# Rotational Dynamics - moment of inertia finding

1. May 24, 2010

### pat666

1. The problem statement, all variables and given/known data
2 children, masses 35kg and 40kg sit at opposite ends of a 3.4m seesaw with mass 25kg with the fulcrum at midpoint. with the seesaw horizontal find
a) the net torque
b)angular acceleration

2. Relevant equations

tau=rF=Ialpha

3. The attempt at a solution
i got the right answer for a ----83.385Nm which i know is equal to Ialpha. my problem is finding the moment of inertial i have tried mr^2 which i knew wouldnt work (tried it just to make sure) anyways i need some help

2. May 24, 2010

### Staff: Mentor

Actually mr^2 will work for part of the answer. To find the total moment of inertia, add up the contributions of each part: The two children plus the seesaw itself. Hint: Treat the children as point masses; treat the seesaw as a thin rod.

3. May 24, 2010

### pat666

so for the the total moment of inertia: 40*1.7^2+-35*1.7^2+Irod?

4. May 24, 2010

### Staff: Mentor

Yes. (But get rid of that minus sign--I suspect it's just a typo.)