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Simple harmonic oscillation: uniform rod

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform rod of mass m and length L is freely pivoted at one end. What is the period of its oscillations? Icm for a uniform rod rotating about its centre of mass is 1/12mL2

    (a) √3g/2L
    (b) 2π √3L/2g
    (c) 2π √2L/3g
    (d) 2π √L/g
    (e) none of the above

    2. Relevant equations
    ω2 = mgL/I

    Icm = 1/12mL2
    I = Icm +1/12mL2

    Period: T = 2π/ω

    3. The attempt at a solution
    I'm fairly certain that the answer is "none of the above", but I'd just like to make sure I'm not neglecting anything in what seems to be a simple plug-and-chug question.

    I = Icm +1/12mL2 = 1/12mL2 + mL2 = 13/12 mL2

    ω = √mgL/I= √(mgL)/(13/12)mL2 = √12g/13L

    T = 2π/ω = 2π √13L/12g
     
    Last edited: Apr 22, 2016
  2. jcsd
  3. Apr 22, 2016 #2
    What is the distance between the Centre of mass of rod and the end? Answer this and you'll realize what you did wrong.
     
  4. Apr 22, 2016 #3
    I just realized what I did wrong. The distance is L/2, so the moment of inertia equation turns into: I = 1/3mL2

    So then ω2 = mg(L/2) / (1/3)mL2

    Therefore T = 2π √2L/3g

    Thanks!
     
  5. Apr 23, 2016 #4
    Welcome :)
     
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