Elastic & gravitational potential energy

[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.

 

[tiny-tim]

Welcome to PF!

Hi Set Abominae! Welcome to PF!

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

… 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.

Just ask yourself: will it make the bead go slower?

ie will increasing the height make the bead go slower? will increasing the length of the elastic make the bead go slower?

I know that … elastic energy is (k/2)(h^2 + x^2 - 2hx^2 + x^4)

how did you get that?

and what is the unstretched length of the elastic?

 

[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:

 

[tiny-tim]

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

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)

So that will be add, then?:shy:

i'm not sure what you're adding to what …

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

 

[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...

 

[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...

 

[tiny-tim]

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.).

Are you taking into account the fact that gpe is minus mgh?

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 …

urggh … don't know anythng about bifurcation diagrams

 

[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?

 

[Set Abominae]

Are you taking into account the fact that gpe is minus mgh?

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?

 

[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?

 

[Set Abominae]

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?

f(x)=\frac{k}{2}(x^{2}+(h-x^{2})^{2}+mgx^{2}<br /> \Rightarrow f&#039;(x)=2kx^{3}+kx-2hkx+2mgx
Solve f&#039;(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&gt;\frac{mg}{k}+\frac{1}{2}

Now, f&#039;&#039;(x)=6kx^{2}+k+2mg-2hk<br /> h&lt;\frac{mg}{k}+\frac{1}{2} \Rightarrow f&#039;&#039;(0)=2mg+k-2hk&lt;0 \Rightarrow stable.
(Note, h=\frac{mg}{k}+\frac{1}{2} \Rightarrow f&#039;&#039;(0)=0 \Rightarrow system is structurally unstable.)

h&gt;\frac{mg}{k}+\frac{1}{2} \Rightarrow f&#039;&#039;(0)&gt;0 \Rightarrow unstable, and
f&#039;&#039;(\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}})=4hk-4mg-2k&gt;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?

 

[mirabella]

[QUOTE f''(0)=2mg+k-2hk
h&gt;\frac{mg}{k}+\frac{1}{2} \Rightarrow f&#039;&#039;(0)&gt;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?

 

[mirabella]

f''(\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}})=4hk-4mg-2k>4k(\frac{mg}{k}+\frac{1}{2})-4mg-2k=0 \Rightarrow[/tex] unstable.

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?

 

[Set Abominae]

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?

Yes, I now have:

1 unstable fixed point x=0 for h&lt;\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&gt;\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...