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The Oscillation Period of A Rod Pivoted at One End

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    The figure shows a 230 g uniform rod pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. K=3.0N/m and the length of the rod is 0.20m.

    What is the rod's oscillation period? You can assume that the rod's angle from vertical is always small.


    2. Relevant equations
    Restorative force=F=-kΔx
    Torque=Fd=force * length of lever arm
    Moment of inertia for a rod pivoted about one end: I=(1/3)mL2
    Angular frequency(w)=2π/T

    3. The attempt at a solution
    -kΔx=I*-w2Θ
    <=>
    -kr2Θ=I*-w2Θ
    <=>
    -kr2Θ/I=-w2Θ
    Substitute I into the equation, L=r in this case.
    -3K/M*Θ=-w2Θ

    Therfore w=sqt(3K/M)
    <=>
    2π/T=sqt(3K/M)
    <=>
    T=2π/sqt(M/3K)=1.00s

    But, I don't really know if my calculations are correct.
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Sorry, I don't know how to do this problem. But check
    -kΔx=I*-w2Θ
    Looks like kg m/s^2 on the left and kg m^2/s^2 on the right.
     
  4. Jan 25, 2009 #3
    If the spring were not there, you would have a simple pendulum. It does not look to me like you have taken this aspect of the problem into your formulation at all. What you have in this problem is a simple pendulum with additional stiffness added. I suggest that you take a second look at the problem (start over).
     
  5. Jan 26, 2009 #4

    Are you sure? It rotates around an axle..that's why I thought it was a physical pendulum. See: http://session.masteringphysics.com/problemAsset/1070632/3/14.CP80.jpg
     
  6. Jan 26, 2009 #5
    OK, pardon the error in termnology. It is a physical pendulum, you are correct. That said, it is just a physical pendulum with a spring added to increase stiffness. Now, shall we start from the beginning?
     
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