Why can the potential energy at any point be chosen to have any value?

[rudransh verma]

Homework Statement

A chain of mass M is placed on a smooth table with 1/3 of its length L hanging from the top. The work done I’m pulling the chain back to surface is?

Relevant Equations

W=F.d

The force is -Mg and distance is L/3. So W=-MgL/3. Not right!
Maybe it’s done using COM but there is no additional information is given.

 

[Orodruin]

Homework Statement:: A chain of mass M is placed on a smooth table with 1/3 of its length L hanging from the top. The work done I’m pulling the chain back to surface is?
Relevant Equations:: W=F.d

The force is -Mg and distance is L/3.

Is the force -Mg during the entire pull?

Als, is it -Mg from the beginning even? If most of the chain is on the table that part will be supported by the table.

 

[rudransh verma]

Is the force -Mg during the entire pull?

O yes you are right. Do we have to use integral?

 

[Orodruin]

O yes you are right. Do we have to use integral?

You could, but you certainly do not have to. What is the potential gravitational energy before/after the pull?

 

[rudransh verma]

What is the potential gravitational energy before/after the pull?

That is not easy to tell. Sum of all the PE of elements of 1/3 chain will be initial PE of the 1/3 part and after the chain is on the table its Mg(height of the table)/3.
Is it Mg(height of the table)/3?

 

[Orodruin]

The height of the table is irrelevant since the L/3 of the chain is hanging freely. I suggest drawing a picture of the before and after situations if you are having trouble visualizing.

 

[rudransh verma]

The height of the table is irrelevant since the L/3 of the chain is hanging freely.

You asked about PE. So its measured from the ground. Height is important. But I don't think its going to solve like that since height is not given.

 

[rudransh verma]

The height of the table is irrelevant since the L/3 of the chain is hanging freely. I suggest drawing a picture of the before and after situations if you are having trouble visualizing.

MgL/18 by taking COM raising it a length L/6.

 

[Delta2]

You cannot use the COM logic here because the chain is not a rigid body, it changes shape as it moves.

 

[rudransh verma]

You cannot use the COM logic here because the chain is not a rigid body, it changes shape as it moves.

How can we do it without the COM concept?

 

[Delta2]

Well @Orodruin is trying to guide you into it. I don't know is I should interfere because I have another way in my mind

Oh well here it goes:
Work does only the weight of the part of the chain that is hanging (which part is different in length as time passes) (why is that can't you think of a reason? More specifically why the supported part (the part that is on the table) does no work with its weight?)

Assume $x$ is the length of the chain that is hanging (at a time t). What is the infinitesimal work $dW$ that the weight of this length $x$ does as it moves $dx$?

Integrate $dW$ (w.r.t $x$) to get the total work $W$

 

[Orodruin]

You cannot use the COM logic here because the chain is not a rigid body, it changes shape as it moves.

I disagree.

You asked about PE. So its measured from the ground. Height is important. But I don't think its going to solve like that since height is not given.

No, it is not relevant. Potential gravitational energy can be measured relative to any reference point you choose. This will not affect potential differences, which is all that is actually physical.

MgL/18 by taking COM raising it a length L/6.

The COM of what?

 

[Delta2]

I disagree.

Yes well it depends on what you disagree. The chain is not a rigid body (I hope you don't disagree on that) but yes maybe we can use the COM logic. I wasn't very sure when I was writing that that anyway my way doesn't use COM logic for this problem.

 

[Orodruin]

Yes well it depends on what you disagree. The chain is not a rigid body (I hope you don't disagree on that) but yes maybe we can use the COM logic. I wasn't very sure when I was writing that that anyway my way doesn't use COM logic for this problem.

Well, the chain is clearly not a rigid body. However, COM methods work perfectly fine.

 

[Delta2]

I get $mgL/18$ with my way too.

 

[rudransh verma]

Potential gravitational energy can be measured relative to any reference point you choose.

let's take the potential energy of COM before the pull as zero. Then the final potential energy is MgL/18 which is the work done to lift it up i think.

 

[rudransh verma]

This will not affect potential differences, which is all that is actually physical.

What do you mean?

 

[rudransh verma]

Assume x is the length of the chain that is hanging (at a time t). What is the infinitesimal work dW that the weight of this length x does as it moves dx?

Integrate dW (w.r.t x) to get the total work

$dW=-Mgdx/3$
Now integrating from 0 to L/6 we get
$\int_{0}^{L/6} -Mg/3 \, dx$
$W=-MgL/18$

 

[phinds]

What do you mean?

If you are standing on the ground and you lift a 10lb barbell up 2 feet over your head, you've done a certain amount of work. Call it W.

If you are standing at the top of the Eiffel Tower and you lift the same barbell the same amount over your head, how much work do you have to do?

 

[rudransh verma]

how much work do you have to do?

W

 

[phinds]

Right. So I assume that you now see why the height of the table is irrelevant in this problem, yes?

I notice a pattern in your questions and answers. You seem intent on blindly fitting things into formulae without trying to figure out what the formulae MEAN

 

[Delta2]

$dW=-Mgdx/3$
Now integrating from 0 to L/6 we get
$\int_{0}^{L/6} -Mg/3 \, dx$
$W=-MgL/18$

Nope you have gone wrong here. The formula you give for dW is wrong and also the boundaries of integration are wrong.

Lets start from the easy stuff. Remember $x$ is the length of chain hanging (NOT the position of the COM) and the chain initially hanging is $L/3$ and at the final position is 0, what do you think the boundaries should be?

 

[rudransh verma]

Right. So I assume that you now see why the height of the table is irrelevant in this problem, yes?

So there is no place where PE is actually zero. I thought on ground its zero.
I think you are right since if we place the ball on ground and dig under it it will fall into the hole which means its PE was not actually zero. It has the potential to fall continuously to the core of the earth.Right?

I notice a pattern in your questions and answers. You seem intent on blindly fitting things into formulae without trying to figure out what the formulae MEAN

Like when?

 

[rudransh verma]

Nope you have gone wrong here. The formula you give for dW is wrong and also the boundaries of integration are wrong.

Why is my way wrong? I took the COM position as zero and final position as L/6. I applied your concept. Now I think we don't have to apply Integration at all since the force does not vary. Just take the initial PE=0 and take the difference of PE. (Mg(L/6-0))/3

 

[kuruman]

So there is no place where PE is actually zero. I thought on ground its zero.
I think you are right since if we place the ball on ground and dig under it it will fall into the hole which means its PE was not actually zero. It has the potential to fall continuously to the core of the earth.Right?

Not quite right. You can choose the potential to be zero anywhere you want. That does not affect the difference in potential energy when an object is displaced by a certain amount. Perhaps you are under the impression that potential energy can only be positive. It can be positive or negative.

Example: You have a well of depth $h$ below the ground. You have a rock at ground level which you drop into the well.
Case I: Choose the zero of potential energy at the bottom of the well.
Initial potential energy: $U_i=+mgh$
Final potential energy: $U_{\!f}=0$
Change in potential energy = $\Delta U=U_{\!f}-U_i=0-mgh=-mgh.$

Case II: Choose the zero of potential energy at ground level.
Initial potential energy: $U_i=0$
Final potential energy: $U_{\!f}=-mgh$
Change in potential energy = $\Delta U=U_{\!f}-U_i=-mgh-0=-mgh.$

It's the change that matters.

 

[Delta2]

Why is my way wrong? I took the COM position as zero and final position as L/6. I applied your concept. Now I think we don't have to apply Integration at all since the force does not vary. Just take the initial PE=0 and take the difference of PE. (Mg(L/6-0))/3

You have blend in your mind my method and the COM method. My method doesn't use COM position, forget COM at all. $x$ is the length of the hanging part which varies from L/3 to 0 as the chain moves. Read again post #11 more carefully please.

 

[rudransh verma]

x is the length of the hanging part which varies from L/3 to 0 as the chain move

dW=-Mgdx/3 but the mass M is varying with respect to x. I think we need the function M(x). What do you say?

 

[Delta2]

Much better, It is actually M(x)gdx, now what do you think M(x) is? We can assume that the mass of the chain is homogeneously distributed.

 

[rudransh verma]

Much better, It is actually M(x)gdx, now what do you think M(x) is? We can assume that the mass of the chain is homogeneously distributed.

$m(x)=Mx/L$
$dW=-Mgxdx/L$
Taking limit from -L/3 to 0
we get $W=MgL/18$

 

[Delta2]

That's great you did it fine. Now one last question remains. Why do you think only the weight of the hanging chain does work (and not the weight of the part of the chain that is supported by the table)?

 

[rudransh verma]

That's great you did it fine. Now one last question remains. Why do you think only the weight of the hanging chain does work (and not the weight of the part of the chain that is supported by the table)?

How can it ? The part of chain is at rest.

In the solution above there is no -ve sign. It should have.

 

[Delta2]

The part of chain is at rest

No it isn't at rest, it moves horizontally, while the hanging part moves vertically...

It is a convention if we should take the work of weight as positive or negative.

 

[rudransh verma]

No it isn't at rest, it moves horizontally, while the hanging part moves vertically...

Actually I took the hanging end and placed it on top of the table. That’s how I did work.

 

[Delta2]

That's another way of putting the chain back at the table than I had in my mind but ok the work of weight should remain the same as my way since weight is a conservative force.

 

[Delta2]

Now that I think of it, that's the advantage of the COM/potential energy method, over my method, that there you can say that the work only depends on the potential energy of the initial and final position and not in the way that you put the chain back in table.

 

[rudransh verma]

is a convention if we should take the work of weight as positive or negative.

Answer should be with -ve sign. It’s not precisely right.
Also we can choose any reference point and it’s the change of PE that is physical and not the PE itself. Someone said this.Can you explain?
Is it because there is no actual place where PE=0 other than when the two bodies stick to each other. Like if we say PE=0 on ground, then if we dig beneath a ball placed on ground it will fall into hole. So it’s PE is actually not zero.

 

[russ_watters]

Also we can choose any reference point and it’s the change of PE that is physical and not the PE itself. Someone said this.Can you explain?
Is it because there is no actual place where PE=0 other than when the two bodies stick to each other. Like if we say PE=0 on ground, then if we dig beneath a ball placed on ground it will fall into hole. So it’s PE is actually not zero.

It's because no matter what reference you use, you will always get the same answer to the problem. So you might as well pick a reference that makes the math easier.

 

[Eberhard]

Perhaps MgL/18 was not understood.
My explanation for this is:

given:
L length of the chain
1 third hangs down
M Mass of the chain

searched:
Work for stroke of chain on table

Formula:
W=mgh

The thickness of the chain is neglected, otherwise thickness/2 would have to be added to the lifting height.
1 third hangs down, its center of gravity must be raised by half of L/3, lifting height is therefore L/6.
The mass of this part is also 1 third of M.

So is W=mgh=(M/3)*g*(L/6)=Mg/18

 

[Delta2]

Perhaps MgL/18 was not understood.
My explanation for this is:

given:
L length of the chain
1 third hangs down
M Mass of the chain

searched:
Work for stroke of chain on table

Formula:
W=mgh

The thickness of the chain is neglected, otherwise thickness/2 would have to be added to the lifting height.
1 third hangs down, its center of gravity must be raised by half of L/3, lifting height is therefore L/6.
The mass of this part is also 1 third of M.

So is W=mgh=(M/3)*g*(L/6)=Mg/18

Yes and no. My objection is where exactly the formula $W=mgh$ comes from. Because if we think like my post #11 then the formula for work becomes more complicated.

You should have mention that because the weight is a conservative force, its work equals to the change of potential energy and that would justify the formula $W=mgh$ where $h$ the difference in height of the initial and final position of the COM (COM comes into play as an easy way of calculating the potential energy of a body with dimensions, that is which is not a single point).

 

[Delta2]

Answer should be with -ve sign. It’s not precisely right.
Also we can choose any reference point and it’s the change of PE that is physical and not the PE itself. Someone said this.Can you explain?
Is it because there is no actual place where PE=0 other than when the two bodies stick to each other. Like if we say PE=0 on ground, then if we dig beneath a ball placed on ground it will fall into hole. So it’s PE is actually not zero.

First of all the theorem of classical mechanics we use is that $$W=-\Delta U=-(U_{final}-U_{initial})=U_i-U_f$$
(it holds for all conservative forces, it is a consequence of the gradient theorem of vector calculus)

Now the usual convention for the potential energy in the gravitational field of Earth is that we take it to be positive, so according to this convention the work will come out negative.

According to another convention though, when the forces in play are attractive we take the potential energy to be negative, and according to this convention the work of weight will be considered positive.

 

[rudransh verma]

Now the usual convention for the potential energy in the gravitational field of Earth is that we take it to be positive, so according to this convention the work will come out negative.

According to another convention though, when the forces in play are attractive we take the potential energy to be negative, and according to this convention the work of weight will be considered positive.

Yes! It has to do with where we take our x=0. And when we take the limit x=0 to x=L/3, I get the correct -ve work.
I like to follow one convention.
Now my question is is there a place where potential energy is actually zero because I have been solving problems taking grounds PE=0.
Also I took x=0 where the chain starts and not from the ground. Is this right?

 

[Delta2]

You can take the origin of the coordinate system anywhere you want and the zeroth level of PE anywhere you want also but you got to keep them the same through the whole calculations (not for example calculate the initial PE in one origin and the final PE in another origin).

The way I understand it is that the PE is actually zero only when there is no gravitational field at all.

 

[rudransh verma]

You can take the origin of the coordinate system anywhere you want

Ok! Right.

the zeroth level of PE anywhere you want

The way I understand it is that the PE is actually zero only when there is no gravitational field at all.

Doesn’t both contradict each other?

 

[Delta2]

Doesn’t both contradict each other?

No, when there is no field at all, the PE is zero everywhere, when we choose PE to be zero somewhere but there is field, the PE isn't zero everywhere.

 

[rudransh verma]

when we choose PE to be zero somewhere but there is field,

Why will you choose PE=0 in a field?

 

[tech99]

You asked about PE. So its measured from the ground. Height is important. But I don't think its going to solve like that since height is not given.

PE in this case is measured down from the table.

 

[Delta2]

Why will you choose PE=0 in a field?

Because according to the theorem $W=-\Delta U$ it is the difference in PE that matters when calculating work when moving through the field and not the absolute value of PE.

 

[rudransh verma]

Because according to the theorem $W=-\Delta U$ it is the difference in PE that matters when calculating work when moving through the field and not the absolute value of PE.

But if we take PE=0 where it actually is it tells us that the body doesn’t have potential to move. Isn’t this useful somehow?

 

[Delta2]

But if we take PE=0 where it actually is it tells us that the body doesn’t have potential to move. Isn’t this useful somehow?

The body doesn't have potential to move, that is acceleration (or equivalently a force that is applied to it), when $F=-\nabla U=\text{(in the case of 1-dimensional motion)}-\frac{dU}{dx}=0$ and not when $U(x)=0$. So to know where $U(x)=0$ isn't useful at all. Except that it makes easier the calculations for work around that point.

 

[rudransh verma]

The body doesn't have potential to move, that is acceleration (or equivalently a force that is applied to it), when $F=-\nabla U=\text{(in the case of 1-dimensional motion)}-\frac{dU}{dx}=0$ and not when $U(x)=0$. So to know where $U(x)=0$ isn't useful at all. Except that it makes easier the calculations for work around that point.

When we drop a ball from a height h the body moves and the body possesses KE. Where does the KE come from? PE. The ball has potential to do action. It has stored energy.