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Pendulum on a cart.

  1. Feb 9, 2009 #1
    Please help; I think my professor might be wrong,

    A cart travels toward an inelastic barrier at a constant speed v. On the cart is a pendulum that is not oscillating before the collision. (It is hanging straight down and traveling at the same speed as the cart.)

    The cart then collides (perfectly inelastically) with the barrier and comes to a complete stop instantaneously. The pendulum support is fixed to the cart and comes to a complete stop, but the pendulum is free to swing.

    Question: What is the maximum angle through which the pendulum can swing for an arbitrary v?

    The puzzle:
    My professor contends that there is a maximum angle, and I contend that there is not (the pendulum will swing in circles for a high enough v.)

    If you think my professor is wrong, I agree, but that's not the question. If you think I am wrong, please be specific about the physics. The question is what train of thought (or physical approximation) would lead to his conclusion that there is a maximum angle to which the pendulum can swing? I cannot ask the professor, as he is trying so hard to be coy.

    Thanks.
     
  2. jcsd
  3. Feb 10, 2009 #2
    I'm with you.

    Assuming the usual ideal conditions for these kinds of pendulum questions (including the use of a rigid massless rod), the total energy of the pendulum's mass at the instant of collision is E=mv^2/2. Of course, v is the velocity of the cart/pendulum at impact and m is the mass of the bob. If the length of the rod is L then the maximum gravitational potential energy V that can be realised is 2mgL. If E<V the pendulum will swing up to some angle less than pi and proceed to swing back and forth. If E=V the bob will swing up through an angle of pi and come to rest balanced directly above the pivot point. If E>V the mass will rotate about the pivot point. [Unless I'm missing something here...]
     
  4. Feb 14, 2009 #3
    The point is, you can calculate a maximum angle. For a high enough v it will go over, but you don't know beforehand if it's a high enough v. So you should find the formula for the angle from energy conservation. I suppose that's all the prof wants.
     
  5. Feb 14, 2009 #4

    arildno

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    Your professor is dead wrong.
    If the parameter [tex]\frac{v^{2}}{2gl}\geq{1}[/tex], the pendulum will swing in circles.
    (g is the acceleration due to gravity, l the length of the pendulum, v being the initial velocity)
     
  6. Feb 14, 2009 #5

    Doc Al

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    I don't see any implication in the professor's question that he thinks the pendulum cannot swing in a circle if v is large enough. He merely asks you to do the calculation for arbitrary v (note that v may or may not be large enough to swing over the top). He's being "coy" because he probably wants you to realize on your own that for speeds above some minimum the pendulum will make a complete swing. Specify that minimum speed and, for speeds below that limit, specify the maximum angle.
     
  7. Feb 14, 2009 #6

    Doc Al

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    How did you arrive at that relationship?
     
  8. Feb 14, 2009 #7

    arildno

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    Well, I made a quick idealization:
    After the collision (conceived as instantaneous), the fulcrum is at rest. Since the collision was perceived as instantaneous, no change in the position of the (mathematical) pendulum has occurred, i.e, it is hanging straight down.
    Therefore, the bob of mass m
    has retained its velocity v, since it has experienced no horizontal force during the collision.
    (The pendulum string could, and have, only have transmitted forces in the vertical direction during the collision).


    The relation follows easily from that, by conservation of mechanical energy, and the requirement that the value of, say, the cosine function, must be less than 1.
     
  9. Feb 14, 2009 #8

    Doc Al

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    Looks to me like you set the initial KE equal to just enough for the pendulum bob to reach the height of the support, not necessarily enough to swing through a complete circle.
     
  10. Feb 14, 2009 #9

    arildno

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    Oh dear, you are absolutely right. My mistake.

    The critical condition should have an extra factor of 2 in its denominator.
     
  11. Feb 14, 2009 #10
    ragidandy:
    Your prof's use of the term "maximum angle" probably just refers to the fact that a real pendulum would not swing quite so much, so your calc would be a maximum possible angle for a given v.
     
  12. Feb 16, 2009 #11
    Thank you all.

    I think you must be right in that my prof. was looking for the angle as a function of v when v is below the critical value. Such trouble I make for myself!
     
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