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Moment of inertia of two perpendicular rods

  1. Jan 13, 2006 #1
    Hi, can someone help me out?

    In the question there are two straight rods which are connected at their ends in such a way that they are perpendicular. (As a visual aid, the situation can be seen by picturing the usual right handed xyz system of coordinates, with each of the two connected rods along the x and y axes).

    The two rods are each of mass m and length 2l. I am asked to produce the expression for the moment of inertia of the element (presumably the connected rods) about the axis through its centre of gravity and perpendicular to both rods.

    Answer: I = (5/3)ml^2

    I've tried a whole bunch of things but it doesn't seem to be getting me anywhere. The first thing that came to mind is that I need to find the location of the axis of rotation. I know that the centre of mass of a single rod with constant density is at its midpoint and my calculations verified that. But I don't know how to find the centre of mass of the lement formed when two rods are connected as described previously.

    Since I don't know where the COM is I'll just assume for now, that it is at the point of connection between the two rods. So on this premise and the visual aid I mentioned before, the axis of rotation is the z-axis.

    Basically, what I then went on to do was to calculate the moment of inertia of of a single rod about an axis through it's COM. Then I used the parallel axis theorem to find the moment of inertia of a single rod about one of its ends. The result came out to be (4/3)ml^2 and this is consistent with the result that for a rod of length L (L=2l in this case), the moment of interia about an axis at one of the ends is (1/3)mL^2.

    I doubled the result and of course it came out incorrectly as (8/3)ml^2 instead of the answer os (5/3)ml^2.

    Those are my attempts. I'm basically unsure about where to start. I've tried some random things but not much seems to work. Any help would be good thanks.
  2. jcsd
  3. Jan 13, 2006 #2


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    Dearly Missed

    Okay, so you know that the C.M's of the two rods lie at the midpoint of each rod.
    Since the masses of the rods are the same, the COMMON C.M for the system will lie midway on the connecting line segment between the two rod C.M's
  4. Jan 13, 2006 #3
    Thanks for the help arildno, I've now managed to obtain the correct answer.
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