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Change of the angular velocity vector

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data
    I don't know how to solve part(b).
    See the photo Q3 for the question.


    2. Relevant equations



    3. The attempt at a solution
    See the photo 2010Q3

    I found the torque and hence the new angular momentum vector. But I wonder how could this help me to find the angle between the new and old angular velocity vector.
     

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  2. jcsd
  3. Dec 14, 2012 #2

    TSny

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    Your result for the moment of inertia tensor looks good to me.

    In your expression for the torque, I don't see where the factor of 1/2 comes from when writing R/2. [EDIT: Otherwise, I think your expression for the final angular momentum is correct].

    Are you familiar with Euler's equations for rigid body rotation? I think they will lead to the answer fairly quickly. If not, then you will need to relate the final angular momentum to the final angular velocity components using the (new) final components of the moment of inertia tensor in the fixed inertial frame.
     
    Last edited: Dec 14, 2012
  4. Dec 14, 2012 #3
    Ha...That was a stupid mistake. That should be R not R/2.
     
  5. Dec 14, 2012 #4
    I have learnt Euler equation. But I am not sure what values to be put for those ω.
    See my photo attached below
     

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  6. Dec 14, 2012 #5

    TSny

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    Maybe a sign error for the last term of your first equation.

    Use these equations at time t = 0 to find the rate of change of each component of ω at time t = 0.
     
  7. Dec 14, 2012 #6
    I used both method to solve this problem.
    But I discovered that the new angular velocitys (only the x component) obtained from 2 methods are not the same.
    Which version is wrong?
     

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  8. Dec 14, 2012 #7

    TSny

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    First version looks good.

    For the second version, note that ##\tau = \dot L## holds only in an inertial frame. At the end of the small time interval the point mass will have rotated out of the yz plane in the fixed inertial frame. So, the moment of inertia tensor will no longer be diagonal in the inertial frame.
     
  9. Dec 14, 2012 #8
    Thank you very much!
     
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