Where the Angular Velocity Vector Coming From?

AI Thread Summary
The discussion focuses on understanding the angular velocity vector, denoted as **ω**, particularly its direction in relation to a rigid body rotating clockwise. The key point is that **ω** is perpendicular to the plane formed by the position vector **r** and the velocity vector **v**, and its direction can be determined using the right-hand rule. The equation **ω = (1/r^2)(v × r)** is highlighted, illustrating how to compute **ω** based on the vectors involved. Additionally, it is clarified that while **ω = dθ/dt** provides the magnitude, the direction requires knowledge of the rotation plane. Overall, the discussion emphasizes the relationship between angular velocity, rotation direction, and the vectors involved in circular motion.
Arman777
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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##
 
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Arman777 said:
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 \vec{\omega} is the axis of rotation. If \vec{v} is the velocity of a point on the body, and \vec{r} is the vector to the center of rotation, then you can compute \omega via:

\vec{\omega} = \frac{1}{r^2} (\vec{v} \times \vec{r})

So it's a vector that is perpendicular to the plane containing both \vec{v} and \vec{r}.
 
stevendaryl said:
The direction of \vec{\omega} is the axis of rotation. If \vec{v} is the velocity of a point on the body, and \vec{r} is the vector to the center of rotation, then you can compute \omega via:

\vec{\omega} = \frac{1}{r^2} (\vec{v} \times \vec{r})

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

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

I see...Is there any equation that contains ##\vec w##, ##dθ## and ##dt## ?
 
Arman777 said:
why there's ##\frac {1} {r^2}## ?

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

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

So the magnitude of \vec{\omega} is just \frac{d\theta}{dt}

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

stevendaryl said:
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
 
Arman777 said:
##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.
 
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