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Rods and springs

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end,on both pulling in opposite directions. The springs have constant k and at equilibrium, their pull is perpendicular to the rod. Find the frequency of small oscillations.

    2. Relevant equations

    [tex] \tau = R \times F[/tex]

    3. The attempt at a solution

    Consider the origin to be the pivot point.

    So :
    [tex]\tau = R \times F [/tex]
    [tex] =\frac{l}{2}\cdot (kx)cos(\theta) - l \cdot kx + l/2 \cdot cos(\theta) + mg \cdot sin(\theta)[/tex]
    I am a little unsure here.
    First theta is the angle between the vertical and the rod.
    I am unsure because clearly when the spring in the middle is stretched by x, the one at the far end must be compressed twice as much, but I'm not sure if the factor of 2 is taken care of by the "R" for the torque or if I would have to apply it twice, once for the "R" and once for "x" from kx.

    [tex] \tau = I \alpha [/tex]
    [tex] = \frac{1}{3} M \cdot l^2 \alpha[/tex]
    [tex] = \frac {1}{3} M \cdot l^2 \frac{d}{d\theta}(0.5 \omega^2)[/tex]
    [tex]\tau = \frac {1}{6} M \cdot l \frac{d}{d\theta}( \omega^2) = -kxcos(\theta) + l/2mgsin(\theta)[/tex]

    After that, I'm not sure what to do :S

    I would integrate with respect to theta to get omega, but x is a function of theta?
    And suppose that x was not, then omega^2 would be negative for certain values k and M.
  2. jcsd
  3. Dec 5, 2008 #2
    I didn't check your earlier work, but based on your last line: Maybe do Taylor's expansions of cos and sin? You'll have to play with how far to keep higher orders, so you get something solvable yet not trivial.
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