The derivatives of a vector in two rotating frames S and S' are related by equation 9.30, and when the vector is fixed in the S frame, the derivative in the S' frame is equal to the cross product of the angular velocity of S' and the vector in the S' frame.
Step-by-step explanation:
In a rotating frame, the derivatives of a vector Q in two frames S and S' are related by the equation (9.30) which states that the derivative in the S' frame is equal to the derivative in the S frame plus the cross product of the angular velocity of the S' frame and the vector Q in the S' frame.
For the special case where Q is fixed in the S frame, the derivative of Q in the S frame is equal to zero. Therefore, the derivative of Q in the S' frame is just the cross product of the angular velocity of S' and Q in the S' frame.
Comparing this with the answer to part (a), we can see that the derivative of Q in the S' frame is the same in both cases.
The probable question can be: Time Derivatives in a Rotating Frame 9.7 * (a) Explain the relation (9.30) between the derivatives of a vector Q in two frames S, and 8 for the special case that Q is fixe 8o and compare with your answer to part (a) d in the frame 8. (b) Do the same for a vector Q that is fixed in the frame [Eq. 9.30