Simple elasticity question

Hey, can someone explain where I'm going wrong here? I'm getting different answers with different methods. I hoped I would find my problem during the writeup but I haven't spotted it. Sorry for the bad formatting.

A particle, of mass 5kg, is attatched to one end of a light elastic string of natural length 0.6 metres and modulus of elasticity 150N. The other end of the string is fixed to a point O.

Find the extension of the string when it hangs in equilibrium at point A below O.

By forces:

At A,

m g = \frac{\lambda e}{l}

(5) (9.8) = \frac{150 e}{0.6}

e = 0.196m

By energy:

At O, taking A as the zero-line for GPE:

Energy = GPE = m g h = 49 (0.6+e)

At A:

Energy = EPE = 49 (0.6+e) = \frac{\lambda e^{2}}{2 l}

49 (0.6+e) = \frac{150 e^{2}}{1.2}

29.4+49e-125e^{2} = 0

125e^{2}-49e-29.4 = 0

e = \frac{49 + (49^{2} - (4) (125) (-29.4))^{0.5}}{250} = 0.719m

Reply 1

Stonebridge

Original post by Xdaamno

m g = \frac{\lambda e}{l}

(5) (9.8) = \frac{150 e}{0.6}

e = 0.196m

By energy:

At O, taking A as the zero-line for GPE:

Energy = GPE = m g h = 49 (0.6+e)

At A:

Energy = EPE = 49 (0.6+e) = \frac{\lambda e^{2}}{2 l}

49 (0.6+e) = \frac{150 e^{2}}{1.2}

29.4+49e-125e^{2} = 0

125e^{2}-49e-29.4 = 0

e = \frac{49 + (49^{2} - (4) (125) (-29.4))^{0.5}}{250} = 0.719m

Where do you get the equation

m g = \frac{\lambda e}{l}

from?

I assume the "modulus" value of 150N should have been 150Nm^-1 and is the Hooke's Law constant (not the Young Modulus which would have units Nm^-2)
In which case the formula is

F = ke where k is the Hooke's Law constant and e the extension.

Reply 2

jaroc

Original post by Xdaamno

By energy:

At O, taking A as the zero-line for GPE:

Energy = GPE = m g h = 49 (0.6+e)

At A:

Energy = EPE = 49 (0.6+e) = \frac{\lambda e^{2}}{2 l}

49 (0.6+e) = \frac{150 e^{2}}{1.2}

29.4+49e-125e^{2} = 0

125e^{2}-49e-29.4 = 0

e = \frac{49 + (49^{2} - (4) (125) (-29.4))^{0.5}}{250} = 0.719m

Well, I believe it is possible to define Hooke's law constant as

k\equiv \lambda /l_0

. But it's a minor problem here I think.

If you want to do this by considering energy, you can't say that equilibrium is at a point where GPE=EPE. What's more, it doesn't really make sense to calculate GPE, because there is no privileged height to which you could refer.

If you want to do this by energy, you need to know that equilibrium is at such a point where the total energy of the system is minimum. So you would need to add GPE+EPE and find such a value of

e

that the derivative of GPE+EPE is 0. Remember that the more the rope is extended, the less gravitational potential energy is there, so that

GPE=mg\left( H_0-e\right)

where

H_0

is the height of the end of the rope when it is not extended (no mass attached to its end).

(edited 13 years ago)

Reply 3

Xdaamno

Thanks for the reply. This is actually a mechanics question in the maths a-level, and the modulus of elasticity is taught in that as basically being defined by that formula. The differentiated-with-respect-to-e formula gives the elastic potential energy at an extension. I'm on my iPod and can't give any good links to back this up, but I'm certain I've got those formulas right!

Reply 4

Xdaamno

Thanks jaroc, I'll reply tomorrow.

Reply 5

Stonebridge

Original post by Xdaamno

As this is a physics forum I assumed it was meant to be one of the usual definitions from physics A level.
Jaroc has now answered the query. Follow his advice.

Reply 6

Xdaamno

Original post by jaroc

Well, I believe it is possible to define Hooke's law constant as

k\equiv \lambda /l_0

e

that the derivative of GPE+EPE is 0. Remember that the more the rope is extended, the less gravitational potential energy is there, so that

GPE=mg\left( H_0-e\right)

where

H_0

is the height of the end of the rope when it is not extended (no mass attached to its end).

I have a few queries about your explanation, but I think might have solved it anyways.

I understand that there is no unique height at which GPE must be calculated - for this reason I arbitrarily chose point A as a baseline to simplify the calculations, which is possible because GPE scales linearly with height. Do you not agree?

Also, surely, for idealised springs and the like in this setup, the spring's GPE+EPE+KE will stay constant? Finding the minimum of the GPE+EPE does nothing but maximize the KE, which is clearly not finding the equilibrium position.

I just occurred to me that the problem in the second argument is probably that, since the non-frictional spring will never come to rest at it's equilibrium from being dropped at O, the EPE+GPE at O and the EPE+GPE at A cannot be equated. I'll put it in the numbers later when I have time, but I think that should work.

Reply 7

jaroc

Original post by Xdaamno

Agreed

In my original response I chose this point to be the point of original height of the bottom end of the spring.

Also, surely, for idealised springs and the like in this setup, the spring's GPE+EPE+KE will stay constant? Finding the minimum of the GPE+EPE does nothing but maximize the KE, which is clearly not finding the equilibrium position.

Ok, GPE+EPE+KE will remain constant when the situation is as follows: you attach mass to the spring at such height that the spring is not extended, and then you let the mass fall down, and in effect swing up and down about the point of equilibrium. Of course to say that GPE+EPE+KE stays constant it requires to assume no friction.

However, here we have a bit different situation. We attach mass to the spring and then we move it very slowly to the point of equilibrium. We move it so slowly that we can say that its KE is 0. We want to let the mass hang freely at the point of equilibrium, that is at the point where the total energy of the system is minimum (remember that KE is 0).

I just occurred to me that the problem in the second argument is probably that, since the non-frictional spring will never come to rest at it's equilibrium from being dropped at O, the EPE+GPE at O and the EPE+GPE at A cannot be equated. I'll put it in the numbers later when I have time, but I think that should work.

That's right, but the thing is that we do not drop it, but move it!

Try to think about it this way: when the mass is at some arbitrary point A, it has some potential energy GPE+EPE (we assume that at the beginning its KE is 0). If it can move to the point where the potential energy is smaller, it will. In the process the mass will gain KE (recall the law of conservation of energy: (change in PE)+(change in KE)=0). Equilibrium is at such a point at which - if you let the mass go - it will remain steady. That means that equilibrium is at the point where the potential energy is minimum.

It may be helpful to look at the definition of potential energy - this is the work that is done by conservative (like gravitational or elastic) forces

F

to move the body from some arbitrary point X to point x=0 (the body gains KE in the process):

\text{PE}=\displaystyle \int_X^0 \mathbf{F}\: \mathrm{d}\mathbf{x}=- \int_0^X \mathbf{F}\: \mathrm{d}\mathbf{x}

.

At equilibrium conservative forces cancel out so that

F=0

. So to find this point we can differentiate PE with regard to x and equate it to 0:

Unparseable latex formula:

F=0 \quad \Longleftrightarrow \quad \dfrac{\mathrm{d}}{\mathrm{d}x} \left( - \text{PE} \right) =\math \dfrac{\mathrm{d}}{\mathrm{d}x} \displaystyle \int F\: \mathrm{d}x =0

.

So as you can see KE does not affect the position of equilibrium in any way. What's important is only PE.

Is it clearer now?

(edited 13 years ago)

Reply 8

Xdaamno

Ah I see, you're saying that using potential energy over conservation of energy is a better way of tackling the question. Thank you for the detail in the argument there, it makes good sense. That the equilibrium position is independant of the kinetic energy of the spring at any given moment is obvious, I guess. From there the potential energy argument checks out intuitively to me.

One question out of interest: Why is it important that we are moving the mass "very slowly" (at dv right?) towards the minimum potential energy? That doesn't seem to be necessary to the argument since you're just considering the potential energy of the system. Is that just part of the transition between my argument and yours, in that you are making KE=0 just because you can? :smile:

Reply 9

jaroc

Original post by Xdaamno

Well, one technical issue,

\mathrm{d}v

means a very small change in velocity (just like

\Delta v

, which is usually bigger), not a small velocity.

You're quite right, we're moving the mass very slowly because it's convenient for the idea of considering the total energy of the system. This way we ensure that it has no KE - and then we can consider the total energy of the system (as we made it equal to just PE). However, as you have noticed, actually we do not have to really do that if we know that only PE is important. So it's just a matter whether you want to be considering the total energy of the system or not. But in general, considering the total energy of a system in order to find its equilibrium is incorrect (unless it happens that potential energy is total energy, i.e. KE is 0), it's better to consider just the PE. I just wanted to stress that it does not make sense to find the minimum of PE+KE. But now, as we've come to the point that only PE matters, we can just find the minimum of PE*.

There was one mistake in my previous post, now I've corrected it. I said that "equilibrium is at the point where the potential energy is 0." Of course the correct version is "where PE is at minimum*".

*To be exact, the points of equilibrium can be at any extremum of PE (that is at such a point where

\tfrac{\mathrm{d}PE}{\mathrm{d}x}=0

). If it's a minimum (

\tfrac{\mathrm{d}^2 PE}{\mathrm{d}x^2}>0

), it's a stable equilibrium. If it's a maximum (

\tfrac{\mathrm{d}^2 PE}{\mathrm{d}x^2}<0

), it's an unstable equilibrium. At stable equilibrium if you move the body by a very small distance it will return to the point of equilibrium (although it may start to swing about this point). At unstable equilibrium, i.e. at the maximum of PE, if you move the body by a very small distance, it will start to accelerate and move away from the equilibrium point as it will move towards lower PE.

In the case of the question you originally posted in the thread there is no maximum of potential energy, and there is only one minimum - the point of stable equilibrium.