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

[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 $2\pi$?

 

[PeroK]

It does look like a job for polar coordinates!

 

[haruspex]

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

 

[Ebby]

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]

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.

 

[Chestermiller]

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

 

[jbriggs444]

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 $curl$.

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.

Also, that may help you resolve any sign-error(s) in your Post #4 working if you wanted.

 

[Ebby]

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?

 

[PeroK]

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.

 

[Ebby]

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

,

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 $\hat \theta$?

 

[PeroK]

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

 

[Ebby]

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?

 

[haruspex]

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. $\vec r=r\hat r$.

 

[Ebby]

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):

 

[haruspex]

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 $\vec r$.

 

[Ebby]

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:

 

[haruspex]

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

View attachment 330150

I previously read your $F_x$ etc. as $\hat r_x$, etc. Please try to use LaTeX.
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]

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.

 

[Ebby]

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:$$W = \int_0^{2\pi} \vec F \cdot R \hat \theta \, d \theta$$
$$= R^2 \int_0^{2\pi} \, d \theta$$
$$= 2 \pi R^2$$

 

[wrobel]

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!

 

[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!

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?

 

[wrobel]

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.

 

[Ebby]

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

 

[PeroK]

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

Yes.