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

1. Jul 23, 2015

### Nathanael

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.

2. Jul 23, 2015

### Staff: Mentor

Are you letting it unroll of its own accord, so that the remaining roll gains KE?

3. Jul 23, 2015

### Nathanael

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. Jul 23, 2015

### Staff: Mentor

The carpet is not a rigid body. Some energy is lost to deforming it. (Assuming the math was correct).

5. Jul 23, 2015

### Nathanael

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. Jul 23, 2015

### Staff: Mentor

I'm not too terribly familiar with integrals. Where did d happen to go?

7. Jul 23, 2015

### nasu

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.

8. Jul 23, 2015

### Staff: Mentor

It also doesn't seem to account for the delta PE of the unrolled section.

9. Jul 23, 2015

### Nathanael

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.

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?

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.

Last edited: Jul 23, 2015
10. Jul 23, 2015

### Staff: Mentor

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. Jul 23, 2015

### Staff: Mentor

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. Jul 23, 2015

### Nathanael

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.

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. Jul 23, 2015

### Staff: Mentor

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. Jul 23, 2015

### Nathanael

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.

Last edited: Jul 23, 2015
15. Jul 23, 2015

### Staff: Mentor

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. Jul 23, 2015

### Nathanael

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. Jul 23, 2015

### Staff: Mentor

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. Jul 23, 2015

### Nathanael

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. Jul 23, 2015

### Staff: Mentor

Are you unrolling this horizontally or vertically?

20. Jul 23, 2015

### Nathanael

Horizontally (should have mentioned that in the OP, sorry).