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Author Topic: What is the simplest explanation of gyroscopic precession?  (Read 35093 times)

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #25 on: 11/04/2011 18:34:35 »
Because of this, the axis resists forces that tend to change the direction of the axis.

Therefore, it requires force to alter the plane of the rotating mass, and, in the case of a flywheel, that force has to be applied as a torque that changes the direction of the axis.

OK, this seems like a good step. But to get a full explanation up to the level of angular momentum we need to be able to take a step in the right direction  :P, that is there has to be a good explanation for why the velocity vector for precession is derived from the cross product of angular momentum and torque.
People seem to be fighting shy of the whole story. In angular momentum terms, why is v not parallel to T?

If you are new to the thread note that a full explanation of gyroscopic precession has been achieved in terms of Newton's laws of motion, v=u+at, s=ut+at˛/2 and the other one *whatever*. The challenge now is to explain an understand it in terms of angular momentum. I'm betting we'll end up going back to Newton's laws again as they underpin angular momentum. Others seem to think not.
 

Offline Geezer

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What is the simplest explanation of gyroscopic precession?
« Reply #26 on: 11/04/2011 19:08:00 »
That's why I was trying to avoid using "angular momentum" and talked about "planar inertia" instead ;D

I was thinking that might allow us to simplify the analysis by completely ignoring that fact that the flywheel is actually rotating. Mind you, it was a bit late in the evening......
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #27 on: 11/04/2011 19:41:21 »
Because of this, the axis resists forces that tend to change the direction of the axis.

Therefore, it requires force to alter the plane of the rotating mass, and, in the case of a flywheel, that force has to be applied as a torque that changes the direction of the axis.

OK, this seems like a good step. But to get a full explanation up to the level of angular momentum we need to be able to take a step in the right direction  :P, that is there has to be a good explanation for why the velocity vector for precession is derived from the cross product of angular momentum and torque.
People seem to be fighting shy of the whole story. In angular momentum terms, why is v not parallel to T?

I admit, my post is very long and probably confusing, but  muddled in there is the whole story of why the right-hand rule is needed?  There's two steps that you need to make to get to the right hand rule definition of torque. 

If you have Geezer's rotating flywheel going at a constant rate, and you want to assign a vector to describe it's constant angular velocity, in which direction would you choose it to point?  Since the angular velocity is constant, it would be good if this vector always pointed in the same direction as this wheel rotates, so it wouldn't make sense for it to point along the tangent to the wheel or radially along a spoke, so the most sensible direction is along the axle--into or out of the plane of rotation. 

Since you have a choice of clockwise or counterclockwise rotation as you look towards the wheel, you can arbitrarily choose a vector pointing towards or away from you to represent each.  To make sure everyone's using the same standard, someone invented the right hand rule to tell you which way the angular velocity vector points.

And once we've agreed that angular velocity points either into or out of the plane, we also have to agree that angular acceleration--increasing or decreasing that velocity, has to also point into or out of that plane, since it points in the same or opposite direction of the velocity. 

I won't go ahead and get to torques and angular momentum yet.  That requires one more step, and without understanding why angular velocity and acceleration are chosen this way, the vector nature of torques and angular momentum won't make sense.  If you do agree that these make sense, then I can post again about how to go from this point to angular momentum and torques.
 

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #28 on: 11/04/2011 20:14:01 »
I admit, my post is very long and probably confusing, but  muddled in there is the whole story of why the right-hand rule is needed?  There's two steps that you need to make to get to the right hand rule definition of torque. 
I did read it - honest. (Could have been shorter).

If you have Geezer's rotating flywheel going at a constant rate, and you want to assign a vector to describe it's constant angular velocity, in which direction would you choose it to point?  Since the angular velocity is constant, it would be good if this vector always pointed in the same direction as this wheel rotates, so it wouldn't make sense for it to point along the tangent to the wheel or radially along a spoke, so the most sensible direction is along the axle--into or out of the plane of rotation. 

Since you have a choice of clockwise or counterclockwise rotation as you look towards the wheel, you can arbitrarily choose a vector pointing towards or away from you to represent each.  To make sure everyone's using the same standard, someone invented the right hand rule to tell you which way the angular velocity vector points.
I am good with this

And once we've agreed that angular velocity points either into or out of the plane, we also have to agree that angular acceleration--increasing or decreasing that velocity, has to also point into or out of that plane, since it points in the same or opposite direction of the velocity. 
I am also good with this

I won't go ahead and get to torques and angular momentum yet.  That requires one more step, and without understanding why angular velocity and acceleration are chosen this way, the vector nature of torques and angular momentum won't make sense.  If you do agree that these make sense, then I can post again about how to go from this point to angular momentum and torques.

shame because it's the

τ=ωpXL            ...http://en.wikipedia.org/wiki/Gyroscope

equation that is really the heart of the problem. It doesn't seem to have an intuitive explanation or derivation that would qualify as "understanding". I have to decompose the vectors back to what it is they really represent to get that.


« Last Edit: 11/04/2011 20:16:14 by JMLCarter »
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #29 on: 11/04/2011 21:17:18 »
We'll hopefully get to the intuitive explanation now.  Keep in the back of your mind the idea that angular acceleration needs to be perpendicular to the plane of rotation.  We know intuitively that a force applied along one of the spokes of the wheel won't make it rotate any faster or slower.  A force applied along it's axle won't make it rotate any faster or slower.  Only a force applied tangentially along the wheel's surface makes it rotate faster or slower. 

So let's say we have a force applied tangentially to the wheel so that it speeds up the wheel.  When the wheel speeds up, this means it has a positive angular acceleration.  This means that somehow this force which was applied in the plane of the wheel has created an angular acceleration which is directed perpendicular to the plane of the wheel.  We want some other quantity which takes into account the force and the point on the wheel at which it's applied and tells us the resulting angular acceleration due to that force.  This quantity has to somehow "know" that only components of the force directed tangentially along the wheel contribute to it's rotation and it's resulting direction should be along the axle, just as the angular acceleration is: this way you can equate this quantity with angular acceleration.  The quantity that does this is the cross product, which is why torque is equal to the cross product between a vector pointing from the axis to the point of contact of the force and the force vector itself. 

Now you might argue that this is unnecessarily complicated or unintuitive.  That's certainly true for the case of a single flywheel rotating on it's axis.  You can do the entire analysis using linear acceleration of points on the wheel without needing right hand rules and torques.  But when you end up with much more complicated systems with multiple degrees of freedom for both rotational and translational motion, torques and angular velocities/accelerations are incredibly useful.  And they're like many things physics--the concept is not intuitive at first, but once you understand it (usually through practice using it), it becomes intuitive.  Once it's intuitive to you, it's a simpler way of dealing with even simple problems like the flywheel or gyroscope.  It's a lot easier to me at least to justify precession in terms of torques and angular momentum than it is to justify it by breaking it into tiny chunks of mass and doing F=ma on each piece.  And if you want someone to actually compute precession rates, it's going to be far, far more difficult without going to torques and angular momentum.

So if you've followed all that, the final little bit is angular momentum.  After all this work and defining cross products, you end up with the quite elegant equation τ=Iα, where τ is the torque vector, I is the moment of inertia about a particular axis of rotation and α is the angular acceleration vector.  If there are no torques applied, then angular acceleration is zero, which means that the angular velocity is constant.  If you construct the quantity L=Iω, where ω is the angular velocity, then this quantity is only changed when a torque is applied.  So you can state this as a law: that L is conserved unless the system is acted upon by an outside torque.  This L is called angular momentum.  (It's called that because linear momentum is arrived at in the exact same way using linear velocity, acceleration, and Newton's second law.)  Again, since angular velocity, acceleration and torques are vectors which have to point (for simplicity) along the axis of rotation, angular momentum has to as well, since it points along the direction of angular velocity.

Sorry this is long-winded, but rotational kinematics usually takes weeks in an introductory physics course. 

If you have followed all of this, then the real payoff is the elegance of the expressions you get out: if θ is rotation angle, ω is angular velocity and α is angular acceleration, τ is torque, I is moment of inertia, L is angular momentum and t is time then:

θ(t)=ω t+1/2 α t2 tells you how far the wheel's rotated,

τ=Iα tells you how the wheel's acceleration and velocity change with applied torques,

dL/dt=τ tells you that angular momentum only changes over time along the direction of an applied torque.

This is all analogous to the linear equations which are usually considered much more intuitive.  Here x is position, v is velocity, a is acceleration, F is force, m is mass, t is time and p is momentum.

x(t)=vt+1/2 at2
F=ma
dp/dt=F.
 

Offline moonstroller

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What is the simplest explanation of gyroscopic precession?
« Reply #30 on: 13/04/2011 03:39:17 »
This is interesting:

"...Gyroscopes would do nothing in outer space.  With no gravity to exert the
torque, there would be no reason for angular momentum to change direction.
The spinning gyroscope would not turn..." ~ask a Scientist.



 

Offline yor_on

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What is the simplest explanation of gyroscopic precession?
« Reply #31 on: 13/04/2011 07:07:10 »
how about this then. You have two forces acting on that toy spinning at every frozen moment of time. One is 'gravity' directed downward along its axis, the other is the momentum created by its mass and spin, directed in a straight line horizontally in the direction of its rotation. if we flicker the motion picture forward until its spin have created a total turnaround we, at every moment we freeze it, will find those two forces acting, with the horizontal straight line a little moved.

The downward 'force' (gravity) anchors the toy where it is, the horizontal tells it where it want to 'move' but as the rotation is around its own axis those horizontal 'nudges' to move in a straight line takes over from the toys urge to 'fall' on its side, induced by the shape/mass of it, relative the gravity.

The momentum you create in that constantly changing 'straight line' is stronger than the gravity acting on it, as long as it spins fast enough. It's the same as when you turn on your bike, you lean in in the direction you want to turn and the wheels spinning acts as a linear 'force' now pointing slightly in the direction you want to turn and 'voila' the bike turn. Trust me, I have a bike, and it works, mostly.. :)

I've only talked about one horizontal straight line here, but if you think of it the whole toys rim will 'consist' of such straight lines, created by its rotation around its own axis. All of them simultaneously wanting to move straight out, horizontally, following the tops rotation. Those 'lines' take each other out, and so you will find that spinning top to be very stable, resisting you, when you try to nudge the top into leaning to a side.
==

Eh, the reason you turn is naturally that you changed your mass distribution, relative the gravity's direction, by leaning to some side. You can see a similar effect if you ever seen those guys riding a bike up on those rounded walls, you know like doing it inside a really big barrel :) their bikes constantly want to go through the wall and continue in a straight line, until gravity pulls them down. But that (linear) force counteract the gravity acting on them, gluing them to the wall, hopefully so :)
==

One more thing, if the speed around its axis is fast enough it doesn't matter how you place the the 'gyroscope' relative gravity. Those constant urges to go away in straight lines, induced by its spin (angular momentum) will be stronger than the 'gravity' acting on it. and that's what precession is as i understands it.  For a cool video of precession look here. there you can see almost see how those 'straight lines' induced by its momentum, counteract the gravity's tries to make the wheel act 'normal'.
« Last Edit: 13/04/2011 08:18:35 by yor_on »
 

Offline yor_on

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What is the simplest explanation of gyroscopic precession?
« Reply #32 on: 13/04/2011 09:03:21 »
Very cool explanation JP. I started to think of it at the first page :) Should have waited to the second maybe. The funny thing is that your math is making sense to me. And that's a good sign of clarity I think. yours I mean, not mine :)
 

Online syhprum

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What is the simplest explanation of gyroscopic precession?
« Reply #33 on: 13/04/2011 15:47:04 »
This has got all unnecessarily complex, if you tie a string to both ends of the axle so that an equal weight is carried on both bearings no precession occurs.
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #34 on: 13/04/2011 16:54:09 »
This has got all unnecessarily complex, if you tie a string to both ends of the axle so that an equal weight is carried on both bearings no precession occurs.

I'm still not quite following your argument about why the bearings are critical. Let's say it's just a spinning top instead of a gyroscope, like this one:


The only place you could remotely consider a "bearing" is the point of contact with the table, but precession still occurs in this case if it tips every so slightly off the vertical while spinning (which it inevitably does).  How does the bearings explanation come into play here?

A toy gyroscope, like this

is essentially just a top put into a housing of some sort so that you can pick it up and play with it without stopping it's rotation.  The explanation of it's precession is basically the same as for a top. 
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #35 on: 13/04/2011 16:59:31 »
In the end, it's complex because I believe you have two choices of how to explain this:

1) In terms of inertia and rotating masses.  This is what Graham and Geezer pointed out.  In this case you end up applying Newton's laws to each point on the rotating gyroscope, which is intuitive, but very messy if you want to do the mathematics.  Even if it's not precessing, each point on the gyroscope is rotating, so it's not following a straight line and is constantly accelerating around a circle due to centripetal force.  This makes applying Newton's laws a bit of a headache.

2) In terms of torques and angular momentum.  In this case, you have to do a lot of up front work to define angular quantities as well as torque and angular momentum, and it's less immediately intuitive how this all works unless you understand how these equations are related to more intuitive concepts, like Newton's laws.  However, if you have to do computations, these are far easier, especially if you end up with a system more complex than a precessing gyroscope.  Also, these equations can be more intuitive than the above method, I think, but only once you've used them enough to really understand their physical meaning.
« Last Edit: 13/04/2011 17:03:42 by JP »
 

Offline burning

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What is the simplest explanation of gyroscopic precession?
« Reply #36 on: 13/04/2011 17:21:36 »
This has got all unnecessarily complex, if you tie a string to both ends of the axle so that an equal weight is carried on both bearings no precession occurs.

If you are saying that you can design a gyroscope so that the bearings will exert no net torque on the gyroscope, I don't think anyone disputes that.  However, it doesn't answer the OP's question.

If you are saying that you can design a gyroscope so that even when you deliberately exert a net nonzero torque on the gyroscope (at a nonzero angle to the axis of rotation) that the gyroscope won't precess, that is simply not true.
 

Online syhprum

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What is the simplest explanation of gyroscopic precession?
« Reply #37 on: 13/04/2011 19:31:29 »
JP
The spinning top is not of course spinning on a point but on a circular track as soon as it tilts over this is where the force to create the precession is injected into the system.
The sharper the point of contact on which it balances the less the precession but of course it can never be zero. 
« Last Edit: 13/04/2011 19:34:13 by syhprum »
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #38 on: 13/04/2011 21:22:30 »
JP
The spinning top is not of course spinning on a point but on a circular track as soon as it tilts over this is where the force to create the precession is injected into the system.
The sharper the point of contact on which it balances the less the precession but of course it can never be zero. 

Ok, let's try this.  Will a top precess if it's placed on a frictionless surface?
 

Offline Geezer

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What is the simplest explanation of gyroscopic precession?
« Reply #39 on: 13/04/2011 21:29:31 »
JP
The spinning top is not of course spinning on a point but on a circular track as soon as it tilts over this is where the force to create the precession is injected into the system.
The sharper the point of contact on which it balances the less the precession but of course it can never be zero. 

Ok, let's try this.  Will a top precess if it's placed on a frictionless surface?


Well, that's difficult to answer, because there is no such thing as a frictionless surface  :D

You could ask if a top precesses in outer space though. I kind of doubt that it will.
 

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #40 on: 13/04/2011 21:33:57 »
This is interesting:

"...Gyroscopes would do nothing in outer space.  With no gravity to exert the
torque, there would be no reason for angular momentum to change direction.
The spinning gyroscope would not turn..." ~ask a Scientist.


Gyroscopes are used to orient satellites in space. You don't need gravity to apply a torque - it can be applied another way (including through friction, I suppose).
Torque -> precession.

The axis of the earth precesses, but I'm not sure what is applying the torque. I assume it is due to moving round the sun with an inclined axis.
« Last Edit: 13/04/2011 21:36:38 by JMLCarter »
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #41 on: 13/04/2011 21:38:47 »
JP
The spinning top is not of course spinning on a point but on a circular track as soon as it tilts over this is where the force to create the precession is injected into the system.
The sharper the point of contact on which it balances the less the precession but of course it can never be zero. 

Ok, let's try this.  Will a top precess if it's placed on a frictionless surface?


Well, that's difficult to answer, because there is no such thing as a frictionless surface  :D

You could ask if a top precesses in outer space though. I kind of doubt that it will.

Ok Mr. SmartyGeez.  We can either introduce a theoretically frictionless surface or put the top on a series of surfaces with decreasing amounts of friction and see what happens.

We could also put it in space, but you'd need to still act on it with gravity, and it would begin accelerating in the direction of gravity.  The same question holds in space, though.  :)
 

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #42 on: 13/04/2011 22:10:42 »
We'll hopefully get to the intuitive explanation now.  Keep in the back of your mind the idea that angular acceleration needs to be perpendicular to the plane of rotation.  We know intuitively that a force applied along one of the spokes of the wheel won't make it rotate any faster or slower.  A force applied along it's axle won't make it rotate any faster or slower.  Only a force applied tangentially along the wheel's surface makes it rotate faster or slower. 
Sure thing

So let's say we have a force applied tangentially to the wheel so that it speeds up the wheel.  When the wheel speeds up, this means it has a positive angular acceleration.  This means that somehow this force which was applied in the plane of the wheel has created an angular acceleration which is directed perpendicular to the plane of the wheel.  We want some other quantity which takes into account the force and the point on the wheel at which it's applied and tells us the resulting angular acceleration due to that force.  This quantity has to somehow "know" that only components of the force directed tangentially along the wheel contribute to it's rotation and it's resulting direction should be along the axle, just as the angular acceleration is: this way you can equate this quantity with angular acceleration.
err and it also has to somehow "know" how far the force is applied from the axis of rotation. torque=force*distance
Oh hang on, you I'll assume you meant Torque not force. Makes sense.

The quantity that does this is the cross product, which is why torque is equal to the cross product between a vector pointing from the axis to the point of contact of the force and the force vector itself.
So that would be the cross product of the Torque and the normalised/unitary angular momentum? Which is the change in angular momentum. OK I'm still aboard

Now you might argue that this is unnecessarily complicated or unintuitive.
Perhaps, but I do get it. Hopefully you are going to go on to precession...

That's certainly true for the case of a single flywheel rotating on it's axis.  You can do the entire analysis using linear acceleration of points on the wheel without needing right hand rules and torques.  But when you end up with much more complicated systems with multiple degrees of freedom for both rotational and translational motion, torques and angular velocities/accelerations are incredibly useful. 
This is my experience also. Try working out how two "nested" gyroscopes behave using finite elements - but with angular momentum you just add the vectors and treat it as a single one. Whereas in the past I might have felt that that operation was a bit of a trick, you've explained it well enough for me.


And they're like many things physics--the concept is not intuitive at first, but once you understand it (usually through practice using it), it becomes intuitive.  Once it's intuitive to you, it's a simpler way of dealing with even simple problems like the flywheel or gyroscope.  It's a lot easier to me at least to justify precession in terms of torques and angular momentum than it is to justify it by breaking it into tiny chunks of mass and doing F=ma on each piece.  And if you want someone to actually compute precession rates, it's going to be far, far more difficult without going to torques and angular momentum.
OK seems to work that way for me also.

So if you've followed all that, the final little bit is angular momentum.  After all this work and defining cross products, you end up with the quite elegant equation τ=Iα, where τ is the torque vector, I is the moment of inertia about a particular axis of rotation and α is the angular acceleration vector.  If there are no torques applied, then angular acceleration is zero, which means that the angular velocity is constant.  If you construct the quantity L=Iω, where ω is the angular velocity, then this quantity is only changed when a torque is applied. 
...by integration of τ=Iα, yes. I'm no stranger to calculus.

So you can state this as a law: that L is conserved unless the system is acted upon by an outside torque.  This L is called angular momentum.  (It's called that because linear momentum is arrived at in the exact same way using linear velocity, acceleration, and Newton's second law.)  Again, since angular velocity, acceleration and torques are vectors which have to point (for simplicity) along the axis of rotation, angular momentum has to as well, since it points along the direction of angular velocity.

Sorry this is long-winded, but rotational kinematics usually takes weeks in an introductory physics course. 
Well, good recap on foundations I feel less rusty, but you haven't got to the bit that was my question yet. Which is about precession.


If you have followed all of this, then the real payoff is the elegance of the expressions you get out: if θ is rotation angle, ω is angular velocity and α is angular acceleration, τ is torque, I is moment of inertia, L is angular momentum and t is time then:

θ(t)=ω t+1/2 α t2 tells you how far the wheel's rotated,

τ=Iα tells you how the wheel's acceleration and velocity change with applied torques,

dL/dt=τ tells you that angular momentum only changes over time along the direction of an applied torque.

This is all analogous to the linear equations which are usually considered much more intuitive.  Here x is position, v is velocity, a is acceleration, F is force, m is mass, t is time and p is momentum.

x(t)=vt+1/2 at2
F=ma
dp/dt=F.

Which is a great recap of the area of the subject. I like the bit at the start about tangential forces (or Torques) as it is the sort of terms I am looking for in the promised explanation of

τ=ωpXL

using all these vector quantities we have discussed. I hope the next post will be able to do that - but it is significantly more difficult.
« Last Edit: 13/04/2011 22:15:13 by JMLCarter »
 

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #43 on: 13/04/2011 22:34:41 »
In a toy gyroscope or toy spinning top the torque that causes the precession is due to gravity in one direction and the force of support from the floor (through the axle and, yes, any bearings) in the other. (Gravity pulls the whole body in one the same direction).

The non-vertical initial alignment of the toy, and asymmetries in its mass mean that it can never run perfectly vertical.

Friction slows the angular velocity of the spinning part (as does air resistance). These are torque vectors parallel to the angular momentum. There is no mechanism by which they can drive/cause precession.

Many examples are given above of precession in the absence of friction (but not in the absence of cross-axial torque of some form).
 

Offline Geezer

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What is the simplest explanation of gyroscopic precession?
« Reply #44 on: 13/04/2011 22:48:27 »
Now I'm totally confused (OK - you lot in the peanut gallery can save your cheap comments for later)

Would a bicycle wheel spinning in outer space precess or not?
 

Offline JMLCarter

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What is the simplest explanation of gyroscopic precession?
« Reply #45 on: 13/04/2011 23:04:53 »
If it was subject to a cross axial-torque it would.
If it was not subject to a cross axial-torque it would not.


Why is anyone talking about friction, that's what confuses me.
« Last Edit: 13/04/2011 23:15:10 by JMLCarter »
 

Offline Geezer

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What is the simplest explanation of gyroscopic precession?
« Reply #46 on: 14/04/2011 00:04:22 »
If it was subject to a cross axial-torque it would.
If it was not subject to a cross axial-torque it would not.


Why is anyone talking about friction, that's what confuses me.

By "cross axial-torque" I'm assuming you mean a torque that tends to alter the alignment of the axis of rotation? (I'd prefer to call this a force that tends to alter the plane of the planar inertia, but that's just because I like to think of it that way.)

In outer space, I don't think there would be any such torque or force.

The reason friction is important is because, if Mr "Know-it-all-theoretical-physicist-JP" was actually able to find a frictionless surface and position it normal to the force of gravity AND spin a spherical spinning-top (of uniform density) on that surface, the spinning-top wouldn't precess.
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #47 on: 14/04/2011 00:38:48 »
The reason friction is important is because, if Mr "Know-it-all-theoretical-physicist-JP" was actually able to find a frictionless surface and position it normal to the force of gravity AND spin a spherical spinning-top (of uniform density) on that surface, the spinning-top wouldn't precess.

I'm very confident it would.  I can show it with math, but I don't know if you'd trust that without a frictionless surface to back it up.  You can experimentally check it by testing a gyroscope on a series of surfaces with decreasing coefficients of friction.  The precession rate should decrease towards zero as you reduce the friction.  Of course, without friction, it will never slow down, and slowing down makes the precession more evident, so you'd have to start it off at a large angle with respect to the surface to begin with...

If I can find my toy gyroscope around, I might give this experiment a shot.  In an oiled glass pan, I should see very little precession.
 

Offline Geezer

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What is the simplest explanation of gyroscopic precession?
« Reply #48 on: 14/04/2011 00:50:27 »
The reason friction is important is because, if Mr "Know-it-all-theoretical-physicist-JP" was actually able to find a frictionless surface and position it normal to the force of gravity AND spin a spherical spinning-top (of uniform density) on that surface, the spinning-top wouldn't precess.

I'm very confident it would.  I can show it with math, but I don't know if you'd trust that without a frictionless surface to back it up.  You can experimentally check it by testing a gyroscope on a series of surfaces with decreasing coefficients of friction.  The precession rate should decrease towards zero as you reduce the friction.  Of course, without friction, it will never slow down, and slowing down makes the precession more evident, so you'd have to start it off at a large angle with respect to the surface to begin with...

If I can find my toy gyroscope around, I might give this experiment a shot.  In an oiled glass pan, I should see very little precession.

Ah! But you may have overlooked a couple of teensy details. The top is completely spherical, and it has uniform density. Therefore, there is no couple.
 

Offline JP

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What is the simplest explanation of gyroscopic precession?
« Reply #49 on: 14/04/2011 00:54:46 »
. . . I am looking for in the promised explanation of

τ=ωpXL

using all these vector quantities we have discussed. I hope the next post will be able to do that - but it is significantly more difficult.

Ok, that's a bit harder, but it isn't that bad if you understand calculus.  I haven't figured out a way to do it in one easy step, though, so I'd hardly call that one intuitive.  :p

First, consider that I've drawn a nice picture of a top.  Or better yet, let someone with art skills do it, as they already have here: http://hyperphysics.phy-astr.gsu.edu/hbase/top.html
That way, I can refer to angles and you'll know what I'm talking about.

Consider that we're working in the center-of-mass rest frame so that the only non-zero torque is generated by the table pushing up on the base of the top.  This will tend to cause it to rotate counter-clockwise about it's center of mass.  You need to start from one of the equations I gave you above and play some calculus tricks:
τ=dL/dt=dL/dθ dθ/dt.

Now θ is the angle around that big circle at the top: it's the angle of precession.  When θ=2π, the top has precessed all the way back to it's starting point.  dθ/dt is proportional to the angular velocity of precession, or ωp.  So we have:
τ=dL/dθ ωp.

The next step is the tricky one, and it took me a bit of thought.  dL is a small change in L as the precession angle changes by a small amount dθ.  If you work out the geometry in that picture, you'll see that dL=L sinφ dθ.  Therefore, dL/dθ=L sinφ, where φ is the angle between a vertical line and the rotation axis of the top.  You'll notice I dropped the vectors here.  They'll return shortly.  But even without vectors, we know that the equation still holds in magnituide:
τ=dL/dθ ωppL sinφ

To get the vectors, you also have to work out the direction of this change in L from the figure.  It points out of the page, or in other words, it points in the direction of the torque which also happens to be the direction of wp x L. This is good, since if the vector directions on both sides of the equation didn't match up, we'd have made a mistake somewhere.  :)

The next point is to recall that cross products have a magnitude given by the product of the vector magnitudes multiplied by the sine of the angle between them.  In other words, the magnitude of  wp x L is wp L sinφ.  This finally means that the cross product  wp x L  matches in both direction and magnitude with our above equation, and so...

τ=wp x L.
 

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What is the simplest explanation of gyroscopic precession?
« Reply #49 on: 14/04/2011 00:54:46 »

 

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