Calculating moment of inertia of a system of balls

haki

I have a bit hard time calculating moment of inertia of a a system of balls. The system is defined as follows: 3 balls that are the same are connected with thin rods soo that the center of each ball is in the corners of a equilateral triangle with length of side to be 0.3 m. Radius of each ball is 0,08 m and mass of each ball is 1,2 kg. Now the fun part: Calculate moment of inertia of the system if the axis of rotation goes trught the center of mass of the equilateral triangle. I of ball is 2/5*m*r^2.
I am puzzled. I know that the center of mass for a equilateral triangle is in 2/3 of its height. Soo each ball is exacly 1/3*Sqrt(3)*a -> 1/3*Sqrt(3)*0,3m away from the axis of rotation. I would simply say that the moment of inertia I = Sigma(m*r^2). I of each ball is I of b = m*(1/3*Sqrt(3)*a)^2 = 1.2 kg * 1/9 * 3 * 0.09 m^2. And then the moment of inertia of the system is simply I of system = 3 times I of ball. I am missing something. Since the radious is also given. I would apprichiate greately if somebody would explain the reasoning behind this problem and also if possible the path to solution.

mukundpa

Go through the parallel axis theorem of M.I.
The ball is not a point mass.

haki

Ah...
The Moment of inertia of the ball is I = 2/5*mass*radius^2 + mass*distance from the axis of the system^2. And the moment of inertia of the system if 3 times of that of single ball?

mukundpa

looks good !

haki

Ok, Thanks!