1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular momentum

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data
    A flywheek with a cavity in its upper surface is constrained to rotate about a fixed vertical axis. Its moment of intertia about the axis is I and the cvity is a distance a from the axis. It is initially rotating with angular velocity w1. Into the cavity is dropped a sphere of mass M and moment of inertia i which is spinning with angular velocity w2 about a vertical axis. At the moment it is dropped into the cavity it has no horizontal motion. Derive an expression for the angular velocity of the composite system when the sphere and the flywheel are at relative rest.

    2. Relevant equations
    Parallel axis theorem.
    L=Iw

    3. The attempt at a solution
    I think I'm struggling with this problem because the sphere and the flywheel are rotating around different axis.

    The moment of inertia of the sphere about the flwheel's axis of rotation= i + Ma^2, so using the conservation of angular momentum if the sphere wasn't rotating initially the systems angular velocity would be straightforward to find.

    However the sphere is rotating, and also the flywheel is constrained so I'm not sure how this affects the system.

    A push in the right direction would be appreciated.
    Thanks
     
  2. jcsd
  3. Jan 5, 2007 #2

    radou

    User Avatar
    Homework Helper

    That's correct, unless I'm missing something huge here. The sphere's angular velocity relative to the flywheel's is given, you found the moment of inertia of the sphere relative to the flywheel's axis, and you should, as mentioned, use conservation of momentum now. Note that the flywheel's moment of inertia also changes.
     
  4. Jan 6, 2007 #3
     
  5. Jan 6, 2007 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, as the flywheel is constrained it will continue to rotate about its own axis. The sphere is spinning about it's own central axis and rotating about the same axis as the fly wheel.
     
  6. Jan 6, 2007 #5
    Does the phrase "the sphere and the flywheel are at relative rest" just mean the centre of mass of the sphere is at rest relative to the flywheel?

    Also I'm not completly sure what radou meant when he said " Note that the flywheel's moment of inertia also changes." Does this mean the moment of inertia of the system about the flywheels axis isn't I+i+Ma^2?

    Thanks for your help.
     
  7. Jan 6, 2007 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes
    I think what he means is you now have to include the sphere in your moment of inertia for the fly wheel. However, this
    is not quite correct (almost there). Note that the sphere is spinning about its own axis but is rotating about the fly wheel's axis (think of a planet orbiting the sun in a circular orbit). Note also, that the sphere need not be spinning with the same angular velocity as it is rotating about the fly wheel's axis. Does that make sense?
     
  8. Jan 6, 2007 #7
    I understand what is going on qualitatively, but I'm not sure how it quantativly changes the system.

    Sorry for being a bit slow.
     
  9. Jan 6, 2007 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No problem, that's what we're here for :smile: . Okay ignore the fact that there sphere is spinning for the moment. Now, if the sphere is just dropped into the fly wheel (and is not spinning), what is the new moment of inertia of the fly wheel and sphere system?
     
  10. Jan 6, 2007 #9
    So then total moment of inertia= I + i + Ma^2
    using the parallel axis theorem
     
  11. Jan 6, 2007 #10

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can treat the sphere as a point mass hence the new moment of inertia of the fly wheel and non-rotating sphere is I + Ma^2
     
  12. Jan 6, 2007 #11
    Oh ok, how would this change if the sphere is spinning?
    Thanks
     
  13. Jan 6, 2007 #12

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay so, now let's write an expression for the conservation of angular momentum;

    [tex]I\omega_{1} + i\omega_{2} = \left(I+Ma^2\right)\omega_{1'} + i\omega_{2'}[/tex]

    Now, after some consideration I believe that the phrase "the sphere and the flywheel are at relative rest" means the sphere stops spinning (my apologies :redface:), thus negated the final term on the RHS.
     
  14. Jan 6, 2007 #13
    So even though the angular momentum vectors before are through different points you can add them like scalars to give the total angular momentum (presumably because they are parallel) ?
    thanks
     
  15. Jan 6, 2007 #14

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, as I understand it.
     
  16. Jan 6, 2007 #15
    Thank you,
     
  17. Jan 6, 2007 #16

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Twas a pleasure :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Angular momentum
  1. Angular momentume (Replies: 7)

  2. Angular momentum (Replies: 1)

Loading...