Is the Work Done by a Variable Force Non-Conservative?

[Steve4Physics]

View attachment 317296

Hi @MatinSAR. Can I point out a fundamental mistake you are making? I don’t think anyone has mentioned it yet (apologies if they already have).

The final expression for W can not contain x or y (or θ). These are variables which change as you move along the path. They must disappear when the definite integral is evaluated.

You made the same mistake in your Post #1 solution, where your final expression for W contained x and y.

The expression for W can contains only given constants (here only the 'a') and pure numbers like 5, π and √2 (just picking random values as examples).

For example, if you determine the formula for the circumference of a circle, x²+y²=a², it wouldn’t make sense to have x and/or y in the final formula.

 

[MatinSAR]

I'm confused. Isn’t that what you already did in post #32? Except that you did not perform the integral.

Yes. It was unfinished.

The integration is very simple. It just involves the use of trigonometric identities.$$(\sin{\theta}\cos{\theta})^2=\frac{\sin^2{2\theta}}{4}=\frac{1-\cos{4\theta}}{8}$$

Thank you.

Hi @MatinSAR. Can I point out a fundamental mistake you are making? I don’t think anyone has mentioned it yet (apologies if they already have).

The final expression for W can not contain x or y (or θ). These are variables which change as you move along the path. They must disappear when the definite integral is evaluated.

You made the same mistake in your Post #1 solution, where your final expression for W contained x and y.

The expression for W can contains only given constants (here only the 'a') and pure numbers like 5, π and √2 (just picking random values as examples).

For example, if you determine the formula for the circumference of a circle, x²+y²=a², it wouldn’t make sense to have x and/or y in the final formula.

Thank you ... I didn't know about this.

 

[MatinSAR]

I hope it's finally true ...

The force is conserative so If I choose another path the final answer shouldn't change, Is it true ?!

 

[haruspex]

The force is conserative

Is it? What do you get if you integrate from 0 to 2π?

I agree with your answer.

 

[MatinSAR]

Is it?

It was mentioned in question.

I agree with your answer.

Thank you.

 

[MatinSAR]

.

.

.

.

.

.

I hope I haven't forgotten anyone.

Thank you for your help and time.

 

[haruspex]

It was mentioned in question.

Try the path along the axes. Isn't F always zero there?

 

[MatinSAR]

Try the path along the axes. Isn't F always zero there?

No it's not.

 

[Steve4Physics]

No it's not.

Oh yes it is! (If you familiar with British pantomime.)

Can I expand on what @haruspex said in Post #57?

Referring to the Post #21 diagram, suppose you take the following route:
- from A(0,a), move along the y-axis to the origin (0,0);
- from the origin (0,0) move along the x-axis to C(2a,0).

$\vec F = x^2y~\hat i + xy^2 \hat~j$

While moving along the y-axis (x=0); both components of $\vec F$ are zero.
Similarly while moving along the x-axis.

The total work done along this path is therefore zero. This is different to your calculated value for the original route from A to C.

So you can see (without doing any maths) that the force is not conservative.

 

[MatinSAR]

Referring to the Post #21 diagram, suppose you take the following route:
- from A(0,a), move along the y-axis to the origin (0,0);
- from the origin (0,0) move along the x-axis to C(2a,0).

Great !
So it's not conserative ...
Thank you.