Moment of inertia of ring with 3 masses on

[j3dwards]

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.

 

[SteamKing]

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.

 

[j3dwards]

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

 

[SteamKing]

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.

 

[j3dwards]

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

 

[SteamKing]

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.

 

[j3dwards]

That looks much better.

Thank you so much!