Work, Power and Energy — Lifting a coiled rope up off a surface

[Differentiate it]

Homework Statement

(view the image)

Relevant Equations

W = ∆KE

The solution said that Average power= net change in energy/time, but why is that true? If you have a different way of solving it, it would be helpful too

 

[PeroK]

Homework Statement:: (view the image)
Relevant Equations:: W = ∆KE

The solution said that Average power= net change in energy/time, but why is that true?

That the definition of power: energy per unit time.

 

[Steve4Physics]

To add to what @PeroK said, 'average' is used because when a quantity Q changes by an amount ΔQ over time-interval Δt, the average rate of change of Q with respect to time during the interval is ΔQ/Δt. That's what average means.

E.g. you travel 10m in 1s, then 30m in 4s. Your average speed during the 5s is (10m+30m)/(1s+4s) = 8m/s.

In your question, you need to ask yourself:
- what is the increase in potential energy?
- what is the increase in kinetic energy?
- what is the time taken, expressed in terms of v and l?

 

[Differentiate it]

To add to what @PeroK said, 'average' is used because when a quantity Q changes by an amount ΔQ over time-interval Δt, the average rate of change of Q with respect to time during the interval is ΔQ/Δt. That's what average means.

E.g. you travel 10m in 1s, then 30m in 4s. Your average speed during the 5s is (10m+30m)/(1s+4s) = 8m/s.

In your question, you need to ask yourself:
- what is the increase in potential energy?
- what is the increase in kinetic energy?
- what is the time taken, expressed in terms of v and l?

Is the work done by the normal force zero here? If yes,Why?

 

[PeroK]

Is the work done by the normal force zero here? If yes,Why?

The normal force acts through no distance.

 

[erobz]

I'm getting something very close to, but not exactly matching any of those answers...

 

[PeroK]

I'm getting something very close to, but not exactly matching any of those answers...

Try harder!

 

[erobz]

Try harder!

Mind If I PM you what I've done?

 

[PeroK]

Mind If I PM you what I've done?

Go ahead!

 

[haruspex]

This is one of a class of rope/chain problems where using conservation of energy and conservation of momentum appear to give conflicting results.
My attitude at one time was that it is safer to trust momentum conservation, but I have come to realise these problems are rather subtle. I now believe the true result depends on detailed properties of the material involved.
The issue is the flexural springiness of the rope. If we treat it as having none, momentum conservation should be the go. At the opposite extreme, energy conservation may be closer, but still not correct because some of the energy will persist in the form of lateral oscillation and/or rotation.
In other instances of these problems, energy conservation can get quite close to the experimental answer. E.g. a rope initially suspended by both ends at one point, then one end released.
The chain fountain is also a member of this class.

 

[erobz]

This is one of a class of rope/chain problems where using conservation of energy and conservation of momentum appear to give conflicting results.
My attitude at one time was that it is safer to trust momentum conservation, but I have come to realise these problems are rather subtle. I now believe the true result depends on detailed properties of the material involved.
The issue is the flexural springiness of the rope. If we treat it as having none, momentum conservation should be the go. At the opposite extreme, energy conservation may be closer, but still not correct because some of the energy will persist in the form of lateral oscillation and/or rotation.
In other instances of these problems, energy conservation can get quite close to the experimental answer. E.g. a rope initially suspended by both ends at one point, then one end released.
The chain fountain is also a member of this class.

I remember discussing this before in a thread with you (It was someone else HW problem), and I can't find it now. But Yeah, using conservation of energy and conservation of momentum yield different results. I used momentum - its not one of the options, so it wasn't the authors intention, but still...

 

[PeroK]

I remember discussing this before in a thread with you (It was someone else HW problem), and I can't find it now. But Yeah, using conservation of energy and conservation of momentum yield different result ( I used momentum - its not one of the options).

We might as well talk about this here @PeroK You think applying momentum conservation is unphysical and not actually Newtons formulation and that is the issue behind the differing results? I'm not trying to cause a ruckus but this whole thing is suspicious.

The energy solution is obviously valid. Given the assumption that there is not additional energy in the rope. Additionally internal energy could always be added as an extension of the solution.

The momentum solution involves an unphysical interpretation of Newton's second law and produces a result which cannot be fully analysed in the way the energy solution can.

 

[erobz]

The energy solution is obviously valid. Given the assumption that there is not additional energy in the rope. That could always be added.

The momentum solution involves an unphysical interpretation of Newton's second law and produces a result which cannot be fully analysed in the way the energy solution can.

So basically, you believe energy (not momentum), has the final say. It isn't clear to me why $dm ~ v$ is unphysical though, but $m ~dv$ is fine.

 

[PeroK]

So basically, you believe energy (not momentum), has the final say.

It's not a question of the final say. It's a question of verification of the answer. The end state has a given energy and momentum. Unless you postulate additional internal energy there can only be one answer. Not two.

In any case, if you rewrite Newton's second law as $F = v\frac{dm}{dt}$ then you can't complain if you get a seemingly contradictory answer!

 

[erobz]

It's not a question of the final say. It's a question of verification of the answer. The end state has a given energy and momentum. Unless you postulate additional internal energy there can only be one answer. Not two.

In any case, if you rewrite Newton's second law as $F = v\frac{dm}{dt}$ then you can't complain if you get a seemingly contradictory answer!

But that's the whole premise for the derivation of the Rocket Equation.

 

[PeroK]

But that's the whole premise for the derivation of the Rocket Equation.

Not this again! You have to be really careful where it's valid and where it isn't.

It's not generally valid to have a changing mass. See, for example:

https://www.physicsforums.com/threads/what-is-three-force.1046341/post-6810372

 

[Lnewqban]

Homework Statement:: (view the image)
Relevant Equations:: W = ∆KE

The solution said that Average power= net change in energy/time, but why is that true? If you have a different way of solving it, it would be helpful too

The net change in energy is the net amount of work done by the pulling force between the full coiled position and the full extension of the rope, until it just takes off from the ground (without considering internal forces in the rope explained by @haruspex above).

Both components of work, force and height, are constantly increasing from a zero value up to the maximum force of the weight of the rope off the ground, which should be $F_{max}=mg=\lambda lg$

 

[erobz]

Not this again! You have to be really careful where it's valid and where it isn't.

It's not generally valid to have a changing mass. See, for example:

https://www.physicsforums.com/threads/what-is-three-force.1046341/post-6810372

I don't like being really careful.

 

[haruspex]

So basically, you believe energy (not momentum), has the final say. It isn't clear to me why $dm ~ v$ is unphysical though, but $m ~dv$ is fine.

Because mass is neither created nor destroyed. It can enter or leave the system under consideration, but it may well bring momentum or take momentum away with it. You then have to account for that gained or lost momentum by regarding it as a virtual force.
E.g., moving cart leaking water. There appears to be no net force on it, but its momentum is not conserved.

 

[erobz]

Because mass is neither created nor destroyed. It can enter or leave the system under consideration, but it may well bring momentum or take momentum away with it. You then have to account for that gained or lost momentum by regarding it as a virtual force.
E.g., moving cart leaking water. There appears to be no net force on it, but its momentum is not conserved.

I just can't figure out how we calculate the force required to lift the chain at $v$.

I can reverse engineer it from Work/energy, but how does it follow from Newtons Laws?

 

[haruspex]

The energy solution is obviously valid.

Not so fast. If the momentum solution is wrong we need an argument for why. It is relatively easy to find reasons why work might not be conserved.

As I wrote, it depends on the properties of the rope (and the details of the setup).
In this particular problem, we need to assume it is being pulled up from a great height so that we can take the lifted portion as effectively vertical at all times. If rising at speed v then the point of lift-off is also moving at speed v around the coil. Thus, each portion of the rope, once lifted, is rising at speed $v\sqrt 2$ helically.
Further, there may be losses in the flexing and unflexing of the rope as it transitions from horizontal to vertical.

Soapbox: be careful with idealisations; they should be treated as limits. Nothing is completely inelastic nor completely inextensible/incompressible. If we take a spring in general to have a spring constant $k_s$ when strain is being increased (compression or extension) and $k_r<k_s$ when being relaxed then:

• inelastic is the limit $k_r\rightarrow 0$.
• inextensible/incompressible is the limit $k_s\rightarrow\infty$
• perfectly elastic is the limit $k_r/k_s\rightarrow 1$.

In principle, some problems might require one to specify the relative rates at which simultaneous limits are approached.

First, consider a completely floppy, inelastic rope. It takes no work to flex it or stretch it, but no work is recovered after. As each small segment is lifted it undergoes at least a tiny stretch to accelerate quickly from rest to v, or whatever. Momentum conservation would seem to apply.

Next, suppose it has perfect longitudinal elasticity. The stretch that occurs as one segment is lifted can store energy which is then used to assist the rise of the next segment. Its speed may transiently exceed v, explaining the apparent momentum violation. Even so, it does seem likely that some of this energy will persist as vertical oscillations in the rope, not doing any useful work.

Now allow some flexural elasticity. As a segment is lifted, it stores energy in its bending, and the release of that as it leaves the ground can give it an extra vertical impulse.

In short, problems like these cannot be answered, and should not be asked, without quite detailed specification of properties, and even then may be very challenging to solve.

 

[erobz]

Another thing. If variable mass is a problem in using conservation of momentum, why is it not an issue with the energies? All of those expressions for energy are derived from a constant mass assumption as well.

For instance the kinetic energy:

$$F = m \frac{dv}{dt} = m \frac{dv}{dx}\frac{dx}{dt} = m \frac{dv}{dx} v $$

The assumption of constant mass $m$ gives:

$$ \int F~ dx = m \int v ~dv$$

The limits of integration don't make any sense for this application. The LHS is identically zero for this problem.

And in this problem the mass $m$ is a function of $x$, So do we actually have:

$$\int \frac{F(x)}{ \lambda x } ~ dx = 0 $$

?

 

[haruspex]

Another thing. If variable mass is a problem in using conservation of momentum, why is it not an issue with the energies? All of those expressions for energy are derived from a constant mass assumption as well.

For instance the kinetic energy:

$$F = m \frac{dv}{dt} = m \frac{dv}{dx}\frac{dx}{dt} = m \frac{dv}{dx} v $$

The assumption of constant mass $m$ gives:

$$ \int F~ dx = m \int v ~dv$$

The limits of integration don't make any sense for this application. The LHS is identically zero for this problem.

There's no issue with either momentum or energy so long as you keep track of any that leaves or enters the system along with the mass. In post #17 the system is the entire rope.

 

[erobz]

There's no issue with either momentum or energy so long as you keep track of any that leaves or enters the system along with the mass. In post #17 the system is the entire rope.

We’ll, I’m hearing the impression that because they yield different results there is an issue. Only one should emerge victorious, or they should agree.

Also, I feel like I’ve made a valid point in #22 that wasn’t addressed. The derivations for energy we use are having issues in this application. The mass is actually a function of $x$. See the edit. Something is going haywire there.

 

[haruspex]

The LHS is identically zero for this problem.

I don't see why. Please explain what F and x are defined to be in that context.

in this problem the mass m is a function of x,

If you define the system as the portion of rope no longer on the table, that is true. But it is not a problem if we define the table as the zero for potential energy, since that will mean the mass being added adds no energy.

We’ll, I’m hearing the impression that because they yield different results there is an issue. Only one should emerge victorious, or they should agree.

Don’t confuse two issues.

My beef with questions like these is that they are essentially unanswerable without a lot more detail. Without that one cannot know how well work is conserved or whether there might be a bit of kick-up from the table to upset the momentum equation.

But in posts #13 to #16 there was discussed the side issue of using momentum conservation on a variable mass system, as one sometimes sees in treatments of the rocket equation. The popular $F=\dot p=m\dot v+\dot m v$ can lead to the wrong result if carelessly applied.

 

[PeroK]

Not so fast. If the momentum solution is wrong we need an argument for why.

@erobz forgot the normal force.

I think we are talking at cross purposes as @erobz messaged me a solution that effectively gives the final KE of the rope as $mv^2$. There must, therefore, be an error in that calculation.

The error was not in the restatement of Newton's law, but forgetting the normal force.

This is the problem with @erboz hijacking another user's homework.

 

[haruspex]

@erobz forgot the normal force.

I think we are talking at cross purposes as @erobz messaged me a solution that effectively gives the final KE of the rope as $mv^2$. There must, therefore, be an error in that calculation.

The error was not in the restatement of Newton's law, but forgetting the normal force.

This is the problem with @erboz hijacking another user's homework.

Pax.

 

[erobz]

This is the problem with @erboz hijacking another user's homework.

I’m always the bad guy.

 

[erobz]

I don't see why. Please explain what F and x are defined to be in that context.

In light of recent revelations, It’s probably better if the chain is understood to be horizontal being pulled on a frictionless surface.

The Integral on the RHS is integrated from $v$ to $v$, (we are applying the force at a constant velocity) so it is 0.

 

[erobz]

Imagine a chain/rope being pulled horizontally out of a loose pile $x \rightarrow^+$ by a force $F$ at constant velocity $v$ (no friction). What does Work Energy give us?

Classically with a constant mass (not a chain coming out of a pile):

$$\int_{0}^{l} F ~ dx = m \int_{v}^{v} v ~dv = \frac{1}{2} m v^2 = 0 $$

The force $F$ is not doing Work. Seems like everything I was taught.

Now on to the chain:

I feel like we should be able to argue that the mass moving at velocity $v$ is actually a function of $x$ i.e. $m(x) = \lambda x$ and incorporate it into the integral on the LHS and perhaps resolve the issue of no work being done by the force $F$...

$$ \int_{0}^{l} \frac{F(x)}{\lambda x} dx = \int_{v}^{v} v ~dv = 0 $$

We still get no work?

Lets ignore the fact that the mass being pulled at velocity $v$ is a function of $x$ ( as far as integration is concerned) and leave it on the RHS.

$$ \int_{0}^{l} F(x)~dx = \lambda x \int_{v}^{v} v ~dv = 0 $$

Still no work?

I feel like this is a problem because in order to solve the OP via energy methods, we are basically assuming (for this simplified problem and for the OP) seemingly against all interpretations that:

$$ W = \int_{0}^{l} F ~ dx = \frac{1}{2} \lambda l v^2 \neq 0 $$

 

[erobz]

To me it seems like to be consistent we always should be writing the following:

$$ W = \int F ~dx = \int \left( m \frac{dv}{dt} + v \frac{dm}{dt} \right) dx $$

Then we can see that for the chain at constant velocity $v$ problem:

$$ W = \int F ~dx = \int \left( \cancel{ m \frac{dv}{dt}} + v \frac{dm}{dt} \right) dx $$

$$ \implies \int F ~dx = \int v \frac{dm}{dt} dx = \int v \frac{dm}{dx} \frac{dx}{dt} dx = \int v \frac{dm}{dx} v dx = v^2 \int_{0}^{l} \lambda ~dx = \lambda l v^2 $$

However, that doesn't really play nicely either. That factor of $\frac{1}{2}$ out front is missing, or should it not actually be there?

 

[erobz]

Another thing that is bothering me.

@erobz forgot the normal force.

The normal force is then doing work that is changing the outcome?

Is the work done by the normal force zero here? If yes,Why?

The normal force acts through no distance.

 

[PeroK]

Okay, I think I've got the issue. I divided the rope up into a large number of segments, then looked at what was happening to the rope already moving upwards at $v$ and the next segment being accelerated to $v$. Here's the issue. The first of these forces acts through a greater distance as the rope is already moving at $v$. The average speed of the accelerating segment is only $\frac v 2$.

So, we cannot use $P = Fv$. We must use $P = F_1v + F_2\frac v 2$.

If we have an infinite acceleration to $v$, then that's where the power calculation goes wrong.

With this approach both methods, energy and momentum yield the same answer.

 

[Differentiate it]

Could I please get help with this one too?

 

[erobz]

Could I please get help with this one too?

Could finish the last one so everyone can get on the same page with what the answer should be (or is intended to be according to the author)?

 

[PeroK]

Could I please get help with this one too?

Given the state of this thread, please open a new thread. You still have to solve this one, as no one has posted a solution.

 

[Steve4Physics]

To me it seems like to be consistent we always should be writing the following:

$$ W = \int F ~dx = \int \left( m \frac{dv}{dt} + v \frac{dm}{dt} \right) dx $$

Then we can see that for the chain at constant velocity $v$ problem:

$$ W = \int F ~dx = \int \left( \cancel{ m \frac{dv}{dt}} + v \frac{dm}{dt} \right) dx $$

$$ \implies \int F ~dx = \int v \frac{dm}{dt} dx = \int v \frac{dm}{dx} \frac{dx}{dt} dx = \int v \frac{dm}{dx} v dx = v^2 \int_{0}^{l} \lambda ~dx = \lambda l v^2 $$

However, that doesn't really play nicely either. That factor of $\frac{1}{2}$ out front is missing, or should it not actually be there?

I see @PeroK has beaten me to it, but since I've already written it ...

I don’t entirely follow your reasoning but this might help.

Consider a simple ‘ideal’ case where one end of the pile of rope is pulled horizontally and at constant velocity. The force ($F$) is then necessarily constant. Successive elementary ‘slices’ of rope are each accelerated from 0 to $v$ in some (albeit arbitrarily small) time-interval with constant acceleration ($a$).

Using simple kinematics, during a slice’s acceleration, its displacement ($s$) is given by $s= \frac {v^2}{2a}$. The work done is therefore $Fs = \frac {Fv^2}{2a}$. That’s essentially where the factor $\frac 12$ (in $\frac 12 mv^2$) comes from.

Your analysis seems to ignore the process of acceleration and assumes each slice is instantaneously accelerated from 0 to $v$; as a consequence, it will give the work done accelerating each slice as $\frac {Fv^2}a$ which is twice the correct value.

For a more rigorous explanation, take a look at how change in kinetic energy is derived from change in momentum here: https://en.wikipedia.org/wiki/Kinetic_energy#With_vectors_and_calculus

Edited - see strikethrough.

 

[PeroK]

Your analysis seems to ignore the process of acceleration and assumes each slice is instantaneously accelerated from 0 to $v$; as a consequence, it will give the work done accelerating each slice as $\frac {Fv^2}a$ which is twice the correct value.

So the bastardisation of Newton's second law was the root of the problem after all:
$$Fdt = vdm \Rightarrow KE = Fdx = Fdt\frac{dx}{dt} = v^2dm$$

 

[Lnewqban]

Could I please get help with this one too?

Just refer to diagram shown in post #17 above.

 

[erobz]

That makes sense now. Its hard to separate onself from the everyday experience of pulling a hose out of a pile. When you said $a$ is constant and hence $F$ is constant my instinct was to claim, that's absurd. BUT in the absence of friction, it does make sense. You're not accelerating the entirety of the rope that's been pulled, just the mass of it coming off the pile, and there are no forces that are growing in proportion to the mass being dragged, without friction.

I can buy it.

 

[Steve4Physics]

So the bastardisation of Newton's second law was the root of the problem after all:
$$Fdt = vdm \Rightarrow KE = Fdx = Fdt\frac{dx}{dt} = v^2dm$$

Just to add that if $x$ is the position of the end of the rope, then the average velocity of the mass element $dm$ during its acceleration is $\frac 1 2 (\frac {dx}{dt})$.

Then$$Fdt = vdm \Rightarrow KE = Fdx = Fdt\frac{dx}{dt} = v^2dm$$becomes$$Fdt = vdm \Rightarrow KE = Fdx = Fdt {\frac 12}(\frac {dx}{dt}) = \frac 12 v^2dm$$ and things look sensible.

 

[erobz]

So the bastardisation of Newton's second law was the root of the problem after all:
$$Fdt = vdm \Rightarrow KE = Fdx = Fdt\frac{dx}{dt} = v^2dm$$

So it seems. One of my Physics Profs. declared to us that Newtons 2^nd Law was actually:

$$\sum \vec F = \frac{d}{dt} \Big( m \vec v \Big)$$

What was the point of that...

 

[PeroK]

So it seems. One of my Physics Profs. declared to us that Newtons 2^nd Law was actually:

$$\sum \vec F = \frac{d}{dt} \Big( m \vec v \Big)$$

What was the point of that...

If $m$ is taken to be constant, then it's a minor restatement of the law in terms of momentum.

It doesn't apply if mass is entering or leaving the system. See, for example:

https://physics.stackexchange.com/questions/53980/second-law-of-Newton-for-variable-mass-systems

https://en.m.wikipedia.org/wiki/Variable-mass_system

 

[Lnewqban]

Just to add that if $x$ is the position of the end of the rope, then the average velocity of the mass element $dm$ during its acceleration is $\frac 1 2 (\frac {dx}{dt})$.

Then$$Fdt = vdm \Rightarrow KE = Fdx = Fdt\frac{dx}{dt} = v^2dm$$becomes$$Fdt = vdm \Rightarrow KE = Fdx = Fdt {\frac 12}(\frac {dx}{dt}) = \frac 12 v^2dm$$ and things look sensible.

Could we then consider only the location of the CoM for each phase of the lifting?
That location, and the acceleration and velocity of each CoM (of the lifted mass only at each instant) will always be half the value of the vertically moving end of the rope.

 

[Steve4Physics]

Could we then consider only the location of the CoM for each phase of the lifting?
That location, and the acceleration and velocity of each CoM (of the lifted mass only at each instant) will always be half the value of the vertically moving end of the rope.

Not sure I fully understand.

For the ideal/simplified situation (as probably intended in the original question) the top of the rope (of total length $l$) is rising vertically at constant speed $v$.

Every bit of rope in the vertical section will have the same speed ($v$) because the rope is inextensible.

But the length of the vertical section of rope is increasing linearly with time. As a result, the speed of the whole rope's CoM will gradually increase from 0 (at the start of uncoiling) to $v$ (at the end of uncoiling).

Note that the accelerations of the uncoiled rope and the vertical section are zero- except for each rope-element during the (arbitrary small) time-interval when it accelerates from 0 to $v$.

Of course, when the whole rope is completely vertical, the height of the top will be $l$ the height of the CoM will be $\frac l2$ (which allows us to find the gain in gravitational potential energy).

For the gain in kinetic energy, we can simply use $\frac 1 2 Mv^2$ where M is the total mass.

 

[haruspex]

The average speed of the accelerating segment is only v2.

Sure, but to make that work you have to assume perfect elasticity and take the rope to have been originally in a vertical stack of zero height, i.e. a spring of zero relaxed length.
(Even then, could there be persistent vertical oscillations? Analysis as a massive spring might settle this.)
Otherwise there will be some sideways motion, as I mentioned in post #21. And in the specific case of this thread, that will take as much KE as the vertical motion.

 

[haruspex]

If $m$ is taken to be constant, then it's a minor restatement of the law in terms of momentum.

It doesn't apply if mass is entering or leaving the system.

Umm… no, that equation still applies. It does represent the momentum of that system. What does not work is then expecting the momentum to be conserved, unless we represent the momentum carried away by (or brought in with) the mass change as an external impulse.

 

[PeroK]

Sure, but to make that work you have to assume perfect elasticity and take the rope to have been originally in a vertical stack of zero height, i.e. a spring of zero relaxed length.
(Even then, could there be persistent vertical oscillations? Analysis as a massive spring might settle this.)

I was trying to figure out how the energy ended up different using a force based approach. The detailed analysis of internal energy within the rope is outside the scope of the this homework.

 

[PeroK]

Umm… no, that equation still applies.

I've provided several references that show that it does not apply in general.

 

[haruspex]

I've provided several references that show that it does not apply in general.

Sorry, for some reason I completely misread the referenced equation as being $\dot{ \vec p}=\frac d{dt}(m\vec v)$. Bizarre.
So what I should have written is that you can make the equation work if you count the rate at which momentum is brought in or out of the system by the mass change as an external force.