Rotational Dynamics - moment of inertia finding

  • Thread starter pat666
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  • #1
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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 wouldnt work (tried it just to make sure) anyways i need some help
 

Answers and Replies

  • #2
Doc Al
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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
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
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so for the the total moment of inertia: 40*1.7^2+-35*1.7^2+Irod?
 
  • #4
Doc Al
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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.)
 

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