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

In summary, the vector ##\vec \theta## is normal to the circle at the point where ##\vec r## is tangent.f

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##

Yes, with ##\vec r## constant that works.

##\vec s = \vec \theta \times \vec r##
What do ##\vec s##, ##\vec r## and ##\vec \theta## represent in this equation?

nasu, vanhees71 and Delta2
What do ##\vec s##, ##\vec r## and ##\vec \theta## represent in this equation?

What do ##\vec s##, ##\vec r## and ##\vec \theta## represent in this equation?
##\vec v## is tangencial speed, in direction of ##\vec s##.

Is the vector ##\vec r## rotating? If so, then ##\vec r## is not a constant vector. In the diagram, ##s## is an arc length. How is ##s## related to the vector ##\vec s##?

Delta2
Is the vector ##\vec r## rotating? If so, then ##\vec r## is not a constant vector. In the diagram, ##s## is an arc length. How is ##s## related to the vector ##\vec s##?
##\vec r## have constant magnitude.
##\theta = \frac {s}{r}##

##\vec r## have constant magnitude.
But in your derivation in the first post, you assumed that the vector ##\vec r## is constant. This assumption would mean that the vector ##\vec r## has constant magnitude and also constant direction. But it looks to me like you are considering a scenario where the vector ##\vec r## is changing direction.

##\theta = \frac {s}{r}##
What is the direction of the vector ##\vec \theta## that you used in your derivation?
What would the vector ##\vec s## look like in your drawing? Can you include it in your drawing?

But in your derivation in the first post, you assumed that the vector ##\vec r## is constant. This assumption would mean that the vector ##\vec r## has constant magnitude and also constant direction. But it looks to me like you are considering a scenario where the vector ##\vec r## is changing direction.

What is the direction of the vector ##\vec \theta## that you used in your derivation?
What would the vector ##\vec s## look like in your drawing? Can you include it in your drawing?
You are right.
##\vec r## need to have constant magnitude and constant direction.
I will continue this tomorrow.

Delta2
The derivation in the link doesn't look correct to me. I'm still uncertain about the vector ##\vec s##. You can define ##\vec s## as $$\vec s = \vec \theta \times \vec r$$ Then ##\vec s## is a vector that is tangent to the circle at the point located at the tip of ##\vec r##. The magnitude of ##\vec s## is the arc length traced out by the tip of ##\vec r## when ##\vec r## rotates by ##\theta##.

With this definition of ##\vec s##, we would have $$\frac{d \vec s}{dt} =\dot{\vec \theta} \times \vec r + \vec \theta \times \dot {\vec r}$$ The last term on the right is not zero. Note that ##\dot {\vec r}## is the velocity ##\vec v## of the particle moving on the circle! So the last term in the expression for ##\large \frac{d \vec s}{dt}## above is ##\vec \theta \times \vec v## and this is a vector that points toward the center of the circle (centripetal). The first term in the expression for ##\large \frac{d \vec s}{dt}## does in fact equal the velocity of the particle. So, $$\frac{d \vec s}{dt} =\vec v+\vec \theta \times \vec v$$ The last term is the odd centripetal term.

So, if ##\vec s## is defined to be ##\vec \theta \times \vec r##, then the time derivative of ##\vec s## does not equal the velocity of the particle due to the presence of the centripetal term.

Maybe I'm misinterpreting what the textbook is saying.

I found what might be an updated version of the textbook. See here.

Note that they have removed the "derivation" which involves taking the time derivative of ##\vec s##. Personally, I would like to see them remove any mention of the vector ##\vec s##. I don't see any use for it.

I tried to get normal and tangential acceleration form ##\vec s = \vec \theta \times \vec r##, but fail. XD
I think it is possible, I just made mistake somewhere.There is a lot to do, so i will not do it again.
##\vec \theta \times \frac {\vec r}{dt}## is definitely normal component of velocity.

Note that they have removed the "derivation" which involves taking the time derivative of ##\vec s##. Personally, I would like to see them remove any mention of the vector ##\vec s##. I don't see any use for it.
I agree. The book defines the vector then never uses it again, so it's a little strange that they left it in at all.

I tried to get normal and tangential acceleration form ##\vec s = \vec \theta \times \vec r##, but fail. XD
I think it is possible, I just made mistake somewhere.There is a lot to do, so i will not do it again.
##\vec \theta \times \frac {\vec r}{dt}## is definitely normal component of velocity.
The problem is the whole approach. It's simpler and more straightforward to start with ##\vec r = r \,\hat r## and calculate its derivatives, using the fact that ##\dot{\hat r} = \dot\theta\,\hat\theta## and ##\dot{\hat \theta} = -\dot\theta\,\hat r## to derive the formulas.

malawi_glenn
##\vec \theta \times \frac {\vec r}{dt}## is definitely normal component of velocity.

In general, the velocity of a particle is always tangent to the trajectory of the particle. So, if a particle moves in a circle the velocity of the particle is tangent to the circle. The normal component of velocity is always zero if "normal component" refers to a direction that is perpendicular to the tangential direction.

The vector ##\vec \theta \times \frac {d \vec r}{dt}## is the same as the vector ##\vec \theta \times \vec v## since ##\vec v = \frac {d \vec r}{dt}## by definition. As mentioned before, the vector ##\vec \theta \times \vec v## has a direction that is centripetal. But ##\vec \theta \times \vec v## is not a component of the velocity ##\vec v##.