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

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?