How Does Angular Momentum Affect Gyroscope Movements and Moon Orbits?

[bobie]

the centripetal force exerted inwards...restricting its movement away from the center of the circle. This force does zero work because it is perpendicular to the velocity of the ball.

The force does zero work not because it is perpendicular but because the ball cannot move, the same as when you are holding up a weight. But, if you lower your arm and let the weight drop a little, G does some work.
In our case the wheel moves, gets velocity, no matter in what direction:
if we push the spinning wheel on the normal plane zx or zy it will start to spin also in that direction, if you think that is not possible, there are videos on the web.

 

[A.T.]

The force does zero work not beacase it is perpendicular but because the ball cannot move,

What are you talking about? The ball IS MOVING. That's why it is called uniform circular MOTION.

 

[Doc Al]

The force does zero work not because it is perpendicular but because the ball cannot move, the same as where you are holding up a weight.

The ball's in uniform circular motion. It's moving.

 

[bobie]

In the case of a gyroscope you are not starting with a non-spinning football. You are starting with a football spinning on one axis and ending with a football spinning on a different axis. It has the same KE before and after. So no work is done.

Thanks for your explanations, jbriggs, I hope you forgive my inaccurate wording, of course Ke is a scalar, I used it instead of velocity.
Before I comment your statements, please let me know if you understood the problem I am posing and if you, too, are excluding that a gyroscope can spin at the same time on two planes/ axes.

For an already-spinning 10-ton wheel, I suspect that you will have problems changing the spin axis

The wheel is spinning (on xy), and we shoot a bullet to a point on its circunference (on y or x), the wheel starts spinning also on the normal plane (zy or zx) is this a real case?

 

[bobie]

What are you talking about? The ball IS MOVING. That's why it is called uniform circular MOTION.

The ball's in uniform circular motion. It's moving.

The ball cannot move perpendicularly, it is of course moving tangentially, you can hold up a weight and run.
The ball is not allowed to fly off at a tangent by the spring, the same as you are not allowing the weight to fall to the ground. Is there any difference?

 

[A.T.]

The ball cannot move perpendicularly,

Of course it could move radially towards the center. The string only prevents it from moving away radially.

But that is completely irrelevant to work done. Work done on the ball doesn't depend on what the ball can do. It depends on what the ball is actually doing: If it moves perpendicular to the force, the force does no work. That follows directly from the dot product in the definition of work:

http://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

W = \int_C \mathbf{F} \cdot d\mathbf{s} = Fs\cos\theta.

In the notable case of a force applied to a body always at an angle of 90 degrees from the velocity vector (as when a body moves in a circle under a central force), no work is done at all

 

[bobie]

Work done on the ball doesn't depend on what the ball can do.

. You are just changing the rotation axis, while preserving KE.

I'd rather concentrate on the issue at hand, that will help me understand theoretical issues. After I grasp this I might comment on your explanations, if you are willing .

I am not changing the rotation axis I am adding a rotation, in this video you can see what I mean
https://www.youtube.com/watch?feature=player_detailpage&v=sy5NY-Dqdys#t=300
at 4:35 mark, the wheel keeps spinnning on its plane, while the man rotates it on a normal plane, here it is only by 90° but you can set it spinning round and round. That does not change the rotation axis, right?

I hope now it's at last clear what I am talking about.
Thanks a lot

 

[Nugatory]

The wheel is spinning (on xy), and we shoot a bullet to a point on its circunference (on y or x), the wheel starts spinning also on the normal plane (zy or zx) is this a real case?

(If you haven't yet learned how to use vectors to represent torque and angular momentum, it's time)

The gyroscope is rotating counterclockwise, with axis of rotation along the z-axis. Thus, its angular momentum vector points in the direction of positive z.

You grab the frame of the gyroscope at the top and the bottom, and you try pulling the top towards you (negative y direction) and pushing the bottom away (positive y direction)- you're trying to rotate the gyroscope in the yz plane to get the axis of rotation to line up with the y-axis with the angular momentum vector pointing in the negative y direction. You're applying a torque, and the torque vector points in the direction of positive x.

The gyroscope frame and the axis of the gyroscope does not just rotate in the yz plane as you'd expect. Instead it precesses, rotating in the xz plane as well. The greater the spin angular momentum of the gyroscope, the more the gyroscope precesses in the xz plane and the less it moves in the yz plane. (You can see this by considering the effect of the torque vector in the positive x direction on the angular momentum vector pointing in the z direction - the resultant angular momentum vector has to lie in the xz plane).

How much work are you doing when you apply the torque and the gyroscope precesses? We have $W=\tau\theta$ where $tau$ is the torque and $theta$ is the angle over which it is applied. But if the axis of rotation stays in the xy plane, $theta$ is zero (you haven't succeeded in tipping the axis of rotation towards you) and the work done is zero.

 

[bobie]

The gyroscope frame and the axis of the gyroscope does not just rotate in the yz plane as you'd expect. Instead it precesses,

Have you watched the video at 4:35?
Anyway, isn't what you call precession just a shifting of the direction of v by 90°?
Have you ever tried to push a beachball downward on the sea? it just flies off horizontally on the surface, shifting the push and the motion by 90°. Can we say it precesses?
It is a phenomenon that is not limited to gyroscopes.

 

[A.T.]

I'd rather concentrate on the issue at hand, that will help me understand theoretical issues.

You should first try to understand linear dynamics (work done by forces), before you move on to gyroscopes (work done by torques).

 

[bobie]

You should first try to understand linear dynamics (work done by forces), before you move on to gyroscopes (work done by torques).

I suppose I should start a new thread on that, but I would appreciate id someone told me what happens if we star the wheel spinning on a normal plane (like the partial one in the video at 4:35 )
Thanks for your help

 

[bobie]

The gyroscope frame and the axis of the gyroscope does not just rotate in the yz plane as you'd expect. Instead it precesses,..

I managed at last to retrieve a video (section 18 here : http://www.gyroscopes.org/1974lecture.asp)where you can see what I mean:
the wheel is spinning on the vertical plane and then it is given a flip on the horizontal plane, it starts and keeps spinning on the horizontal plane. If the wheel is not solid and is allowed to fluctuate , it just leans a few degrees on the vertical, if it were solid it would stay put.
Now the wheel has motion, v (and Ke) on two axes. We added rotational velocity on a new different axis so we must have done work, by a jerk.
Suppose now the wheel is the one shown in section 17, we have done a huge amount of work, not arbitrarily small , by a jerk, haven't we?.

Thanks for your help.

 

[Nugatory]

Have you ever tried to push a beachball downward on the sea? it just flies off horizontally on the surface, shifting the push and the motion by 90°. Can we say it precesses?

No, that phenomenon is not precession; it has completely different cause unrelated to torque and angular momentum.

 

[bobie]

No, that phenomenon is not precession;

That is surely not precession,
"Precession is a change in the orientation of the rotational axis of a rotating body. ... In other words, the axis of rotation of a precessing body itself rotates around another axis."
I was comparing it to what happens in the quoted video (https://www.youtube.com/watch?feature=player_detailpage&v=sy5NY-Dqdys#t=300 at the beginning (from 2:50 on), when the axis of rotation does not rotate around another axis. When you (or G) try to push down the gyroscope and change the axis/plane of rotation , it does not change orientation downward nor rotates around another axis like a spinning top (this is precession: http://en.wikipedia.org/wiki/File:Gyroscope_precession.gif), but it gets a tangential v and moves straight ahead. (In Prof. Lewin video the axis is longer and you can see clearly the tangential motion). Can you call this precession?
This seems also to contradict what many maintained so far, that you do no work when you try to change the plane of rotation. If G does no work, where does the KE of tangential motion come from?
...
Likewise in the second part of the same video (from 4:35 on) when the man changes the plane of rotation, there is no precession but the chair starts moving tangentially in the opposite direction. Where does that energy come from if not from the work done by the man pushing on the axis of rotation? Can you call this precession?.
I regret I cannot retrieve a video in which a Prof states that that is not conservation of angular momentum, but 3rd law of motion. It makes more sense, if you reflect just a second.
...
Likewise in the 18 section quoted in my previous post, where do you spot precession?

 

[A.T.]

This seems also to contradict what many maintained so far, that you do no work when you try to change the plane of rotation.

That "try to change" is too vague. You can change the plane of rotation without doing net work. But you also can "try to do it" and end up doing work.

You do no work, if you apply a torque perpendicular to angular velocity. But if the wheel spins while the wheel axis is rotating around some other axis, then the net angular velocity is not parallel to the wheel axis. So while holding the axis bearings, you can apply torques that have a component parallel to angular velocity, and thus do work.

 

[jbriggs444]

Likewise in the second part of the same video (from 4:35 on) when the man changes the plane of rotation, there is no precession but the chair starts moving tangentially in the opposite direction.

Where does that energy come from if not from the work done by the man pushing on the axis of rotation? Can you call this precession?.

There is precession in that section of the video. It is just not obvious.

The man is sitting in [the equivalent of] a swivel chair with the bicycle wheel held by its axles, one end of the axle in either hand. The man appears to be giving a hard twist to the axle over the span of about one second, bringing it to a vertical orientation. It looks like a clockwise or counter-clockwise twist (he does it several times in each direction).

In fact, the imparted force is not clockwise or counter-clockwise. It is out at the top and in at the bottom (or out at the bottom and in at the top). The applied torque is actually at right angles to what you think you see in the video.

As the axle of the bicycle wheel rotates, this applied torque shifts from being out-at-the-top, in-at-the-bottom to out-at-the-right, in-at-the left. That amounts to an unbalanced torque on the man in his chair. As a result of this torque, the man in the chair begins rotating around a vertical axis.

There is no work done on the bicycle wheel as a result of this. It ends up spinning at the same rate that it started (with respect to the non-rotating room).

There is work done on the man. He starts out stationary and ends up spinning. That is because he extends his right arm (or left -- depending on which event we're talking about) and retracts his left and his body moves as this happens. Force multiplied by body movement = work done on his body.

I regret I cannot retrieve a video in which a Prof states that that is not conservation of angular momentum, but 3rd law of motion.

Conservation of angular momentum applies here. As does Newton's third law. It's not an either-or proposition.

 

[bobie]

That "try to change" is too vague. You can change the plane of rotation without doing net work. But you also can "try to do it" and end up doing work.
.

he extends his right arm (or left -- depending on which event we're talking about) and retracts his left .

Thanks for your explanations, in the second part it is arguable if an arm is extented, it rather seems rotating at same distance. (In this video the girl clearly starts with fully stretched arms, so there is no chance of changing angular momentum, only direction is rotated by 90°)
But, to make things easier :
in the first part nobody is interfering with the wheel (apart from G), where does the KE it acquires come from?
of course from gravity, but
-the wheel remains at the same height from the ground ( no acceleration, no work done, no KE gained)
- the axis of rotation does not move, does not precess (no work done ), and even if it moved (as many said) that requires / can do no work.

 

[bobie]

applies here. As does Newton's third law. It's not an either-or proposition.

What is the aspect where 3rd law is relevant? do you mean the rotation of the platform?

 

[bobie]

The moon orbits the Earth in a [nearly] circular orbit. It is constantly changing direction under a force at right angles to its motion. This force does not change the speed of the moon. .

Could you please explain that?
There are two possibilities:

- either the moon moves a little tangentially an then g pulls it back into the track of the orbit (making sortof saw-tooth)
- or the moon is always on track and olny changes the direction of its vector, (describing a perfect curve circle/ellipse)

Which is the mainstream point of view?
Thanks