- #1
Eemu
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Problem: I have to find out the power that is required to set the balanced seesaw from equilibrium to an oscillation that changes direction in an inclination of 30°.
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.
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.