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Angular Momentum Parking Gate

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data



    x2bfH7J.png
    2. Relevant equations

    Torque = r X F

    3. The attempt at a solution

    r is 2.5m since that's the length of the red (minus 0.5), and F should be the weight, so ma? That gives something like (5 * 2.5/3) * 9.8 * 2.5, which doesn't give me an answer at all :(
     
  2. jcsd
  3. Apr 16, 2015 #2

    haruspex

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    You've included the weight of the short end, but as though it is at the mass centre of the long end.
     
  4. Apr 16, 2015 #3
    Do I need to calculate them individually? The correct answer here is 49, and I think my approach as a whole is wrong.
     
  5. Apr 16, 2015 #4

    haruspex

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    Yes, your error is more fundamental, I realised just after posting. You seem to be calculating moment of inertia, which is not what is asked for.
     
  6. Apr 16, 2015 #5
    Can you point out where? I thought I was calculating torque.
     
  7. Apr 16, 2015 #6

    haruspex

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    You squared the distance and divided by 3. That happens the formula for the moment of inertia of a bar, but it is not in the (correct) equation you quoted.
     
  8. Apr 16, 2015 #7
    I'm sorry, I don't understand. Are you saying I SHOULD be using inertia of a bar, I = 1/3 M r^2 ? Because right now I am literally just doing f = ma.
    And if I need to use inertia, should I use the torque = Inertia * angular acceleration equation?
     
  9. Apr 16, 2015 #8

    haruspex

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    No, I said the equation you quoted is correct:
    But that's not what you did here:
    Yiu have two factors of 2.5 and a divisor of 3. That looks like a moment of inertia calculation to me.
     
  10. Apr 16, 2015 #9
    Oh!

    It wasn't, that was a straight up r x F calculation. See, I was confused by the part where the bar is attached "0.5m from the end". So I figured that the actual mass would not be 5kg, but rather 5 * (2.5/3), since that would be the fraction of the bar that I'm actually looking at.
     
  11. Apr 16, 2015 #10

    haruspex

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    Ok. In that case, your error is that you have treated the short end as simply not being there. It will produce a counter torque.
    The easiest way is just to treat all the mass as being at the mass centre (which is valid for finding the effective weight).
     
  12. Apr 16, 2015 #11
    I see, how clever!

    The mass center would be 1m away from the pole-thingy.

    So torque = 1 * 5 * 9.8

    = 49.

    Thanks so much!
     
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