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Classical Mechanics - Find angular velocity of two rods

  1. Jan 22, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello!
    I apologize for my poor setup, first post.
    I am given a system of two rods, hope you can see my image.http://file:///C:/Users/Mikkel/Downloads/Mek2_exam_Jan2016_final.pdf [Broken]
    One along the x-axis with mass = 2m and length = 2l
    Another perpendicular with the other with mass = m and length = l

    The system rotate around a point O

    Find the angular velocity at the point when the systems longest rod is vertical.

    2. Relevant equations

    I have calculated the center of mass rcm = ( (4/3)*l , (1/6)*l )

    and the distance from pivot point to the center of mass = (1/6) * sqrt(65) * l

    Also the moment of inertia for the system = 7*m*l2

    3. The attempt at a solution

    I have tried the following so far:

    I tried to use conservation of energy to solve it:

    m*g*h + 0 = 0 + (1/2)*I*w2

    I am thinking that I have a total of mass 3
    and the height has to be the x-coordinate for center of mass (I think this is where I might be wrong?)

    3*m*g*(4/3)*l = (1/2)*7*m*l2 * ω2

    after a bit of calculations....

    ω = sqrt((8*g)/(7*l))

    However my facit says ω = sqrt((9*g)/(7*l))

    I can't figure it out!

    Appreciate any tips ! :)
     

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    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Jan 22, 2017 #2

    haruspex

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    You have omitted a contribution to the change in height of the mass centre.
     
  4. Jan 22, 2017 #3
    Hmm I don't see how. I was thinking that, when the rod is vertical the height is zero.
     
  5. Jan 22, 2017 #4

    TSny

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    Draw a diagram and mark the location of the center of mass at the initial and final positions.
     
  6. Jan 23, 2017 #5
    Ah you are a god damn genius sir!

    So the center of mass changes to (3/4) * l when it is vertical and then it becomes (9g/7l)
    Thank you!
     
  7. Jan 23, 2017 #6

    haruspex

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    It is rather simpler if you do not bother finding the mass centre of the L shape. Just work out and sum the two moments of inertia, and likewise the two changes in PE.
     
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