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10:53
@JohnRennie: Hi sir :-)
@GuruVishnu hi :-)
Are you free now sir?
I'm asked to find the work done by the force F as a particle moves from A to B:
The force F always points towards B throughout the particles journey.
That should be straightforward. Use the angle as the variable to integrate
11:01
I tried to correlate the angle made by the force F with the horizontal to the vertical angle made by the position vector about the centre. But with no improvement. IS there any alternate method?
@JohnRennie If we use the angle made with the horizontal, will that not vary non-uniformly? It varies faster near A and slower near B, am I right, sir?
The work is $\mathbf F \cdot d \mathbf r$
Ok sir. I'll try this again. A different doubt but related to this, how do we tell whether this force is conservative or non-conservative?
And that is $F dr \cos\phi$, where $\phi$ is the angle between the force and dr. Yes?
@JohnRennie Yes sir.
$F$ is a constant and $dr = Rd\theta$
11:04
Yes sir.
so $dW = FR \cos\phi d\theta$
Yes sir. I arrived at this. But was unable to find a relation between $\phi$ and $d\theta$, maybe I need to think some more time.
And $\phi$ will be a function of $\theta$ though I'll need to think exactly what that function is, so you'll get $dW = FR f(\theta) d\theta$
where $f(\theta) = \cos\phi$
Ok sir. No problem. I'll do that part :-)
Could you tell how to determine whether this force is conservative or not?
Integrate that from $\theta =0$ to $\theta = \pi/2$ and you have the answer.
Does that make sense?
This is all off the top of my head so I cannot guarantee it's validity! :-)
11:07
@JohnRennie If I figured out the relation between $\theta$ and $\phi$, I'm sure, I'll get the answer. I'm not sure how to arrive at it now.
Oh. Did you ask that for the nature of force?
@GuruVishnu a force $\mathbf F$ is conservative if and only if $\nabla \times \mathbf F = 0$
I don't think it would be easy to calculate the curl for the force in your question.
That's where the problem arises. I haven't learnt 'curl' in much detail. I'm just using the classic, simple definitions of conservative and non-conservative force.
Your thinking you could just draw the line AB, then the work is length(AB) times F?
If it's conservative that would be correct ...
But here F is not a constant vector. Or is it fine to use the same here too?
You are told F always points towards B. Yes?
11:10
Yes sir. So its direction always changes as the particle moves from A to B.
And if you travel along the line AB then your direction of travel always points towards B.
So the force and your travel are in the same direction, which makes the integral easy.
AHA!
Oh. Yes. So as per your method, we're trying to find the work along the path AB and trying to check whether it matches the initial result.
If the force always points towards B then its curl will be zero so it is conservative.
So the answer is $W = 20\sqrt{6^2 + 6^2}$
@JohnRennie I agree sir. But if we're not sure that the force is conservative, how can we make use of this method to assure that the work along AB is same as the work along the arc?
@GuruVishnu if you don't know whether the force is conservative or not you have no option but to integrate along the path.
But this force is conservative
11:14
@JohnRennie Ok sir. Did you figure this fact using the 'curl' concept? Or any other logic?
I'm not much into vector calculus and I'm think it's time to learn that.
The curl of a radially symmetric field is always zero.
Curl is basically a rotation. If a field has a non-zero curl then there will be field lines that form loops.
Like magnetic fields?
In this case the field lines are straight lines through B so there cannot be any looped field lines and hence the curl must be zero.
By 'radially symmetric' do you refer to a 'central' force which always acts along the line joining two points, sir?
@GuruVishnu yes, exactly. Magnetic fields are the archetypal example of a non-conservative field because the filed lines form loops.
@GuruVishnu yes. We are told the force always points towards B. Yes?
11:17
Yes sir.
So the field lines are straight lines radiating outwards from the point B.
Just like the electric field from a point charge, only this force doesn't change with distance.
Ok sir. I can understand this. This is similar to opposite charges placed in the system with a constrain that one must move only along the curved path.
Have you got the answer?
I understood this 'conservative' part based on electrostatic forces. But to fully understand, I think, I need to understand vector calculus. For the first part, I need to do the integration after our discussion.
Do the integration and see what the result is ...
11:21
Fun fact: I've studied about vectors as well as calculus. But I don't know vector calculus.
Could you recommend any good sources where I can get the basics of vector calculus? I'm seeing it more often now while reading answers on the main site.
I get $\phi = 45 - \theta/2$
Having found the force is conservative, as you told, the work along AB, the straight line is $120\sqrt 2$ which is correct. I need to do this using the formal (long) approach.
Is curl just a property of shape of the field rather than its magnitude sir? Here the force is uniform, but in our electrostatic analogy, it has an inverse square relationship.
Yes, it is a property of the shape not the value. Any field that has field lines forming loops will have a non-zero curl regardless of how the field varies along the field line.
I need to go now. I may be around later.
Ok sir. Thank you for your time and help. Good bye :-)
Bye :-)

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