Where the Angular Velocity Vector Coming From?

[Arman777]

I know that If we have a rigid body rotating clokwise direction,the angular velocity vector should be in the into the screen.But also I know that

##w=dθ/dt## so..Whats the equation that tells us that ##\vec w## is into the screen ?

Is it coming from some vector product ? Or we know that its ##\vec R x\vec v=\vec w##

 

[stevendaryl]

I know that If we have a rigid body rotating clokwise direction,the angular velocity vector should be in the into the screen.But also I know that

##w=dθ/dt## so..Whats the equation that tells us that ##\vec w## is into the screen ?
I thought ##\vec R x \vec {Δθ}=\vec w## (##\vec R## is in the upward direction).But sounds wrong...

The direction of [itex]\vec{\omega}[/itex] is the axis of rotation. If [itex]\vec{v}[/itex] is the velocity of a point on the body, and [itex]\vec{r}[/itex] is the vector to the center of rotation, then you can compute [itex]\omega[/itex] via:

[itex]\vec{\omega} = \frac{1}{r^2} (\vec{v} \times \vec{r})[/itex]

So it's a vector that is perpendicular to the plane containing both [itex]\vec{v}[/itex] and [itex]\vec{r}[/itex].

 

[Arman777]

The direction of [itex]\vec{\omega}[/itex] is the axis of rotation. If [itex]\vec{v}[/itex] is the velocity of a point on the body, and [itex]\vec{r}[/itex] is the vector to the center of rotation, then you can compute [itex]\omega[/itex] via:

[itex]\vec{\omega} = \frac{1}{r^2} (\vec{v} \times \vec{r})[/itex]

So it's a vector that is perpendicular to the plane containing both [itex]\vec{v}[/itex] and [itex]\vec{r}[/itex].

why there's ##\frac {1} {r^2}## ?

I see...Is there any equation that contains ##\vec w##, ##dθ## and ##dt## ?

 

[stevendaryl]

why there's ##\frac {1} {r^2}## ?

I see...Is there any equation that contains ##vec w##, ##dθ## and ##dt## ?

Well, the [itex]\frac{1}{r^2}[/itex] is intuitively correct from the following reasoning: If you have circular motion, then [itex]v = \frac{2 \pi r}{T}[/itex] where [itex]T[/itex] is the time for a complete circle. So the magnitude of [itex]\vec{v} \times \vec{r}[/itex] is [itex]\frac{2 \pi r^2}{T}[/itex]. Dividing by [itex]r^2[/itex] gives [itex]\frac{2\pi}{T}[/itex] which is the same as [itex]\frac{d\theta}{dt}[/itex].

So the magnitude of [itex]\vec{\omega}[/itex] is just [itex]\frac{d\theta}{dt}[/itex]

You can't get the direction of [itex]\vec{\omega}[/itex] from [itex]d\theta[/itex] and [itex]dt[/itex]. You need to know what plane the rotation is in.

 

[Arman777]

I understand now.We know that ##\vec w x \vec R= \vec v##
so ##(\vec w x \vec R) x \vec R= \vec v x \vec R##

Which its ##\vec w (\vec R.\vec R)-\vec R (\vec w.\vec R)=\vec v x \vec R##
##\vec w R^2-0=\vec v x \vec R##
##\vec w=\frac {1} {r^2} (\vec v x \vec R)##

I just should know that ##\vec w## is founding by right hand rule and ##\vec w x \vec R= \vec v##

You can't get the direction of ⃗ωω→\vec{\omega} from dθdθd\theta and dtdtdt. You need to know what plane the rotation is in.

I see ok

 

[Cutter Ketch]

##w=dθ/dt## so..Whats the equation that tells us that ##\vec w## is into the screen ?

This 1D equation is a simplification of the 3D description. In making this simplification θ is defined about an axis with a direction, so the direction is already implicitly in there.