equation, in scalars, is $\tau=\Omega_p L$. You know $\tau$ and want $\Omega_p$.
What is the angular momentum, L?

Could you please comment my results? Your hints are very helpfull for me. It seems you have to be very good teacher.

 

[haruspex]

The angular momentum of the disc? $L=J\omega=\frac{1}{2}mR^2\omega$.
So $\tau=N_n$
$\Omega_p \frac{1}{2}mR^2 \sin{\theta}=-K\omega \sin{\theta}$
$\Omega_p=-\frac{2K}{mR^2}\omega$
But I'm bit confused now ...

Getting there.
The precession is about a horizontal axis, so is at right angles to the spin axis. Lose the sin(θ) in the cross product.
Next, use a small angle approximation for the remaining sin(θ).

 

[Vrbic]

Getting there.
The precession is about a horizontal axis, so is at right angles to the spin axis. Lose the sin(θ) in the cross product.
Next, use a small angle approximation for the remaining sin(θ).

Ohh I add sin(θ) on LHS I don't know why.
So than I suppose small angle so sin(θ) -> θ.
But there is probably mistake because torque $\tau=\frac{dL}{dt}=I \frac{\Omega_p}{dt}$ no?
Than I expect both omegas are same and I have differential eq.:
$\phi''+\frac{2K\phi}{mR^2}\phi'=0$...it is not nice :) I would expect some other typ...for dumping oscilations.

 

[haruspex]

But there is probably mistake because torque $\tau=\frac{dL}{dt}=I \frac{\Omega_p}{dt}$ no?

The first equality is true (in terms of magnitudes, but not as a vector equation), but not the second.
Thefrictional torque is largely aligned with the angular momentum, so most of it is just slowing ω. Only the component of torque orthogonal to L leads to precession. Go back to the equation you had in post #27, but droppong the sn θ on the left and substituting θ for sin θ on the right.
What is the differential relationship between Ω_p and θ? I.e., what is the consequence of Ω_p for θ?

 

[Vrbic]

The first equality is true (in terms of magnitudes, but not as a vector equation), but not the second.
Thefrictional torque is largely aligned with the angular momentum, so most of it is just slowing ω. Only the component of torque orthogonal to L leads to precession. Go back to the equation you had in post #27, but droppong the sn θ on the left and substituting θ for sin θ on the right.
What is the differential relationship between Ω_p and θ? I.e., what is the consequence of Ω_p for θ?

Ok, I have to summary it again:
There are two movements done on start:
1) rotating about vertical axis z
2) for my better intuition if $\theta=\pi/2$ and no rotation about vertical axis z, in special case it would roll on the ground in circle. Is it same motion?
Or similar when you throw a coin, before definitely fall down it wobble without rotation ...is it same motion?

 

[haruspex]

2) for my better intuition if θ=π/2 and no rotation about vertical axis z, in special case it would roll on the ground in circle. Is it same motion

Not sure what you mean. If you roll a coin along the ground, as it starts to fall over, gravity creates a torque orthogonal to the rotation This causes the coin to precess in such a way that it turns towards the direction of fall. The tendency to fall is thereby inhibited, causing the coin to continue in a wide arc instead of simply falling flat.
The Frisbee problem is different in that there is no centripetal acceleration.

So, can you answer my question in post #31: What effect does the precession, Ω_p, have on the tilt angle θ?
Think carefully about the direction of the precession. Remember that its axis is orthogonal to both the torque and the vertical spin axis.

 

[Vrbic]

Not sure what you mean. If you roll a coin along the ground, as it starts to fall over, gravity creates a torque orthogonal to the rotation This causes the coin to precess in such a way that it turns towards the direction of fall. The tendency to fall is thereby inhibited, causing the coin to continue in a wide arc instead of simply falling flat.
The Frisbee problem is different in that there is no centripetal acceleration.

So, can you answer my question in post #31: What effect does the precession, Ω_p, have on the tilt angle θ?
Think carefully about the direction of the precession. Remember that its axis is orthogonal to both the torque and the vertical spin axis.

OK
$tau=N_n$
$mR^2\Omega_p/2=-K\omega \sin{\theta}$
$\Omega_p=-\frac{2K\omega\sin{\theta}}{mR^2}$ , $\sin{\theta}->\theta$ for small $\theta$.
$\Omega_p=-\frac{2K\omega}{mR^2}\theta$
ok?

 

[haruspex]

OK
$tau=N_n$
$mR^2\Omega_p/2=-K\omega \sin{\theta}$
$\Omega_p=-\frac{2K\omega\sin{\theta}}{mR^2}$ , $\sin{\theta}->\theta$ for small $\theta$.
$\Omega_p=-\frac{2K\omega}{mR^2}\theta$
ok?

Yes, that's ok up to there, but it's not what I am asking now. That equation tells you the precession rate that arises for a given θ. I am now asking how that precession then alters θ. Combining these facts will give you a differential equation to solve.