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Angular acceleration caused by mass over a pulley

  1. Mar 30, 2013 #1
    icmurt.png

    The method I'm using in the image is the one that makes the most sense to me, but I've tried others and still cannot get at the stated answer, which is 2.65 rad/s^2
     
  2. jcsd
  3. Mar 30, 2013 #2

    Doc Al

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    Do not assume that the tension in the rope equals the weight of the hanging mass. Set up two equations, one for the post and one for the hanging mass. Then you can solve for the acceleration.
     
  4. Mar 30, 2013 #3
    Sorry, how do I go about that? If the tension in the string isn't 5g, I don't know how to do it.
     
  5. Mar 30, 2013 #4

    Doc Al

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    Don't guess, solve! Call the tension T and set up your equations. That will be one of the unknowns. Luckily, you'll have two equations and two unknowns.
     
  6. Mar 30, 2013 #5
    I'm not even guessing; I just don't know how to set up the equations. T - 5g = 5a, I think? Then I am 100% lost at the post.
     
  7. Mar 30, 2013 #6

    Doc Al

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    Good! Since the mass falls, I would choose positive as down and let "a" stand for the magnitude of the acceleration.

    Now write an equation for the post.
     
  8. Mar 30, 2013 #7
    I'm not sure how I can be any more direct: I am 100% lost at the post. It's not for lack of thinking, I just don't know what the equation ought to be?
     
  9. Mar 30, 2013 #8

    Doc Al

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    You already set up the equation for the post in your first post! Your mistake was assuming that the tension was equal to mg. Set the unknown tension equal to T and write out that equation.
     
  10. Mar 30, 2013 #9
    If you mean [itex]rF=I/alpha[/itex], then isn't that a different acceleration to the acceleration for the block?
     
    Last edited: Mar 30, 2013
  11. Mar 30, 2013 #10

    Doc Al

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    Yes, that's what I mean. Almost. It should be rF = I*alpha.

    Yes, one is an angular acceleration and the other is a linear acceleration. How are they related? (The rope is attached to the post.)
     
  12. Mar 30, 2013 #11
    Yeah, that's what I meant - LaTeX troubles :P

    Well, regardless, I still get the wrong answer:

    5jW8JuM.png

    And I checked, in case I mixed my signs up, adding 6.25 rather than subtracting, which gives 2.84, still not the correct answer.
     
  13. Mar 30, 2013 #12

    Doc Al

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    You need to fix the sign of "a" in your equation for the hanging mass.

    And how does alpha relate to "a"?
     
  14. Mar 30, 2013 #13
    Which doing will give me 2.84 as the final answer, which is still incorrect? =/
     
  15. Mar 30, 2013 #14

    Doc Al

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    Write your equations. Tell me how alpha relates to "a".

    Don't solve them, just write them.

    (I solved it and got the correct answer, so we need to see where you are going wrong.)
     
  16. Mar 30, 2013 #15
    I don't know how it relates it general, but at that instant, they are the same?

    5g - T = 5a
    [itex]\tau = rF=r(5g-5a) = 61.3 - 6.25a = Ia, a(I+6.25) = 61.3[/itex]
    [itex]\frac{61.3}{\frac{1}{3}*15*1.75^2 + 6.25} = a[/itex]
    a = 2.84
     
  17. Mar 30, 2013 #16

    Doc Al

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    How can an angular acceleration (in rad/s^2) be the same as a linear acceleration (in m/s^2)?

    But they are related. Consider the point of attachment of the rope with the post. How does the linear acceleration of that point relate to the angular acceleration of the post?

    You're almost there.
     
  18. Mar 30, 2013 #17
    Ok, [itex]a=r\alpha[/itex]

    Solving from there is just basic algebra. Prior to googling, I had not seen this, nor thought about a concrete relationship between them, which must've been why I could not see how to do the problem. Thank you for putting up with me :P
     
  19. Mar 30, 2013 #18

    Doc Al

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    Now you've got it. :approve: You should be able to the the correct answer now.

    And you're welcome.
     
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