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

[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.

 

[Hall]

@rudransh verma Hello pal! Assume the chain to be having uniform density. So, we have $d = \frac{dM}{dx}= M/L$. I have assumed the chain to be one dimensional.

Now, try to understand that the part which is hanging down is made of up little tiny rings (infinitesimal rings) that are joined together to from a chain, the ring closest to the table need not to be l
lifted up, and the ring at the bottom need to be lifted up by a distance of L/3. Consider the ring which is at a distance $x$ from the table, its mass is $dM$ (infinitesimal ring) and it has to be lifted up by a distance of $x$, so, work done for that particular ring is
$$
dM g~ x$$
Now, if we were to analyze this for all the rings and sum it up to get the total work, we would get
$$
\int_{0}^{L/3} dM g x$$
$$
g \int_{0}^{L/3} M/L dx x $$
In the above, I replaced $dM$ by but what I wrote in the beginning.
$$
g ~M/L~ \frac{x^2}{2} \big|_{0}^{L/3}$$

Can you proceed on?

 

[Delta2]

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.

I am not sure about this argument but I think it is actually the changes in the PE energy as the ball moves that make the work and hence the KE. According to the theorem if there are no changes in PE, there is no work, it doesn't matter if $U_i=U_f=0$ or $U_i=U_f=150$, the work will be zero.

 

[rudransh verma]

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.

I am not sure about this argument b

No you are right! PE is not a absolute quantity. @John Rennie from physics stack exchange told me that. It can be chosen as zero where ever we want. It is the difference W that matters. We can choose PE as zero on ground by assuming we are at inifinity from the ground and are calculating W.

 

[jbriggs444]

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.

The reference PE associated with the zero field is just as arbitrary as the reference PE associated with any other conservative field.

The gradient of the constant field with value x is just as zero as the gradient of the constant field with value zero.

 

[kuruman]

No you are right! PE is not a absolute quantity. @John Rennie from physics stack exchange told me that. It can be chosen as zero where ever we want. It is the difference W that matters. We can choose PE as zero on ground by assuming we are at inifinity from the ground and are calculating W.

That same idea was expressed, directly or indirectly, on this your thread in posts #6 and #12 (@Orodruin), #19 (@phinds), #25 (@kuruman) and #37 (@russ_watters). Yet you seem to have ignored all of the above and now you quote somebody else from somewhere else as if it were a piece of information that you see for the first time.

My question to you is what kind of sense does it make to you to ask for our guidance, which we offer freely, if you then turn around and ignore it? Never mind us, what's in it for you?

 

[phinds]

My question to you is what kind of sense does it make to you to ask for our guidance, which we offer freely, if you then turn around and ignore it?

 

[Delta2]

That same idea was expressed, directly or indirectly, on this your thread in posts #6 and #12 (@Orodruin), #19 (@phinds), #25 (@kuruman) and #37 (@russ_watters). Yet you seem to have ignored all of the above and now you quote somebody else from somewhere else as if it were a piece of information that you see for the first time.

My question to you is what kind of sense does it make to you to ask for our guidance, which we offer freely, if you then turn around and ignore it? Never mind us, what's in it for you?

Yes it seems like something stroke him and he finally saw the light. Maybe he took the words from John Rennie at stack exchange more seriously than ours here at PF.

 

[kuruman]

Maybe he took the words from John Rennie at stack exchange more seriously than ours here at PF.

I googled John Rennie and I got this Wikipedia article. I guess Rennie's word carries more weight because he has a Wikipedia article and we don't. Mystery solved.

On edit: For the record, a PM from @caz informed me that the Wikipedia John Rennie is not the stack exchange John Rennie. A profile of the latter can be found here. I apologize for any embarrassment my inadvertent mistake may have caused. The mystery remains unsolved.

 

[rudransh verma]

That same idea was expressed, directly or indirectly, on this your thread in posts #6 and #12 (@Orodruin), #19 (@phinds), #25 (@kuruman) and #37 (@russ_watters). Yet you seem to have ignored all of the above and now you quote somebody else from somewhere else as if it were a piece of information that you see for the first time.

I would never ignore anyone if it’s a fact but sometimes it’s difficult for me to understand something that has been said indirectly. I don’t mean to prioritise anyone over the other. I just want to understand the reality ie physics. And in this no one is more right than the other. It’s the science that is right always. Sometimes a simple “Yes” is enough than pouring information because sometimes it gets too much overwhelming. And I like that about John. (He is a Colloid scientist from England).

 

[rudransh verma]

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.

Oh! Yes! Sorry I missed that.