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Deriving angular velocity vector algebra?

  1. Mar 16, 2010 #1
    where x represents cross product

    currently if i forget i figure these out using the right hand rule, but how do you get each equation visa versa using vector algebra

    i started with w = rxv

    how do you derive that v = wxr

    i got up to this

    w = rxv
    w= -(vxr)
    rxw = -rx(vxr)
    rxw = v(r.r) - r(r.v)
    rxw= v -r(r.v)

    but can u assume r.v are perpendicular?
    is this the right approach? and you even relate the 2 equations this way??
  2. jcsd
  3. Mar 16, 2010 #2


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    This is not true in general.

    We know that w=rxv mus be orthogonal to both r and v by definition.

    If v=wxr, then v would be orthogonal to both w and r. We know v is orthogonal to w by hypothesis, but v is not necessarily orthogonal to r.

    Bottom line, you would have to assume orthogonality of r and v to make any kind of assertion like this.
  4. May 10, 2010 #3
    You were wrong from the start I'm afraid.

    For v_t is tangential velocity (theta_hat component)

    v_t=|v_t|=|rhat x v|
    sometimes but not always more useful:
    rhat=r/r and
    v_t=|(r/r) x v|
    in vector form then:
    w= (rhat/r) x v=(r/r^2) x v

    wr= rhat x v

    in absolute value this is:

    Of course in absolute value you could have just skipped all the vector stuff,
    so presumably it is the vector result that you are interested in.
    Last edited: May 11, 2010
  5. May 11, 2010 #4
    BTW it is true that:
    where v_t=v_t theta_that=rhatxv

    Since all three are now orthogonal , proof of that comes from unit vector cross product rules, basically the right hand rule anyway, except it will work for a right or left handed rule since w depends in the first place on which rule you're using. That's why if one side of an equal is an axial vector, the other side also should also , and also why, as in the case here, an axial vector crossed with a vector, is a vector. The result of that cross product doesn't depend on the rule.
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