Euler-Lagrange Equation Q: Where Does 2mr˙r˙θ Come From?

[volican]

On the following post, where it says q=θ for the Euler-Lagrange equation where does the 2mr˙r˙θ come from?

https://www.physicsforums.com/threads/variable-length-pendulum.204840/

 

[BvU]

Hello Volican,

It comes from the time derivative of $\partial {\mathcal L}\over \partial \dot\theta$ $${\partial {\mathcal L}\over \partial \dot\theta }= mr^2\dot \theta $$

 

[volican]

Thanks for writting back, much appreciated. I can see where that comes from and also the -mgrsinθ but can't see how they got the 2mr˙r˙θ. Any ideas? I am thinking that I can just completley ignore the dr/dt as it is not in the euler lagrange equation? Is this wrong?

 

[BvU]

$$
{d\over dt} \left ( {\partial {\mathcal L}\over \partial \dot\theta } \right ) = {d\over dt} mr^2\dot \theta = m {d\over dt} \left ( r^2\right ) \dot \theta + mr^2 {d\over dt} \dot \theta = 2mr\dot r\dot\theta + mr^2\ddot\theta$$

 

[volican]

Aewsome! to differentiate r^2 did you use the chain rule?

 

[BvU]

Yes, the chain rule is awesome $$
{d\over dt} r^2 = {d\over dt} \left ( r \right ) r + r {d\over dt} \left ( r \right ) = 2 r\dot r $$

Make sure you speak this jargon fluently when pursuing further studies !