Why is the work done by gravity not equal to the -ΔGPE? (unrolling carpet)

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  • #1
Nathanael
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Suppose we have a carpet rolled (along it's length) into the shape of a cylinder (of radius R0). Now we allow the carpet to unroll to a radius Rf < R0.
We can model the mass of the cylindrical part of the carpet as M = kR2

The differential work done by gravity would be dW = -MgdR = -kR2gdR
This leads to the total work done by gravity W = ∫dW = -(kg/3)Δ(R3)

But if we consider just the initial and final states: We start with mass M0=kR02 a distance R0 above the ground and end with Mf=kRf2 a distance Rf above the ground.
So the (negative of the) change in gravitational potential energy is M0gR0-MfgRf = kg(R03-Rf3) = -kgΔ(R3) which is three times what the work done by gravity was.
 

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  • #2
NascentOxygen
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Are you letting it unroll of its own accord, so that the remaining roll gains KE?
 
  • #3
Nathanael
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Are you letting it unroll of its own accord, so that the remaining roll gains KE?
Yes. We can also assume it unrolls without slipping.

Seems irrelevant; shouldn't the work done by gravity equal the -ΔGPE regardless?

If we let a particle fall freely a distance, or if we apply an upwards force so that it falls with a constant velocity, the work done by gravity will still equal the -ΔGPE either way.
 
  • #4
Dale
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The carpet is not a rigid body. Some energy is lost to deforming it. (Assuming the math was correct).
 
  • #5
Nathanael
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The carpet is not a rigid body. Some energy is lost to deforming it. (Assuming the math was correct).
I am not arguing that energy is conserved.

The work done by gravity should be equal to the -ΔGPE regardless of if energy is lost. This is basically the definition of GPE.
 
  • #6
Drakkith
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This leads to the total work done by gravity W = ∫dW = -(kg/3)Δ(R3)

I'm not too terribly familiar with integrals. Where did d happen to go?
 
  • #7
nasu
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How do you figure that formula for work? If you unroll it by a small angle, some parts go up, some go down. You seem to assume that the whole mass M goes down by dR.
 
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  • #8
Dale
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It also doesn't seem to account for the delta PE of the unrolled section.
 
  • #9
Nathanael
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How do you figure that formula for work? If you unroll it by a small angle, some parts go up, some go down. You seem to assume that the whole mass M goes down by dR.
Sorry, I don't see your point: If we were to account for every piece of mass then we should find that only the displacement of the center of mass is relevant, right?
If a ball is rotating and falling, some parts go up, some go down... but the CoM movement still gives the work done by gravity (as that is where gravity is effectively acting).

I am implicitly splitting it into two objects at each instant, the cylindrical part of the carpet and the flat part of the carpet. I don't see why this is wrong, though.

It also doesn't seem to account for the delta PE of the unrolled section.
I'm not sure what you mean. By unrolled section you mean the cylindrical part? By "it also" you are talking about my work calculation? Can you explain?


I'm not too terribly familiar with integrals. Where did d happen to go?
The "d" is not a variable or value, it's like the "d" in dy/dx (if you know about derivatives). So "dR" means a differential change in R.
 
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  • #10
Dale
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and end with Mf=kRf2 a distance Rf above the ground.
I mean that this formula is incorrect. It ignores the mass which has been unrolled.

By unrolled section I mean the part which is not in the cylinder any more.
 
  • #11
russ_watters
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Looks to me like you set up dW incorrectly: The mass is a variable too, since as you unroll it, it loses mass. Your dW appears to use a constant mass (at each "step") and only a differential of height.

In essence/from the other side (as the others were saying), you are leaving mass on the floor, so that mass is dropped the entire distance of the radius, not just the incremental distance.
 
  • #12
Nathanael
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I mean that this formula is incorrect. It ignores the mass which has been unrolled.

By unrolled section I mean the part which is not in the cylinder any more.
The total work on the carpet is the work on the cylindrical part plus the work on the flat part, right? The work on the flat part is zero, which is why I left it out. Sorry if I missed your point.

Looks to me like you set up dW incorrectly: The mass is a variable too, since as you unroll it, it loses mass. Your dW appears to use a constant mass (at each "step") and only a differential of height.

In essence/from the other side (as the others were saying), you are leaving mass on the floor, so that mass is dropped the entire distance of the radius, not just the incremental distance.
But if we consider how much of the carpet is left on the floor (at each "step") it would be Rdθ (which we could express in terms of dR and the thickness if we wanted) which is only a differential amount.
Since only a differential amount of mass is left behind at each "step," it will have a negligible effect on the location of center of mass of the cylindrical part.

It is because of this effect that I took the mass to be variable (w.r.t. R). How else would I quantify this effect?
 
  • #13
russ_watters
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But if we consider how much of the carpet is left on the floor (at each "step") it would be Rdθ (which we could express in terms of dR and the thickness if we wanted) which is only a differential amount.
Since only a differential amount of mass is left behind at each "step," it will have a negligible effect on the location of center of mass of the cylindrical part.
If differentials we're truly negligible, we wouldn't use them at all. That one, when you sum it, accounts for all of the mass of the carpet if you unroll the whole thing!
 
  • #14
Nathanael
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Edited;

Okay, I was thinking it just leads to higher-order-differentials but I was doing it wrong, sorry.

Using the formula RCoM = ΣiRimi/(Σimi) we get:
##dy_{CoM} = R-\frac{k(R-dr)^2(R-dr)}{kR^2} = 3dR##
Whereas in my OP I was using ##dy_{CoM} = dR## which is why it was small by a factor of 3

Thanks.
 
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  • #15
Drakkith
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The "d" is not a variable or value, it's like the "d" in dy/dx (if you know about derivatives). So "dR" means a differential change in R.

Whoops, I meant the d in MgdR. You have it up until you do the integral, at which point it disappears. Just wondering why.
 
  • #16
Nathanael
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Whoops, I meant the d in MgdR. You have it up until you do the integral, at which point it disappears. Just wondering why.
Just to be clear, it is (M)*(g)*(dR). In other words the "dR" is a single term.

The reason it disappears is because I integrated it: ##\int -MgdR = \int -kgR^2dR = -kg \int R^2dR = -kg\frac{R_f^3-R_0^3}{3} = -\frac{kg}{3}\Delta (R^3)##

(Sorry for the confusion, I left out this step in the OP)
 
  • #17
Dale
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The total work on the carpet is the work on the cylindrical part plus the work on the flat part, right? The work on the flat part is zero, which is why I left it out. Sorry if I missed your point.
I am talking about your expression for the energy in the initial and final state. That expression fails to account for the energy of the flat part.

Edit: I didn't check the math but it looks like you may have figured out the problem in post 14
 
  • #18
Nathanael
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I am talking about your expression for the energy in the initial and final state. That expression fails to account for the energy of the flat part.
I can write it like this if you'd like:

##[PE_{initial}]-[PE_{final}] = [M_0g(R+C)]-[M_fg(R+C)+(M_0-M_f)gC]##

In my OP I took the ground to be the zero potential (hence C = 0) because it doesn't make a difference.
 
  • #19
Dale
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Are you unrolling this horizontally or vertically?
 
  • #20
Nathanael
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Are you unrolling this horizontally or vertically?
Horizontally (should have mentioned that in the OP, sorry).
 
  • #21
Dale
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I was envisioning it vertically please ignore my comments. I am not sure if any of them are relevant.
 
  • #22
Drakkith
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Just to be clear, it is (M)*(g)*(dR). In other words the "dR" is a single term.

Oh. I thought d was distance above/below some reference height.
 

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