Relation between force in cartesian, polar.

In summary, Goldstein(3rd) discusses generalized potentials and how they are defined. He shows that the component of the forces in the two coordinate systems (cartesin, spherical polar) are related to each other as Q_{j}=F_{i}\cdot \frac{\partial r_{i}} {\partial q_{j}} \cdots
  • #1
merrypark3
30
0
Goldstein(3rd) 1.15

Generalized potential, U as follows.

[tex] U( \stackrel{\rightarrow}{r} ,\stackrel{\rightarrow}{v})=V(r)+\sigma\cdot L[/tex]

L is angular momentum and [tex]\sigma[/tex] is a fixed vector.


(b) show thate the component of the forces in the two coordinate systems(cartesin, spherical polar) are related to each other as
[tex]Q_{j}=F_{i}\cdot \frac{\partial r_{i}} {\partial q_{j}} \cdots (a) [/tex]

So I did,
[tex] Q_{j}= - \frac{\partial U}{\partial q_{j}} + \frac{d}{dt} (\frac{\partial U}{\partial \dot q_{j}})[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial \dot x_{k}}{\partial \dot q_{j}} \frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}+ \frac{\partial x_{k}}{\partial q_{j}} \frac{d}{dt}(\frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]=\frac{\partial x_{k}}{\partial q_{j}}(
- \frac{\partial U}{\partial x_{k}} + \frac{d}{dt} (\frac{\partial U}{\partial \dot x_{k}}))+ \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}
[/tex]

[tex]=\frac{\partial x_{k}}{\partial q_{j}} (
F_{k})+ \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}
[/tex]



If the last term in the last line vanishes, [tex] Q_{j} [/tex] and [tex] F_{k} [/tex] satisfies the relation (a), but it DOESN't vanish. What's my problem??
I've evaluated the last term in this condition, but It doesn't...
 
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  • #2
Replace x with r:

[tex]\frac{d}{dt}(\frac{\partial r_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot r_{k}}[/tex]

Does U change with respect to r dot?
 
  • #3
chrisk said:
Replace x with r:

[tex]\frac{d}{dt}(\frac{\partial r_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot r_{k}}[/tex]

Does U change with respect to r dot?

Yes, There's L(angular momentum) in the U
 
  • #4
Explicitly write the angular momentum terms and see if any have r dot terms.
 
  • #5
chrisk said:
Explicitly write the angular momentum terms and see if any have r dot terms.

Isn't x the component cartesian coordinate?
 
  • #6
You could have started the problem as this:

[tex] Q_{r}= - \frac{\partial U}{\partial r} + \frac{d}{dt} (\frac{\partial U}{\partial \dot r})[/tex]

and expressed the generalized potential as this:

[tex] U( \stackrel{\rightarrow}{r} ,\stackrel{\rightarrow}{v})=V(r)+\sigma_{\theta}\L_{\theta}+\sigma_{\phi}\L_{\phi}[/tex]

Express the L terms explicitly in terms of m, r, theta, phi, and sigma and the second term on the right side of the generalized force equation will equal zero.
 
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FAQ: Relation between force in cartesian, polar.

What is the difference between force in cartesian and polar coordinates?

The main difference between force in cartesian and polar coordinates is the way in which the force is represented. In cartesian coordinates, force is represented by a vector with x and y components, while in polar coordinates, force is represented by a magnitude and direction angle.

How do you convert between force in cartesian and polar coordinates?

To convert force from cartesian to polar coordinates, you can use the following equations:

  • Magnitude: F = √(Fx^2 + Fy^2)
  • Direction angle: θ = tan^-1(Fy/Fx)
To convert from polar to cartesian coordinates, you can use the following equations:
  • x-component: Fx = Fcos(θ)
  • y-component: Fy = Fsin(θ)

How is force affected by the coordinate system used?

The coordinate system used does not affect the magnitude or direction of the force. However, the representation of the force may differ in cartesian and polar coordinates.

Can force be added or subtracted in both cartesian and polar coordinates?

Yes, force can be added or subtracted in both cartesian and polar coordinates. However, the method of addition/subtraction may differ depending on the coordinate system used.

Which coordinate system is better for representing force?

This depends on the situation and personal preference. Cartesian coordinates are often preferred for their simplicity and ease of use in calculations, while polar coordinates are preferred for their ability to easily represent circular motion. It is important to understand and be proficient in both coordinate systems as a scientist.

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