Gravitational potential energy of a tilted teeter toy

[Lis]

Homework Statement

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.

Homework Equations

U=mgh

The Attempt at a Solution

U(θ) = mg(L-Lcos(θ))

 

[kuruman]

U(θ) = mg(L-Lcos(θ))

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

 

[Lis]

 

[kuruman]

Thanks for the drawing.

U(θ) = mg(L-Lcos(θ))

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 = 90^o, the potential energy is mgL. Does that look right? What exactly finds itself at distance L above your reference when the toy is tipped 90^o? 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.)

 

[Lis]

I don't think i understand you quite. Do you mean " Therefore we can view gravity as acting on the length of the two masses"?

 

[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?

 

[Lis]

on the top off L?

 

[kuruman]

Please read this
http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html
and see if it rings a bell.

 

[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.