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Seesaw physics and center of mass

  1. Sep 24, 2010 #1
    I want to write a program that basically mimics the physics of seesaws. Part of a project I'm working on. I understand how equilibrium works and everything, but let's say that I have a mass on each end of the seesaw, and both masses are equal distance from the pivot. If both masses have the same mass then the seesaw would be in equilibrium and the seesaw would be horizontal. But, what if one of the masses is heavier than the other? The center of mass would be shifted towards the heavier mass and the seesaw would start to rotate since the center of mass applies a torque. I'm just trying to find an equation that will tell me how fast it's rotating depending on the masses. I broke out my old physics book, but the equations I'm coming up with aren't working. Maybe there's just something I'm missing. Are there 3 torques from the two ends and center of mass or just the center of mass torque? can I use the moment of inertia here or is it not necessary? Any help would be greatly appreciated.
  2. jcsd
  3. Sep 29, 2010 #2
    Assuming the see-saw is anchored in the middle, apply Newtons second law for circular motion. That is the total rotating moment( force x arm) = polar moment of inertia x angular acceleration. The net force is weight(gravity force with direction taken correctly) x arm . Integrating this gives you angular velocity and integrating again gives you angular displacement . This works for any masses and is a standard problem in mechanics. You can include the torque due to friction in the anchor and due to moving in air if you want too.
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