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Conservation of angular momentum

  1. Nov 25, 2008 #1
    I'm tasked with the design of a variable inertia flywheel. I've concentrated on the change in location of mass about an axis and varying 'r'. I'm trying to analyze what happens without any change in torque (I.E. the flywheel free spins). My HS physics tells me that the angular momentum must remain constant- Lo=Le. But also, kinetic energy must be conserved - Eko=Eke. As I run trials with re=ro/2, the results substantiate what I would expect, since L is a function of w and Ek is a function of w2 (I can calculate a new w to conserve one (e.g. L) but not both).

    I know that I can adjust r without adding torque to the axis of rotation and it seems that the work required should not enter into the conservation question (if I increase r, centrifugal force would make work a negative value). Where am I missing the conservation question? Can it be both angular momentum and kinetic energy and if not, which is conserved? And what happened to the other?

    I've searched this forum as well as other internet sites and have found references to this dichotomy, but each leaves the fundamental question unanswered.
  2. jcsd
  3. Nov 26, 2008 #2
    I think your reasoning is Ok. It's just that Kinetic Energy is not conserved. As you indicated, if you increase r, that's negative work. It shows up as a reduction in Kinetic Energy. The angular velocity is less. If you can do this without applying a torque to the object, the Angular Momentum should remain the same.

    That's a challenging puzzle.
  4. Nov 26, 2008 #3
    I've attached a sample of the basic calculations in a PDF form. I saw a form of this question in an earlier post regarding a physics class demonstration: a puck with a string attached is rotated on a table. The string is pulled from the axis such that the puck moves in toward the axis until the radius of rotation is 1/2 the original. Most of the responses suggested that the angualr velocity would double, however, in order to maintain a constant angular momentum, the velcocity must increase by a factor of 4.

    Essentially, attached spreadsheet indicates the same factor for an increase in kinetic energy, which is what you would expect. But why?

    Attached Files:

  5. Nov 26, 2008 #4

    It appears to me that the other responses for the puck example are assuming Kinetic Energy is conserved. That would result in the conclusion that angular velocity that is only doubled. As you have identified, changing the location of the puck involves work, positive if r is reduced, negative if r is increased. Work being energy has to go someplace. It can go into heat or mechanical deformation. But in this case, it goes into rotational motion. That's why its not surprising to see the angular velocity in the puck example increases by a factor of 4, not 2.

    Total Energy is conserved. Kinetic Energy by itself is not.

    I am curious. Can you provide a link to the earlier post?
    Last edited: Nov 26, 2008
  6. Dec 1, 2008 #5
    ML - Thanks for your reply. I've been off in the mountains with no internet- scary feeling.

    I suspect there's a more direct way to give you the link I referred to but I'm not that well versed in the forum features, so you might try this link:
    Otherwise, search for "spinning puck".

    I've decided that the work involved in changing the radius not only must be included, it's also a major factor. I don't know yet where that might take me. For example, when you move a mass that is at radius=1m to radius=0.8m, did you move the mass 0.2m, or did you move it the distance that it actually moved (I.E. a spiral depending on the ratio of angular velocity to linear velocity)? I think probably not because that is factored in when the force required is due primary to centripetal force as opposed to the actual inertia of the mass being moved. But.....?
  7. Dec 1, 2008 #6
    Welcome back from the mountains. With regard to the puck problem, I would say you have already solved it in your pdf link. Using the consept of conservation of momentum, reducing the radius by a half increases the angular velocity by a factor of 4.

    What else are you trying to calculate? The work done? The new Kenetic Energy?
  8. Dec 1, 2008 #7
    The formula for kinetic energy is: K=1/2*I*w^2. This yields a kinetic energy that is 4 times the the original. In the series of equations to get this result, the only gray area where additional work could be applied is in the simple assumption that r suddenly becomes r/2. But my problem is trying to understand how r to r/2 makes energy increase 4 times. I've attempted to use centripetal force to determine the work required to move the mass from r to r/2 (F+dt), but this work turns out substantially exceeding the kinetic energy gain.
  9. Dec 1, 2008 #8
    The difference between the new and old Kinetic Energies must be exactly the work done. It would be hard to calculate the work directly. Remember, as you pull the puck in, the rotation increases but the centrifugal force goes down. So FORCE times the increment of DISTANCE gets smaller as the radius gets smaller. I could try to spend some time coming up with a formula if you really want corroboration. You really want to nail this down don't you.
  10. Dec 1, 2008 #9
    My calculus is more than a little rusty. I calculated the centripetal force for r and for r/2 using the respective velocities and then used an average (Fe+Fi/2)+Fi times distance. The answer yields work that is about 1.6 times higher than the delta kinetic energy. I'm sure the average is not going to get the job done.

    I really appreciate your feedback. I can definitely use help. As it turns out, I actually need to produce this apparatus and predicting forces and work involved is pretty important. Thanks again
  11. Dec 1, 2008 #10
    One thing I failed to mention: since velocity increases by 4 times (assuming r to r/2), and force is proportional to v^2, my numbers show centrifugal force increasing rather substantially.
  12. Dec 2, 2008 #11
    You are right. I jumped too fast on that one. Here is a link to a web page that does the Work calculation using Integration.

  13. Dec 2, 2008 #12
    I believe you have it. Thank you so much. I shutter to think how long it woulkd have taken me to get there (or if). If I reduce this to pratice may I email you with a status?
  14. Dec 2, 2008 #13
    I'm trying to keep my emails to a minimum. But I think you could post an update anytime on this thread. I would be interested in reading about your progress.
  15. Dec 2, 2008 #14
    Mike - the result from your integration looks good but I have been trying to reconcile one aspect that has me baffled. You may have explained it in your statement regarding the refernce frame, but it still confuses me:

    Your equation for centrifugal force is: F=m*w^2*r but it is normally F = m*w^2/r
    It seems that force would in any case be inversely proportional to r. Can this be right?
  16. Dec 2, 2008 #15
    You are thinking radial acceleration for a rotating object is "v^2/r".

    Yes but, when you use angular velocity its "omega squared times r".
  17. Dec 3, 2008 #16
    Thanks - I got it. I see why
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