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Moment of inertia of a rod

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data

    "Three identical thin rods, each of length L and mass m, are welded perpendicular to one another. The assembly is rotated about an axis that passes through the end of one rod and is parallel to anotther, determine the moment of inertia of this structure."

    2. Relevant equations

    I(for a long thin rod with a rotation axis through it's end) = 1/3 ML^2

    3. The attempt at a solution

    I(for the whole system) = I (rod 1) + I (rod 2) + I (rod 3)

    For the rod where the rotation axis passes through the end of the rod, i'm thinking that its moment of inertia should be 1/3 ML^2.

    For the rod which is parallel to the axis of rotation, I have been told by my prof that the monent of inertia is zero. Could someone breifly explain why this is.

    For the last rod, i'm not sure what it moment of inertia is.
    If any of this is right, please let me know and any other help is appreciated.
     
  2. jcsd
  3. Jan 15, 2007 #2

    Doc Al

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    Staff: Mentor

    How are they welded together? At their centers? (A diagram would be helpful.)
    Good.

    Good.

    Sounds good.

    That depends on where the rod is compared to the axis of rotation. If the axis of rotation is parallel to the rod and passes through its center, then the distance of the rod to the axis is zero--and so is the moment of inertia about that axis. But if the rod is parallel to the axis but at some distance from it, then the moment of inertia is not zero. (Again, a diagram would help.)

    Consider the parallel axis theorem.
     
  4. Jan 15, 2007 #3
    the rods are welded together at their centers and the paralell rod does have a translated distance from the axis of rotation. Sorry, no diagram.
     
  5. Jan 15, 2007 #4
    Considering the parallel axis theorem: I_z = I_cm + MD^2.
    Is I_cm(center mass) the moment of inertia the object would have if the rotational axis were through the rods center?
     
  6. Jan 15, 2007 #5
    or is I_cm the moment of inertia of the three rods together?
     
  7. Jan 15, 2007 #6

    Doc Al

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    No problem. So, since the parallel rod has some distance from the axis of rotation, its moment of inertia about that axis is not zero. (Since the entire mass has the same distance from that axis, you can consider that parallel rod as being a point mass for the purpose of calculating its moment of inertia.)

    Exactly. For a thin rod, what's the moment of inertia about its center? Given that, you can find the rod's moment of inertia about any (parallel) axis.
     
  8. Jan 15, 2007 #7
    Does this mean that this rod will also involve using the parallel axis theorem?

    I_cm(Of a thin rod about its center) = 1/12 ML^2, but since it is parallel, the rotational axis would cut through it horizontally, not vertically( which i think it needs to be to use that equation).
     
  9. Jan 15, 2007 #8

    Doc Al

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    You certainly could use the parallel axis theorem, but you should know how to find the moment of inertia of a point mass a given distance from an axis.

    If you're talking about finding the moment of inertia of the parallel rod, then you'd need to start with its moment of inertia parallel to its axis (and through its center).

    But I thought you were talking about one of the other rods. To keep from going nuts, let's say the three rods are aligned with the x, y, & z axes (centered at 0,0,0). And that the axis of rotation that we are using is parallel to the x-axis and is defined by y = -L/2, z = 0. So you need the moments of inertia of all three rods about that axis.

    The rod along the y-axis is rotating about one end--so you know its moment of inertia. The rod about the x-axis is the one parallel to the axis of rotation, so we've discussed that one a bit. And the rod about the z-axis is perpendicular to the axis of rotation--that's the one that I'd use the parallel axis theorem for.

    Make sense?
     
  10. Jan 15, 2007 #9
    Thanks for your help. Using the axis helps clarify things alot.
     
  11. Nov 20, 2011 #10
    I would like to take this question to the end. I understand that the moment of inertia for the rod aligned with the y-axis is 1/3*M*L^2. Couldn't I model the other two rods as a particle, distance L/2 from the axis of rotation, using the formula I = M*(L/2)^2 ?
     
    Last edited: Nov 20, 2011
  12. Nov 20, 2011 #11
    If so, then

    I = I_cm + M*D^2 = 1/3*M*L^2 + 2*M*(L/2)^2 = 5/6*M*L^2
     

    Attached Files:

  13. Nov 20, 2011 #12

    Doc Al

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    Looks to me like your axis of rotation is parallel to the y-axis. Is that the problem you are addressing?
     
  14. Nov 20, 2011 #13
    Yes, my picture shows rotation parallel to the y-axis. In that case, isn't the moment of inertia for the rod aligned with the x-axis 1/3*M*L^2 ? Couldn't I model the other two as a particle, distance L/2 from the axis of rotation, as I suggested above? Then the moment of inertia for the system would be the sum of the moments of inertia for each rod, i.e. the rod aligned with the x-axis and the two particles.
     
  15. Nov 20, 2011 #14

    Doc Al

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    Yes.
    That would work for the rod aligned with the y-axis, since all points are equidistant from the axis of rotation. But not for the rod aligned with the z-axis. For that one, use the parallel axis theorem.
     
  16. Nov 20, 2011 #15
    *Light bulb*

    Haha, thank you! I guess it may have been easier to use the parallel axis theorem from the get-go. I_cm would then be 2(M/12*L^2) and the total mass would be 3M.
     
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