Conceptual Difficulty with Lagrange's Eqn

[H_man]

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.

$$T = \frac{1}{2}m( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)$$

And

$$T' = \frac{1}{2}( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)$$

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

Now, as far as I understand the above equation is $$a_\theta$$. That is, the $$\theta$$ component of acceleration.

However, it seems I am wrong.

My book tells me I have to divide the expression by $$h_\theta$$

where

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

Producing $$a_\theta = r \ddot \theta + 2 \dot r \dot \theta - r \dot \phi^2 sin \theta cos \theta$$

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 $$F_\theta$$??

:uhh:

 

[Dr Transport]

A quick check of the units in equation (3) would show that it isn't an aceleration, you have $m^{2} s^{-2}$ not $m s^{-2}$.

 

[H_man]

Oops. Yep, you is right. Just had a look at the derivation again and I see where my confusion arose.

School-boy error

Thanks muchly!