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Oscillating piece of wire

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform piece of wire is bent into an upside down V shape with angle [itex]\theta[/itex] between two legs of length L. The wire is placed over a pivot. Show that the angular frequency of small-amplitude oscillations about the equilibrium is
    [tex]\omega=\sqrt{\frac{3g\cos(\frac{\theta}{2})}{2L}}[/tex]


    2. Relevant equations
    maybe centre of mass of a rod comes into it
    maybe [itex]\omega=\sqrt{\frac{k}{I}}[/itex]


    3. The attempt at a solution
    Don't know how to start this problem
     
  2. jcsd
  3. Jul 12, 2009 #2

    Doc Al

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    Staff: Mentor

    Hint: Treat it as a physical pendulum. (Which is what it is.)
     
  4. Jul 12, 2009 #3
    the shape of the thing makes it a bit tricky because i think i have to find out the centre of mass
    perhaps the centre of mass could be found by
    [tex]\cos(\frac{\theta}{2})}=\frac{x}{\frac{L}{2}}[/tex]
    where x is the vertical distance as measured from the pivot point
    so
    [tex]x=\frac{L}{2}\cos(\frac{\theta}{2})}[/tex]
    and let the mass of the wire be m
    then
    [tex]I\alpha=mgx\sin\phi[/tex]
    [tex]I=mx^2[/tex]
    where
    [tex]\alpha=\frac{d^2\phi}{dt^2}[/tex]
    [tex]mx^2\frac{d^2\phi}{dt^2}=mgx\sin\phi[/tex]
    [tex]x\frac{d^2\phi}{dt^2}=g\sin\phi[/tex]
    [tex]\frac{L}{2}\cos(\frac{\theta}{2})}\frac{d^2\phi}{dt^2}\approx g\phi[/tex]
    not sure where a 3 will pop in but thats my best effort
     
  5. Jul 12, 2009 #4

    diazona

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    Homework Helper

    It's not true for this case that [itex]I = mx^2[/itex], because it's not a point mass, it's a rod. (Actually it's like two rods)
     
  6. Jul 13, 2009 #5
    ok i think i have it
    [tex]I=\frac{2}{3}mL^2[/tex]
    [tex]I\alpha\approx 2mgx\phi[/tex]
    and then omega can be calculated from this
    [tex]\frac{1}{3}L^2\alpha\approx gx\phi[/tex]
     
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