Work done by a tangential force shouldn't be zero

[etotheipi]

Homework Statement

N.B. It's hard to write out the problem so here's a link as well: https://isaacphysics.org/questions/work_done3?board=399e042f-90e1-49a9-bace-29ed242057ff
It's part three that I'm stuck on - the answers to the first two are F0 and 90 respectively.

A particle is moving in the (𝑥,𝑦)-plane in a circle of radius 𝑟 centred on the origin i.e. such that 𝑟^2=𝑥^2+𝑦^2.

Its displacement from the origin at any time is given by 𝑟=𝑥𝑖 +𝑦𝑗. The force on the particle is given by

𝐹 =𝐹0/r (𝑦𝑖−𝑥𝑗)

with i and j as the basis vectors.

How much work is done by the force as the particle moves round? (Work done is given by 𝐹.𝑟)

Relevant Equations

The question gives the statement "Work done is given by 𝐹.𝑟" but I think this should be F . delta r instead.

Applying the dot product to the position vector and force vectors yields the result that the force vector is always at 90 degrees to the position vector, namely that it is directed tangentially.

The mark scheme gives the work done as 0, however I don't agree. I believe they got this by finding the dot product of the force and position vector from the origin giving 0, whilst this seems completely invalid because the equation for work involves the force dotted with a displacement vector along which the particle is moving (albeit an infinitesimal in this case).

It seems that to work this out we need to do some sort of much more complicated integral of the form

$$W = \int \vec{F} . d\vec{r}$$

I worked out that this is not an exact differential (by computing d(Fx)/dy and d(Fy)/dx and comparing), so the force is not conservative, so we will not necessarily do no work once we go around the circle.

Would I be right in thinking that such a tangential force would actually do work on the particle (instead of 0)? This question has confused me a lot because I would say the work done is the integral of all of the force dot ds contributions around the circle, but the mark scheme appears to be dotting the force with the position vector, which makes no sense?

Sorry if I'm missing something!

 

[BvU]

I worked out

Dear $e^\pi$, show us what you worked out !

 

[haruspex]

Homework Statement: N.B. It's hard to write out the problem so here's a link as well: https://isaacphysics.org/questions/work_done3?board=399e042f-90e1-49a9-bace-29ed242057ff
It's part three that I'm stuck on - the answers to the first two are F0 and 90 respectively.

A particle is moving in the (𝑥,𝑦)-plane in a circle of radius 𝑟 centred on the origin i.e. such that 𝑟^2=𝑥^2+𝑦^2.

Its displacement from the origin at any time is given by 𝑟=𝑥𝑖 +𝑦𝑗. The force on the particle is given by

𝐹 =𝐹0/r (𝑦𝑖−𝑥𝑗)

with i and j as the basis vectors.

How much work is done by the force as the particle moves round? (Work done is given by 𝐹.𝑟)
Homework Equations: The question gives the statement "Work done is given by 𝐹.𝑟" but I think this should be F . delta r instead.

Applying the dot product to the position vector and force vectors yields the result that the force vector is always at 90 degrees to the position vector, namely that it is directed tangentially.

The mark scheme gives the work done as 0, however I don't agree. I believe they got this by finding the dot product of the force and position vector from the origin giving 0, whilst this seems completely invalid because the equation for work involves the force dotted with a displacement vector along which the particle is moving (albeit an infinitesimal in this case).

It seems that to work this out we need to do some sort of much more complicated integral of the form

$$W = \int \vec{F} . d\vec{r}$$

I worked out that this is not an exact differential (by computing d(Fx)/dy and d(Fy)/dx and comparing), so the force is not conservative, so we will not necessarily do no work once we go around the circle.

Would I be right in thinking that such a tangential force would actually do work on the particle (instead of 0)? This question has confused me a lot because I would say the work done is the integral of all of the force dot ds contributions around the circle, but the mark scheme appears to be dotting the force with the position vector, which makes no sense?

Sorry if I'm missing something!

I agree with your analysis. It's a blunder.

 

[BvU]

I can't read the problem statement: the link gives a green turning thingy that turns forever

Still would like to see $\vec F\cdot d\vec r$

 

[jbriggs444]

I worked out that this is not an exact differential (by computing d(Fx)/dy and d(Fy)/dx and comparing), so the force is not conservative, so we will not necessarily do no work once we go around the circle.

What makes you think that the force is not conservative?

As I see it the x component of the force is directly proportional to the x component of the displacement from the origin. This gives rise to what is obviously a conservative field. Same for the y component. The one field amounts to a repulsion and the other field amounts to an attraction, but since both are conservative, so is their sum.

 

[etotheipi]

What makes you think that the force is not conservative?

As I see it the x component of the force is directly proportional to the x component of the displacement from the origin. This gives rise to what is obviously a conservative field. Same for the y component. The one field amounts to a repulsion and the other field amounts to an attraction, but since both are conservative, so is their sum.

I've written it down in a weird way, though I think the x component of the force is proportional to the y displacement and vice versa.

 

[ZapperZ]

Why can't you solve this in plane polar coordinates? Rewrite F in terms of polar coordinates, and apply that work integral to it, where dr = dr r^ + r dθ θ^.

Zz.

 

[jbriggs444]

I've written it down in a weird way, though I think the x component of the force is proportional to the y displacement and vice versa.

Ahhh, my mistake then.

 

[Delta2]

I agree that the force isn't conservative but still the work around a particular path could be zero. Show us your work on how you calculated the work of the force around 1 complete circle.

EDIT: After looking at the plot of the force field at wolframalpha I agree that the work shouldn't be zero.
https://www.wolframalpha.com/input/?i=plot+vector+field+(y,-x)
I think there must be some typo in the formula for the force, since the particle is doing circular motion there must be a component of the force normal to the velocity, that is in the radial direction in this case, and the force as given has no radial component.

 

[etotheipi]

I've tried the polar integration but have no idea if I've done it correctly. Here's my work:

$$x= r\cos{\theta}, y = r\sin{\theta}$$ and consequently $$dx = -r\sin{\theta}d \theta, dy = r\cos{\theta}d \theta$$So$$\vec{F} = (F_{0}\sin{\theta}) \vec{i} - (F_{0}\cos{\theta}) \vec{j}$$Now considering the particle moves in a clockwise direction

$$W = \int_{2\pi}^{0} -F_{0}r\sin^{2}{\theta}d \theta - F_{0}r\cos^{2}{\theta}d \theta$$
$$W = -F_{0}r \int_{2\pi}^{0} d \theta = 2\pi r F_{0}$$
This answer makes some sense to me because it is the circumference multiplied by the magnitude of the force, though I don't know for sure. Is this the right way of solving the problem?

Why can't you solve this in plane polar coordinates? Rewrite F in terms of polar coordinates, and apply that work integral to it, where dr = dr r^ + r dθ θ^.

Zz.

 

[etotheipi]

I think there must be some typo in the formula for the force, since the particle is doing circular motion there must be a component of the force normal to the velocity, that is in the radial direction in this case, and the force as given has no radial component.

Yeah this confused me as well. I suspect since the question intended some sort of situation with an inward force and tangential displacement, for which we would instantly be able to see that W = 0.

 

[haruspex]

there must be some typo in the formula for the force, since the particle is doing circular motion there must be a component of the force normal to the velocity, that is in the radial direction in this case, and the force as given has no radial component.

I see no suggestion that the particle's motion is determined by the given force. There may be other forces present.

 

[rmwhitehill7918]

I agree with your statement, a way to check if a force is conservative is to check whether its curl is zero, but in this case, it obviously cannot be zero since the force is tangential to a circular path at all points along the path, implying that it "swirls" about the origin. Also, by checking we can see that $\vec{\nabla} \times \vec{F} = -2\frac{F_0}{r} \hat{z}$.

I worked out the integral by parameterizing $x$ and $y$ in terms of $\theta$ and found the differential path increment $d\vec{r}$ with that parametrization, and I then assumed that the particle was traveling in a counterclockwise direction and obtained the result $W = -\frac{F_0}{r}2\pi$. If the particle were traveling the other direction, this would be a positive value. Either way, as a qualitative assessment, since the force is tangential to the path either in the same direction or in the opposite direction, the force must do work on the particle.