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Falling Plate

  1. Aug 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform square plate ABCD has mass 0.8 kg and side length .8 m. The square is pivoted at vertex D and initially held at rest so that sides AB and CD are horizontal (see the diagram). After it is released, the plate swings downward, rotating about the pivot point. Resistive force can be neglected. The acceleration due to gravity is g = 9.8 m/s2. The rotational inertia of a square plate of side d relative to the axis perpendicular to the plate and passing through the center of mass is md^2/6.

    Find the rotational inertia of the plate relative to the axis of rotation.
    Find the angular speed of the plate at the moment when BD is horizontal.
    Find the linear speed of B at the moment when BD is horizontal.
    Find the linear speed of B at the moment when B is at the bottom position.

    2. Relevant equations

    3. The attempt at a solution
    I used the parallel axis theorem to find the moment of inertia around the axis of rotation. I got
    1/6(.8)(1.2)^2+.8(sqrt(2)/2)^2=.592 but that isn't correct.
     

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    Last edited: Aug 10, 2013
  2. jcsd
  3. Aug 10, 2013 #2

    TSny

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    What does the quantity √(2)/2 represent?
     
  4. Aug 10, 2013 #3
    never mind
     
    Last edited: Aug 10, 2013
  5. Aug 10, 2013 #4
    √(2)/2 represents the distance from point D to the center of mass.
     
  6. Aug 10, 2013 #5

    TSny

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    Shouldn't the 1.2 m come into play here?
     
  7. Aug 10, 2013 #6
    Oh ok, that makes sense. I got .768, is that right?
     
  8. Aug 10, 2013 #7

    TSny

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    I believe that's correct. (Add units of course.)
     
  9. Aug 10, 2013 #8
    Ok, then how do I start solving part B?
     
  10. Aug 10, 2013 #9

    TSny

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    Is anything conserved while the plate swings down?
     
  11. Aug 10, 2013 #10
    Energy,right?
     
  12. Aug 10, 2013 #11

    TSny

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    Right. Give it a shot.
     
  13. Aug 10, 2013 #12
    Ok I got 4.937. Is that right?
     
  14. Aug 10, 2013 #13

    TSny

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    I don't know, I haven't worked it out. Please show your work and we can see if you are setting it up correctly.
     
  15. Aug 10, 2013 #14
    m*g*h=.5*I*w^2

    m*g*h=.5*.768*m*r^2*(v/r)^2

    g*h=.384*v^2

    9.8*.6=.384*v^2

    v=3.91 (not 4.937 due to arithmetic error)
     
  16. Aug 10, 2013 #15

    TSny

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    Your set up looks good!

    But note that in the second question, you are asked to find the angular speed rather than the linear speed.

    Also, I don't agree with the value that you used for h in the calculation. How far does the center of mass of the square move downward for this question?
     
  17. Aug 10, 2013 #16
    I still think that its .6. Can you tell me why I'm wrong?
     
  18. Aug 10, 2013 #17

    TSny

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    I'm a little confused about the value of the side length of the square plate. I thought it was 1.2 m, but now when I look at your original post, it appears to be 0.8 m.

    But either way, h would not be 0.6 m for the second question. Did you draw a picture showing the orientation of the square at the moment of release and then another picture for when BD is horizontal? Be sure to mark the center of mass in each case.

    Since a diagram wasn't included, I am interpreting the geometry as best I can from the wording of the problem.
     
    Last edited: Aug 10, 2013
  19. Aug 10, 2013 #18

    TSny

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    OK, now I see the picture. I had misinterpreted it. I thought that initially point B was directly above D so that BD would be a side of the square. But, BD is a diagonal of the square. So, if the length of a side is 1.2 m, then I agree that h would be 0.6 m for the second question. Sorry about that.
     
  20. Aug 10, 2013 #19
    For parts 3 and 4 is h= to .8 and .8(2+sqrt(2)/2) respectively?
     
  21. Aug 10, 2013 #20

    TSny

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    h for part 3 should be the same as for part 2.

    Can we clarify if the length of the side of the square is 1.2 m or is it 0.8 m?
     
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