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Moment on door vehicle

  1. Nov 20, 2009 #1
    Validity of Einstein's Elevator Experiment

    I solved the problem below using Einstein's http://en.wikipedia.org/wiki/Equivalence_principle" [Broken], or "elevator experiment," and I easily find an answer. I used traditional dynamics analysis and I get a different answer! Where is the mistake in my solution(s) or assumptions? I'm pretty good with dynamics, but I'm having trouble resolving this. Any help greatly appreciated!

    A door is hinged to a vehicle at pt. O and is open at angle [tex]\theta[/tex]. The vehicle's acceleration is a[tex]_{}o[/tex]. I need to find the moment on the door caused by the vehicle's acceleration.

    Answer from Solution 1: M = m*a[tex]_{}o[/tex]*r*sin([tex]\theta[/tex])
    Answer from Solution 2: M = a[tex]_{}o[/tex]*r*sin([tex]\theta[/tex])/(1/m+r^2/I)

    a[tex]_{}o[/tex] is the acceleration of the vehicle in the x direction.
    [tex]\theta[/tex] is the angle of the door away from the x axis towards the y axis.
    m is the mass of the door.
    r is the distance from the hinge to the door's center of mass.
    I is the door's moment of inertia.


    SOLUTION 1
    ----------

    Based on my interpretation of Einstein's "elevator experiment" or "equivalence principle," instead of saying that the vehicle is accelerating, I can say that the vehicle is stationary with the door being exposed to a gravity field in the -x direction with magnitude a[tex]_{}o[/tex]. This makes it a very easy pendulum problem.

    Force is applied at center of gravity
    F = m*a[tex]_{}o[/tex]

    Moment is force times moment arm
    M = F*r*sin([tex]\theta[/tex])
    M = m*a[tex]_{}o[/tex]*r*sin([tex]\theta[/tex])


    SOLUTION 2
    ----------

    I know that the vehicle pushes on the door at pt. O (but I don't know magnitude or direction):

    F = force vector applied to pt. O

    F has magnitude f in the [tex]\theta[/tex]F direction. To see what this force does to the door, I move F to the center of gravity (pt. cg). The moment caused by moving this vector is:

    M = R x F

    R is the vector from pt. cg to pt. O. "x" is the cross product.

    Now I know how the body will move for a given force vector.

    A_CG = F/m
    ALPHA = M/I

    ALPHA is the radial acceleration vector, and A_CG is the acceleration vector of pt. cg.

    I want to figure out what force vector will cause the door to move such that the hinge point accelerates in the x direction with magnitude a[tex]_{}o[/tex]. I can then calculate the moment.

    To find acceleration of pt. O:

    A_O = A_CG + OMEGA x ( OMEGA x R ) + ALPHA x R

    My vectors ([x,y,z]' indicates a column vector):

    OMEGA = omega*[0,0,1]'
    R = r*[-cos([tex]\theta[/tex]),-sin([tex]\theta[/tex]),0]'
    F = f*[cos([tex]\theta[/tex]F),sin([tex]\theta[/tex]F),0]'
    ALPHA = M/I = 1/I * R X F

    From here out I'll use:

    c[tex]\theta[/tex] = cos([tex]\theta[/tex]), st = sin([tex]\theta[/tex])
    c[tex]\theta[/tex]f = cos([tex]\theta[/tex]F), stf = sin([tex]\theta[/tex]F)

    My x direction and y direction equations come out to:

    a[tex]_{}o[/tex] = f/m*c[tex]\theta[/tex]f + omega^2*r*c[tex]\theta[/tex] - r^2*f/I*s[tex]\theta[/tex]*(c[tex]\theta[/tex]*s[tex]\theta[/tex]f-s[tex]\theta[/tex]*c[tex]\theta[/tex]f)
    0 = f/m*s[tex]\theta[/tex]f + omega^2*r*s[tex]\theta[/tex] + r^2*f/I*c[tex]\theta[/tex]*(c[tex]\theta[/tex]*s[tex]\theta[/tex]f-s[tex]\theta[/tex]*c[tex]\theta[/tex]f)

    I can combine the two equations to get rid of the omega^2*r terms. Divide the first equation by c[tex]\theta[/tex] then subtract the second equation divided by s[tex]\theta[/tex].

    a[tex]_{}o[/tex]/c[tex]\theta[/tex] = f/m*(c[tex]\theta[/tex]f/c[tex]\theta[/tex]-s[tex]\theta[/tex]f/s[tex]\theta[/tex]) + r^2*f/I*(c[tex]\theta[/tex]*s[tex]\theta[/tex]f-s[tex]\theta[/tex]*c[tex]\theta[/tex]f)*(-s[tex]\theta[/tex]/c[tex]\theta[/tex]-c[tex]\theta[/tex]/s[tex]\theta[/tex])

    Multiply both sides by st*ct

    a[tex]_{}o[/tex]*s[tex]\theta[/tex] = f/m*(s[tex]\theta[/tex]*c[tex]\theta[/tex]f-c[tex]\theta[/tex]*s[tex]\theta[/tex]f) + r^2*f/I*(s[tex]\theta[/tex]*c[tex]\theta[/tex]f-c[tex]\theta[/tex]*s[tex]\theta[/tex]f)
    a[tex]_{}o[/tex]*s[tex]\theta[/tex] = f*(s[tex]\theta[/tex]*c[tex]\theta[/tex]f-c[tex]\theta[/tex]*s[tex]\theta[/tex]f)*(1/m+r^2/I)

    f = a[tex]_{}o[/tex]*s[tex]\theta[/tex]/((s[tex]\theta[/tex]*c[tex]\theta[/tex]f-c[tex]\theta[/tex]*s[tex]\theta[/tex]f)*(1/m+r^2/I))
    M = R x F = r*f*(s[tex]\theta[/tex]*c[tex]\theta[/tex]f-c[tex]\theta[/tex]*s[tex]\theta[/tex]f) (magnitude of M)

    M = a[tex]_{}o[/tex]*r*sin([tex]\theta[/tex])/(1/m+r^2/I)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
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