Rotational Dynamics - moment of inertia finding

[pat666]

Homework Statement

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

Homework Equations

tau=rF=Ialpha

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 wouldn't work (tried it just to make sure) anyways i need some help

 

[Doc Al]

my problem is finding the moment of inertial i have tried mr^2 which i knew wouldn't work (tried it just to make sure) anyways i need some help

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.

 

[pat666]

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

 

[Doc Al]

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

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