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Find centre of mass

  1. Jan 29, 2017 #1

    Kajan thana

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    1. The problem statement, all variables and given/known data
    The lamina is free to rotate about a fixed smooth horizontal axis, perpendicular to the plane of the lamina, passing through the point A. and hangs in equilibrium.
    Find the angle between AC and the horizontal.

    I have attached a the solution to the question, it seems like there is no line to the question and theta there are showing in the working out.

    2. Relevant equations
    I think the answer is 9.46 degree as tanφ= 1/6

    3. The attempt at a solution.
    I have attached screenshot to this thread. Is the solution wrong or have I been thinking in a wrong way? What are looking at that theta shown in the screenshot?
     

    Attached Files:

    Last edited by a moderator: Jan 29, 2017
  2. jcsd
  3. Jan 29, 2017 #2

    Nidum

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    More like thinking upside down . If you were holding the lamina in the position shown what would happen when you let go of it ?
     
  4. Jan 29, 2017 #3

    Kajan thana

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    lamina will turn so the centre of mass is below the pivot.
     
  5. Jan 29, 2017 #4

    Nidum

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    That's it .
     
  6. Jan 29, 2017 #5

    Kajan thana

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    yes but my question why have they goy theata there when they ask for the theta between the horizontal and AC
     
    Last edited: Jan 29, 2017
  7. Jan 29, 2017 #6

    Kajan thana

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    i thought about it. It makes sense what they have done.
     
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