Deriving angular velocity vector algebra?

  • #1

Main Question or Discussion Point

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??
 

Answers and Replies

  • #2
jav
35
0
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.
 
  • #3
2
0
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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  • #4
2
0
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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