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Homework Help: Rotational Kinematics - Torque

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data

    https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/spring09/homework/09/pivoting_stick/5.gif [Broken]
    A stick of uniform density with mass M = 8.7 kg and length L = 0.6 m is pivoted about an axle which is perpendicular to its length and located 0.14 m from one end. Ignore any friction between the stick and the axle.
    The stick is held horizontal and then released.
    What is its angular acceleration as it passes through the vertical position?

    2. Relevant equations
    Torque = I*(angular acceleration)

    Torque = r X F



    3. The attempt at a solution

    I need to find the net torque of the system to solve this problem.
    But when i draw a free-body diagram, i can't find the tangential force to the stick which should be used in the torque equation.(I find only horizontal forces which make no sense...)

    Please help me out here.....

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 9, 2010 #2

    Doc Al

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    Staff: Mentor

    What forces act on the stick? It seems you are leaving out an obvious one.
     
  4. Jan 9, 2010 #3
    You mean when it's vertical?
    I think you meant the "gravity" i guess..?
    Otherwise,Could you explain..?
     
  5. Jan 9, 2010 #4

    Doc Al

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    Staff: Mentor

    Yes, gravity is what I was thinking of. But I misread the problem... I thought it was let go from the vertical position. D'oh!

    In any case, do any horizontal forces act on the stick (forces that can exert a torque, that is) when it is vertical? What's the torque on the stick at that point?
     
  6. Jan 9, 2010 #5
    No horizontal forces when it's vertical, i suppose?

    Then there is no torque at that point, hence, angular acceleration is zero..

    OMG....... That's the answer.....
    I thought this was a tricky question.. never considered this would be the answer...


    Thanks a lot !!!!
     
  7. Jan 9, 2010 #6
    Can i ask a tiny additional question ?
    The question is
    "What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical? "

    And i have got all the numerical things.
    I thought the vertical component should be "ma - mg" at vertical position. (here, 'a' is centripetal acceleration)

    But it turns out that it should be "ma + mg" must be the vertical component..

    I think centripetal acceleration points to the pivot(+y direction) and mg points down(-y direction) at vertical position..
    Then, how come the ADDITION happens here...?


    Please help me out here.....

    Thanks a lot.
     
  8. Jan 9, 2010 #7

    Doc Al

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    Staff: Mentor

    Right.
    Just apply Newton's 2nd law.

    In general: ΣF = ma

    Applying that to this particular problem: Fp - mg = +ma
    Thus: Fp = mg + ma

    Make sense?
     
  9. Jan 9, 2010 #8
    It totally makes sense...!!
    Thanks a lot !!

    God bless you !
     
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