What are the fundamentals of motion in a non-inertial reference frame? First of all, note that I understand what is considered off and on topic here. I'm posting this because I likely lack some sort of conceptual understanding of motion in non-inertial reference frames, and I genuinely wish to understand where my reasoning has failed.

In this problem we are given that $$\omega_{1}$$=0.25rad/s and $$\omega_{2}$$=0.4rad/s, the length of the crane boom, denoted as $$x_{B/A}$$, is 20ft. Additionally, all angular velocity is constant, i.e. $$\alpha$$=0. Similarly, u=1.5ft/s and $$\frac{du}{dt}$$=0.

I'm trying to find out what the acceleration of point B with respect to A is in this instant.

I began by using given equations for motion in a non-inertial reference frame: $$a_{B/A}$$=$$\alpha \times r_{B/A}$$ + $$\omega \times (\omega \times r_{B/A})$$ + $$2\omega \times (\frac{dr_{B/A}}{dt})_{Oxy}$$ + $$(\frac{d^{2}r_{B/A}}{dt^{2}})_{Oxy}$$

This yields the following:
$$a_{B/A}$$=0$$\times \begin{bmatrix} 0 \\ 20\sin{30} \\ 20\cos{30}\end{bmatrix}$$ + $$\begin{bmatrix}0.4 \\ 0.25 \\ 0\end{bmatrix} \times \left(\begin{bmatrix}0.4 \\ 0.25 \\ 0\end{bmatrix} \times \begin{bmatrix} 0 \\ 20\sin{30}\\20\cos{30}\end{bmatrix}\right)$$+$$2\begin{bmatrix}0.4\\0.25\\0\end{bmatrix}\times\begin{bmatrix} 0 \\ 1.5\sin{30} \\1.5\cos{30}\end{bmatrix}$$+0=$$\begin{bmatrix}1.65 \\ -2.64\\-3.25\end{bmatrix}$$

Curiously, the y and z components of my answer are correct, but the x component is off by 1: it should be 2.65.

I'm not entirely sure where I've gone wrong here - I think there is something incorrect about my conceptual understanding of motion in rotating frames.

From my understanding, the Coriolis force is created from the linear component of the velocity relative to the rotating frame. I.e., if u = 1.5ft/s, that should imply that the Coriolis acceleration of point b is $$2\begin{bmatrix}0.4\\0.25\\0\end{bmatrix}\times\begin{bmatrix} 0 \\ 1.5\sin{30} \\1.5\cos{30}\end{bmatrix}$$. Otherwise, if u were 0, then that should imply that the coriolis acceleration is 0. This matches up with my conceptual understanding of Coriolis acceleration. If you are standing still on Earth's surface, there is no Coriolis acceleration, but the second you start moving independently of Earth's surface, say, by running and jumping, you begin to develop a small Coriolis acceleration.

• You wrote 15 in text and 20 in the vector, which is it? Also, for what time are you computing? $t=1s$? – Emil Dec 9 '18 at 0:34
• My bad - the length of the boom was supposed to be 20ft. I'm trying to find out what the relative acceleration of point B is at this instant. I'll fix and clarify the question. – – Daniel Agramonte Dec 9 '18 at 1:32
• Re "Fundamentals of motion in a non-inertial reference frame": I just want to point out that the way you are using terms are at odds with each other. Fundamentally speaking Newtonian motion is understood from an inertial framework. – Mozibur Ullah Dec 9 '18 at 3:29
• Noted. Took away the newtonian mechanics tag – Daniel Agramonte Dec 9 '18 at 3:34