Homework Help: Moment of inertia of rug beater

1. The problem statement, all variables and given/known data
There are 2 models of rugbeaters. Model A has a 1m long handle and a 40cm edge length square. The handles mass is 1kg and the squares mass is 0.5kg. Model B has a 0.75m long handle and a square that has 30cm edge length. The mass of the handle is 1.5kg and the mass of the square is 0.6kg. Which model is easier to use?

2. Relevant equations
[tex]I=\Sigma m_i r^2_i=\int r^2 dm[/tex]
the rod has inertia of
[tex]I=\frac{1}{3}ML^2[/tex]


3. The attempt at a solution
I break the problem into 2 parts: the handle and the square.
The handle can be found using [tex]I=\frac{1}{3}ML^2[/tex]
I'm not sure how to find the inertia of the square part, it's distance is growing longer away from the axis of rotation. I'm guessing I'm going to have to calculate an integral.

 

how would using the parallel axis theorem help? I don't know what u mean about the google search.