Precession is not due to inefficient bearings -
Geezer/Syphrum, you seem to agree on something but from what I understood of the question, I tend to agree with JMLCarter. I think there must be some misunderstanding somewhere.
If all this angular momentum stuff seems odd to you, it's possible to analyze it from Newton's laws, but that requires chopping the top up into tiny pieces and analyzing the forces on each piece. Angular momentum is basically a very elegant short method to take care of all of this.
friction [xx(]... ...anyway...If all this angular momentum stuff seems odd to you, it's possible to analyze it from Newton's laws, but that requires chopping the top up into tiny pieces and analyzing the forces on each piece. Angular momentum is basically a very elegant short method to take care of all of this.
I never liked angular momentum - well, as a tool it is great, but I think the derivation of its properties from "simpler" laws of motion does create a better explanation. (Which I did in the earlier post - how would you do it?).
Perhaps also consider a gyroscope in free space, (such as used to orient satellites). I think it's simpler than the classic toy or spinning top in which one is examining motion that is polluted/constrained by the floor or little tower thing.
JP
I agree with/understand all you have said about angular momentum. But you stopped just short of the bit I find tricky. Which is to "prove/understand" using angular momentum principles that a torque force on the axis results in a fixed angular velocity perpendicular to torque and angular momentum.
Perhaps I need to re-visit vector dot and cross-products.
Do you understand why the angular momentum vector due to spin points out the top of the gyroscope, and why the one due to gravity pulling it sideways points at 90 degrees to that?
I thought that the question was how does the force that cause the precession of a rotating disc arise.
My explanation is that it arises in the bearings on the shaft that is used to tilt the rotating disc.
Could anyone please explain how they tilt the disk without using a shaft and bearings as all counter arguments seem to ignore the existence these things.
Any more on the vectors? Why is a cross product the right operation to use? I can't see how proof of that can come from angular momentum, it seems necessary to look below at what the angular momentum represents.
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.
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.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.I am good with this
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 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.
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.
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.
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.
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?
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.
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.
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
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.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.
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.
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.
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.
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.
. . . 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.
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.
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.
That's where your mistake is, Geezer! Perfect spheres are called cows (http://en.wikipedia.org/wiki/Spherical_cow), not tops!
Well, I tried it out on an oiled glass surface, a table cloth and a wooden cutting board and it precessed on all of them noticeably, and didn't seem to be significantly less on the glass. I guess putting it in a Teflon pan would be best, since that has a coefficient of static friction of 0.04, but I'd rather not scratch up my cookware with a gyroscope. ;)
Well, my gyroscope isn't quite spherical, but maybe we can neglect any deviation from sphericity as being minor. Then precession is experimental error!
If not, then perhaps syphrum would find this useful. :)
Well, my gyroscope isn't quite spherical, but maybe we can neglect any deviation from sphericity as being minor. Then precession is experimental error!
If not, then perhaps syphrum would find this useful. :)
If you had this magical "frictionless surface," what you'd expect to see is that the center of mass of the top stays in one spot, while the body of the top moves around due to precession and spin. The point of contact with the table would freely slide around. This is because there is a net torque about the center of mass, due to the normal force, but no net force. No net force means the center of mass doesn't move. The net torque leads to precession.