- #26

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

The equation, in scalars, is ##\tau=\Omega_p L##. You know ##\tau## and want ##\Omega_p##.

What is the angular momentum, L?

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- #26

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

The equation, in scalars, is ##\tau=\Omega_p L##. You know ##\tau## and want ##\Omega_p##.

What is the angular momentum, L?

- #27

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The angular momentum of the disc? ##L=J\omega=\frac{1}{2}mR^2\omega##.No.

The equation, in scalars, is ##\tau=\Omega_p L##. You know ##\tau## and want ##\Omega_p##.

What is the angular momentum, L?

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 ...

- #28

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Could you please comment my results? Your hints are very helpfull for me. It seems you have to be very good teacher.No.

The equation, in scalars, is ##\tau=\Omega_p L##. You know ##\tau## and want ##\Omega_p##.

What is the angular momentum, L?

- #29

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Getting there.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 ...

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).

- #30

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Ohh I add sin(θ) on LHS I don't know why.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).

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.

- #31

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The first equality is true (in terms of magnitudes, but not as a vector equation), but not the second.But there is probably mistake because torque ##\tau=\frac{dL}{dt}=I \frac{\Omega_p}{dt}## no?

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

- #32

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Ok, I have to summary it again: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?

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?

- #33

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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.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

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

Think carefully about the direction of the precession. Remember that its axis is orthogonal to both the torque and the vertical spin axis.

- #34

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OKNot 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.

##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?

- #35

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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.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?

- #36

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So do you mean that ##\Omega_p=\frac{d\theta}{dt}##?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.

- #37

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Yes, except you have to get the sign right.So do you mean that ##\Omega_p=\frac{d\theta}{dt}##?

- #38

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Now I'm confused, why ##\Omega_p=-\frac{d\theta}{dt}##? Is it because this precession acts against wobble which is in opposite direction?Yes, except you have to get the sign right.

- #39

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To deduce it, you need to use the right hand rule and the correct vector product equations. But I am content to take a shortcut: we know from experience that the wobble tends to lessen, not grow.Now I'm confused, why ##\Omega_p=-\frac{d\theta}{dt}##? Is it because this precession acts against wobble which is in opposite direction?

Ok, that's a bit lazy, so here's an attempt.

Let the frisbee be spinning anticlockwise when viewed from above. The spin vector ω, and hence the angular momentum vector, therefore point upwards. Viewing the frisbee from the side, let it be tilted clockwise by angle θ. The frictional torque vector points down to the left at θ to the vertical. The component leading to the precession points horizontally to the left.

The torque component is the cross product, precession vector x angular momentum. Using the right hand rule, to get the torque component pointing left the precession vector must be pointing towards you. That means the precession is anticlockwise, thus acts to reduce θ.

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- #40

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I'm sorry I have still problem with describing of movement. If I will understand 100% I will be able to answer anything I hope...but I'm still not sure. So here is definition of Euler's angles. https://en.wikipedia.org/wiki/Euler_angles#/media/File:Eulerangles.svg What is happening in my case? What is constant and what angles aren't constant in 2 cases:To deduce it, you need to use the right hand rule and the correct vector product equations. But I am content to take a shortcut: we know from experience that the wobble tends to lessen, not grow.

Ok, that's a bit lazy, so here's an attempt.

Let the frisbee be spinning anticlockwise when viewed from above. The spin vector ω, and hence the angular momentum vector, therefore point upwards. Viewing the frisbee from the side, let it be tilted clockwise by angle θ. The frictional torque vector points down to the left at θ to the vertical. The component leading to the precession points horizontally to the left.

The torque component is the cross product, precession vector x angular momentum. Using the right hand rule, to get the torque component pointing left the precession vector must be pointing towards you. That means the precession is anticlockwise, thus acts to reduce θ.

1) Without friction

2) With friction

My attempt:

1) Im sure that ##\gamma## is not constant and ##\dot{\gamma}=\omega## is spin of the disk (in our case). ##\beta## is our ##\theta##...but is it constant (without friction) and ##\alpha## is not constant?

2) Same as in 1) but torque of friction causes that ##\beta## is not constant and it acts in opposite direction of ##\beta##?

- #41

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In respect of the picture at that link, taking the red circle to represent the frisbee:I'm sorry I have still problem with describing of movement. If I will understand 100% I will be able to answer anything I hope...but I'm still not sure. So here is definition of Euler's angles. https://en.wikipedia.org/wiki/Euler_angles#/media/File:Eulerangles.svg What is happening in my case? What is constant and what angles aren't constant in 2 cases:

1) Without friction

2) With friction

My attempt:

1) Im sure that ##\gamma## is not constant and ##\dot{\gamma}=\omega## is spin of the disk (in our case). ##\beta## is our ##\theta##...but is it constant (without friction) and ##\alpha## is not constant?

2) Same as in 1) but torque of friction causes that ##\beta## is not constant and it acts in opposite direction of ##\beta##?

Yes, β there is our θ.

ω is the rotation rate about the vertically up (blue Z) axis.

The frictional torque vector points down along the negative red Z axis.

The component of frictional torque that causes precession is orthogonal to the spin axis, so points along the positive blue Y axis.

The precession itself is orthogonal to both of those, so points along the green axis. The signs say it is pointing away from you in the picture, so will tend to reduce β.

- #42

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Ok...here is the snag. I thought that the spin about rotation axis is different movement than wobbling. That there is (with respect to the picture) non zero ##\beta## (but constant). The spin rate about red Z axis ##\dot{\gamma}##. And the wobbling, rotation about blue Z axis. Is possible throw the disks (frisbee) in such way? Or is it nonsense?In respect of the picture at that link, taking the red circle to represent the frisbee:

Yes, β there is our θ.

ω is the rotation rate about the vertically up (blue Z) axis.

The frictional torque vector points down along the negative red Z axis.

The component of frictional torque that causes precession is orthogonal to the spin axis, so points along the positive blue Y axis.

The precession itself is orthogonal to both of those, so points along the green axis. The signs say it is pointing away from you in the picture, so will tend to reduce β.

- #43

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Sorry I thought I explained in post #14. The wobble we see in the video is simply the fact that the frisbee is not horizontal, but is rotating about a vertical axis. The blue dot just goes around the vertical axis, not up and down.Ok...here is the snag. I thought that the spin about rotation axis is different movement than wobbling. That there is (with respect to the picture) non zero ##\beta## (but constant). The spin rate about red Z axis ##\dot{\gamma}##. And the wobbling, rotation about blue Z axis. Is possible throw the disks (frisbee) in such way? Or is it nonsense?

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Yes you are right. We do. I'm sorry I only ask if it is possible in real world to throw it in this way. And if it is possible to solve.Sorry I thought I explained in post #14. The wobble we see in the video is simply the fact that the frisbee is not horizontal, but is rotating about a vertical axis. The blue dot just goes around the vertical axis, not up and down.

- #45

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If you mean such that it wobbles at a different rate from its spin, no. There is no force that would alternate like that.I only ask if it is possible in real world to throw it in this way

So, using the equations in posts 34 to 38, can you solve it now?

- #46

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Definitely. Generally it is something like ##c_1e^{-c_2t}## ... It means, it decreases exponentially.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.

Big big thanks for your patience. I know it was hard with me :) But on the other hand, you led me very clearly.

Thank you very much.

- #47

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You are welcome. It was an interesting problem.Definitely. Generally it is something like ##c_1e^{-c_2t}## ... It means, it decreases exponentially.

Big big thanks for your patience. I know it was hard with me :) But on the other hand, you led me very clearly.

Thank you very much.

- #48

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Really thank you. I will try think about a problem if the rotations are a bit different as I thought first. Can I write you in here my attempt and discuss again?You are welcome. It was an interesting problem.

- #49

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Of course, but start a new thread for each problem.Really thank you. I will try think about a problem if the rotations are a bit different as I thought first. Can I write you in here my attempt and discuss again?

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