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Moment of Inertia of rods

  1. Nov 17, 2006 #1
    I'm a physics major with a quick HW question for a Calc-based physics class. I'll present the problem along with my work. Any help would be appreciated:

    HW Problem:

    Consider 3 uniform thin rods all perpendicular to one another in such a manner that they align with the x, y, and z axii respectively. The centers of mass for the rods coincide with the origin. Each rod has a mass M and a length L. Imagine an axis of rotation parallel to the y axis running through the very end of the rod along the x-axis. Determine the moment of inertia for the system in terms of M and L.

    We are looking for an equation that identifies the moment of inertia of the object in terms of M and L.

    My Approach:

    This appears to be a problem in which the moment of inertia of the system is the sum of the moments of inertia of the individual rods (with respect to the axis of rotation).

    For the rod along the y-axis, this is simply calculated from the standard equation of a long thin rod with aor at one end I=1/3 ML^2.

    For the rod along the z-axis, this also is an application of the parallel axis theorem. We use the standard equation for the moment of inertia of a long thing rod rotating about an axis passing through its center of mass (perpendicular to the rod's length). This gives I=1/12 M (L/2)^2 +M(L/2)^2. Here, the aor for the entire object is a distance L/2 away from the z-axis (half the distance of a rod).

    So far, we have the total inertia of the system to be 1/3 ML^2 + 1/12 M (L/2)^2 +M(L/2)^2

    For the rod along the y, no standard form is given, but we know that a single discrete particle has a moment of inertia of MR^2, where R is the radius from axis of rotation. The length of this particle makes little difference if the particle is small. The moment of inertia for a hoop or thin cylindrical shell (different from a hollow cylinder) based on the particle premise is MR^2, where R is the radius from axis of rotation. Summing all our equations:

    Total I = y + z + x
    Total I = 1/3 ML^2 + 1/12 ML^2 + M(L/2)^2 + M(L/2)^2


    Algebraically, this reduces to 11/12 ML^2. Is my approach wrong? Apparently, the answer is incorrect.
     
    Last edited: Nov 17, 2006
  2. jcsd
  3. Nov 17, 2006 #2

    OlderDan

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    The approach is correct, but I think I see a problem highlighted in red in the quote.
     
  4. Nov 17, 2006 #3
    I believe I erred in posting the equations from copy / paste. The posted equation has been corrected to the appropriate value:

    Total I = 1/3 ML^2 + 1/12 ML^2 + M(L/2)^2 + M(L/2)^2

    This is the equation I used to reduce to the 11/12 ML^2 and was told the answer was incorrect.
     
  5. Nov 18, 2006 #4
    I believe this to be mathematically similar. But to be sure, let's show it to be so...

    Let's assume the object rotates around the y-axis then translate this according to the parallel axis theorem.

    Since the aor is the y-axis, the moment of inertia of the system includes the "arms" that point along the x- and z- axis.

    The moment of inertia for each "arm" is given by the standard equation for a long thin rod rotating about its center of mass 1/12 ML^2. So the moment of inertia of the object before the parallel axis theorem is

    I = x + z
    I = 1/12 ML^2 + 1/12 ML^2 = 2/12 ML^2

    Now, we choose an axis of rotation parallel to y that is a distance L/2 away. According to the theorem our new moment of inertia I is given by:

    I = I + MD^2

    here M is the mass of all 3 rods or 3M and D is L/2. Substitution gives

    I = 2/12 ML^2 + 3M(L/2)^2 = 11/12 ML^2

    Here, we do have to make the assumption that the radius of each rod is negligible compared to its length, thus the rod along y contributes no moment of inertia to the system before application of the parallel axis theorem.
     
    Last edited: Nov 18, 2006
  6. Nov 18, 2006 #5

    OlderDan

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    You must have read something between your last two posts that I deleted. Sorry about that. I realized I had gone astray and that the results are in fact the same if you do the CM calculation first, as I had thought in the first place.

    I'm not seeing anything wrong with what you did using either approach, and as you have shown they do in fact lead to the same result.
     
    Last edited: Nov 18, 2006
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