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Rigid solid

  1. Dec 5, 2004 #1
    Let's say we have a lorry and it's rear door is wide open (Pi angle). All of a sudden the lorry accelerates. How much time will it take the door to close?

    I think I need elliptic integrals to solve this problem. Is it that complicated? How would you solve it?


    THanks for your help!
     
  2. jcsd
  3. Dec 5, 2004 #2
    Infinity?

    Just guessing here, but given that it's the rear door (at the back side of the lorry, right?), I suspect it wouldn't close at all...
     
  4. Dec 5, 2004 #3
    The answer is t=3.03(L/a)^(1/2) where L is the length of the door and a the acceleration. But I don't have the foggiest how to solve it.

    I send a picture of the lorry seen from above. Elliptic integrals have to be used. I'd be grateful if you could give me a hint!
     

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  5. Dec 6, 2004 #4
    Is it the conservation principle what I have to aplly here? Or just the movement equations of the rigid solid?
     
  6. Dec 6, 2004 #5

    arildno

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    What's the problem?
    To set up the differential equations leading to the elliptic integral, or the evaluation of that integral?
     
  7. Dec 6, 2004 #6
    TO set up the differential equation. I don't really know what properties of the rigid solid to use. I know that elliptic integrals are to be used because the teacher has said it so.

    Thanks for your help
     
  8. Dec 6, 2004 #7

    arildno

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    All right:
    1)It's easiest to do this in the rest frame of the lorry.
    This is a NON-inertial reference frame, and if the acceleration of the lorry, measured in an inertial frame, is [tex]a\vec{i}[/tex] the door experiences, in the lorry's reference frame, a pseudo-force [tex]-ma\vec{i}[/tex] acting at the C.M of the lorry.
    m is the mass of the door).

    2) Consider now the torque produced by the pseudo-force about the hinge.
    (Choosing the hinge as our point eliminates the need to estimate the forces acting on the door AT the hinge!)

    Let the position vector of the C.M of the door, measured from the hinge be:
    [tex]\vec{r}=\frac{L}{2}(\cos\theta(t)\vec{j}-\sin\theta(t)\vec{i})[/tex]
    That is, at t=0, we set the angular displacement equal to zero, and L is the length of the door.
    When the angle equals [tex]\pi[/tex] the door has closed.

    The torque of the pseudo-force about the hinge is therefore:
    [tex]\vec{r}\times(-ma\vec{i})=\frac{La}{2}\cos\theta(t)\vec{k}[/tex]

    3. The moment-of momentum equation may therefor be written as follows, in scalar form:
    [tex]m\frac{La}{2}\cos\theta=\frac{1}{3}mL^{2}\frac{d^{2}\theta}{dt^{2}}[/tex]
    Or:
    [tex]\frac{d^{2}\theta}{dt^{2}}=\frac{3a}{2L}\cos\theta[/tex]

    4.
    Our initial conditions reads:
    [tex]\theta(0)=0,\dot{\theta}(0)=0[/tex]
    ("dot"-notation means temporal derivative)
    We multiply our moment-of momentum equation with [tex]\dot{\theta}[/tex], integrate the result from t=0 to an arbitrary t-value, and makeuse of the initial conditions:
    [tex]\frac{1}{2}(\dot{\theta}(t))^{2}=\frac{3a}{2L}\sin\theta(t)[/tex]
    Or, since we're only interested in the time period where [tex]\theta[/tex] increases up to [tex]\pi[/tex] we have:
    [tex]\dot{\theta}=\sqrt{\frac{3a}{L}\sin\theta}[/tex]
    We make a change of variables, and find that the period T must fulfill:
    [tex]T=\int_{0}^{\pi}\sqrt{\frac{L}{3a}}\frac{d\theta}{\sqrt{\sin\theta}}}[/tex]
     
  9. Dec 6, 2004 #8

    Clausius2

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    Well, that comes up with the demonstration that Arildno must be a gurú of something. I don't know what, but he should be one.

    Just a doubt: why do you have not considerated the aerodinamic drag?. In fact I would think that is the predominant force that enhances the door closing.

    Ooops, I see, the lorry is first at rest. So I see, it will be logical to think that inertia forces close the door. Hmmm. But I'm not sure....
     
  10. Dec 6, 2004 #9

    arildno

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    You're right about the drag, Clausius:
    But do you REALLY think you'd end up with a nice little integral in that case?
    I think drag is to be assumed negligible, in order to end up with a comparatively nice expression (not an uncommon feature with exercises..)

    (It's basically the same equation you'd get with a pendulum swinging in a constant g-field)
    Congratulations with the Engineer Award, BTW.
     
  11. Dec 6, 2004 #10

    Clausius2

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    Sure it will be more difficult. You're right. Well, maybe Feynmanfan wants to spend this afternoon solving a pretty equation. He is spanish too, and today here is the Constitution Day (festive), so he won't have nothing to do apart of this... :rofl:
     
  12. Dec 6, 2004 #11
    Gracias a todos!
    THanks for the nice explanation. And you're right, CLausius. Today is the Constitution Day and here I am solving (trying to solve) problems.
     
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