- #1

Eemu

- 2

- 0

I also have to know the power required to eventually stop that motion and bring the seesaw back to equilibrium.

I have calculated the angular velocity at the equilibrium point (when dropped from 30°) to 0.88 rad/s. Answer to my first question ought to be that the power is the same that is required to set the seesaw in 0.88 rad/s, but I just can't see the equation.

Moment of inertia for a seesaw = (1/12)mL^2

Many thanks for your time!

Eemu from Gothenburg, Sweden.