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Derive the Lagrangian for the system

  1. Apr 25, 2009 #1
    Investigating how a car bounces with the framework of an idealized model. Let the chassis be a rigid, square plate, of side a and mass M, whose corners are supported by massless springs, with spring constants K,K,K and k < K (the faulty one). The springs are confined so they stretch and compress vertically, with unperturbed length L. The density of the plate is uniform.

    (a) explain why the system has 3 generalized co-ordinates to be described completely.

    1 translational ( up and down ) and 2 rotational.

    (b) show your coordinates.

    http://img11.imageshack.us/img11/1859/coordsd.jpg [Broken]

    (c) Derive the Lagrangian for the system assuming that all motions are small and the four springs are identical.

    This is where I'm stuck.....

    L = KE + PE

    KE = .5mv^2 + .5I(phi dot)^2 + .5I(theta dot)^2

    where v = velocity of the centre of mass.
    phi dot is the derivative of phi (does this give you the angular velocity?)
    theta dot is the derivative of theta
    I is the moment of inertia of a square plate (Ma^2)/12

    PE = 4.(.5kz^2) = 2kz^2

    Is this even remotely correct? Any help as always is greatly appreciated.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 25, 2009 #2

    Doc Al

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    Re: Lagrangian

    For one thing, L = KE - PE.

    This is OK. I would use z as the position of the center of mass, thus v = z dot. (z = 0 is the equilibrium position.)

    Careful here. You have three coordinates: z, theta, and phi. Express the height of each corner in terms of those variables. (z is just the position of the center of mass.)

    And don't forget gravity.
     
  4. Apr 25, 2009 #3
    Re: Lagrangian

    I have no idea on how to express the height of each corner in terms of phi theta and z :S I mean I can presume they involve sin and cos of phi and theta but anything more then that and I'm lost.

    Also for gravity is that just mgh? where h = z?
     
  5. Apr 25, 2009 #4

    Doc Al

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    Re: Lagrangian

    Yes, it will involve sinθ, but for small angles sinθ ≈ θ. Start from the center of mass and figure out how having the plate tilted at some angle θ affects the height of the corner. Hint: If one side goes up, the other goes down.

    Give it a shot.

    Yep.
     
  6. Apr 25, 2009 #5
    Re: Lagrangian

    So for the springs is this even remotely correct...

    PE = 4 * .5 * K * ( (2/a)sin(theta) + (2/a)sin(phi) + z )^2 + mgz

    if so, does sin(theta and sin(phi) just become theta and phi?
     
  7. Apr 26, 2009 #6

    Doc Al

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    Re: Lagrangian

    You're getting warmer. First specify the vertical position of each corner. They can't all be the same (if one side tilts up, the other must tilt down):

    z1 = z + (a/2)sin(theta) + (a/2)sin(phi)
    z2 = z + (a/2)sin(theta) - (a/2)sin(phi)

    and so on, for all four corners.

    Yes.
     
  8. Apr 26, 2009 #7
    Re: Lagrangian

    z1 = z + (a/2)sin(theta) + (a/2)sin(phi)
    z2 = z + (a/2)sin(theta) - (a/2)sin(phi)
    z3 = z - (a/2)sin(theta) - (a/2)sin(phi)
    z4 = z - (a/2)sin(theta) + (a/2)sin(phi)

    which become..

    z1 = z + (a/2)(theta) + (a/2)(phi)
    z2 = z + (a/2)(theta) - (a/2)(phi)
    z3 = z - (a/2)(theta) - (a/2)(phi)
    z4 = z - (a/2)(theta) + (a/2)(phi)

    so...

    PE = .5K(z1^2) + .5K(z2^2) + .5K(z3^2) + .5K(z3^2) + mgz

    (as for the 2/a, what a careless mistake. I'll just say it was late >.<)

    and now if this is correct my L = KE - PE

    and then i use the lagrange equation

    ae9c64c6d51a4dac2002030f436866fd.png

    to solve for phi, theta, z, phi dot, theta dot and z dot.

    please tell me im on the right track. I'm still struggling to see how this is an improvement on F = ma :P

    Thanks for all the help thus far Doc Al truly appreciated.
     
  9. Apr 26, 2009 #8

    Doc Al

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    Re: Lagrangian

    Yes, you're on track now. :wink:

    The power of the Euler-Lagrange method will become clear when you have problems with a few more forces and constraints. Sometimes the forces are very hard to identify and keep straight, especially their directions, so applying F = ma directly would be too hard. But the Lagrangian is usually straightforward to write down immediately. (With practice!) It might take you an hour just to try and figure out how the forces are acting in a problem (if you can do it at all)--but using E-L you'll have the equations of motion in 5 minutes. (Of course, you still have to solve the equations...)
     
  10. Apr 26, 2009 #9
    Re: Lagrangian

    I see.... Well i might see how i go solving this... Just quickly though seeing the the body is a square (Symmetric) won't that mean the solution for phi, phidot and theta, thetadot will be the same?
     
  11. Apr 26, 2009 #10

    Doc Al

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    Re: Lagrangian

    Yes, the equations for phi and theta will be identical.
     
  12. Apr 29, 2009 #11
    Re: Lagrangian

    I've solved the equations and ended up with 3 modes of oscillation

    w1 = 2((K/M)^.5)

    and

    w2 = w3 = 2((3K/M)^.5)

    But the following question isn't making too much sense.

    By making sensible guesstimates of how far a car sink when you sit in it, estimate K and hence the bounce frequency w1. does w1 conform with common experience.

    So the M (mass of chassis/car) ~ 1500kg

    K = F/x

    F ~ 800 newtons (80 * 10)
    x ~ 2cm (I really have no idea on this >_>)

    so K = 40000

    I get a value of ~ 10.3 for w1. But how do I interpret this (Are the units radians per second?)?


    Also the following has me stumped...

    Estimate roughly how far apart the corrugations on a road must be to induce maximum bouncing?

    does this have to do with w1 matching/lining up with the distance/frequency between the corrugations in the road?

    Thanks again!
     
  13. Apr 29, 2009 #12

    Doc Al

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    Re: Lagrangian

    Looks good.

    Yes, the units of ω are radians per second. And your estimates are as good a guess as any.


    Sounds right to me. You'll have to estimate the typical car speed.
     
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