# Moment of inertia of a three particle system

1. Jun 3, 2012

### Avatrin

1. The problem statement, all variables and given/known data
I have three particles of mass m at (-a,-a), (a,-a) and (0,a)

Find the moment of inertia I_z around the center of mass of the system for an axis along the z-axis.

2. Relevant equations
Center of mass:
CM = Ʃmr

Moment of inertia:
I = Ʃmp^2

m = mass
r = position
p = distance from axis

3. The attempt at a solution
I found that the center of mass is CM = (0,-a/3);
CM_x = (-am + am + 0m)/3m = 0
CM_y = (-a-a+a)m/3m = -a/3
This is something I know is correct..

The real problem starts with the moment of inertia:
p_1 = 4a/3
p_2 = p_3 = √(a^2 + (2a/3)^2)

(p_1)^2 = (16/9)a^2
(p_2)^2 = (a^2 + (2a/3)^2) = (13/9)a^2

Ʃmp^2 = m(16/9)a^2 + m(13/9)a^2 + m(13/9)a^2
= m((16/9)a^2 + (26/9)a^2) ≈ 4.67ma^2

The problem is that the book says that the solution is 4.01ma^2. Where have I made a mistake?

2. Jun 3, 2012

### Sleepy_time

I got the answer that you had as well. Make sure that you've read the question correctly.

3. Jun 3, 2012

### Avatrin

Last edited: Jun 3, 2012