Why is linear velocity represented as cross product of angular velocity of the particle and its position vector? Why not vice versa? (Consider rigid body rotation)

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    $\begingroup$ Are you asking why velocity is given by $v = \omega \times r$ rather than $v = r \times \omega$? $\endgroup$ – John Rennie Mar 19 '15 at 7:18
  • $\begingroup$ yes!if you know the answer please let me know! $\endgroup$ – user74370 Mar 19 '15 at 7:23
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    $\begingroup$ I think it's just because we conventionally take anti-clockwise angles to be positive. I can't think of any deeper reason. $\endgroup$ – John Rennie Mar 19 '15 at 7:45

OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes: This is UCM with respect to a stationary axis.

Let the particle rotate about the axis OO' ... Within time interval dt let its motion be represented by the vector whose direction is along axis obeying right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ.

Now, if elementary displacement of particle at a be specified by radius vector r,

From the diagram, it is easy to see that, for infinitesimal rotation, dr= dφΧr ... 1 (crossproduct)

By definition, ω = dφ/dt

Thus taking the elementary time interval as dt, all given equations surely hold!

Thus we can divide both sides of equation 1 by dt which is corresponding time interval!

So we get dr/dt = dφ/dt X r of course r value won't change WRT the particle and axis, so r/dt is essentially r!

So result is, v = ωΧr

  • $\begingroup$ And just so u know, if u tried v = rXω, that would be fundamentally wrong, bcuz of how the direction of ω us assumed by the right hand rule convention!, then of course you would obtain an illogical result, anyway , that's if u go by convention!! $\endgroup$ – sugatasen Mar 19 '15 at 7:46
  • $\begingroup$ Ok then basically i've been questioning the right hand screw rule itself..and do you know a proof for the right hand screw rule? Btw this is a good approach :-) $\endgroup$ – user74370 Mar 19 '15 at 7:56
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    $\begingroup$ The right hand screw rule is just convention; we could define the cross product the opposite direction and it would be fine as long as the same new version was applied universally $\endgroup$ – danimal Mar 19 '15 at 10:27

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