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Rotational Motion About a Fixed Axis

  1. Oct 26, 2004 #1
    Hello again.

    I thought I had this problem figured out but once again, I did not.

    Here's the problem: A solid circular disk has a mass of 1.2kg and a radius of 0.12m. Each of three identical thin rods has a mass of 0.12kg. 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 it's center. (Hint: When considering the moment of inertia of each rod, note that all of the ass of each rod is located at the same perpendicular distance from the axis.)

    So....I drew a picture of this stool. The radius of the top is 0.12m and each leg/rod would also have the same radius. My thinking in this is because they are placed at the outside edge of the disk. The axis I drew was perpendicular to the top in it's center, making all three legs/rods the same distance away from the axis which I thought to be 0.12m.

    So I used this equation to find the answer for the whole system/stool but was wrong with my answer ----> [itex]I = m_1r_1^2 + m_2r_2^2[/itex]

    But now I just started thinking that perhaps this is what I should be doing instead.

    That above equation equates to this equation -----> [itex]I = mL^2[/itex] Which would find the inertia of the rod only. Again this is just what I'm thinking. So what I should be doing is using my first equation above to find the inertia of the rods, and then find the inertia of the disk and then add those two to get my final anwer.

    I know to find the inertia of the disk -----> [itex]I = \frac{1}{2}MR^2[/itex]

    The thing I'm a little unsure of is, do I need to find the inertia for all three rods, or can I just find it for one of them and multiply that number by 3, or will one just do....

    Any help provided will be greatly appreciated.
     
  2. jcsd
  3. Oct 26, 2004 #2

    arildno

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    Moments of inertia are additive.
    What values did you try out?
    (I thought the m1,r1,m2,r2 notation was rather cryptic..)
     
  4. Oct 26, 2004 #3
    Ok, I think you answered my question then. I should be finding the moment of inertia for each rod and add that to the moment of inertia for the disk...

    I didn't think the m1r1 notation was cryptic...m1 being the mass of the rod : .12kg and r1 being the radius : .12m. My other question was could I just calulate the inertia for one rod and then multiply by three or do I need to calculate them individually then add?
     
  5. Oct 27, 2004 #4

    arildno

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    Your notation was rather cryptic to me, since you never specified what the terms where supposed to mean..

    Just multiply the moment of inertia for one rod with 3 and add that to the moment of inertia of the disk.
     
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