Physics challenge: freshman physics edition

[Kuroneko]

You pick up the end of a flexible rope wound around a coil, of uniform density 0.5 kg/m, and walk away at 2 m/s. Assume that the coil remains stationary and has negligible friction. If the rope is 10 m long,
(a) find the kinetic energy of the rope when it finishes unwinding,
(b) the horizontal force the rope exerted on you before it did, and
(c) explain the relationship between (a) and (b).
Please use spoiler tags.

Despite initial appearances, there is actually a physically interesting point to this very simple situation, but first let's solve it.

[Bakustra]

Here's my solution: Spoiler

I presume that if the coil remains stationary, that I must walk in an increasing spiral around the coil to unwind the rope, so:
a. The KE would be 10 J.
b. The horizontal force would be 2 N.
c. (a) and (b) are related through the length of the uncoiled rope. The length determines the KE through the mass of the rope, and the radius of the spiral. As r increases, m will effectively increase, causing F and KE to increase. Or have I misinterpreted the problem?

[Darth Holbytlan]

Spoiler

Let D be the "density" of the rope, v be the velocity which you travel, L be the current length of rope behind you, and t be the time you've spent dragging the rope (where L=0 at t=0). Then

L=vt
m=DL=Dvt is the mass of the rope being dragged behind you.
p=mv=Dv2t is the momentum of the rope.

The force you apply to the rope is dp/dt=dDv2t/dt=Dv2, therefore the force the rope applies to you is -Dv2.

The kinetic energy of the rope is (1/2)mv2 = (1/2)Dv3t = (1/2)Dv2L. The relationship between the two is clearly the ratio -L/2.

Plugging in the maximum length of the rope and the other givens produces force = -2N and kinetic energy = 10J.

OK, so why is the relationship -L/2? The kinetic energy of the rope is just the work done on the rope, or ∫F⋅d distance. Since the force F is a constant, the ratio is just ∫d distance = distance. But while applying the force, the rope only travels half the distance you travel (L) on average, or L/2.

[Feil]

Do you mean the horizontal impulse? Force is incalculable given the information presented.

[Feil]

Spoiler

a) 10m*0.5kg/m = m = 5kg
v=2m/s
T=.5mv^2=10J

assume person walking on a surface and exerting such (nonuniform) force as is needed to maintain uniform velocity v=2m/s, rope initially stationary
b) impulse I=m(v1)-m(v0)=10kgm/s

c) I=integral[dT/dx)dt]
T=integral[(dI/dt)dx]

[Feil]

Spoiler

Plugging in the maximum length of the rope and the other givens produces force = -2N and kinetic energy = 10J

Spoiler

b. The horizontal force would be 2 N.

Spoiler

Not going to work. In fact, we can state with certainty that whatever the force is, it is not constant over the time period in question (the time it takes the rope to spool out)

let t=t1-t0
suppose towards a contradiction that force is uniform
F=T/(t)
T=.5*5kg*(v_final)^2=.5*5kg*(2v_avg)^2=10kg*100m^2/t^2 => F = 10kg*100m^2/t^3
clearly F and T share a maximum and that maximum occurs when t is at a minimum

T_max = 10J (the rope must stay behind you, and so cannot have a higher velocity than you)
T_min approaches 0 as t=> infinity
ergo t_min=T_max/sqrt(10kg*100m^2)=10s

(as you predicted, F_max=2N)

define a new term, h=the length your arm has to stretch to accomplish this
h_min=t_min*(v_you)-10m=10m
h_max=t_max*(v_you)-10m approaches infinity
your arms are not at least 10m long
ergo force is not uniform over the time period in question

[Bakustra]

Spoiler

Plugging in the maximum length of the rope and the other givens produces force = -2N and kinetic energy = 10J

Spoiler

b. The horizontal force would be 2 N.

Spoiler

Not going to work. In fact, we can state with certainty that whatever the force is, it is not constant over the time period in question (the time it takes the rope to spool out)

let t=t1-t0
suppose towards a contradiction that force is uniform
F=T/(t)
T=.5*5kg*(v_final)^2=.5*5kg*(2v_avg)^2=10kg*100m^2/t^2 => F = 10kg*100m^2/t^3
clearly F and T share a maximum and that maximum occurs when t is at a minimum

T_max = 10J (the rope must stay behind you, and so cannot have a higher velocity than you)
T_min approaches 0 as t=> infinity
ergo t_min=T_max/sqrt(10kg*100m^2)=10s

(as you predicted, F_max=2N)

define a new term, h=the length your arm has to stretch to accomplish this
h_min=t_min*(v_you)-10m=10m
h_max=t_max*(v_you)-10m approaches infinity
your arms are not at least 10m long
ergo force is not uniform over the time period in question

Spoiler

I presumed, for my answer, that I would be walking in an increasing spiral to unwind the rope. I took "assume the coil remains stationary" to preclude any pivoting of the coil. I also took the wording to imply the force at the moment that the rope finishes unwinding, which would be Fmax, as the force wouldn't be constant over any time period while the rope is unwinding. In this case, I think we may need some clarification from Kuroneko on this matter.

[Kuroneko]

You're meant to walk away in a constant direction, with you dragging an increasing length of rope (although pretty much any way that this is accomplished is fine, I believe). Apologies if this wasn't clear--it is about the inertia of the dragged rope only.

Do you mean the horizontal impulse? Force is incalculable given the information presented.

No, I meant force, and yes, it is calculable.

[Bakustra]

You're meant to walk away in a constant direction, with you dragging an increasing length of rope (although pretty much any way that this is accomplished is fine, I believe). Apologies if this wasn't clear--it is about the inertia of the dragged rope only.

Do you mean the horizontal impulse? Force is incalculable given the information presented.

No, I meant force, and yes, it is calculable.

Oh. Whoops!

[Glass Pearl Player]

Spoiler

The rope has a mass of 5 kg overall, and will move at 2 m/s when I've uncoiled it after 5 seconds and 10 m. That gives, with E=0.5 m v^2, a kinetic energy of 10 J. That happened over a space of 10 m, so a force of 1 N acted on it.
The impulse of the rope is m(t)*v, where m(t) is the mass of the moving part at time t. Over 5 seconds, it goes linearly from zero to 10 kg m/s. Force is change of impulse over time, so a force of 2 N acted on it. that happened over the space of 10 m, so since work is force times distance, we get an energy of 20 J.
What's going on here?
My guess: The value of 10 J for kinetic energy is correct, and 1 N for the force the rope exerts on me. The other Newton of force is internal to the rope, and the remaining 10 J are stored in the tension of the rope. They should reappear as the end of the rope flying into my back.

[Feil]

Ghetto edit - my third post on this thread states that 2N is correct if one assumes linear acceleration. This is not the case - I neglected to divide mv^2 by 2 in my head. F_max=1N in that series of calculations.

I am interested to see how you show F to be calculable, Kuroneko.

[Darth Holbytlan]

Spoiler

Not going to work. In fact, we can state with certainty that whatever the force is, it is not constant over the time period in question (the time it takes the rope to spool out)

let t=t1-t0
suppose towards a contradiction that force is uniform
F=T/(t)

Spoiler

This equation (F=T/(t)) is not correct. It doesn't even produce the right units for force, which should be kg m/s2, not kg m2/s3 as you have below:

Spoiler

T=.5*5kg*(v_final)^2=.5*5kg*(2v_avg)^2=10kg*100m^2/t^2 => F = 10kg*100m^2/t^3
clearly F and T share a maximum and that maximum occurs when t is at a minimum

Spoiler

You shift from Tfinal to Tt at the third "=", and your formula for the latter is wrong. You cannot average the velocity of an object to compute the kinetic energy. You have to do it piecewise. Your formula has a Tt→∞ as t→0, which is clearly nonsense.

Spoiler

T_max = 10J (the rope must stay behind you, and so cannot have a higher velocity than you)
T_min approaches 0 as t=> infinity
ergo t_min=T_max/sqrt(10kg*100m^2)=10s

Spoiler

Kinetic energy is 0 only at t=0. Even if the equation were correct, you cannot apply it outside the time where the rope is uncoiling. After that there is no more stationary rope to be accelerated and no more work being done. You could even let go and the rope will still keep traveling at 2m/s forever, thanks to us ignoring friction. Therefore it has a constant, non-zero kinetic energy ever after.

Spoiler

(as you predicted, F_max=2N)

define a new term, h=the length your arm has to stretch to accomplish this
h_min=t_min*(v_you)-10m=10m
h_max=t_max*(v_you)-10m approaches infinity
your arms are not at least 10m long
ergo force is not uniform over the time period in question

Spoiler

Here's why the force is constant: At any point in time while the rope is uncoiling, you have a part that is trailing behind you, traveling at 2m/s. At any point in time, only the tiny part of the rope between that and the coil is being accelerated, from 0 to 2m/s. The coiled rope has no effect and the trailing rope only acts to transmit your force to that accelerating part—it contributes nothing as well. This process is identical no matter little or much of the rope has uncoiled, so the forces should be identical as well.

Edit: For clarity.

[Feil]

Spoiler

This equation (F=T/(t)) is not correct. It doesn't even produce the right units for force, which should be kg m/s2, not kg m2/s3 as you have below:

Spoiler

You're right. I confused energy with impulse.

Spoiler

You shift from Tfinal to Tt at the third "=", and your formula for the latter is wrong. You cannot average the velocity of an object to compute the kinetic energy. You have to do it piecewise. Your formula has a Tt→∞ as t→0, which is clearly nonsense.

Spoiler

You are wrong here. Specifically, you're ignoring the definition of t that I stated at the start of that post, which is the time between when the coil starts unwinding and when the coil finishes unwinding. Similarly, T is the final kinetic energy, not the instantaneous kinetic energy, as specified in the opening post, and - with a presumption of linear acceleration - can be expressed in terms of the average velocity because the average velocity exists in fixed ratio to the final velocity.

As before, the last part of what I wrote - F = 10kg*100m^2/t^3 - is incorrect, for the reasons you noted.

Spoiler

Kinetic energy is 0 only at t=0. Even if the equation were correct, you cannot apply it outside the time where the rope is uncoiling. After that there is no more stationary rope to be accelerated and no more work being done. You could even let go and the rope will still keep traveling at 2m/s forever, thanks to us ignoring friction. Therefore it has a constant, non-zero kinetic energy ever after.

Spoiler

Again, you are using a different t. Surely you will agree that if it takes infinite time for the rope to leave the coil, then the rope is not in fact moving, and its kinetic energy is zero.

Spoiler

Here's why the force is constant: At any point in time while the rope is uncoiling, you have a part that is trailing behind you, traveling at 2m/s. At any point in time, only the tiny part of the rope between that and the coil is being accelerated, from 0 to 2m/s. The coiled rope has no effect and the trailing rope only acts to transmit your force to that accelerating part—it contributes nothing as well. This process is identical no matter little or much of the rope has uncoiled, so the forces should be identical as well.

Spoiler

How do you propose getting the rope off the coil at a rate of 2m/s if the rope is not processing around the coil at a rate of 2m/s?

[Wyrm]

Spoiler

You are wrong here. Specifically, you're ignoring the definition of t that I stated at the start of that post, which is the time between when the coil starts unwinding and when the coil finishes unwinding.

Spoiler

DH was already using t as a free time variable. You then stomped on that notation by redefining t as the duration of time the coil spends unwinding, customarily denoted ∆t. It's very bad form to stomp on someone's notation when critiquing it.

Spoiler

How do you propose getting the rope off the coil at a rate of 2m/s if the rope is not processing around the coil at a rate of 2m/s?

Spoiler

We're not given the size of the spindle, or the method of unwinding, so the problem would be unidentified and ill-specified if we had to consider it. Convention for these "freshman physics problems" is that such complications are to be assumed ignorable—otherwise, the problem is just to damned complicated for a freshman physics problem and you're not given enough informatino to properly deal with it anyway.

For instance, having to draw the rope out from a small spindle with a rusty axis would add an unknown but significant amount of force to the draw.

[Darth Holbytlan]

Spoiler

You are wrong here. Specifically, you're ignoring the definition of t that I stated at the start of that post, which is the time between when the coil starts unwinding and when the coil finishes unwinding. Similarly, T is the final kinetic energy, not the instantaneous kinetic energy, as specified in the opening post, and - with a presumption of linear acceleration - can be expressed in terms of the average velocity because the average velocity exists in fixed ratio to the final velocity.

Spoiler

It took me some effort to parse out any meaning to what you wrote at all, so I'm not too surprised that I misunderstood that. But see Wrym's point, above.

Also, look further on at this bit:

T_min approaches 0 as t=> infinity

Now, all of the sudden, t is a variable rather than a constant =5s. You provided absolutely no notice of the shift in meaning, so it's not too surprising I thought it was in a different place. In fact, you didn't explain how t is varying at all.

Spoiler

Again, you are using a different t. Surely you will agree that if it takes infinite time for the rope to leave the coil, then the rope is not in fact moving, and its kinetic energy is zero.

Spoiler

OK, so at this point you were actually varying vfinal in the original problem, resulting in an increase in t. Got it. Yes, I agree that as vfinal→0 (and therefore t→∞), the kinetic energy also approaches 0. But I don't see what relevance it has.

Spoiler

How do you propose getting the rope off the coil at a rate of 2m/s if the rope is not processing around the coil at a rate of 2m/s?

Spoiler

Again, see Wrym's point. The conventions in these sorts of problems are weird, but basically come down to "If it wasn't specified, it can be ignored". In this case the problem specified a coil, so we assume it is just lying motionless on the ground, positioned so that the top part can be pulled off without disturbing the rest. No diameter was specified for the coil, so we treat it as 0—effectively we have a point mass from which we can pull 10m of rope.

[Kuroneko]

Perhaps the problem would have been improved if no coil at all mentioned, and the rope was simply in a small bundle on the ground, but not tangled, so that it could be pulled smoothly, with the length and mass of the dragged rope increasing in time, instead of over the top of the coil. Sorry about that, Feil.

Anyway, the final kinetic energy is 10J, and the horizontal force is 2N in the appropriate direction. I'm going to pick on GPP a bit because he concisely illustrated the apparent incogruity:

The rope has a mass of 5 kg overall, and will move at 2 m/s when I've uncoiled it after 5 seconds and 10 m. That gives, with E=0.5 m v^2, a kinetic energy of 10 J. That happened over a space of 10 m, so a force of 1 N acted on it.
The impulse of the rope is m(t)*v, where m(t) is the mass of the moving part at time t. Over 5 seconds, it goes linearly from zero to 10 kg m/s. Force is change of impulse over time, so a force of 2 N acted on it. that happened over the space of 10 m, so since work is force times distance, we get an energy of 20 J.
What's going on here?

So the relationship between the kinetic energy and force is not the usual one. A direct answer:

But while applying the force, the rope only travels half the distance you travel (L) on average, or L/2.

But we might as well try to account for it in detail. Substituting Newton's second law into the infinitesimal amount of work done,
[1] F dx = dx d(mv)/dt = dx[ v(dm/dt) + m(dv/dt) ] = v²dm + m d(v²/2).
The latter term would normally give the kinetic energy, but in this case it is the former that's relevant, and integrating it shows the work done is exactly twice the kinetic energy: mv² rather than mv²/2. Thus, you applied 2N of force on the rope (and hence it did an equal and opposite on you) and did 20J of work in dragging it, even though only 10J went to its kinetic energy.

As a toy model, one can think of two particles of mass m each coupled by a massless spring. If one imparts one of them with velocity v parallel to the spring, its kinetic energy is T = mv²/2. But conservation of momentum requires that the system, having total mass 2m, travels at half the velocity, and hence its kinetic energy will be m(v/2)² = T/2. The other half must therefore be in the oscillation of the masses about the center of mass. The rope, modeled as a chain of similarly coupled masses, this means that half of the work you do turns into the vibratory motion of the atoms: heat.

Even without friction, half of your work is wasted. It is a consequence of the rope's inertia only.

[Darth Holbytlan]

That's neat, Kuroneko. I really wasn't expecting a heat term to drop out of a toy problem like this. I was so focused on the kinetic energy of the string that I didn't really process why the problem asked about the force the rope exerts on you, or that there was an energy discrepancy to account for. Cool.