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

[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.

 

[rudransh verma]

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.

Sorry guys.

 

[phinds]

I would never ignore anyone if it’s a fact

So, you think we've just been giving you opinions instead of facts? Why are you here if you don't believe we know what we're talking about?

 

[rudransh verma]

So, you think we've just been giving you opinions instead of facts? Why are you here if you don't believe we know what we're talking about?

I do believe you are right but let it make sense to me.
Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

 

[russ_watters]

Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

It isn't true of the spring either. As I said before, it is common to choose easy/convenient zero reference, but what is easy/convenient can vary.

A common alternative scenario for a spring is when it is constrained or pre-loaded. That's what the "tare" button on a scale is for.

One of the problems with your posture here is that you have blinders on. You are locked into preconceived logic/intuition and thus unable to really listen to/absorb what people are telling you. It would help you if you could generalize the things you are learning and get off wayward paths faster.

 

[rudransh verma]

One of the problems with your posture here is that you have blinders on. You are locked into preconceived logic/intuition and thus unable to really listen to/absorb what people are telling you. It would help you if you could generalize the things you are learning and get off wayward paths faster.

Good point. I will try.

 

[kuruman]

I do believe you are right but let it make sense to me.
Actually in case of spring it has PE=0 in relaxed state and its PE increases no matter where it moves. So I thought this must be the case for ball Earth system too. Somewhere I strongly believed in the back of my mind that this applies everywhere.

What applies everywhere is that the potential energy at any point can have any value you choose because you can choose the zero of potential energy anywhere you choose. This is true even for a spring. Here is how it works for a spring which I assumed has one end fixed whilst the other is allowed to move and stretch or compress the spring.

When you write ##PE=\frac{1}{2}kx^2## the zero of potential energy is at the free end of the spring when the spring is relaxed. Also zero is the position of the free end of the spring. In other words, there are two zeroes implicitly defined at the end of the spring: (a) the zero of potential energy and (b) the origin of coordinates. The latter choice makes it possible to make the position of the free end equal to the displacement of the spring from the relaxed position. This is useful because the change in potential energy of the spring (remember the change is what counts) which is always ##\Delta U = \frac{1}{2}k(\Delta x)^2## can be written as ##\Delta U = \frac{1}{2}k x^2## because we chose the origin of coordinates so that ##\Delta x=x.##

To help you digest this, consider what would happen if you measured coordinate ##x## from the fixed end and kept the zero of potential energy at the free end of the relaxed spring. Let ##L## be the relaxed end of the spring. Then the force exerted by the spring is written as ##F=-k(x-L)## and the force is negative when the spring is stretched (##x>L##) and positive when the spring is compressed (##x<L##).

First I will find the change in potential energy when the free end of the spring moves from ##x_1## to ##x_2##. That change is the negative of the work done by the spring force: $$\Delta U=U_2-U_1=-\int_{x_1}^{x_2}Fdx=+\int_{x_1}^{x_2}k(x-L)dx=\frac{1}{2}k\left(x_2^2-x_1^2\right)-kL(x_2-x_1).$$ This, I repeat, is the change in potential energy when the spring moves from ##x_1## to ##x_2.##

One can get the potential energy function ##U(x)## by choosing the potential energy to be zero at the starting or reference point ##x_1=x_{\text{ref}}## in which case ##U_1=0## in the above equation. Also the end point in the equation can be any ##x## because the subscript ##2## is no longer necessary. Then $$\Delta U=U_2-0= U(x)=\frac{1}{2}k\left(x^2-x_{\text{ref}}^2\right)-kL(x-x_{\text{ref}}).$$This is the most general expression for the potential energy of a spring of length ##L## where ##x## is the position of the free end relative to the fixed end. Note that it is zero at ##x=x_{\text{ref}}##.

If you choose the potential to be zero at the free end, i.e. ##x_{\text{ref}}=L##, then the potential simplifies to $$U(x)=\frac{1}{2}k\left(x^2-L^2\right)-kL(x-L)=\frac{1}{2}k(x-L)^2=\frac{1}{2}k(\Delta x)^2.$$Finally, if you choose the zero of coordinates to be at the free end, ##\Delta x=x## and you end up with the familiar $$U(x)=\frac{1}{2}kx^2.$$Please read this carefully and digest it. See how you can adjust the zeroes to get different values for the potential energy at a given point. It will liberate your mind.