Moment of inertia of rug beater

[man_in_motion]

Homework Statement

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?

Homework Equations

I=\Sigma m_i r^2_i=\int r^2 dm
the rod has inertia of
I=\frac{1}{3}ML^2

The Attempt at a Solution

I break the problem into 2 parts: the handle and the square.
The handle can be found using I=\frac{1}{3}ML^2
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.

 

[benhou]

You can use the parallel axis theorem. By the way, how did you make it show up in the google search?

 

[man_in_motion]

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

 

[benhou]

First find the rotational axis through the centre of mass, make sure the axis is parallel to the actually axis of rotation. And then use the theorem.

By google search, I meant this

http://www.google.com/search?um=1&h...uare rug beater&aql=&oq=&ie=UTF-8&sa=N&tab=iw

 

[benhou]

Hey, I saw your post in Physics Help Forum as well. Are you still stuck with the question? If you had shown that you tried for the square part like you did in that Forum, I would have followed up.
Anyway, the rotational axis is at the end of handle as we know. Before we shift the rotational axis to the centre of mass and apply the parallel axis theorem, it's important to know the orientation of the square relative to the rotational axis. If we shift it, the axis goes through two thin edges. If you can figure out the moment of inertia at that axis, you can apply the theorem. Now try it.