Masses on a rod inertia problem

[Staerke]

Homework Statement

Three balls are attached to a rod. Ball one has mass 0.614 kg and is 14.0 cm from the pivot. Ball two has mass 0.634 kg and is 25.0 cm from the pivot. Ball three has mass 0.729 kg and is 35.0 cm from the pivot. The rod has length 86.0 cm and mass 0.486 kg. Calculate the rotational inertia of the system (the size of the balls may be neglected).

THEN

If the system starts to spin around its pivot point with an angular speed of 5.45 radians per second, what is the kinetic energy of the system?

Homework Equations

I'm stuck on the first part, I=(1/3)ML^2

The Attempt at a Solution

I know since it's a pivoting rod it's (1/3)ML^2
What I've done is this:
(1/3*.14/86*.25/.86*.35/.86) = .00642
Basically adding up the fractional lengths for each mass.

Next I add up the masses including the mass of the rod and multiply it by the number above and the length squared:
.00642*((.614+.634+.729+.486)*(.86^2) and get .011694 which is obviously wrong.

I have absolutely no idea what to do here, so I'd really appreciate a breakdown of how to do this (without calculus, we're not using calculus in my introductory physics class and honestly it will just confuse me)
Thanks for any assistance.

 

[clementc]

Without calculus =(
Well, in a dodgy series notation kind of thing...
I=\sum mr^2=m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 +...
In other words, the moment of inertia of a body is just the sum of all the individual moment of inertias of every point on it.
For example , if I had a rod of negligible mass with a 2kg ball 3m from the pivot and 4kg ball 5m from the pivot, the moment of inertia of the whole thing would be 2(3²) + 4(⁵) = whatever I cbb reach for calculator LOL

So what you want to do in this question is add up the moment of inertias to get the total moment of inertia. You have to also add the moment of inertia of the rod as its mass using 1/3ML^2 is not negligible.