Conceptual question on Modulus of elasticity

[Abhishekdas]

Homework Statement

Lets say a wire has Youngs modulus Y, Initial length L area A and is supporting some external load/force of magnitude F...its extension is dl..We say that the Potential energy of wire is 1/2(YA/L)(dl²)...But i am having a problen with the energy changes...

Homework Equations

The Attempt at a Solution

If its hanging down the work done by external force (gravity) is mgh...
So we want the potential energy of the wire...I was trying to derive it by using the external force...Now work done by external force should be negative of work done by spring (from work energy theorem here change in KE is zero as initially and finally its at rest)...
So is 1/2(YA/L)(dl²) = mg*dl? IF so how?

 

[collinsmark]

Homework Statement

Lets say a wire has Youngs modulus Y, Initial length L area A and is supporting some external load/force of magnitude F...its extension is dl..We say that the Potential energy of wire is 1/2(YA/L)(dl²)...But i am having a problen with the energy changes...

Homework Equations

The Attempt at a Solution

If its hanging down the work done by external force (gravity) is mgh...
So we want the potential energy of the wire...I was trying to derive it by using the external force...Now work done by external force should be negative of work done by spring (from work energy theorem here change in KE is zero as initially and finally its at rest)...
So is 1/2(YA/L)(dl²) = mg*dl? IF so how?

In your specific example, kinetic and potential energy are not conserved. Some of the original, gravitational potential energy (mgΔh) will inevitably be lost to heat or to some other external force.

Let me describe it another way. Suppose you held the load in your hand such that the wire is just barely taught, but not under any stress. Then you released the load. The load will "fall" a little bit as the wire stretches. But once it reaches the lowest point, it doesn't stop there! The wire pulls the load back up to where the load was originally, and the process repeats. The load/wire system has gone into simple harmonic motion. At this point, you could stop the oscillations yourself, with your hands for example, which is an external force. Or you could let the oscillations die out on their own, but that in that case, some of the energy gets lost to heat.

If you want to figure out where this potential energy equation comes from, don't use conservation of energy. Instead, use the definition of work. And yes, you are correct to treat the wire as a spring.

W = ∫F·dx
F = -kx

I'll let you take it from there.

 

[Abhishekdas]

Hi...collinsmark...thanks for the reply...

I don't get how energy is converted to heat...I mean in case of a normal shm energy remains constant but why not in this case...? And why will the oscillations die out in absence of any air resistence? Isnt stretching of a string exactly similar to a spring? But with a very high value of k(force constant)?

 

[collinsmark]

Hi...collinsmark...thanks for the reply...

I don't get how energy is converted to heat...I mean in case of a normal shm energy remains constant but why not in this case...? And why will the oscillations die out in absence of any air resistence? Isnt stretching of a string exactly similar to a spring? But with a very high value of k(force constant)?

In the absence of any resistance, the oscillations wouldn't die out. But then you'd have another type of energy to deal with: the mass's kinetic energy.

All I'm saying is that the total energy of the system is in two parts, ignoring friction: The wire's potential energy and the mass' kinetic energy. If you introduce friction, the mass' kinetic energy will be converted to heat over time.

My original point is that you don't want to take the total energy of the system and assume that all of it is the potential energy of the wire.

But what you can do is use the work energy theorem, and the definition of force, to derive the wire's potential energy equation.

Give it a try. You know

$$\vec{F} = \frac{YA}{L} \vec x$$

where x is the displacement from stretching, and

$$W = \int \vec F \cdot \vec{dx}$$

And from the work energy theorem, you can evaluate the work as the wire's potential energy.

And yes, it's exactly the same thing as a spring.

 

[Abhishekdas]

hi...collinsmark...
i got 1/2(YA/L)(dl^2) using your relation and integrating it...but still...

thanks for explaining so much...but believe me i am trying to understand but i am still not getting my original doubt cleared...
You are saying that the total energy is divided into PE of string and KE of mass...but what is that supposed to imply...i mean i am still confused...

Ok...Work don't by external force in stretching the wire is F.dl...here dl is the extension(not the differential element)...now this F is constant and has a magnitude of YAdl/L...so work done by it is (YA*dl^2)/L...now half of it is potential energy of spring ...where does the other half go...?

 

[collinsmark]

Ok...Work don't by external force in stretching the wire is F.dl...here dl is the extension(not the differential element)...now this F is constant and has a magnitude of YAdl/L...so work done by it is (YA*dl^2)/L...now half of it is potential energy of spring ...where does the other half go...?

In your scenario, there are two forces involved. There is the constant, external force, and there is the spring-like force that is proportional the amount of stretch.

The two forces are not equal (except at one particular length of stretching -- the future center of oscillation). That means that the net force is not equal to zero.

Newton's second law says ma = Σ F_i. And since the sum of all forces is nonzero, the mass accelerates. And it keeps accelerating until the specific point where the spring-like force is equal in magnitude and opposite in direction to the constant force. When the mass reaches this point, the instantaneous acceleration is zero, but its velocity is at maximum. Then the mass begins to decelerate because the magnitude of the spring-like force overtakes the strength of the constant force. Then the mass comes to a stop and then goes back in reverse. The process repeats indefinitely, unless there is friction, in which case the mass eventually comes to rest at the very point where the magnitude of spring-like force equals the magnitude of the constant force.

Maybe it's worthwhile to go back and examine the first part again. And let's ignore friction for the moment. When the mass initially reaches the point where the magnitude of the spring-like force equals the magnitude of the constant force, that's where the mass's kinetic energy is at maximum.

So, for a given stretch dl, where dl is the exact amount of stretch required for the the spring-like force to be equal in magnitude to the constant force, the constant force has potential energy of -(YA*dl^2)/L. The potential energy of the wire is 1/2(YA/L)(dl^2). So where did the rest of the energy go? The mass's kinetic energy. Remember, the mass's kinetic energy is maximum at that point.

 

[Abhishekdas]

Ya...perfect...it goes into the KE of the mass...hmmm...didnt think of that...You really explained it in a very detailed mannner...the way i wanted...thanks man...

But there is one more confusion which has been created...SO it turns out that the mass does beyond the length dl...isnt it...(actually it is 2*dl as we know from Simple harmonic motion)...So in ideal situations doesn't this happen? I mean does the mass stop as soon as it the forces become equal? And if we are neglecting the thing that it goes beyond dl we are neglecting a large chunk of extension...But i guess it doesn't affect the equations of PE or stress...or Youngs Modulus...or whatever...I mean maybe it doesn't affect the maths of the analysis of the situation...What do you say?

 

[collinsmark]

But there is one more confusion which has been created...SO it turns out that the mass does beyond the length dl...isnt it...(actually it is 2*dl as we know from Simple harmonic motion)...So in ideal situations doesn't this happen? I mean does the mass stop as soon as it the forces become equal?

Be careful. The answer is no. When the forces become equal (in magnitude), the mass's acceleration goes to zero. But it still has a non-zero velocity. (All of this comes directly from Newton's laws -- and it is the ideal situation, the way we're treating it.) So it goes right past the point that we're calling 'dl'. Once it gets past that point it starts slowing down.

It eventually comes to a stop (for an instant) at 2 times dl, if we ignore friction.

If we introduce friction, the oscillations will eventually die out, and in time, the mass will finally come to rest at dl. And in that case, half of the original energy of the system ended up getting converted to heat.

Or, instead of friction, you could use your hand to gently lower the mass from its original position to length dl. In that case, the mass comes to rest at the position dl. But now there are three forces to consider. The spring like force, the force of gravity, and the force from your hand. And in this situation, half of the original energy of the system goes into your hand (the system does work on your hand).

 

[Abhishekdas]

Ya...i know...in ideal situation it shud stop at 2dl because KE isn't zero at dl ...rather its maximum so it should go to 2dl...Thats why i raised the question...But i am not getting how friction will make the mass stop at dl...Its ok that half of the energy is converted to heat but how does friction make it stop at dl?

 

[Abhishekdas]

And coming to the explanation using the hand...If we are using our hand and its pulled down gently then force due to kx=force of hand at every time so wd by the is equal and opposite nad that explains the energy considerations...I guess...So there is no extra energy to worry about...what do you say? And using the hand can be a nice method to stop it at dl...And i didnt get what you meant by system doing work on my hand...Could you please explain that...?

 

[Dadface]

Hi Abhishekdas when the mass eventually comes to rest it must be at a place where the resultant force equals zero. As collinsmark explained this is at an extension of dl.It is similar to a simple pendulum where there is a continual interchange between gravitational PE and KE.The KE is a maximum at the lowest point of the swing where the resultant force is zero and where the pendulum bob eventually comes to rest.
With the spring there's an infinite number of ways that the mass can be lowered towards the final extension of dl but the final result is always the same which is that although the elastic PE gained is half of the GPE lost the total energy is conserved.

 

[Abhishekdas]

With the spring there's an infinite number of ways that the mass can be lowered towards the final extension of dl but the final result is always the same which is that although the elastic PE gained is half of the GPE lost the total energy is conserved.

Hi Dadface...can yu explain this part i did not get this...
And you the final conclusion is unles there is friction or we are using out hand to move it infinitely slowly the extension of the spring is twice the standard dl (here dl is the point at which external force is equal to kx)...Isnt it?

 

[Dadface]

Let me give a slightly different example.Imagine that a mass M is lifted through a height h in a uniform gravitational field of field strength g.There's an infinite number methods that can be used and routes that can be taken to lift the mass.For some methods it is easy to calculate the energy conversions but for others it becomes very difficult if not impossible.
Whatever method is used,however,we can be fairly confident that energy is conserved and that there has been an increase in GPE of Mgh.
With the mass on the spring there's an infinite number of ways that the mass can be lowered to its final position and whatever method is used we can be fairly confident that the loss of GPE is Mgh but that energy has been conserved.

 

[Abhishekdas]

Ok...i got your example but i did not get how it will help in in understanding the situation in the spring...