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A simple harmonic problem

  1. May 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A L = 1.48 m long metal rod is pivoted from a point one third of the way along its length. The rod has a mass of 0.208 kg. What period will the rod oscillate with when released from a small angle?

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    2. Relevant equations
    [itex] T = 2 \pi \sqrt{\frac{L}{g}}[/itex]

    3. The attempt at a solution
    should I think the rod as a point mass about it's centre mass? and use centre mass to calculate the period by the formula?
    This is my approach:

    [itex] \frac{1}{2}L-\frac{1}{3}L [/itex] is the length of the string
    [itex] T = 2 \pi \sqrt{\frac{L}{g}} = 2 \pi \sqrt{\frac{\frac{L}{6}}{g}} = 2 \pi \sqrt{0.025} = 0.997 s[/itex]
    Could someone give me some advice please?
     
  2. jcsd
  3. May 19, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    That approach is too simple. Consider a rod pivoted very close to the center of mass - your formula would give a very short period, but the actual period would be very long.

    Think of torque and angular momentum.
     
  4. May 20, 2015 #3
    Thank you for your reply, and this is my new approach:
    distance from pivot to centre mass is L/6
    torque(mg) = sin(theta)*(L/6)*mg
    also torque = I*angular acceleration.
    I am confused here, as w in simple harmonic motion is angular frequency and in the torque formula it is angular velocity? are they the same?
     
  5. May 21, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    They are not, and I don't see where an ##\omega## would appear in the formula for torque. In general: if you would get the same symbol for different meanings, use a different symbol or indices (point 4 here).
     
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