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Equivalence of Lagrangians through modified g field.

  1. Jan 6, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Write down the Lagrangian of a simple pendulum in terms of it's angle θ to the vertical suspended from a pivot attached to a moving carriage at constant velocity ##v##. Suppose that the carriage is now moving at a velocity ##v(t)=at## so it is accelerating uniformly. Show that the Lagrangian is equivalent to that of a pendulum in a modified gravitational field and deduce the strength and direction of this field.

    2. Relevant equations
    The lagrangian for the pendulum with carriage at constant speed v (1)
    The lagrangian for the pendulum with carriage at variable speed v(t). (2)

    3. The attempt at a solution
    (1) is $$L = \frac{m}{2}\ell^2 \dot{\theta}^2 + mg\ell \cos \theta + \frac{m}{2}v^2 + mv\ell \cos \theta \dot{\theta}$$
    For purposes of finding the E.O.M the third term here can be neglected.

    (2) is $$L' = \frac{m}{2}\ell^2 \dot{\theta}^2 + mg\ell \cos \theta + \frac{m}{2}(at)^2 + m (at)\ell \cos \theta \dot{\theta}$$
    Since ##\theta## is the generalized coordinate, I think I can neglect the third term again. (It will vanish when I take derivatives), but I do not see why physically. The quantity ##at## will have a different value at each point in time so it affects the Lagrangian numerically. Is it because the 'information' that is useful is encoded in the functional form of the Lagrangian through it's dependence on ##\theta##, ##\dot{\theta}## and ##t##? So, in general, does that mean if there is a term in the Lagrangian that is independent of all the generalized coords/ velocities and time, we can neglect it?

    As for the question, there is a hint to use the derivative ##\frac{d}{dt} (t \sin \theta)## in the calculation and I can see where that is going, however, I was wondering if there was another way to solve this without requiring this 'trick'.

    Many thanks.
     
  2. jcsd
  3. Jan 6, 2014 #2
    You reasoning is correct. Any additive term in a Lagrangian that does not depend on the generalized coordinates or their time derivatives can be ignored, because it does not affect equations of motion. Note that dependence on time in an additive term per se is ignorable.

    As for the question, what equations of motion do you obtain from the second Lagrangian? How are they different from the equations for a stationary pendulum in a vertical field of gravity?
     
  4. Jan 6, 2014 #3
    There is something that can help you
    cartForcesOnPendulum.png

    http://blogs.mathworks.com/seth/2008/10/09/challenge-metronome-and-cart-equations-of-motion/
     
  5. Jan 6, 2014 #4

    CAF123

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    Hi voko,
    The equations differ by a term ##\frac{a}{\ell} \cos \theta## on the LHS. I can then write the E.O.M for the second Lagrangian as $$\ddot{\theta} + \frac{\sqrt{g^2 + a^2}}{\ell} \sin(\theta + \phi) = 0,$$where ##\tan \phi = \frac{a}{g}## is the offset and ##|g_{eff}| = \sqrt{a^2 + g^2}##
    How do we see the direction of the acceleration ##a## in these equations?

    @Malverin, thanks for the link.
     
  6. Jan 6, 2014 #5
    What do you think of angle ##\phi##?
     
  7. Jan 6, 2014 #6

    CAF123

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    Angle ##\phi## is the angle the pendulum makes with the vertical initially. From the equation for ##\phi##, this means that ##a## points horizontally. If the x axis points rightwards, (I measured ##\theta## in the direction of motion, assuming motion of carriage rightwards) how do we see that ##\vec{a}## points in the negative x direction?
     
  8. Jan 6, 2014 #7
    I am not sure what "initially" means here.

    More importantly, where is the stable equilibrium of the system? Would that have any connection with the direction of the effective field of gravity?
     
  9. Jan 6, 2014 #8

    CAF123

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    I was interpreting it as the angle the pendulum makes with the vertical at t=0. (Instantaneously, before the carriage starts moving from zero initial velocity at t=0, the pendulum is at an angle ##\phi## to the vertical).

    In the case of a stationary pendulum in a vertical gravity field, the stable equilibrium is at ##\theta = 0##. The orientation of the pendulum at the stable equilibrium will be parallel to the gravity field. For a moving pendulum in a non vertical gravity field, the stable equilibrium is similarly parallel to the direction of the gravity field.
     
  10. Jan 6, 2014 #9
  11. Jan 6, 2014 #10
    That's an initial condition. It does not get to be encoded in the differential equation.

    Where is the stable equilibrium of the system according to the equation in #4?
     
  12. Jan 6, 2014 #11

    CAF123

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    Possible points of stable equilibrium would be values of ##\theta## where ##\ddot{\theta}## vanishes. So ##\theta## satisfying $$\theta = n\pi - \arctan\left(\frac{a}{g}\right)$$

    This equation will give all points of stable and unstable equilibria. To determine a point of stability, consider the sign of a corresponding potential function ##U = -\omega^2 \cos (\theta + \phi)##, where ##\phi = \arctan(a/g)##. When ##n=0## then ##\theta = -\arctan(a/g)## so the effective field points in the third quadrant. This corresponds to ##U## being minimized. For ##\theta = \pi - \arctan(a/g)##, U is maximized. The shift of ##\pi## means the stable points and unstable points alternate since ##\cos \theta## passes through a cycle of maxima and minima every ##\pi## units.

    Is that analysis correct?

    If one stable equilibrium is in the third quadrant for the given value of ##\phi## then since ##g## is vertical, ##a## is horizontally pointing leftwards. So, ##\vec{g_{\text{eff}}} = \langle -a, g \rangle##.
     
    Last edited: Jan 6, 2014
  13. Jan 6, 2014 #12
    I am not sure why ##a## has to be leftward; other than that, the analysis is solid.
     
  14. Jan 6, 2014 #13
    I think, because in non inertial system there is an inertial force

    F= -m*a
     
  15. Jan 6, 2014 #14

    CAF123

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    My understanding is that for an observer in an accelerating frame, the equivalent statement of NII as measured in the inertial frame is given by $$m\vec{a_{S'}} = F_S + F_{trans, S'} + F_{centrifugal, S'} + F_{Coriolis, S'} + F_{azimuthal,S'}$$where ##S'## is the accelerating frame attached to the carriage and ##S## the inertial one. We are not dealing with a case where the last three terms are present and ##F_{trans, S'}## is of the form ##-m\dot{v(t)} = -ma##.

    Since I was considering the Lagrangian for the pendulum (which included terms relative to the pivot which accelerates at ##a##) I think the above reasoning is ok. More specifically, the Galilean transformation I was dealing with was ##\vec{v} = \vec{u} + \vec{V}##, where ##\vec{u}## is the velocity of the pendulum wrt accelerating frame attached to pivot and ##\vec{V}## is the velocity of the accelerating frame attached to pivot wrt inertial frame. The transformation took place between an inertial frame and the accelerating frame attached to the pivot, so indeed NII is affected by the transformation and the only present fictitious force in the frame is that given above.
     
    Last edited: Jan 6, 2014
  16. Jan 7, 2014 #15
    I do not think that addresses my issue with the direction of ##a## itself. You said that you assumed the carriage was moving rightwards. So that means ##a## is directed rightwards. Then you said ##a## was directed leftwards. This is not an issue of different frames because ##a## is given in the ground frame.

    Another thing that I just realized is that in your expression for potential energy ##g## is a positive constant, while the direction of the acceleration due to gravity is negative: ## \vec g = (0, -g) ##. So I think the formula for the effective field of gravity must be ##(-a, -g)##. This is more consistent with your statement about the third quadrant.
     
  17. Jan 7, 2014 #16

    CAF123

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    Yes, the acceleration of the carriage is to the right.
    What I meant was that for an observer in the frame of the pivot, they would see a single fictitious force directed leftwards which is given by -ma. Reading back through my posts though, I see why I did not make this clear.
    I defined the y axis positive down and x axis positive rightwards with the origin at the pivot of the pendulum.
     
  18. Jan 7, 2014 #17
    I think we are on the same page now.
     
  19. Jan 7, 2014 #18

    CAF123

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    Thanks for all your help!
     
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