Finding work done in spherical coordinates

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Homework Help Overview

The problem involves calculating the work done by a force expressed in spherical coordinates as it moves from the origin to a point along the x-axis. The force is given in terms of spherical coordinates, and the context suggests a focus on the line integral of the force over a specified path.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the force over the specified path and question the implications of the force components at specific angles, particularly at \varphi = 0. There is an exploration of how the force behaves along the path and its impact on the work done.

Discussion Status

Some participants have provided insights into simplifying the integral by examining the force along the path, particularly at \varphi = 0. There is an acknowledgment of the relationship between the force components and the resulting work calculation, with ongoing clarification of interpretations regarding the force's effect on work.

Contextual Notes

There is a noted assumption that the motion occurs solely along the radial direction, with no change in the angular coordinates, which influences the calculation of work. The discussion also highlights the potential for misunderstanding regarding the force's value at specific angles.

HeisenbergW
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1. Find the work done by the force F=r3*cos2[itex]\varphi[/itex]*sin[itex]\varphi[/itex]*[itex]\hat{r}[/itex] + r3*cos[itex]\varphi[/itex]*cos(2[itex]\varphi[/itex]) [itex]\hat{\varphi}[/itex]
from the point (0,0,0) to (2,0,0)

Homework Equations


Work=[itex]\int[/itex] F*dr
where dr= dr[itex]\hat{r}[/itex] + rd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex]

The Attempt at a Solution



When muliplying the line element, dr, by the force, F, I come up with
[itex]\int[/itex] r3*cos2[itex]\varphi[/itex]*sin[itex]\varphi[/itex] dr +[itex]\int[/itex] r4*cos[itex]\varphi[/itex]*cos(2[itex]\varphi[/itex]) d[itex]\varphi[/itex]

I believe the r goes from 0 to 2, and there is no change in [itex]\varphi[/itex]

I end up with 4*cos[itex]^{2}[/itex][itex]\varphi[/itex]*sin[itex]\varphi[/itex]
but then when I plug in 0 for [itex]\varphi[/itex], the answer ends up being zero, which I have a hard time believing since it moves from 0 to 2.
Any help is greatly appreciated
Thank You in advance.
 
Last edited:
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F=r3*cos2[itex]\varphi[/itex]*sin[itex]\varphi[/itex]*[itex]\hat{r}[/itex] + r3*cos[itex]\varphi[/itex]*cos(2[itex]\varphi[/itex]) [itex]\hat{\varphi}[/itex]

What is the force along [itex]\varphi = 0[/itex]?
(This should simplify your line integral.)
 
Check the force along the path you are given (because it's really a line integral through a vector field), and it should become fairly simple to see why that is. Notice that your psi component didn't really change in the integral that mattered.
 
Thanks for the replies
I believe you are saying that the force along [itex]\varphi[/itex]=0 is just zero along the r component, which is the only component that matters, since there is no motion in the other two coordinates. Since my force is actually zero at [itex]\varphi[/itex]=0, it doesn't matter that I went from (0,0,0) to (2,0,0), since no force means no work.

Am I interpreting your comments correctly?
Thanks for the feedback. Always appreciated.
 

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