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Homework Help: Conceptual Difficulty with Lagrange's Eqn

  1. Aug 21, 2005 #1
    I shall give a short example to illustrate where I am confused.

    If we take the K.E. of a particle in spherical polar coords.

    [tex]
    T = \frac{1}{2}m( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)
    [/tex]

    And

    [tex]
    T' = \frac{1}{2}( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)
    [/tex]

    Now plugging this in Lagrange's equation:
    [tex]
    \frac {d} {dt} \frac {\partial T'} {\partial \dot \theta} - \frac {\partial T'} {\partial \theta} = r^2 \ddot \theta + 2r \dot r \dot \theta - r^2 \dot \phi^2 sin \theta cos \theta
    [/tex] --------- Line 3

    Now, as far as I understand the above equation is [tex]a_\theta[/tex]. That is, the [tex]\theta[/tex] component of acceleration.

    However, it seems I am wrong.

    My book tells me I have to divide the expression by [tex]h_\theta[/tex]

    where

    [tex]
    h_\theta = \left ( \left ( \frac {\partial x} {\partial \theta} \right )^2
    + \left ( \frac {\partial y} {\partial \theta} \right )^2
    + \left ( \frac {\partial z} {\partial \theta} \right )^2 \right )^\frac {1}{2} = ( (r cos \theta cos \phi )^2 + ( (r cos \theta sin \phi )^2 + r^2 sin^2 \theta )^\frac {1}{2} = r
    [/tex]

    Producing [tex]a_\theta = r \ddot \theta + 2 \dot r \dot \theta - r \dot \phi^2 sin \theta cos \theta [/tex]

    So making the reasonable assumption that the book is correct and I am not. What does the expression (Line 3) that I thought was the acceleration represent?

    This is especially confusing as I know that if we do not divide T by m then line 3 should produce the force [tex]F_\theta[/tex]??

    :frown: :frown: :uhh:
     
  2. jcsd
  3. Aug 21, 2005 #2

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    A quick check of the units in equation (3) would show that it isn't an aceleration, you have [itex] m^{2} s^{-2} [/itex] not [itex] m s^{-2} [/itex].
     
  4. Aug 21, 2005 #3
    Oops. Yep, you is right. Just had a look at the derivation again and I see where my confusion arose.

    School-boy error :blushing:

    Thanks muchly!!
     
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