- #36
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Thank you ... I have done it with a math editor.
Work done by variable force
- Thread starter MatinSAR
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Yeah, you replace it with whatever variable you want.the equation of path is x=(2a) sinθ and y=(a) cosθWhat is ##\theta## here? It can't be the argument parameter of the polar representation of (x,y) since we would deduce ##\tan^2(\theta)=\frac 12##. So I assume it is an arbitrary parameter unrelated to polar coordinates. But then, where do the integration bounds come from?I see no diagram.f
- #37
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What about Theta ?! Did I specify theta correctly?
I get the factor 6 again ! Let me send a picture ...
- #38
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- #39
erobz
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I'm getting what you are getting.
$$ \int 3 x y^2 dy = \int 3 \cdot y^2 \cdot 2 \sqrt{a^2 - y^2} dy $$
Let:
##\frac{y}{a} = \sin \beta##
## \frac{ \sqrt{a^2 - y^2} }{a} = \cos \beta##
It follows that
$$ y^2 = a^2 \sin^2 \beta$$
$$\sqrt{a^2 - y^2} = a \cos \beta$$
$$ dy = a \cos \beta d \beta$$
$$ \implies \int 3 x y^2 dy = 6 a^4 \int \sin^2 \beta \cos^2 \beta d \beta $$
$$ \int 3 x y^2 dy = \int 3 \cdot y^2 \cdot 2 \sqrt{a^2 - y^2} dy $$
Let:
##\frac{y}{a} = \sin \beta##
## \frac{ \sqrt{a^2 - y^2} }{a} = \cos \beta##
It follows that
$$ y^2 = a^2 \sin^2 \beta$$
$$\sqrt{a^2 - y^2} = a \cos \beta$$
$$ dy = a \cos \beta d \beta$$
$$ \implies \int 3 x y^2 dy = 6 a^4 \int \sin^2 \beta \cos^2 \beta d \beta $$
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- #40
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I also get a factor of 6, although I would never have done the math the way that he did it>
- #41
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Was it true ?!
What was wrong with the way I have done the math
Can you please tell me what is your final answer ?!
- #42
erobz
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I didn't do the integral!Was it true ?!
What was wrong with the way I have done the math
Can you please tell me what is your final answer ?!
- #43
erobz
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Post your work, it will get checked by someone...
- #44
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Yes, my mistake… I should have waited until I had the time to check properly.
- #45
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- #46
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It is a line integral, not a double integral, so you must arrange that there is only one variable in the integrand. If you want to do it as an integral wrt y then you must first replace all the occurrences of x with what x is as a function of y.
But really, the parametric approach using theta is much easier.
- #47
erobz
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Show the work. Explain how you get the last result?
- #48
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So my answer is wrong , isn't it ?!
Thank you ... I will try to do it using parametric approach and I will send a picture of the work ...
Thank you ... I will try to do it using parametrich approach and I will send a picture of the work ...
- #49
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I'm confused. Isn’t that what you already did in post #32? Except that you did not perform the integral.
- #50
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The integration is very simple. It just involves the use of trigonometric identities.$$(\sin{\theta}\cos{\theta})^2=\frac{\sin^2{2\theta}}{4}=\frac{1-\cos{4\theta}}{8}$$
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- #51
Steve4Physics
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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.
- #52
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Yes. It was unfinished.
Thank you.
Thank you ... I didn't know about this.
- #53
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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 ?!
The force is conserative so If I choose another path the final answer shouldn't change, Is it true ?!
- #54
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- #55
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- #56
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I hope I haven't forgotten anyone.
Thank you for your help and time.
- #57
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Try the path along the axes. Isn't F always zero there?
- #58
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No it's not.
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- #59
Steve4Physics
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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.
- #60
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