# Derivative of cross product

1. May 16, 2010

### Lavace

1. The problem statement, all variables and given/known data

http://damtp.cam.ac.uk/user/dt281/dynamics/two.pdf" [Broken]

Looking at page 5, equations (2.19) and (2.20)

3. The attempt at a solution

I cannot understand how they derived the (2.20), at first from comparing the solutions I had assumed r(dot)' had disappeared as we were differentiating with respect to dr'.

I then went about the derivative of the cross product:

Omega X r(dot)' + .... But in the solution we find Omega x (Omega X r').

Last edited by a moderator: May 4, 2017
2. May 16, 2010

### D H

Staff Emeritus
What that paper means by $\partial L/\partial \mathbf r'$ is

$$\frac{\partial L}{\partial{\mathbf r}'} \equiv \frac{\partial L}{\partial x'}\hat x' + \frac{\partial L}{\partial y'}\hat y' + \frac{\partial L}{\partial z'}\hat z'$$

It might be easier for you to see how (2.20) follows from (2.19) by using (2.19) in its first form,

$$L = \frac 1 2 m \Bigl((\dot x - \omega y')^2 + (\dot y + \omega x')^2 + \dot z^2\Bigl)$$

The result in (2.20) is a much more general result. It applies to any rotation vector $\boldsymbol \omega$, not just the pure z rotation used in that example.