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Additive moments of inertia?

  1. Nov 8, 2004 #1
    Hi! Im working on this problem:

    A solid circular disk has a mass of 1.2 kg and a radius of 0.17 m. Each of three identical thin rods has a mass of 0.16 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to form a three-legged stool. Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center. (Hint: When considering the moment of inertia of each rod, note that all of the mass of each rod is located at the same perpendicular distance from the axis.)
    Here is my thinking so far (but im stumped!):
    -Moment of Inertia's are additive
    - The moment of inertia for the circular disk is (1/2)M*R^2
    - I don't know how to get the moment of inertia for the three legs because they dont give me a mass. Im pretty sure the formula is 1/3M*L^2 but I dont know L ??

    Thanks for the help.
  2. jcsd
  3. Nov 8, 2004 #2


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    Re-read the problem - the mass of the rods is stated explicitly. The length of the rods should be unimportant - perhaps you can figure out why.
  4. Nov 10, 2004 #3
    Oh whoops, I didn't mean to put that the mass was not known. So it's been a couple of days and Im still stumped on why the length would not matter.
    Is this close to the right track?

    (1/2*M*R^2)+(1/3m)+(1/3m)+(1/3m) = I

    M= mass of stool top
    m = mass of one leg
  5. Nov 10, 2004 #4


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    The legs are parralel to the axis of rotation right? So, in the plane of rotation (perpendicular to the axis of rotation) they're essentially point masses. (Does this make sense to you?)

    I get:
    [tex]\frac{1}{2}MR^2 + 3 \times (\frac{R}{3})^2m[/tex]
  6. Nov 10, 2004 #5
    could you please explain that for me? it would seem like the legs would matter, since there is a r (distance between them and the axis of rotation). we just covered angular momentum and i have yet to work through the homework.

    thanks, if you've the time!
  7. Nov 10, 2004 #6
    NateTG, why you put R/3? Since the rods can be considered as point masses on the solid disc, I think it should be mR^2 for each, right?
  8. Nov 10, 2004 #7
    well from the lab on angular momentum that we did, two point particles rotating on a bar -- the moment of intertia can be calculated as


    now that i see what he's talking about with that r/3 part, he's doing the same thing. it's like there are 3 of them adding to the moment of inertia. i don't understand what the 1/3 of R is for. it seems to me like it should be 3mR^2.
  9. Nov 11, 2004 #8


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    That was a brain fart. I thought the rods were [tex]\frac{1}{3}[/tex] of the radius out rather than at the edge.
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