Moment of inertia of ring with 3 masses on

  • #1
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Homework Statement


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?

Homework Equations


I = Σma2

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:

Answers and Replies

  • #2
SteamKing
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Homework Statement


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?

Homework Equations


I = Σma2

The Attempt at a Solution


I = 3(2 x 0.6) = 1.728 kgm

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

Thank you.
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.
 
  • #3
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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.

So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2
 
  • #4
SteamKing
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So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2
Don't forget to include the contribution to the MOI from the ring.
 
  • #5
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Don't forget to include the contribution to the MOI from the ring.

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
 
  • #6
SteamKing
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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
That looks much better.
 
  • #7
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