# Elastic & gravitational potential energy

1. Mar 22, 2009

### Set Abominae

Hi there.

I have a bead of mass m, which slides down a frictionless parabolic wire in the form y=x^2, and is attached by elastic to the point (0,h), and I want to write down total energy f(x) (= elastic energy + gravitational potential energy) (no mention of kinetic energy...) of the bead at (x,x^2).

I know that g.p.e is mgx^2, and that elastic energy is (k/2)(h^2 + x^2 - 2hx^2 + x^4), but I'm not sure whether I'm adding the g.p.e. to the elastic energy, or subtracting it, despite the fact that I want f(x) = elastic energy + g.p.e.

Any clarification would be great :)

Thanks.

2. Mar 22, 2009

### tiny-tim

Welcome to PF!

Hi Set Abominae! Welcome to PF!

(try using the X2 tag just above the Reply box )

ie will increasing the height make the bead go slower? will increasing the length of the elastic make the bead go slower?
how did you get that?

and what is the unstretched length of the elastic?

3. Mar 22, 2009

### Set Abominae

The question states that:
When the elastic has length d its elastic energy is $$\frac{k}{2}$$ $$d^{2}$$, where d> 0.....

So when the bead is at $$(x,x^{2})$$, we have that:
$$d = \sqrt{x^{2}+(h-x^{2})^{2}}$$

(It'a a question from a math assignment, so its probably not totally accurate...)

So that will be add, then?:shy:

4. Mar 22, 2009

### tiny-tim

hmm … that's unusual … but if the question says so, i suppose it's ok

(perhaps the elastic starts further back, and goes round a peg at (0,h), and the unstretched length is up to the peg)
i'm not sure what you're adding to what …

anyway, give a reason, so we can see you're not guessing!

5. Mar 22, 2009

### Set Abominae

Actually, thinking about it, I would imagine that I would subtract the g.p.e from the elastic energy to get f(x) (though I'm cautious about doing so, as the question says f(x)=elastic energy + g.p.e.).

Another reason for doing this is that if I sum them to get f(x), when I start drawing my bifurcation diagram later in the question, I get:

A single fixed point $$x=0$$ is stable for $$h<\frac{mg}{k}+\frac{1}{2}$$,

and 3 fixed points

$$x=0, \pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}$$ are all unstable for $$h>\frac{mg}{k}+\frac{1}{2}$$, and I've never seen a system where all 3 fixed points are unstable before...

6. Mar 22, 2009

### Set Abominae

But I also get the same result of 3 unstable fixed points when I subtract the g.p.e. from the elastic energy....

7. Mar 23, 2009

### tiny-tim

Are you taking into account the fact that gpe is minus mgh?
urggh :yuck: … dunno anythng about bifurcation diagrams

8. Mar 23, 2009

### Bob S

Here is a related problem that I. Newton solved in one day back in about 1697. Suppose we have a frictionless bead sliding on a wire from point A at (0,h) to point B at (x,0) by the fastest route. What is the shape of the wire from point A to point B? A straight line is the shortest distance, but a parabola is longer and faster. But a parabola is not the fastest. So what is?

9. Mar 23, 2009

### Set Abominae

I forgot about that! Upon further thought, I would imagine that I would subtract the negative gpe, so be adding it to the elastic energy to give f(x). I say this by considering energy conservation - assuming we don't have any stupidly big oscillations, the higher the bead gets, the larger mgh gets, and the smaller the elastic energy.

Likewise, when the bead is at x=0, it has it least gpe (0 if we take the x-axis as our 'zero height', but the elastic is at maximum extension (again, assuming well-behaved oscillations).

So by energy conservation, I'm inclined to write that
$$f(x) = (elastic energy) + mgx^{2}$$

Thoughts?

10. Mar 26, 2009

### mirabella

I have the same problem for my maths assignment :) check ur calculations carefully. I have a single unstable fixed point at x=0 for h<mg/k+0.05, second derivative is >0 and 3 stable fixed points for h>mg/k+0.05, second derivative<0. Why do u have the opposite?

11. Mar 26, 2009

### Set Abominae

$$f(x)=\frac{k}{2}(x^{2}+(h-x^{2})^{2}+mgx^{2} \Rightarrow f'(x)=2kx^{3}+kx-2hkx+2mgx$$
Solve $$f'(x)=0$$ to give $$x=0$$ for $$h\leq\frac{mg}{k}+\frac{1}{2}$$,
and $$x=0, x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}$$ for $$h>\frac{mg}{k}+\frac{1}{2}$$

Now, $$f''(x)=6kx^{2}+k+2mg-2hk h<\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)=2mg+k-2hk<0 \Rightarrow$$ stable.
(Note, $$h=\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)=0 \Rightarrow$$ system is structurally unstable.)

$$h>\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)>0 \Rightarrow$$ unstable, and
$$f''(\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}})=4hk-4mg-2k>4k(\frac{mg}{k}+\frac{1}{2})-4mg-2k=0 \Rightarrow$$ unstable.

(Sorry for the mess!)

I don't think there's any mistakes here... How did you go about solving it?

12. Mar 27, 2009

### mirabella

[QUOTE f''(0)=2mg+k-2hk
$$h>\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)>0 \Rightarrow$$ unstable QUOTE]

multiply h>mg/k+0.05 by 0.05k ( sign stays the same since k>0). U get 2kh>2mg+k i.e
2mg+k-2kh<0 so f''(0)<0 stable ?? Same for the other two.
I know it's a cusp catastrophe but how do u justify that?
Do u have a symmetry question (with a see-saw) later on in your assignment?

13. Mar 27, 2009

### mirabella

Actually this is ok. I had a wrong sign there, but u still have a mistake for x=0 so check it. We have ether one unstable at x=0 or one stable at x=0 and two unstable at +-sqrt(h-mg-1/2). Bifurcation looks much better now :) Thoughts?

14. Mar 27, 2009

### Set Abominae

Yes, I now have:

1 unstable fixed point $$x=0$$ for $$h<\frac{mg}{k}+\frac{1}{2}$$

1 fixed point $$x=0$$ for $$h=\frac{mg}{k}+\frac{1}{2}$$, and system is structurally unstable.

1 stable fixed point $$x=0$$ and 2 unstable fixed points at
$$x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}$$ for $$h>\frac{mg}{k}+\frac{1}{2}$$

I do indeed have a symmetry question (check your private messages). I wasn't really too sure on whether this was a cusp catastrophe or not. I couldn't use any of the equations for the canonical cusp catastrophe, but was inclined to believe that it's a cusp catastrophe, as we just from a maximum to a mininum and two maximums (basically, the opposite of the canonical cusp catastrophe: just the same situation with all equations made negative). But, that's my only real argument....