# Moment of inertia of ring with 3 masses on

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1. Jun 3, 2015

### j3dwards

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. Jun 3, 2015

### SteamKing

Staff Emeritus
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. Jun 3, 2015

### j3dwards

So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2

4. Jun 3, 2015

### SteamKing

Staff Emeritus
Don't forget to include the contribution to the MOI from the ring.

5. Jun 3, 2015

### j3dwards

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. Jun 3, 2015

### SteamKing

Staff Emeritus
That looks much better.

7. Jun 3, 2015

### j3dwards

Thank you so much!!