# Work done by non-conservative force while particle moves in circle

• Ebby
In summary, the conversation discusses the problem of showing that the force is non-conservative, involving the work done for a round trip. The participants consider different methods, including using Cartesian and polar coordinates, and discussing the direction of the tangent vector and the unit vectors. The final solution is determined to be relatively simple, involving expressing the force vector in terms of unit vectors and using polar coordinates to integrate.
Ebby
Homework Statement
Show that the force is non-conservative
Relevant Equations
F = kx
x^2 + y^2 = R^2

I have to show that the force is non-conservative, i.e. that work done for the round trip ##\neq 0##.

Rearranging the circle equation, I can say that:$$x = \sqrt {R^2 - y^2}$$$$y = \sqrt {R^2 - x^2}$$
Then I have:$$F_x = -\sqrt {R^2 - x^2}$$$$F_y = \sqrt {R^2 - y^2}$$
Now, as I understand it, I must integrate these forces thus:
$$W_x = \int_a^b -\sqrt {R^2 - x^2} \, dx$$$$W_y = \int_a^b \sqrt {R^2 - y^2} \, dy$$
So now I'm not sure what my limits of integration ##a## and ##b## should be. Taking the expression for ##W_x##, for example, I imagine the ##x## coordinate starting at a value of ##R##, going to ##-R## and then back to ##R## again. But this will just mean ##W_x## is zero, which is not what I want.

Should I change the integraton to be over an angle rather than over a displacement to reflect the fact that the path is circular, and use limits of ##0## and ##2\pi##?

It does look like a job for polar coordinates!

The issue with your method is the ambiguity in the signs. ##x=\pm\sqrt{R^2-y^2}## etc.

I think I see. I need to take into account the sign ambiguity. So I can stay with using Cartesian coordinates:$$W_x = \int_R^{-R} -+ \sqrt {R^2 - x^2} \, dx + \int_{-R}^R -- \sqrt {R^2 - x^2} \, dx$$
$$= \int_R^{-R} - \sqrt {R^2 - x^2} \, dx + \int_R^{-R} - \sqrt {R^2 - x^2} \, dx$$
$$= -2 \int_R^{-R} \sqrt {R^2 - x^2} \, dx$$

$$W_y = \int_R^{-R} + \sqrt {R^2 - y^2} \, dy + \int_{-R}^R - \sqrt {R^2 - y^2} \, dy$$
$$= \int_R^{-R} + \sqrt {R^2 - y^2} \, dy + \int_R^{-R} + \sqrt {R^2 - y^2} \, dy$$
$$= 2 \int_R^{-R} \sqrt {R^2 - y^2} \, dy$$

$$W_{total} = W_x + W_y = -2 \int_R^{-R} \sqrt {R^2 - x^2} \, dx + 2 \int_R^{-R} \sqrt {R^2 - y^2} \, dy$$
$$\pi R^2 - \pi R^2 = 0$$

Oh dear.

PeroK
PeroK said:
It does look like a job for polar coordinates!

Ebby
OK, I shall try with polar coordinates, if I can work that out. But I did spot that I'm wrong about ##Wy##. I believe that the Cartesian method will yield ##Wy = 0##, and so ##W_{total} = \pi R^2## by this method.

What is the force vector in terms of unit vectors? What is unit vector in the direction of the tangent to the circular path?

Why do any integrations at all? You are not asked for a quantitative result. You are asked for a qualitative result. Use a little geometric intuition.

First, pick a direction for the path. Clockwise or counterclockwise.

Can you decide whether the ##x## component of the force does positive or negative work over the portion of the path in the first and second quadrants? Can you decide whether the ##x## component of the force does positive or negative work over the portion of the path in the third and fourth quadrants? Can you make the same decisions for the ##y## component in the second and third quadrants and then again in the fourth and first quadrants?

Steve4Physics
For information (since no one has yet mentioned it) the problem can be solved easily/quickly by showing that ##curl(\vec F) \ne 0##. A specific path is not need. But presumably you haven’t yet met ##curl##.

Ebby said:
I believe that the Cartesian method will yield ##Wy = 0##
It won't!

If you follow @jbriggs444 44 suggestions and @Chestermiller ’s hint about the dire tion of the tangent vector, you can show that (going round clockwise) the direction of ##\vec F## is always the same as the direction of motion. That would be sufficient 'proof' IMO.

I think I get it now:

$$F_r = \sqrt {{F_x}^2 + {F_y}^2} = \sqrt {{(-y)}^2 + x^2}$$where:$$x = R\cos(\theta),\, y = R\sin(\theta)$$
So we can form the integral:$$W_{total} = R^2 \int_0^{2 \pi} \sqrt {\sin^2(\theta) + \cos^2(\theta) \, d\theta}$$
$$= 2 \pi R^2$$And it was that simple? Is that right?

Ebby said:
I think I get it now:

$$F_r = \sqrt {{F_x}^2 + {F_y}^2} = \sqrt {{(-y)}^2 + x^2}$$where:$$x = R\cos(\theta),\, y = R\sin(\theta)$$
So we can form the integral:$$W_{total} = R^2 \int_0^{2 \pi} \sqrt {\sin^2(\theta) + \cos^2(\theta) \, d\theta}$$
$$= 2 \pi R^2$$And it was that simple? Is that right?
I don't follow this. You need to look at the unit vectors in polar coordinates:
$$\hat r = (\cos \theta) \hat x + (\sin \theta) \hat y$$$$\hat \theta = (-\sin \theta) \hat x + (\cos \theta) \hat y$$And express ##\vec F## in terms of these unit vectors.

Chestermiller said:
What is the force vector in terms of unit vectors? What is unit vector in the direction of the tangent to the circular path?
,
PeroK said:
I don't follow this. You need to look at the unit vectors in polar coordinates:
$$\hat r = (\cos \theta) \hat x + (\sin \theta) \hat y$$$$\hat \theta = (-\sin \theta) \hat x + (\cos \theta) \hat y$$And express ##\vec F## in terms of these unit vectors.
I made a sketch since I'm confused about the unit vectors. Where is ##\hat \theta##?

What you've drawn there is ##\hat \theta##. Not ##\hat r##, which is in the same direction as the radial position vector.

PeroK said:
What you've drawn there is ##\hat \theta##. Not ##\hat r##, which is in the same direction as the radial position vector.
And there's no confusion here about ##\hat r## being the tangent unit vector (which is what I thought it was) and R the radius?

Ebby said:
And there's no confusion here about ##\hat r## being the tangent unit vector (which is what I thought it was) and R the radius?
##\vec r## is the position vector, r is its magnitude and ##\hat r## is the radial unit vector. ##\vec r=r\hat r##.

MatinSAR
OK, I'll think about that. I think I've gotten way confused here because I had something like the following sketch in mind (which is obviously wrong, but I include it merely to elucidate my state of confusion):

Ebby said:
OK, I'll think about that. I think I've gotten way confused here because I had something like the following sketch in mind (which is obviously wrong, but I include it merely to elucidate my state of confusion):

View attachment 330098
To write ##\hat r=\hat r_x+\hat r_y##, those last two terms must mean "the x component of the unit vector ##\hat r##", etc. So that should turn into ##\cos(\theta)\hat x+\sin(\theta)\hat y##. What you have written would have magnitude r instead of magnitude 1 and is at a right angle to ##\vec r##.

haruspex said:
To write ##\hat r=\hat r_x+\hat r_y##, those last two terms must mean "the x component of the unit vector ##\hat r##", etc. So that should turn into ##\cos(\theta)\hat x+\sin(\theta)\hat y##. What you have written would have magnitude r instead of magnitude 1 and is at a right angle to ##\vec r##.

I drew it out using @PeroK's nomenclature. I still have questions:

Ebby said:
I drew it out using @PeroK's nomenclature. I still have questions:

View attachment 330150
The second line is wrong. ##\hat r=\cos(\theta)\hat x+\sin(\theta)\hat y##; ##\vec r=r\hat r=r\cos(\theta)\hat x+r\sin(\theta)\hat y##.
##\hat\theta## is neither an angle nor a displacement. It is the unit vector in the anticlockwise tangential direction.

PeroK said:
I don't follow this. You need to look at the unit vectors in polar coordinates:
$$\hat r = (\cos \theta) \hat x + (\sin \theta) \hat y$$$$\hat \theta = (-\sin \theta) \hat x + (\cos \theta) \hat y$$And express ##\vec F## in terms of these unit vectors.
See post number #11 for the definition of the unit vectors.

PeroK said:
See post number #11 for the definition of the unit vectors.
Ah that was careless. I did not mean to write that.

Yes I understand the unit vectors now.

Decomposing, this means that ##\vec F = R \hat \theta + 0 \hat r##.

And now to form the work integral. I imagine it's the integral over ##d\theta## of ##\vec F## dotted with something? Perhaps it's:$$\int_0^{2\pi} \vec F \cdot R \hat \theta \, d \theta$$

I'd appreciate completing this!

EDIT:

Well, I'll carry on with it assuming:$$W = \int_0^{2\pi} \vec F \cdot R \hat \theta \, d \theta$$
$$= R^2 \int_0^{2\pi} \, d \theta$$
$$= 2 \pi R^2$$

Last edited:
PeroK
Assume that the force is conservative. By definition this implies that there exists a function ##V=V(x,y)## such that
$$F_x=-\frac{\partial V}{\partial x},\quad F_y=-\frac{\partial V}{\partial y}.$$
From the identity
$$\frac{\partial^2 V}{\partial x\partial y}=\frac{\partial^2 V}{\partial y\partial x}$$
it follows that
$$\frac{\partial F_y}{\partial x}=\frac{\partial F_x}{\partial y}.$$ This is the only relation to check!

nasu
wrobel said:
Assume that the force is conservative. By definition this implies that there exists a function ##V=V(x,y)## such that
$$F_x=-\frac{\partial V}{\partial x},\quad F_y=-\frac{\partial V}{\partial y}.$$
From the identity
$$\frac{\partial^2 V}{\partial x\partial y}=\frac{\partial^2 V}{\partial y\partial x}$$
it follows that
$$\frac{\partial F_y}{\partial x}=\frac{\partial F_x}{\partial y}.$$ This is the only relation to check!
Is that sufficient in the case where the particle is constrained? If we had a non-conservative force that vanished on the given circle, say, then would that not be conservative on the restricted domain of the particle?

PeroK said:
Is that sufficient in the case where the particle is constrained? If we had a non-conservative force that vanished on the given circle, say, then would that not be conservative on the restricted domain of the particle?
This is a good question. Surely not. And that is why I think that the statement of the problem is not good. Two interpretations are possible. The circle may be treated as a constraint, or it may be merely a curve that is suggested for checking that the work done is not zero. These are different statements.
My answer is for the second case. If the circle is a constraint, then one should calculate a generalized force ##Q=Q(\theta)##. This force is locally conservative, but globally not. The corresponding potential is a multivalued function on the circle.

In the general case, we have only a local result: if a differential form ##Q_idq^i## is closed, then it is locally exact.

UPD It is possible that the force in ##\mathbb{R}^2## is not conservative but the corresponding generalized force on the circle is conservative.

Last edited:
In message 21 above, is the integral correct? Thanks.

Ebby said:
In message 21 above, is the integral correct? Thanks.
Yes.

SammyS

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