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Pivoting Rod

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform rod is pivoted at its center and a small weight of mass M = 5.07 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod.


    2. Relevant equations
    Now suppose that the rod has length L = 5.4 m and mass mrod = 10.1 kg. Suppose also that there are no external forces applied (i.e. Fh = Fv = 0). The system is released from rest at the q = 37° angle. What is the angular acceleration just after it is released?
    angular acceleration:1.7437 rad/s^2
    What is the angular velocity when the rod is vertical?

    3. The attempt at a solution
    mgh=.5Iw^2
    where h=sin 53*L/2=2.156
    w=sqrt(2mgh/I)
    this is not the correct answer. Can somebody help me. I dont know what im doing wrong.
     
  2. jcsd
  3. Mar 2, 2009 #2

    Doc Al

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    Staff: Mentor

    OK. (What's I? What's m?)
    What's h the change in height of?
     
  4. Mar 2, 2009 #3
    sorry i dont quite understand what you're asking, I=moment of inertia m=mass w=omega
    the potential energy is mgh and kinetic energy is .5Iw^2
     
  5. Mar 2, 2009 #4

    Doc Al

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    I meant: I = moment of inertia of what? m = mass of what? h = change in height of what?
     
  6. Mar 2, 2009 #5
    I=moment of inertia of rod-weight system=61.50
    m=10.1 kg
    h=change in height of the rod from 37 degrees to 90 degrees
     
  7. Mar 2, 2009 #6

    Doc Al

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    OK.
    That's just the mass of the rod.
    What part of the rod? What's the change in height as it moves from one position to the other?

    How does the gravitational PE of the entire system change? What's the change in PE of the rod? Of the attached weight?
     
  8. Mar 2, 2009 #7
    The part of the rod thats attached to the weight. The change in height is the point from which theta=37 degrees to the height when the rod is vertical at 90 degrees. The potential energy of the rod does not change because the pivot point is the center of mass?
     
  9. Mar 2, 2009 #8

    Doc Al

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    All good. Redo your calculation of the change in height of the attached weight. (It's not L/2*sin(53).)
     
  10. Mar 2, 2009 #9
    im having trouble redoing the calculation for the change in height. is the sin 53 part correct? because if you want the rod to be vertical, the tip of the rod with the weight will be pointing down, thus 90-37=53 degrees?
     
  11. Mar 3, 2009 #10

    Doc Al

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    Do this: Find the initial position of the tip of the rod (measured from the pivot point). Find the final position. Compare.
     
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