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Homework Help: Rod Pendulum Problem

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A rod suspended on its end and acting as a physical pendulum swings with a period of 2.24 s. What is its length? (g = 9.80 m/s2)

    2. Relevant equations
    T=2(pi) * (sqrt)(L/G)

    3. The attempt at a solution
    Okay, so I'm given the period, which is 2.24 seconds.

    I'm also given G, as it's 9.8.

    I plug in the values and the resulting equation is what I get:

    2.24=2pi * (sqrt)(L/9.8)

    I divide 2.24 by 2pi, which leaves me with this:


    I square both sides and get:


    Multiply 9.8 * .127097 should give me L.

    I get L=1.24555. I put it into my online homework assignment, and it's wrong.

    What am I doing incorrectly?
  2. jcsd
  3. Nov 18, 2009 #2
  4. Nov 18, 2009 #3
    im getting the same as you :(
    are you sure those are the correct numbers?
  5. Nov 18, 2009 #4
    Yup, I copy pasted the problem right out of the assignment.
  6. Nov 18, 2009 #5


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    Homework Helper

    Here the length of the physical pendulum is the distance of center of mass from the point of suspension.
  7. Nov 18, 2009 #6
    Don't I need the mass (or masses) to calculate the center of mass for an object?
  8. Nov 18, 2009 #7


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    No. In this case center of mass is the center of gravity. It is the mid point of the rod.
  9. Nov 18, 2009 #8
    I don't know if I understood this right, but the L in the equation must then be doubled?


    2.24=2pi * (sqrt)(2L/9.8)

  10. Nov 18, 2009 #9


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    No. Length should be L/2.
  11. Nov 18, 2009 #10
    It still doesn't work.

    I plugged in (L/2) for the L, so the equation looks like this:

    2.24 = 2pi * (sqrt)((L/2)/9.8)

    Solving for L:

    2.24/(2pi) = (sqrt)((L/2)/9.8) ----> 0.356507 = (sqrt)((L/2)/9.8)

    (0.356507)^2 = ((sqrt)((L/2)/9.8))^2 -----> 0.127097 = (L/2)/9.8

    L/2 = (0.127097)*9.8 -----> L/2 = 1.24555

    L = 1.24555 * 2 -----> L = 2.49111

    This value is also incorrect. Where am I making a mistake?
  12. Nov 18, 2009 #11


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    In the physical pendulum, You have to take in to account the moment of inertia and torque.
    The formula for period of oscillation is given by
    T = 2π*sqrt(I/τ), where I = 1/3*M*L^2 and τ = M*g*L/2
    So T = 2π*sqrt(2L/3g)
    Now solve for L.
  13. Nov 18, 2009 #12
    It worked, finally! Thanks for the help, it was really useful!
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