Deriving angular velocity vector algebra?

AI Thread Summary
The discussion centers on deriving the angular velocity vector equations using vector algebra, particularly the relationships between w = r x v and v = w x r. The participants explore the implications of orthogonality among the vectors involved, noting that w must be orthogonal to both r and v by definition. Concerns are raised about assuming that r and v are perpendicular, which is necessary for certain assertions to hold true. The conversation also touches on the tangential velocity component and its relation to angular velocity, emphasizing the importance of vector representation. Ultimately, the conclusion is that the relationships depend on the orthogonality of the vectors involved, and the cross product rules are crucial for understanding these dynamics.
SpartanG345
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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??
 
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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.
 
You were wrong from the start I'm afraid.

For v_t is tangential velocity (theta_hat component)

w=v_t/r
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

then
wr= rhat x v

in absolute value this is:
wr=v_t

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.
 
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BTW it is true that:
wxr=v_t
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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