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Calculate the rotational inertia of the pendulum

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    The pendulum consists of a uniform disk with radius r=10.0cm and mass 500g attached to a uniform rod with length L=500mm and mass 270g.
    a. Calculate the rotational inertia of the pendulum about the pivot point.
    (answer .205kg m^2)
    b. What is the distance between the pivot point and the center of mass of the pendulum?
    (answer 47.7 cm)
    c. Calculate the period of oscillation.
    (answer 1.5s)

    2. Relevant equations
    Ic for rod is 1/12 x M x L^2
    Ic for disk is 1/2 x M x R^2
    Io is Ic + mh^2, where h is from the point of pivot to the center of mas.
    T=2 x pi x [Io / (mgh)]^(1/2)

    3. The attempt at a solution
    a. Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
    Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk
    Io = Ic + mh^2, I don't know what to do with the two Ic, and cannot find the Io without h, which is to be found in question b. What am I missing?

    b. Io = Ic + mh^2 = .205 kg m^2 = Ic + (.770kg)(h^2) to find h, but cannot find Ic with the two previous number I got. No Idea Need Help!

    c. T= 2 x pi x [Io / (mgh)]^(1/2) =
    2 x pi x [(.205kg m^2) / [(.770kg)(9.81m/s^2)(.477m)]]^(1/2)= 1.5s
    can get with the answers for a and b

    If someone had msn (horseland20@hotmail.com) I will be on if you can help if not post!
    I would appreciate the help.
     
    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 30, 2008 #2
    Any one able to help, I have spent over two hours getting no where. Suggestion may help!
     
  4. Jan 30, 2008 #3

    Doc Al

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    Staff: Mentor

    So far, you've calculated the moment of inertia of each piece about its own center of mass. Now use the parallel axis theorem to find the moment of inertia of each piece about the pivot point.
    You don't need the center of mass of the pendulum to answer this part of the question.
     
  5. Jan 30, 2008 #4
    I don't mean to be dumb:confused:, but i thought I used the parallel axis theorem, because that is where I from the formulas:redface:.
     
  6. Jan 30, 2008 #5

    Doc Al

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    Staff: Mentor

    Where did you use the parallel axis theorem? I just see your calculations for the two Ic's.
     
  7. Jan 30, 2008 #6
    My text book stats the parallel axis theorem for a disk is 1/2 x M x R^2
    and a thin rod 1/12 x M x L^2, maybe I call is it Ic when it is not.
     
  8. Jan 30, 2008 #7

    Doc Al

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    Staff: Mentor

    Please double-check your book--it can't be that bad! :wink:

    Ic means the moment of inertia about the center of mass. Those are just the formulas for calculating the moment inertia of those two shapes about their centers.

    The parallel axis theorem allows you to use Ic to find the momentum of inertia about some other (parallel) axis. That's what you need to use to find the moment of inertia about the pivot point. Read this: Parallel Axis Theorem
     
  9. Jan 30, 2008 #8
    You are correct, I am reading my book wrong. they are Ic formulas.

    So what I am stuck on is what is the d(in the formula and the site) or h in the one I stated Io is Ic + mh^2.

    I have no idea, except is h not what b is asking for?
     
  10. Jan 30, 2008 #9

    Doc Al

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    Staff: Mentor

    For part a, you need the moment of inertia of each piece about the pivot point. So "d" in the parallel axis formula is the distance from the center of each piece to the pivot point.

    In part b, you need the distance from the pivot to the center of mass of the pendulum. (So you would first figure out where the center of mass is.)
     
  11. Jan 30, 2008 #10
    ok, thanks for clearing that up!

    to make sure,

    Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
    Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk

    I= Ic of rod + Md^2 = (.005625 kg m^2) + (.270kg)(.250m)^2 = .0225kg m^2 both mass and I = Ic of disk + MD^2 = (.0025 kg m^2) + (.500kg)(.6)^2 = .1825kg m^2

    .0225kg m^2 + .1825kg m^2 = .205 kg m^2 got it.


    Now help with question b
     
    Last edited: Jan 30, 2008
  12. Jan 30, 2008 #11

    Doc Al

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    Staff: Mentor

    That looks good.
    That looks good.

    No.

    What's the distance between the center of the disk and the pivot? (It depends on whether the rod ends at the edge or the center of the disk.)
     
  13. Jan 30, 2008 #12
    Got it, thank! While I was waiting I reread and got it from what you said before, thank you so much. It feels good to have it!
     
  14. Jan 30, 2008 #13
    I got part b and it feel so good to have got this problem!

    Thanks So Much!
     
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