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The Potential Energy Function in Three-Dimensional Motion

  • Thread starter Terrycho
  • Start date
Problem Statement
A particle is attracted toward the z-axis by a force F vector proportional to the square of its distance from the xy-plane and inversely proportional to its distance from the z-axis. Add an additional force perpendicular to F vector in such a way to make the total force conservative, and find the potential energy.
Relevant Equations
→ → → → →
F = ∇ V, and a conservative force F satisfies ∇ X F = 0
I set the location of the particle (x,y,z); therefore,


the force F_1 is (z^2/root(x^2+y^2) * x/root(x^2+y^2) , z^2/root(x^2+y^2) * y/root(x^2+y^2), 0), since cosΘ is x/root(x^2+y^2).
→ → → →
And also, the force F_1 and the additional force F_2 are perpendicular so, F1 ⋅ F2 =0.

So, I got F_x=-F_y. (I set F_2 as (F_x, F_y, F_z)

The total Force F_tot is ( z^2/root(x^2+y^2) * x/root(x^2+y^2) +F_x , z^2/root(x^2+y^2) * y/root(x^2+y^2) -F_y , F_z )
→ →
Then, I used Del operator ∇ X F = 0, which gave me the following result.

(1) ∂F_z/∂y + ∂F_x/∂z = 2yz/(x^2+y^2)

(2) ∂F_x/∂z - ∂F_z/∂x = 2zx/(x^2+y^2)

(3) ∂F_z/∂y + ∂F_z/∂x = (2yz-2zx)/(x^2+y^2)


I am kind of able to feel by doing something with the equations (1),(2), and (3), you can figure out the F_tot which leads to get to know the potential Energy but... I got stopped here!

I'd really appreciate if you could help me out. I attached some photos below for those of you who got confused with my messed up complicated symbols!
 

Attachments

Sorry for all the confusing arrows... I tried to express those symbols as vectors, so I put the arrows on the upper line with lots of space but they just ignored the space and put all the arrows together...

If you let me know how to compute vectors, I will edit the post right away!! Thanks!
 

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