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I Angular velocity of falling box

  1. Mar 7, 2016 #1
    Hi,

    I'm preparing a computer software to simulate the fall of an object for an academic project. The object is rotating (not rolling) over the circular bottom point. I know that the angular velocity at the horizontal point is ω = √(3g/L). I would like to calculate the angular velocity at every angle (from 90° to 180°). Can you please provide a simple formula for this?

    Thanks
     

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  2. jcsd
  3. Mar 7, 2016 #2

    mfb

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    It is your project.
    Energy conservation should work. For every rotation angle, you can calculate the potential energy, the difference to the original value got converted to kinetic energy.
     
  4. Mar 7, 2016 #3
    So, you are proposing the following:

    The difference in PE is equal to the KE, that is ΔPE = KE.
    Ex: PE1 - PE2 = ½ m⋅v2
    Then I solve for v, but this v is the vector from the COM to the ground. Will then have to calculate ω.
    Or else, use the angular KE?
    Is that right?
     
    Last edited: Mar 7, 2016
  5. Mar 7, 2016 #4

    mfb

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    The kinetic energy of a rotating object is not 1/2 m v^2. You'll have to find the center of rotation and the moment of inertia around it, or sum rotational energy (around a different point) plus lateral motion.
     
  6. Mar 7, 2016 #5
    Yes, my mistake, KE = ½⋅I⋅ω2
    May I ask what you mean by the centre of rotation? Is this the pivot point on the ground?
     
  7. Mar 7, 2016 #6

    mfb

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    If you measure I around a point that does not move, yes. But then I depends on the angle if you want to be precise.
    Yes.
     
  8. Mar 7, 2016 #7
    Is this because of the opposite forces of the rotating body towards the rotation point? Can you clarify this please?
     
  9. Mar 7, 2016 #8

    mfb

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    The point that has contact to the surface changes over time.
     
  10. Mar 10, 2016 #9
    Hi again,
    I did the calculations on my software, but the graph of ω is not what I expect. See series 1 on attached image (x axis =frame no, y axis = ω).
    It should show a slower increase in acceleration at the beginning of the fall. Any ideas?
     

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  11. Mar 10, 2016 #10

    mfb

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    Where is series 1?
    Yes, acceleration should increase over time.
     
  12. Mar 10, 2016 #11
    Yes, series 1, but it shouldn't accelerate at this step. It should look like the attached image. Notice that for the first frames/angles the velocity is not as accelerated as on the later frames/angles. I think something is missing here...
     

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  13. Mar 10, 2016 #12

    mfb

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    I don't see series 1 there. Just 2, 3 and 4.
     
  14. Mar 10, 2016 #13
    Is only series 1 in the graph at post #11
     
  15. Mar 10, 2016 #14

    davenn

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    you are still not understanding mfb's comment

    there is NO series 1 plotted in your graph, only series 2, 3, and 4

    so where is the series 1 data ?


    D
     
  16. Mar 10, 2016 #15
    There IS one series in ejs-graph-jpg
     
  17. Mar 11, 2016 #16

    mfb

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    Yes, but that graph shows the expected result. So where is the problem? The attachment of post 9, where you discussed the problem with "series 1", does not have a series 1.
     
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