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Moment of inertia of ring with 3 masses on

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Three particles, each with mass m = 0.5 kg, are fixed on a uniform circular ring of mass M = 2 kg, radius a = 0.6 m and centre O, to form the corners of an equilateral triangle. What is the moment of inertia of this system?

    2. Relevant equations
    I = Σma2

    3. The attempt at a solution
    I = 3(2 x 0.62) = 2.16 kgm2

    I'm confused because it is either that or zero and I can never tell when it is zero.

    Thank you.
     
    Last edited: Jun 3, 2015
  2. jcsd
  3. Jun 3, 2015 #2

    SteamKing

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    The moment of inertia is never zero, unless the mass is zero.

    Draw a sketch of the ring and the masses. Use the Parallel Axis Theorem to calculate the MOI of this assembly.
     
  4. Jun 3, 2015 #3
    So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2
     
  5. Jun 3, 2015 #4

    SteamKing

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    Don't forget to include the contribution to the MOI from the ring.
     
  6. Jun 3, 2015 #5
    Oh I've been doing this very wrong. So is it: I = Σma2 + MR2 = 3(0.5 x 0.62) + (2 x 0.62) = 1.26 kgm2
     
  7. Jun 3, 2015 #6

    SteamKing

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    That looks much better.
     
  8. Jun 3, 2015 #7
    Thank you so much!!
     
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