I Partial differentiation

1. Aug 4, 2016

Happiness

I believe there is a mistake in the second equation of (5.139).

The equation is obtained from (5.138) using the Euler-Lagrange equation
$\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{\partial L}{\partial\theta}.$

LHS$\,\,=\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{d}{dt}(mr^2\dot{\theta}-mr\dot{x}\cos\theta)$
$=mr^2\ddot{\theta}-mr\dot{x}(-\sin\theta)\dot{\theta}-mr\ddot{x}\cos\theta\,\,\,\,\,\,\,\,\,\,$ (Note that $\dot{r}$ terms are ignored.)

RHS$\,\,=mgr\sin\theta$

Equating LHS and RHS, and dividing by $m$ and $r$, we have
$r\ddot{\theta}+\dot{x}\sin\theta\dot{\theta}-\ddot{x}\cos\theta=g\sin\theta$.

Am I right?

2. Aug 4, 2016

axmls

There's a $\theta$-dependent term that you have forgotten to differentiate inside the parentheses.