# Gravitational potential energy

1. Oct 19, 2016

### Lis

1. The problem statement, all variables and given/known data
A teeter toy is composed of a massless central stick of length L and two massless sticks of length l attached at angles α, each with a mass m at the end (see the figure). We imagine tilting the toy by an angle θ from the upright position.

a) Find an expression for the gravitational potential energy of the whole object, as a function of θ.

2. Relevant equations

U=mgh

3. The attempt at a solution

U(θ) = mg(L-Lcos(θ))

2. Oct 19, 2016

### kuruman

Can you provide your reasoning how you got this equation? A drawing would be helpful.

3. Oct 19, 2016

### Lis

4. Oct 19, 2016

### kuruman

Thanks for the drawing.
Is this correct? Let's see. It says that when θ = 0 (top figure), the potential energy is zero. That's defines your choice of reference. Your expression also says that when θ = 90o, the potential energy is mgL. Does that look right? What exactly finds itself at distance L above your reference when the toy is tipped 90o? To see how to treat the problem sensibly, consider that gravity is an external force acting on a system of two masses. Therefore we can view gravity as acting on the ____ of the two masses. (Fill in the blank.)

5. Oct 19, 2016

### Lis

I dont think i understand you quite. Do you mean " Therefore we can view gravity as acting on the length of the two masses"?

6. Oct 19, 2016

### kuruman

You have a system of two equal masses. If you were to treat this system as if its entire mass of 2m were concentrated at one point, where would that point be?

7. Oct 19, 2016

### Lis

on the top off L?

8. Oct 19, 2016

### kuruman

9. Oct 19, 2016

### arpon

Potential energy, $$U_{total}= m_{total} \cdot g\cdot h_{center~of~mass}$$
Or you can calculate the height for the two masses separately, then calculate their respective potential energy and add them.