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Homework Help: Relation between force in cartesian, polar.

  1. Mar 18, 2010 #1
    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....
     
    Last edited: Mar 19, 2010
  2. jcsd
  3. Mar 19, 2010 #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?
     
  4. Mar 19, 2010 #3
    Yes, There's L(angular momentum) in the U
     
  5. Mar 19, 2010 #4
    Explicitly write the angular momentum terms and see if any have r dot terms.
     
  6. Mar 20, 2010 #5
    Isn't x the component cartesian coordinate?
     
  7. Mar 20, 2010 #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.
     
    Last edited: Mar 20, 2010
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