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Pivoted bar swinging top to bottom

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform bar of mass and length is pivoted at one end and is held vertically as shown. After the bar is released, it swings downward with no friction in the pivot.

    At the instant the bar is at the bottom of the swing, find the magnitudes of the following quantities:

    The angular speed of the bar

    The angular acceleration of the bar

    The linear speed of the free end of the bar

    The horizontal component of the acceleration of the free end of the bar

    The vertical component of the acceleration of the free end of the bar

    The linear speed of the midpoint of the bar

    The horizontal component of the acceleration of the midpoint of the bar

    The vertical component of the acceleration of the midpoint of the bar

    The horizontal component of the force exerted on the bar by the hinge

    The vertical component of the force exerted on the bar by the hinge

    2. Relevant equations



    3. The attempt at a solution

    For the first part I found the MoI and got 1/3ML^2 through parallel axis, then used conservation of energy : mgh=.5 I w*2 and got sqrt(3*g/L) (used L/2 for h), but it's wrong.
     

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    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2

    Delphi51

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    I got that, too! But I think we have the change in height wrong. The center of mass falls more than L/2; it swings down below the pivot point.
     
  4. Aug 11, 2013 #3
    Then what is the change in height?
     
  5. Aug 11, 2013 #4

    Delphi51

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    Initially the center of mass of the swinging bar is L/2 above the pivot.
    After the swing it is L/2 below the pivot.
    At least that is how it reads to me . . . it actually doesn't say what "at the bottom of the swing" means. The arc shown in the diagram suggests it only goes down to the horizontal orientation. Worth a try to see if a change in height of L produces the "right" answer.
     
  6. Aug 11, 2013 #5
    You're right. To find the angular acceleration can I use the formula alpha=T/I, and if so does T=Lg?
     
  7. Aug 11, 2013 #6

    Delphi51

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    T = torque. Yes, that should work. Note that the torque varies with position of the rod as it falls. In fact, the torque appears to be zero initially so I wonder why it falls! Anyway, you will want to sketch a diagram of it in the bottom of the swing position showing the force that causes the torque and its direction so you can see what torque there is.
     
    Last edited: Aug 11, 2013
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