Calculating Angular Velocity & Acceleration of Rotating Frame-B

[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

This leads to:

[(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 , but feel this is a better formulation of the same problem.

 

[Achy47]

Thank you for your thought-provoking question. I am a scientist and I would like to provide some insight on the issue you have raised.

Firstly, let's clarify the notation used in the post. The basis vectors I, J, K represent the fixed reference frame (Frame-F), while the basis vectors i, j, k represent the rotating frame (Frame-B). The angle Θ represents the rotation of Frame-B with respect to Frame-F. The derivatives of these vectors (di/dt, dj/dt) represent the angular velocity of Frame-B with respect to Frame-F.

Now, to answer your question about the lack of terms accounting for change in the axis of rotation, we need to understand the concept of angular acceleration. Angular acceleration is the rate of change of angular velocity with respect to time. In your calculations, you have correctly obtained the angular velocity (ω) of Frame-B with respect to Frame-F. However, in order to calculate the angular acceleration, we need to take the derivative of ω with respect to time.

In your equation, [(dω/dt) w.r.t. Frame-F], you have taken the derivative of ω with respect to Frame-F, which is incorrect. Since ω is a vector quantity, its derivative should be taken with respect to the rotating frame (Frame-B). This means that the correct equation for angular acceleration should be:

[(dω/dt) w.r.t. Frame-B] = (d/dt)[(dΘ/dt)k]

Now, let's simplify this equation using the product rule of differentiation:

[(dω/dt) w.r.t. Frame-B] = (d2Θ/dt2)k + (dΘ/dt)(d/dt)k

The first term on the right-hand side represents the change in magnitude of the angular velocity, while the second term represents the change in direction of the angular velocity (i.e. the axis of rotation). This shows that the terms accounting for the change in axis of rotation are present in the equation, but they have been combined with the term for change in magnitude.

In conclusion, the terms for change in axis of rotation are not missing from the equation, but they are combined with the term for change in magnitude of angular velocity. I hope this helps to clarify the issue.