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Deriving an equation of motion

  1. Jan 24, 2005 #1
    a uniform thin rod of length L and mass m is pivoted at one end the point is attached to the top of a car accelerating at a rate A.

    a) what is the equilibrium angle between the rod and the top of the car?
    b) suppose that the rod is displaced a small angle phi from the equilibrium derive the equation of motion for phi. Is the equilibrium angle stable or unstable?

    I was able to get part a wich using newtons 2nd law for non inertial frames is
    tan(equilibrium angle) = g/A.

    but i am stuck on part b. I could use the lagrangian method to get phi(double dot) but that would be really messy, and was wondering if there would be a better way of getting it. oh and also he gave the hint to ignore air resistance and that torque= I*alpha.

  2. jcsd
  3. Jan 24, 2005 #2


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    Getting the equation isn't hard but solving it might be...

    I'm going to take theta to be the total angle from the horizontal, and phi as the angle from the equilibrium point.

    Calculate the torque acting on the rod... this is in terms of theta. Then write down I * alpha = torque.

    Plug in the value of I in terms of m and L. Plug in [tex]\alpha=\frac{d^2\theta}{dt^2}[/tex]

    Then finally replace [tex]\theta[/tex] with [tex]\phi + tan^{-1}(g/A)[/tex]

    That seems to give you the equation you need. It's a second order diff. equation... I'm not sure of the solution right now. But do you need the solution?
  4. Jan 24, 2005 #3
    thanks for the help, it seems i forgot that alpha= phi(double dot), as soon i understood that the rest of the problem wasnt too hard.
  5. Jan 24, 2005 #4
    The differential equation may be simplified for small [tex]\phi[/tex], since

    [tex]\sin\phi = \phi[/tex] and

    [tex]\cos\phi = 1[/tex] (approximately).

    I don't know how to write ~ in latex.
  6. Jan 24, 2005 #5


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    I'm curious about the final solution to this problem... Did you have to solve the diff. eq? If you did, can you post it?

    It seems stable to me because the second derivative of phi, is zero at the equilibrium point... but I'm not sure if this is the correct reasoning.
  7. Jan 24, 2005 #6


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    This what you're looking for?

  8. Jan 25, 2005 #7
    Thank you Hurkyl, that's what I was looking for.
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