# Moment of inertia of ring with 3 masses on

## 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?

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:

SteamKing
Staff Emeritus
Science Advisor
Homework Helper

## 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?

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.

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
Staff Emeritus
Science Advisor
Homework Helper
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.

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
Staff Emeritus
Science Advisor
Homework Helper
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.

That looks much better.
Thank you so much!!