- #1

- 42

- 14

- Homework Statement
- Can we get ##\vec v = \vec \omega \times \vec r## by deriving ##\vec s = \vec \theta \times \vec r## with respect to time?

- Relevant Equations
- Nothing

My try:

##\vec s = \vec \theta \times \vec r##

##\frac {d}{dt} (\vec s) = \frac {d}{dt} (\vec \theta \times \vec r)##

##\vec v = \frac {d}{dt}(\vec \theta) \times \vec r + \vec \theta \times \frac {d}{dt}(\vec r)##

##\vec v = \vec \omega \times \vec r + \vec \theta \times \vec v##

And I realized while writing this that ##\vec \theta \times \frac {d}{dt}(\vec r) = 0## because ##\vec r## is constant.

And at the end we get just ##\vec v = \vec \omega \times \vec r##

##\vec s = \vec \theta \times \vec r##

##\frac {d}{dt} (\vec s) = \frac {d}{dt} (\vec \theta \times \vec r)##

##\vec v = \frac {d}{dt}(\vec \theta) \times \vec r + \vec \theta \times \frac {d}{dt}(\vec r)##

##\vec v = \vec \omega \times \vec r + \vec \theta \times \vec v##

And I realized while writing this that ##\vec \theta \times \frac {d}{dt}(\vec r) = 0## because ##\vec r## is constant.

And at the end we get just ##\vec v = \vec \omega \times \vec r##