# Vector Derivative of Angular Velocity Doesn't Allow For Change in Rotational Axis

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1. Nov 8, 2014

### CitrusLime

Greetings all,

I have been pondering this for a few days and cannot come to a conclusion. Suppose you have Frame-F with basis vectors I, J, & K. Also suppose you have Frame-B (basis vectors i, j, and k) which is rotating w.r.t. Frame-F. In such a case:

i = (cos Θ)I + (sin Θ)J
j = -(sin Θ)I + (cos Θ)J

[(di/dt) w.r.t. Frame-F] = (dΘ/dt)j
[(dj/dt) w.r.t. Frame-F] = (dΘ/dt)i

Now, you can calculate the angular velocity of Frame-B w.r.t Frame-F
using the following equations:

ω = i x [(di/dt) w.r.t. Frame-F] = i x (dΘ/dt)j = (dΘ/dt)k

Now, we take the derivative of this angular velocity (w.r.t to Frame F), using the formula for the derivative of a cross-product:

(d/dt)[u x v] = [u x (d/dt)v] + [(d/dt)u x v]

So...

[(dω/dt) w.r.t. Frame-F] = {i x (d/dt)[(dΘ/dt)j} + {(di/dt) x (dΘ/dt)j}

Where...

{(di/dt) x (dΘ/dt)j} = (dΘ/dt)j x (dΘ/dt)j = 0

And...

{i x (d/dt)[(dΘ/dt)j} = [i x (d2Θ/dt)j] + [i x -(dΘ/dt)2i]

So...

[(dω/dt) w.r.t. Frame-F] = [i x (d2Θ/dt)j] = (d2Θ/dt)k

The result indicates that if the magnitude of ω is W, then the
derivative of ω is (W)k. Since k is is always parallel to the angular-velocity, the result does not seem to include any terms that allow for change in axis of rotation. Where did these terms go???

NOTE: I had posted a similar question http://physics.stackexchange.com/questions/144894/question-on-using-transport-theorem-to-determine-angular-acceleration-of-a-rotat [Broken], but feel this is a better formulation of the same problem.

Last edited by a moderator: May 7, 2017
2. Nov 13, 2014