How Much Power Is Needed to Oscillate and Stop a Seesaw at 30°?

[Eemu]

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.

 

[HallsofIvy]

Are you sure the question asked about "power" (perhaps it is a translation problem).

You should be able to calculate the total energy in the seesaw (kinetic energy and potential energy change from moment to moment but the total energy stays the say- calculate the potential energy at the extreme angle, when the kinetic energy is 0, relative to the horizontal ) and, since the seesaw in equilibrium had no energy- that is the energy that must have been put in.
I don't see HOW you can calculate "0.88 rad/s" since you haven't said anything about how fast the seesaw is going or what time it takes to go from horizontal to 30 degrees.
Assuming no friction, which you pretty much have to since you are given no way of calculating friction, exactly the same energy is required to stop it.

However, technically (and in English), "power" is "energy per unit time" and you have given no time required for this.

 

[Eemu]

You're right, I failed to give the required information.

The factors that affect the angular acceleration are the length of the board L (20m) and the gravitational acceleration g (9.82m/s^2)

The angular acceleration can be considered to be constant (for small inclinations) and changing direction each time the board passes its equilibrium position.

Angular acceleration for an unloaded seesaw:
a = T / I =(1/2)mg(1/4)L / (1/12)mL^2 = (3/2)g / L = 0.75rad/s^2

Velocity when passing the equilibrium can be calculated from the constant acceleration equation 2as = v^2 - v0^2 (0.88rad/s), and the oscillation period from v = v0 - at (1.18s from 30° to equilibrium)

This is with no friction.

Apologies for my insecure english, this is the first time I speak physics in this language.