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Finding the period

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data

    The pendulum shown below right consists of a uniform disk with radius r = 20.0 cm and mass m = 1.2 kg attached to a uniform rod with length L = 40 cm and mass M = 0.8 kg. What is the period of the motion?

    3508mqs.jpg

    2. Relevant equations
    T = 2∏√(I/mgL)


    3. The attempt at a solution

    I dont know what to plug into what.
     
  2. jcsd
  3. Dec 9, 2012 #2

    Doc Al

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    Staff: Mentor

    To start with, you need to calculate the rotational inertia (I) of the system about the axis of rotation. How would you do that?

    Also, in your formula for T what does L represent? You'll have to find that as well.
     
  4. Dec 9, 2012 #3

    T = 2∏√(I/mgL)


    L represents the distance to the center of mass of the system?
    L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
    = 0.38m?


    using parallel axis theorem:

    I = 1/2MR2 + ML2
    = (0.5)(1.2)(0.2)2 + 0.8(0.2+0.4)2
    =0.312?
     
  5. Dec 9, 2012 #4

    Doc Al

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    Staff: Mentor

    Right.
    What's the distance from the center of the disk to the axis?


    Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
     
  6. Dec 9, 2012 #5
    L = [(1.2*0.6)+(0.8*(0.4/2)]/(1.2+0.8)
    = 0.44m?


    using parallel axis theorem:

    I of disk = 1/2MR2 + ML2= 1/2(1.2)(0.2)2 + (1.2)(0.6)2= 0.456?
    I of thin rod = 1/3ML2 = 1/3(0.8)(0.4)2= 0.0427?

    I total = 0.4987?

    I have an exam in 45 minutes so i really need to get this cleared up before that.
     
    Last edited: Dec 9, 2012
  7. Dec 9, 2012 #6

    Doc Al

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    Staff: Mentor

    Looks good.
     
  8. Dec 9, 2012 #7
    wen i plug m into the formula, T = 2∏√(I/mgL)

    do i use the mass of the entire system? so 1.2 + 0.8 = 2kg?
     
  9. Dec 9, 2012 #8

    Doc Al

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    Staff: Mentor

    Yes.

    For more insight into the physical pendulum and how that equation comes about, review this: Physical Pendulum
     
  10. Dec 9, 2012 #9
    thanks so much for ur help! really appreicate it. glad there are people like you out there.

    but last question, can you explain why it has a negative sign in

    Torque = -mgLsinθ?
     
  11. Dec 9, 2012 #10

    Doc Al

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    Staff: Mentor

    Because the torque is in the direction of decreasing θ. (It's a restoring force, tending to bring the system back to the equilibrium point.)

    Good luck on your exam!
     
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