# Rotational Inertia Seesaw Picot

1. Nov 18, 2009

### premedonna89

CORRECTION ON TITLE: it should be "Rotational Inertia Seesaw Pivot"

1. The problem statement, all variables and given/known data

Alice (20kg) and Bonnie (25kg) sit on seesaw (10m long, 12 kg mass). Pivot of seesaw is in the middle. In order for the seesaw to stay level we determined (in the previous step of the multi-part problem) that Bonnie has to sit 4 m away from the pivot/center.

What is the rotational inertia of this system about the seesaw pivot?

2. Relevant equations

Inet = Ia + Ig + Ib

Ig = 1/12 x mass x length^2^

3. The attempt at a solution

I actually have the answer already (this is a practice exam). But I don't understand how he got it.

Inet = Ia + Ig + Ib

Inet = mass,a x radius,a^2^ + 1/12 x mass x length^2^ + mass,b x radius,b^2^

Inet = 500 + 100 + 25

Inet = 625 kg/m^2^

I don't understand how:

Shouldn't it be: 25 x 4^2^ = 400 ???

I imagine radius, b = 1, therefore 1 square is just 1. Since Bonnie is sitting 4 meters away from the center, and the edge is 5 meters away from the center, so 5 minus 4 = 1. Then multiply that by 25 kg to get 25. However, I can't find anywhere in the textbook section for inertia that says to measure radius from the end of the seesaw (?) rather than from the center/pivot.

Thank you!

2. Nov 18, 2009

### clamtrox

You can of course calculate the moment of inertia with respect to any point (and the result is different for different points) but you're right about this making no sense at all. The distance in your formulas is indeed the distance from the point where you wish to calculate I.