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Rotational Mechanics - Question

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of length "L" is kept vertically on a smooth horizontal surface at a point "O". If it is rotated slightly and released, it falls down on the horizontal surface. To what distance will the lower end of the rod "shift" from point "O" ?

    Note : The answer in the book is given terms of "L"..

    2. Relevant equations

    Formulas of rotations. Check the wikipedia link.


    3. The attempt at a solution

    I can not think of any way. I tried integrating Lcosθ/2 from 0 to pi/2. I am not sure of any logic. How will the lower point shift if the rod is doing pure rotation ? Any hints will do.

    Please help !!

    Thanks in advance...

    :smile:
     
    Last edited: Apr 10, 2013
  2. jcsd
  3. Apr 10, 2013 #2

    Doc Al

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    Consider the forces acting on the rod.
     
  4. Apr 10, 2013 #3

    arildno

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    What external forces, if any, act upon the rod in the HORIZONTAL direction?
     
  5. Apr 10, 2013 #4

    arildno

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    Blaargh, you are so fast, Doc Al.
     
  6. Apr 10, 2013 #5
    The force mg acting along the centre of mass perpendicular to axis of rotation.

    The normal reaction from the floor will have two components: Rsinθ and Rcosθ..

    Rcosθ will balance the weight and Rsinθ will cause the shifting of lower end, right ?

    I assume rod is slightly rotated by angle θ.

    So what should I do after this ?
     
  7. Apr 10, 2013 #6

    arildno

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    The surface is..SMOOTH.
    What does that condition entail?
     
  8. Apr 10, 2013 #7
    It entails that there will be no friction acting on the rod's lower end by the floor. But how will that help me ? (Thanks BTW..)
     
  9. Apr 10, 2013 #8

    Doc Al

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    Apply Newton's 2nd law. What will be the translational acceleration of the rod? (In what direction?)
     
  10. Apr 10, 2013 #9
    Rsinθ=ma , right ?

    R is the reaction force from the floor to the lower end of the rod.
     
  11. Apr 10, 2013 #10

    arildno

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    So, whatever force acts at the contact point, will it be:
    a) Strictly horizontal
    b) Strictly vertical
    c) A mixture?
     
  12. Apr 10, 2013 #11
    A mixture, right ? I already mentioned that in my previous post. I mentioned that normal reaction at contact point will have two components, horizontal which balances the weight and vertical component which causes the acceleration in lower point.
     
  13. Apr 10, 2013 #12

    Doc Al

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    What's the definition of "normal"? And how can a horizontal force balance a weight, which is a vertical force?
     
  14. Apr 10, 2013 #13
    I am not saying that horizontal force balances the weight. If I rotate the rod slightly by angle θ, then by geometry rod will make angle (90-θ) with the floor. Now, horizontal component of normal reaction, Rcos(90-θ)= Rsinθ causes acceleration, and vertical component Rsin(90-θ)=Rcosθ balances the weight of the rod. Yes now that's correct. I mistyped in my previous post. Sorry about that !!

    So now, what can I do after this ?
     
  15. Apr 10, 2013 #14

    Doc Al

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    You seem to think that the normal force of the floor on the rod will be parallel to the rod. No.

    The problem is much easier than you think. Almost a trick question.
     
  16. Apr 11, 2013 #15
    I am sorry if I misunderstood somewhere.
    The normal reaction will be "just" perpendicular to the plane of floor, right ? And it will not have components right ? Then how will there be horizontal acceleration on the lower point of rod ?
     
  17. Apr 11, 2013 #16

    Doc Al

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    Right!

    Right!

    If you divide the rod into pieces, then one piece can certainly exert a horizontal force on the other. But no need to do that. Treat the rod as a whole; no external horizontal forces act on the rod.

    Hint: The net force determines the acceleration of what part of the rod?
     
  18. Apr 11, 2013 #17
    Sorry, but net external force on the rod seems to be producing a "couple" on the rod when the rod gets slightly rotated and had unstable equilibrium, and nothing else.
     
  19. Apr 11, 2013 #18

    Doc Al

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    So you think the net external force is zero?
     
  20. Apr 11, 2013 #19

    arildno

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    The normal force to a surface is simply so large that the object the surface is in contact with does not pentrate into that surface.

    Thus, the normal force is NOT, necessarily, equal to the normally directed component of gravity.
     
  21. Apr 11, 2013 #20
    If Normal Reaction were not equal to force of gravity on centre of mass of the rod, there will be either net force up or net force down. In former case, rod will jump up and in latter case, it will penetrate in the floor. That is why it has to equal force of gravity.

    Doc Al, I can not see any net external force on the rod acting. I can only see net external torque.
     
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