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Equations of motion of a pendulum in a cart.

  1. Dec 31, 2007 #1
    I am looking over an old problem about a cart that has a pendulum on it, and you are supposed to find the equations of motion. The pendulum is made from a mass and a wire. Because wire can only support tensile loads, the tension has to be directed along the direction of the wire. Any force not directed along the axis of the wire will cause it to buckle.

    But consider I replace the wire with a rigid bar. So I now have a rigid pendulum. I want to say that the force in the bar will remain along the axis of the bar, but I cant for the life of me show why. Because it is now a bar, it has to be pin connected to the cart, which means it can only have a reaction force in the (x,y) direction at the pin. Any ideas?
  2. jcsd
  3. Dec 31, 2007 #2
    Consider a coordinate system attached to the point mass at the end of a massless rod of length L. There is no net motion in the axial direction of the rod. This means the net acceleration in the axial direction is zero, and the reaction force on the pin in the axial direction is equal to [tex] T= m (g\cdot sin \theta + \frac{ V^2}{L}) [/tex].

    Now two possible solutions exist for the reaction on the pin in the [tex]e_\theta[/tex] direction. (a) There is no reaction force, and so the component of weight acting in this direction accelerates the mass downard. (b) there is an equal and opposite reaction force of the component of weight in this direction, and as a result it creates a couple that causes pure rotation around the pin axis. Right now, I dont see why this is an invalid solution. It should produce the same effect, no?
    Last edited: Dec 31, 2007
  4. Jan 2, 2008 #3
    Ah, I figured it out.


    Now, I want to figure out what the RxN forces are, and what the internal forces are for a rod of fixed mass m.

    The answer is, the reaction forces are exactly the same as a pendulum of wire or string. So it is not a couple moment of reaction forces at the pin, even though that makes senes conceptually.
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