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

[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.

 

[haruspex]

The detailed analysis of internal energy within the rope is outside the scope of the this homework.

Unfortunately it is not.

You made the force/momentum approach produce the work conservation answer by assuming that a perfectly elastic string of very little extensibility pulled up at a steady rate would not acquire significant persistent vertical oscillations. It is unclear to me whether that is the case.
Even then, to get your answer, the significant lateral KE which would come from pulling up from a horizontal loop (as discussed above) has to be ignored. I have not checked whether including that gives one of the listed answers.

Other valid models can produce other answers.

E.g. consider the model of a chain in which each segment, length L, starts standing vertically. The segments have no width. As each segment is lifted to height L it engages the next segment. If that is a coalescence then the work required is doubled. Even if it is perfectly elastic, would there not be persistent vertical oscillations?
Or start with zero width segments lying horizontally in a stack in a zigzag arrangement. As a segment is lifted it rotates about its lower end. Will the energy that goes into that rotation be recovered in helping to lift the next segment, or will there be residual horizontal wiggles?

Such considerations are only outside the scope of the question if those models are somehow excluded.

 

[Lnewqban]

There is also a twisting result as the rope goes up, unless the free end can rotate.
It is easily visible in construction sites when pulling an extension cord, or wire-rope, or hose that have been resting on the ground as a coil.
Same happens in reverse, when forming the coil on the ground.

 

[haruspex]

More thoughts…
If we model the rope as flexing freely down to some minimum radius of curvature, r, then the portion just above the ground consists of a quadrant of a circle of that radius. The acceleration of each bit of it is towards the centre of curvature, and that must be provided by tension in the rope.
Ignoring gravity, this makes the tension equal along the arc, including where it joins the part that is still horizontal. Specifically, $T=\lambda v^2$.
The rate of doing work in lifting new rope is therefore $\lambda v^3$.

Looking at it from work conservation instead, each dt a mass segment $\lambda\cdot dx$ of rope finishes with a vertical speed v and a horizontal speed v for a total KE of $\lambda dx .v^2=\lambda dt .v^3$, hence the rate of acquiring KE is $\lambda v^3$, agreeing with the result above.