# Gyroscopic precession

by jbar18
Tags: gyroscopic, precession
 P: 53 Hi, I'm having some trouble understanding how this works. I'm trying to put the pieces together but I'm hoping people here can help me. Specifically I'm wondering about a gyroscope positioned so that its axis lies in the horizontal plane and one end sits on top of a rod. So far what I've got is this: There is a torque acting at the centre of mass of the gyroscope. Torque equals the rate of change of angular momentum, and so the angular momentum of the gyroscope must change so long as there is a torque acting. Angular momentum must be conserved. I'm not really sure how that applies here. Maybe this is why I'm having trouble. The main problem I'm having is this: It sounds logical when somebody else says it, but I can't seem to properly understand it myself. If this works on a gyroscope, why doesn't it work on a stationary rod (this still has a torque acting?). In this case the rod would gain angular momentum in the vertical plane I guess, due to gravity. So then if the rod is spun, like a gyroscope, it should precess as well? Maybe it does a little bit, but no matter how thick or heavy or long the rod is, it will always fall pretty quick. So why is this different for a gyroscope? I've been thinking and thinking but I just can't really clear it up. Any help appreciated. Thanks.
 P: 6 Here is my conceptual understanding of the gyroscope: Imagine a spinning gyroscope sitting at an angle on top of a rod. Now freeze time. The lower side of the gyroscope is being pulled DOWN by the torque of gravity around the axis of the gyro, and so that point on the gyro will gain downward momentum. Now let time go for a second until that point is on top. Now the torque of gravity around the axis is pulling it UP, but the point still has most of its downward momentum, the upward torque will cancel the downward momentum, but not be able to pull it any farther upward. This is happening continuously on every point on the gyroscope. Hence, the torque vectors "blur" together, and the gyroscope remains frozen at its angle. The amount of blurring and hence the gyro's ability to stay up depends on the angular momentum of the gyroscope (the speed of blurring times the ability of the object to hold momentum/kinetic energy). See the figure: Notice that the KE vectors bulge in the middle. These bulges aren't quite balanced and create the rotation around the axis. The rotation will diminish as the blurring becomes more ideal (angular momentum approaches infinity). Now, your question: Spinning the rod could work, it is in fact a species of gyroscope. But you would need to get it going ridiculously fast for it to stay up. The reason is that it has a lower Moment of Inertia (wikipedia it). Gyroscopes stay up when they have a high enough angular momentum. Because angular momentum is defined by angular velocity times the moment of inertia, a lower moment of inertia means you need a higher angular velocity to achieve the same angular momentum and staying power. So, your rod is gyroscope, just one that needs to be going really, really fast.
 P: 53 Thank you, mindoftea. Although what you said about blurring kind of went over my head, any information is helpful right now, I need to somehow internalise this. It doesn't really help that I haven't really ever played around with a gyroscope, that would probably help me gain some intuition on the topic. However, I am still puzzled. I'll tell you what I was told, and maybe you can help me figure it out. I will discuss the picture you have shown as an easier way to see what I'm talking about. There are the vertical torques, as you have labelled on your diagram. Now torque gives us the rate of change of angular momentum, and so since there is a net torque, the angular momentum of the gyroscope must change. However, angular momentum must also be conserved, and so as a vector quantity, the angular momentum will stay at the same magnitude but will change direction. Therefore the gyroscope precesses. (I'm not quite sure why it changes direction in the way that it does, but okay, I'll accept it for now) However, let us take a gyroscope that is not spinning. The same torques exist on the gyroscope, the only thing that is different is that the angular momentum vector is null. In this case, the gyroscope will surely fall off from its rod to the ground. What I'm having trouble with, more than anything, is why giving the gyroscope angular momentum keeps it off the floor. And how this doesn't seem to fit very well with the when the gyroscope is spinning. There is still a net torque on a non-spinning gyro, and so there must be some change in it's angular momentum. Again, applying the same rules as before, since angular momentum must be conserved, the vector can only change direction, and so presumably the gyro will fall towards the ground doing somersaults as part of this. However, the direction and magnitude of the torque vector is entirely unchanged. The ONLY difference as far as I can see is that in one scenario, angular momentum is non-zero. Why does this single condition entirely change how the gyroscope moves? Same torque, same everything apart from that the angular momentum is zero in one case. This entirely changes the direction in which the gyroscope will move. I am making an assumption here, that small angular momentums also cause precession. In reality it seems like if the angular momentum was very small, it also wouldn't work, but I have put this down to things like losses of rotational energy as friction and so on, and so the angular momentum would decrease to zero fairly quickly. So basically, the main thing that is still bugging me is how any amount of angular momentum changes the direction of motion from downwards to sideways. Every time I think I've got hold of it I go back and compare it to a non-spinning gyro and it just doesn't seem to hold. I must be missing something. Perhaps conservation of energy plays a bigger role than I've been told? That seems promising, but I'm just having trouble coming up with anything that makes me go "aha! That's what it is." Any help appreciated, and thanks again mindoftea for your reply. The stuff about blurring will probably be more useful once I get over this hurdle though! I'll be sure to read it over a few more times again once I've got it. Thanks :)
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## Gyroscopic precession

 Quote by jbar18 Now torque gives us the rate of change of angular momentum, and so since there is a net torque, the angular momentum of the gyroscope must change. However, angular momentum must also be conserved, and so as a vector quantity, the angular momentum will stay at the same magnitude but will change direction.
There are at least two things wrong with your thinking here.

One is that when angular momentum is conserved, it is the vector quantity, not just the magnitude, that is conserved. The other problem is that angular momentum is not conserved here. There are external torques.

To help clarify things, let's look at linear momentum. An external force acting on an object changes the linear momentum of that object. So doesn't this conflict with the fact that linear momentum is a conserved quantity? Of course not. Linear momentum is a conserved quantity in an isolated system. A object that has external forces acting on it is not an isolated system. Conservation of linear momentum does not apply to this object.

The exact same thing happens with respect to angular momentum when external torques act on some object. The object is not an isolated system. The law of conservation of angular momentum does not apply to this object.
 P: 6 So, I've decided that the way to answer this question is to completely solve the gyroscope problem without using vector operations. This will take quite a while, but should make sense. First I'm going to idealize the gyroscope: it has a stationary massless frame which contacts the pivot on the bottom. Through this frame is a massless, frictionless axle, with the gyro's weight attached. The weight is concentrated in an infinitely thin ring of radius r, called the ring. So in my earlier image, the only part with mass is the gold ring. My plan of action for doing this is as follows: first set up a 1D polar coordinate system around the axis which describes the points on the ring, called Co1. Then, a 2D polar coordinate system around the tip of the pivot in the same plane as a vertical and the axis, called Co2. Images: Now, find a function f(θ1) that gives us the magnitude of the linear acceleration of a point on the ring due to the gyroscope falling in Co2, based on an angle θ1 in Co1:[INDENT] Imagine the gyroscope spinning at an angle. The torque acting on it would be given by: $\tau =mgl\sin \left( \phi \right)$. Where m is the mass of the gyro, g is the gravitational acceleration, phi is the angle of the gyroscope from the horizontal, and l is the distance from the center of mass to the pivot. Now, divide the torque, t, by the moment of inertia of the gyroscope rotated around the pivot, I, to yield the angular acceleration: $\alpha =\frac{\tau }{I}$. The angular acceleration means how fast the gyroscope's angle is accelerating when falling over. Now for a single point on the gyro described by θ1 in Co1: First, we need to find the coordinates of the point in Co2: In rectangular, the y coordinate would be l, the distance from the center of mass to the pivot, and the x coordinate would be the cos(θ1). Convert to polar by taking the angle, θ, as atan(y/x), and then the radius, r, as √(l2+cos2(θ1)). The magnitude of the linear acceleration would then be calculated by: |a|=rα. In full: \left| a \right|=\sqrt{I^{2}+\cos ^{2}\left( \theta _{1} \right)}\frac{mgl\sin \left( \phi \right)}{I} Anyway, because the ring is spinning at a constant rate, θ1 can be written as function of time, namely θ1=ωt. So now we have a function f(ωt) that gives us linear acceleration of a point on the ring in Co2. Now, we integrate f(ωt) w/r/t time to get the magnitude of the linear velocity of that point: $v=\int_{0}^{t}{f\left( \omega t \right)\; dt}$. Now, multiply that number by the mass of the point, m, to get the momentum, and integrate w/r/t velocity to get the KE of a point on the gyro: $k_{e}=\int_{0}^{v}{\frac{m}{\omega }F\left( \omega t \right)\; dv}$. I have to go, and will finish this later, but that is the main idea. Now just take the vertical components and add up the points.