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Moment of Inertia and anemometer

  1. Apr 18, 2004 #1
    An anemometer for measuring wind speed consists of four metal cups, each of mass m = 146 g, mounted on the ends of essentially massless rods of length L = 0.3 m. The rods are at right angles to each other and the structure rigidly rotates at f = 12 rev/s. Treat the cups as point masses.

    a) What is the moment of inertia of the anemometer about the axis of rotation?

    I know moment of inertia is calculated by the sum of masses times distance to axis of rotation. I tried to substitute (L^2)/2 as my distance to the rotational axis, and I know it's 4 times each mass, (I also converted to kg) so I got 0.0827 but it's not correct. Any help would be appreciated
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2004 #2

    Doc Al

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    Staff: Mentor

    Not true. The rotational inertia of a point mass is I = mR2, where R is the distance to the axis.
    I don't know if R = L or L/2, since I don't know how "rod" is defined: are there two rods or four?

    In this case, the total I = 4mR2.
     
  4. Apr 18, 2004 #3
    The picture shows four rods. Yeah sorry I meant MR^2, basically I used pythagorem theorem and I found that R^2 equals 2(L/2)^2 which simplifies to (L^2)/2. I still can't solve the problem though because I don't know how to define the distance from each of the point mass to the axis of rotation.
     
  5. Apr 18, 2004 #4

    Doc Al

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    Is it four rods of length L? Or two rods of length L attached in the middle?
    Why are you using the pythagorean theorem??

    If there are four rods of length L, then the distance to the axis is L.
     
  6. Apr 18, 2004 #5
    That's a really good question... :biggrin: unfortunately the question doesn't make it clear...can you explain to me though how I would go about trying each setup-both 2 rods connected at the center or 4 rods connected. Thank you.
    If there are four rods of length L, then the distance to the axis is L.[/QUOTE]
    I tried just squaring L and plugging it into our I equation (4ML^2) but it still registers as incorrect...sorry for all the grief doc.
     
    Last edited: Apr 18, 2004
  7. Apr 18, 2004 #6

    Doc Al

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    The equation is I = 4MR^2. Try assuming that R = L/2.
     
  8. Apr 18, 2004 #7
    Yup! That worked. Thanks a lot for your patience
     
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