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Writing a dynamical equation

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume a rigid body with a massless rod that pivots about point O. Displacement x is measured from equilibrium position. Assuming x is small, that the weight at the end of the rod is 5N and spring constant is 400n/m, obtain the dynamical equation of the system.

    hw31.jpg


    2. Relevant equations

    [tex]\sum f =ma[/tex]
    f= -kx

    3. The attempt at a solution

    When I look at the system, I see only two forces acting on it. One of which is gravity, and one of which is the force of the spring. So I come up with

    ma=-kx-mg , so

    [tex]m\ddot{x}=-kx-mg[/tex]

    My intuition tells me something is missing though, since the length of the rod is given. Any suggestions?
     
  2. jcsd
  3. Nov 8, 2009 #2

    tiny-tim

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    Hi dankaroll! :smile:
    Yeees :redface:

    you're using x to mean two different distances, aren't you? :wink:

    (and either take moments, or use conservation of energy :smile:)
     
  4. Nov 8, 2009 #3
    Are you talking about finding the actual distance x the ball drops down once the 5N force is applied?

    I would think you take a moment at point O to find the force of the spring the instant after the system is released.. which would give you

    [tex]\sum Mo=0 ; 5N(3a)+(kx)(a)=0[/tex]
    [tex]kx= (5N*3a)/(a) = (10a N)[/tex]

    so from there you can find the displacement of the spring, which is

    [tex]x=(10a N)/(400N/m) = .025a [/tex]

    so then you can find the angle, and then the distance that the ball drops,

    [tex]x= .025(3a) = .075a [/tex]

    which would be the max distance the ball would drop, since over time the system would lose energy. am i right?:biggrin:
     
  5. Nov 8, 2009 #4

    tiny-tim

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    You're asked for the dynamical equation of the system.

    In other words, how does x (or x') depend on t (or how does x' depend on x)?

    Where in your equations is any reference to the speed (x'), or the angular speed?
     
  6. Nov 28, 2009 #5
    my only question is, wouldnt angular velocity introduce a theta term into the dynamical equation? how would you handle that?
     
  7. Nov 28, 2009 #6
    Ok, well I see that angular acceleration is equal to tangental accel/radius

    and angular velocity is equal to tangental velocity / radius

    and angular position is equal to tangental position / radius

    so....

    [tex](1/3)m\ddot{x}=-(1/3)\dot{x}-kx[/tex]

    on to something?
     
  8. Nov 28, 2009 #7

    tiny-tim

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    The question keeps mentioning the displacement x, not the angle, so it clearly wants an answer in terms of x, not the angle.

    If you insist on using angle, you can calculate it from x and a. But the rod is massless, and you can assume the rigid body is a point mass, so just use its ordinary velocity, not its angular velocity.
     
  9. Nov 28, 2009 #8

    tiny-tim

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    I don't really understand what you're doing …

    where is g? where is a? why 1/3 ? :confused:
     
  10. Nov 28, 2009 #9
    I used 1/3 because the conversion from angular to tangential is a/r or v/r.. and the ball is 3a away.

    The equation should have the input force in it though.
     
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