Forces in a Rotating Frame

This post derives newton’s law in a rotating frame of reference using the more generalized framework of coordinate frames and rotation matrices. It has a nice intermediate result which is a general formula for computing the acceleration of a point in a new reference frame. It also has the benefit of starting from a simple and intuitive vector equation: $r^{p0} = r^{10} + r^{p1}$.

Notation

• $\omega_a^{cb}$ : angular velocity of c with respect to b represented in the coordinates of frame a.
• $r_a^{cb}$: vector from b to c represented in the coordinates of frame a.
• $T_{ba}$: matrix that rotates vectors from frame a into frame b

System Setup

This section explains the geometric picture. Define two frames, frame 0 and frame 1, and say frame 0 is an inertial frame and frame 1 is rotating about some vector $\omega$. Next, we define a point p. An observer in frame 1 is watching point p: $r_1^{p1}(t)$. He wants to apply newton’s law to understand it’s motion. He knows that the frame he is in is not inertial, so he cannot just apply $F=ma$.

Figure: Frame 0 is an inertial frame and frame 1 is rotating w.r.t frame 1. An observer in frame 1 is watching point p.

Forces in a Rotating Frame: Coriolis Force, Centrifugal Force, Euler Force

The vector from frame 0 to p equals the vector from frame 0 to 1 plus the vector from frame 1 to p. This is easy to reason and can be geometrically seen in the figure above. At this point, I want to emphasize that there are no assumptions about the motion of frame 0 or frame 1 or p.

$$r^{p0} = r^{10} + r^{p1}$$

We can rewrite the equation in terms of specific basis, specifically in terms of frame 0 since we are after the point’s acceleration in an inertial frame. Notice that we are rotating the vector $r^{p1}$ from frame 1 into frame 0. The motivation for this is because $r_1^{p1}$ is what an observer in frame 1 sees.

$$r_0^{p0} = r_0^{10} + T_{01} r_1^{p1}$$

Next, we take the derivative twice to find the acceleration of point p w.r.t frame 0. This is just calculus.

$$\dot{r}_0^{p0} = \dot{r}_0^{10} + T_{01} \dot{r}_1^{p1} + \omega_0^{10} \times T_{01} r_1^{p1} \\ \ddot{r}_0^{p0} = \ddot{r}_0^{10} + T_{01} \ddot{r}_1^{p1} + \omega_0^{10} \times T_{01} \dot{r}_1^{p1} + \omega_0^{10} \times (T_{01} \dot{r}_1^{p1} + \omega_0^{10} \times T_{01} r_1^{p1}) + \frac{d\omega_0^{10}}{dt} \times T_{01} r_1^{p1} \\ \ddot{r}_0^{p0} = \ddot{r}_0^{10} + T_{01} \ddot{r}_1^{p1} + 2 ( \omega_0^{10} \times T_{01} \dot{r}_1^{p1} ) + \omega_0^{10} \times (\omega_0^{10} \times T_{01} r_1^{p1}) + \frac{d\omega_0^{10}}{dt} \times T_{01} r_1^{p1} \\ \ddot{r}_0^{p0} = \ddot{r}_0^{10} + \ddot{r}_0^{p1} + 2 ( \omega_0^{10} \times \dot{r}_0^{p1} ) + \omega_0^{10} \times (\omega_0^{10} \times r_0^{p1}) + \frac{d\omega_0^{10}}{dt} \times r_0^{p1} \\$$

The last two steps were only expansion and then applying the rotation matrices. Already the familiar forms of the coriolis, centrifugal and euler forces are appearing. A few comments about this equation. First, there was absolutely no assumption made about the motion of the frames or the points. This is a very general equation. Second, the equation is saying that the acceleration of point p w.r.t. frame 0 is given by adding the acceleration of frame 1 w.r.t 0 ($\ddot{r}_0^{10}$) plus the acceleration of the point p w.r.t. frame 1 ($\ddot{r}_0^{p1}$) plus other terms due to the relative rotation.

Now we introduce two assumptions. First, we assume that frame 0 is an inertial reference frame. This means that we can use Newton’s law. Second, we assume that frame 1 is only rotating with respect to frame 0, and it is not moving/accelerating (e.g. $\ddot{r}_0^{10} = 0$).

\[F_0 = m \ddot{r}_0^{p0} \\ F_0 = m \ddot{r}_0^{p1} + 2 m ( \omega_0^{10} \times \dot{r}_0^{p1} ) + m \omega_0^{10} \times (\omega_0^{10} \times r_0^{p1}) + m \frac{d\omega_0^{10}}{dt} \times r_0^{p1} \\]

This equation is true no matter what basis we work in, so we change it to frame 1 because that is the frame the observer is in and he will view and represent the coordinates of p and his own frame’s rotation in the basis of that frame. Mathematically though, you can multiply both sides by $T_{10}$ and use the following fact: if $a$ and $b$ are vectors and $M$ is an invertible matrix, then $(Ma) \times (Mb) = (detM)M^{-T}(a \times b)$.

hint: remember that any rotation matrix $T_{ab} \in SO(3)$, where $SO(3)$ is the special orthogonal group. Look up the properties of this set!

\[F_1 = m \ddot{r}_1^{p1} + 2 m ( \omega_1^{10} \times \dot{r}_1^{p1} ) + m \omega_1^{10} \times (\omega_1^{10} \times r_1^{p1}) + m \frac{d\omega_1^{10}}{dt} \times r_1^{p1} \\]

The last thing to notice is that $m \ddot{r}_1^{p1}$ is mass times the acceleration of point p w.r.t. frame 1. This is what we are after. We want to find a set of forces so Newton’s law holds in frame 1 and we can treat it as an inertial frame. Rearranging for $m \ddot{r}_1^{p1}$:

$$\bbox[2pt,border:1px solid red]{ m \ddot{r}_1^{p1} = F_1 - 2 m ( \omega_1^{10} \times \dot{r}_1^{p1} ) - m \omega_1^{10} \times (\omega_1^{10} \times r_1^{p1}) - m \frac{d\omega_1^{10}}{dt} \times r_1^{p1} \ }$$

This is the final, desired result. Recall the definitions of the coriolis, centrifugal and euler forces:

• Corilois Force: $- 2 m ( \omega_1^{10} \times \dot{r}_1^{p1} )$
• Centrifugal Force: $- m \omega_1^{10} \times (\omega_1^{10} \times r_1^{p1})$
• Euler Force: $- m \frac{d\omega_1^{10}}{dt} \times r_1^{p1}$

This equation tells us that in a rotating frame, mass times acceleration is equal to the applied force plus the three fictitious forces defined above.