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

[Vladimir_Kitanov]

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 $\vec v = \vec \omega \times \vec r$

 

[haruspex]

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

 

[TSny]

$\vec s = \vec \theta \times \vec r$

What do $\vec s$, $\vec r$ and $\vec \theta$ represent in this equation?

 

[Vladimir_Kitanov]

What do $\vec s$, $\vec r$ and $\vec \theta$ represent in this equation?

 

[Vladimir_Kitanov]

What do $\vec s$, $\vec r$ and $\vec \theta$ represent in this equation?

$\vec v$ is tangencial speed, in direction of $\vec s$.

 

[TSny]

View attachment 305181

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 $\vec s$?

 

[Vladimir_Kitanov]

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.
$\theta = \frac {s}{r}$
$\theta$ is angle in radians

 

[TSny]

$\vec r$ 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.

$\theta = \frac {s}{r}$
$\theta$ is angle in radians

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?

 

[Vladimir_Kitanov]

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.

 

[Vladimir_Kitanov]

Direction of $\vec r$ depend on angle $\theta$ but $\frac {d}{dt}(\vec r)$ is 0.
https://www.marianopolis.edu/wp-content/uploads/2019/04/4-Rigid-Bodies.pdf

 

[TSny]

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.

 

[TSny]

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.

 

[Vladimir_Kitanov]

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.

 

[vela]

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.

 

[vela]

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.

 

[TSny]

$\vec \theta \times \frac {\vec r}{dt}$ 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 $\vec v$.