Angular Momentum: Spinning Flywheel on a Rotating Table

1. Jul 21, 2009

Sam_Goldberg

1. The problem statement, all variables and given/known data

Hi guys, this question is from Kleppner and Kolenkow's mechanics book, problem 7.2:

A flywheel of moment of inertia I rotates with angular velocity w at the middle of an axle of length 2L. Each end of the axle is attached to a support by a spring which is stretched to length L and provides tension T. You may assume that T remains constant for small displacements of the axle. The supports are fixed to a table which rotates at constant angular velocity, W, where W << w. The center of mass of the flywheel is directly over the center of rotation of the table. Neglect gravity and assume that the motion is completely uniform so that nutational effects are absent. The problem is to find the direction of the axle with respect to a straight line between the supports.

I'm sorry that I could not provide a picture, but you can find one by searching inside on amazon. The page is 334.

2. Relevant equations

3. The attempt at a solution

Okay, here goes. There will be three axes. The z axis points to the sky; the y axis points along the springs; and the x axis is perpendicular to the springs. All axes meet at the center of the flywheel. I will describe the orientation of the axle which holds the flywheel by two angles, A and B. If A changes wrt time, there is an angular velocity in the x direction, and if B changes, in the z direction. I am not completely sure about this, but from the problem statement, it seems as if the angles A and B are small, that is, they can be dealt with in the first order.

I will now find the torques about the x and z axes and relate them to the changes in angular momentum wrt time. For the x axis:

Torque(x axis) = - 2Lsin(A)T = -2LTA

The change in angular momentum wrt time comes from three sources. First, the flywheel itself can rotate about the x axis, it produces (1/2)IA'', since the moment of inertia about a perpendicular axis is (1/2)I. Second, the angular momentum from the flywheel is a vector, and it can produce an x component due to the rotation of the table at angular velocity W. The value of this is IwW. Finally, the rotation of the axle itself about the z axis changes the vector component; this contribues IwB'. So, we have:

(1/2)IA'' + IwW + IwB' = -2LTA

The torque about the z axis is similar, except for a sign difference for a term:

(1/2)IB'' - IwA' = -2LTB

We have two coupled differential equations, which presents a problem. I could simplify these equations by eliminating a term(s), but I'm not sure which one(s) is too small to be considered. I know that W << w, but that's about it.

Perhaps I made a mistake in coming up with these equations. Let me know if that's the case as well. Thanks in advance.

2. Jul 22, 2009

Sam_Goldberg

Hmm... It looks as if no one has replied yet. Let me try to clarify some points that may be confusing. If the problem and/or my attempt at the solution still does not make sense, by all means let me know; do not hesitate.

Let's first try to clear up the problem. We have a table that is rotating at a constant angular velocity W (remember that I use W for the table and w for the flywheel). At opposite sides of the table are two posts, each of which holds a spring that provides tension to the object of interest in the middle. That object (which I found the torques on) is an axle of length 2L which has a rotating flywheel jammed into it. Note that my special object is floating (quoting the problem, neglect gravity), being supported by the tension of the springs which are constant throughout the problem. If this still does not make sense, I'm sorry that I could not present a picture (physics forums would not accept my file for some reason) but it's definitely on amazon. Just find Kleppner and Kolenkow's book on that site; click on the book's image (it should say around it "search inside!"); and search the number 334 to get the page.

Now let's talk about the attempt at a solution. I will clarify the meaning of my angles A and B. For A, let the axle point along the y axis, and then give it a small tilt such that the screw that tilts it points outward along the x axis. If this happens, then A is changing. I use A' for the first time derivative and A'' for the second time derivative. The angle B is similar, except that the screw points along the z axis.

Finally, the book has not done a differential equation tougher than simple harmonic motion, so I know that it's not making me solve the final two equations that were obtained; that's why I feel that I need to neglect some terms, or scratch these equations for the correct ones.

Please let me know if there is anything more that is unclear; I wil gladly restate it. Thanks.

3. Jul 22, 2009

cepheid

Staff Emeritus
Last edited by a moderator: May 4, 2017
4. Jul 22, 2009

Sam_Goldberg

Thank you cephid for the picture. However, it still appears as if no one has responed. If you come across this post and are unable to completely answer my question, please either tell me that you do not have an immediate answer but have a suggestion on a different way I can look at the problem, or that I did not state everything clearly enough and you would like me to clarify something I said earlier. I'm self-studying the Kleppner and Kolenkow book in preparation for university and I don't have anyone to talk to, so it would be greatly appreciated if someone could give at least a little help.

5. Jul 22, 2009

bodensee9

Imagine that your springs make an angle theta with the horizontal. So then your torque with respect to center is then 2lsin(theta) = 2l(theta). So since you have 2 contributions, you have total 4ltheta. Torque = dL/dt = IwW, where W = capital omega. So then you can find theta and that will give you the direction.

6. Jul 22, 2009

Sam_Goldberg

Hi bodensee,

Thanks for the post. However, I am still unsure. Aren't there two degrees of freedom, that is, two angles, A and B, that we must take into account? Also, looking at my very first post, I have for the contributions to the change of angular momentum wrt time two extra terms, (1/2)IA'' and IwB' that I believe should be included. (or maybe they are too small as I was mentioning. If this is the case, why?). Finally, isn't the magnitude of the torque 2LTA, and not 4LTA?

7. Jul 22, 2009

bodensee9

uh, it seems to me that you're double counting with A" and B'? I thought that displacement wrt x would be theta and then B' is then W. In anycase, it's 4TL(theta) because you have 2 springs, and each length to center is 2L. And I'm sure this solution is correct as I submitted this on homework and got full credit.

8. Jul 22, 2009

diazona

There's no need to consider motion in the vertical (z) direction, since there are no vertical forces that would make the flywheel's axis wobble that way. So basically B=0.

I agree with bodensee9's reasoning.

9. Jul 22, 2009

Sam_Goldberg

I think I see what you guys are saying. There is no change of angular momentum about the z direction; it stays constant, B = 0. However, since the flywheel has a component of angular momentum that can change about the x axis, we need a constant torque about the x axis for torque = dL/dt to be satisfied. And it appears this is where A comes into play. A stays constant so that the change in angular momentum wrt time about the x axis may be constant. Bodensee and diazona, thank you very much.