Statics: Potential Energy

In summary, the conversation discusses a physics problem involving a spring and a rod, and the process of finding the equilibrium position and determining stability using the second derivative of the potential energy equation. The conversation concludes with the realization that the equilibrium position is stable at 0 degrees and unstable at 31 degrees, contrary to the initial intuition.
  • #1
epheterson
22
0

Homework Statement


Includes picture, download question http://docs.google.com/Doc?id=ajccwmhrc3rx_37gs2ht7g9" [Broken].

Homework Equations


The potential energy equation ends up being
[tex]V = 1/2 k x^2 + W y[/tex]

The Attempt at a Solution


xspring = 2 sin([tex]\Theta[/tex])
ym = 2 cos([tex]\Theta[/tex])
[tex]V = 1/2*40 (2 * sin(\Theta))^2 + 7 * 2 * g cos \Theta[/tex]
[tex]dV/d\Theta = 160sin(\Theta)cos(\Theta)-14g*sin(\Theta) = 0[/tex]

Solving returns theta = 0, 0.5404057 rad (31 degrees)

That sounds about right, so now I look for stability

[tex]d^2V/d\Theta^2 = 160cos(\Theta)^2-160sin(\Theta)^2-14g*cos(\Theta)[/tex]

This is where I'm stuck. Using intuition, I feel the system should be UNstable at theta = 0 and find stability at theta = 31 degrees.

Solving the second derivative, I get [tex]d^2V/d\Theta^2(0) > 0[/tex] and [tex]d^2V/d\Theta^2(.54) < 0[/tex]

Is the math wrong or is my understanding of stability wrong?
 
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  • #2
The math looks OK.

Did you sketch the potential? Try drawing it for [tex]-\pi/2 \le \theta \le \pi/2[/tex] (radians).
How does the graph indicate whether the extremum is stable or unstable?
 
  • #3
I found out I did it exactly right. When the second derivative is negative, the system is unstable and when it's positive, the system is stable.

What's going on is at the top position the spring is pulling on the rod and keeping it in place, so without excess force pushing on the rod, it will tend to stay in place, even if you push a fair amount, the spring will bounce back into place and the rod will be vertical again.

If you find the equilibrium position at 31 degrees, it will stay but be quite fragile. if you push down, the entire system will give and the rod will lay flat. If you push the rod to the right, the spring will snap it back to the vertical position.

So there we have it, after all my math was correct and my understanding of stability was wrong.

My teacher was as blown away as I was.
 
  • #4
As I said, nothing was wrong with your math, just with your intuition :smile:

Attached is a graph of V(theta) zoomed in a bit (from -pi/4 to pi/4). You can clearly see here that the minimum at 0 is stable (imagine putting a ball close to it, it will roll down to 0). The maximum is clearly unstable, if you put a ball exactly there it will stay, but if you move it a little it will fall down.
 

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1. What is the definition of potential energy in statics?

Potential energy in statics refers to the energy stored in a system due to its position or configuration. It is the energy that an object has because of its potential to do work.

2. How is potential energy calculated in statics?

Potential energy in statics is calculated by multiplying the force acting on an object by its displacement. The formula for potential energy is PE = F * d, where PE is potential energy, F is the force, and d is the displacement.

3. What are the different types of potential energy in statics?

The different types of potential energy in statics include gravitational potential energy, elastic potential energy, and electric potential energy. Gravitational potential energy is the energy stored in an object due to its position in a gravitational field, elastic potential energy is the energy stored in a stretched or compressed spring, and electric potential energy is the energy stored in an electric field.

4. How does potential energy affect the stability of a system in statics?

Potential energy can affect the stability of a system in statics by determining the equilibrium point of the system. If the potential energy of a system is at a minimum, the system is in a stable equilibrium. However, if the potential energy is at a maximum, the system is in an unstable equilibrium and any small disturbance can cause the system to collapse.

5. What is the relationship between potential energy and kinetic energy in statics?

In statics, potential energy and kinetic energy are interchangeable. As an object moves, its potential energy will decrease as its kinetic energy increases, and vice versa. This is known as the principle of conservation of energy, where the total energy of a system remains constant.

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