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I Where the Angular Velocity Vector Coming From?

  1. Feb 4, 2017 #1
    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##
     
  2. jcsd
  3. Feb 4, 2017 #2

    stevendaryl

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    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].
     
  4. Feb 4, 2017 #3
    why theres ##\frac {1} {r^2}## ?

    I see...Is there any equation that contains ##\vec w##, ##dθ## and ##dt## ?
     
  5. Feb 4, 2017 #4

    stevendaryl

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    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.
     
  6. Feb 4, 2017 #5
    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##

    I see ok
     
  7. Feb 5, 2017 #6
    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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