Velocity and Acceleration of an Object Moving in a Rotating Coordinate System 📂Classical Mechanics

Velocity and Acceleration of an Object Moving in a Rotating Coordinate System

Formulas

The velocity and acceleration of an object in a rotating coordinate system are as follows.

$$ \mathbf{v} = \mathbf{v}^{\prime} + \boldsymbol{\omega} \times \mathbf{r}^{\prime} +\mathbf{V}_{0} $$

$$ \mathbf{a} = \mathbf{a}^{\prime} + \dot{\boldsymbol{\omega}} \times \mathbf{r}^{\prime}+ 2\boldsymbol{\omega} \times \mathbf{v}^{\prime}+\boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}^{\prime}) + \mathbf{A}_{0} $$

Rotating Coordinate System¹

Imagine a train in motion and a fly buzzing around inside it. For a person inside the train, considering only the movement of the fly simplifies explaining its motion. However, for a person outside the train, factoring in the train’s movement as well makes it significantly more complicated. To simplify, let’s divide the fly’s motion into two components: the movement of the train relative to the ground and the movement of the fly relative to the train. Describing the movement of the fly relative to the train, as seen from inside the train, thus becomes simple. Likewise, describing the movement of the train relative to the ground is not difficult. By dividing complex movements into understandable parts like this, they can be more easily managed. Now consider the motion of an object moving on a rotating coordinate system fixed to the ground, analogous to observing someone moving on a spinning disco ride from outside.

Regardless of the coordinate system, vectors $\mathbf{r}$ and $\mathbf{r}^{\prime}$ are the same. Despite their different names, they are undoubtedly the same vectors.

$$ \begin{align*} \mathbf{r} &= x \mathbf{i}+ y\mathbf{j}+z\mathbf{k} \\ &= x^{\prime} \mathbf{i}^{\prime}+ y^{\prime}\mathbf{j}^{\prime}+z^{\prime}\mathbf{k}^{\prime} \\ &= \mathbf{r}^{\prime} \end{align*} $$

At this point, $\mathbf{i}=\hat{\mathbf{x}}$, $\mathbf{j}=\hat{\mathbf{y}}$, and $\mathbf{k}=\hat{\mathbf{z}}$ apply.

Velocity

Now let’s differentiate $\mathbf{r}$ and $\mathbf{r}^{\prime}$ with respect to time. Differentiating $\mathbf{r}$ is straightforward as always. However, care is needed when differentiating $\mathbf{r}^{\prime}$. $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ do not change over time, hence $\dfrac{d\mathbf{i} }{dt}=0$, but $\mathbf{i}^{\prime}$, $\mathbf{j}^{\prime}$, and $\mathbf{k}^{\prime}$ are rotating, so their rate of change with respect to time is not $0$. Differentiating with this in mind yields

$$ \begin{align*} \dfrac{d \mathbf{r} } {dt} &= \dfrac{dx}{dt} \mathbf{i}+ \dfrac{dy}{dt} \mathbf{j}+\dfrac{dz}{dt} \mathbf{k} \\ &= \dfrac{dx^{\prime}}{dt} \mathbf{i}^{\prime}+x^{\prime}\dfrac{ d\mathbf{i}^{\prime}}{dt} + \dfrac{dy^{\prime}}{dt}\mathbf{j}^{\prime} + y^{\prime}\dfrac{d \mathbf{j}^{\prime} } {dt} + \dfrac{dz^{\prime}}{dt} \mathbf{k}^{\prime} + z^{\prime}\dfrac{d \mathbf{k}^{\prime} }{dt} \\ &=\frac{d\mathbf{r}^{\prime}}{dt} \end{align*} $$

$\dfrac{dx^{\prime}}{dt} \mathbf{i}^{\prime}+\dfrac{dy^{\prime}}{dt}\mathbf{j}^{\prime} +\dfrac{dz^{\prime}}{dt} \mathbf{k}^{\prime}$ is the velocity of the object as seen from a fixed coordinate system, that is, the velocity of a person seen by someone also riding the disco ride. Therefore, summarizing:

$$ \mathbf{v} = \mathbf{v}^{\prime}+x^{\prime}\dfrac{ d\mathbf{i}^{\prime}}{dt}+ y^{\prime}\dfrac{d \mathbf{j}^{\prime} } {dt}+ z^{\prime}\dfrac{d \mathbf{k}^{\prime} }{dt} $$

Now, let’s calculate the derivatives of the three unit vectors.

When angles are sufficiently small, the length of the arc can be approximated to the length of the chord

$$ | \Delta \mathbf{i}^{\prime} | \approx | \mathbf{i}^{\prime} | \sin \phi \Delta \theta = \sin \phi \Delta \theta $$

Therefore,

$$ \begin{align*} \left| \dfrac{d \mathbf{i}^{\prime}} {dt} \right| &= \lim \limits_{\Delta t \rightarrow 0} \left| \dfrac { \Delta \mathbf{i}^{\prime}} {\Delta t} \right| \\ &=\sin \phi \lim \limits_{\Delta \rightarrow 0} \dfrac{\Delta \theta}{\Delta t} \\ &= \sin \phi \dfrac{d\theta}{dt} \\ &= \sin \phi \omega \end{align*} $$

Also, $\Delta \mathbf{i}^{\prime}$ is perpendicular to $\boldsymbol{\omega}$ and $\mathbf{i}$. Hence, it can be expressed by the cross product of the two vectors.

$$ \dfrac{d \mathbf{i}^{\prime} }{dt}=\boldsymbol{\omega} \times \mathbf{i}^{\prime} $$

Similarly, the equations below are valid.

$$ \dfrac{ d \mathbf{j}^{\prime}}{dt}=\boldsymbol{\omega} \times \mathbf{j},\quad \dfrac{ d \mathbf{k}^{\prime} }{dt}\boldsymbol{\omega} \times \mathbf{k}^{\prime} $$

Therefore,

$$ \begin{align*} x^{\prime}\dfrac{ d\mathbf{i}^{\prime}}{dt}+ y^{\prime}\dfrac{d \mathbf{j}^{\prime} } {dt}+ z^{\prime}\dfrac{d \mathbf{k}^{\prime} }{dt} &= \boldsymbol{\omega} \times x^{\prime}\mathbf{i}^{\prime}+ \boldsymbol{\omega} \times y^{\prime} \mathbf{j}^{\prime} + \boldsymbol{\omega} \times z^{\prime} \mathbf{k}^{\prime} \\ &= \boldsymbol{\omega} \times \left( x^{\prime}\mathbf{i}^{\prime}+ y^{\prime} \mathbf{j}^{\prime} + z^{\prime} \mathbf{k}^{\prime} \right) \\ &= \boldsymbol{\omega} \times \mathbf{r}^{\prime} \end{align*} $$

Which means,

$$ \mathbf{v} = \mathbf{v}^{\prime} + \boldsymbol{\omega} \times \mathbf{r}^{\prime} $$

If we specify,

$$ \dfrac{d \mathbf{r}}{dt}_{f} = \frac{ d \mathbf{r}^{\prime} }{ dt }_{r}+\boldsymbol{\omega}\times \mathbf{r}^{\prime}=\left[ \dfrac{d}{dt}_{r}+\boldsymbol{\omega}\times \right] \mathbf{r}^{\prime} $$

The subscript $_{f}$ signifies differentiation with respect to time in a fixed coordinate system, deriving from the initial of “fixed”. The subscript $_{r}$ signifies differentiation with respect to time in a rotating coordinate system, taken from the initial of “rotation”. Thus, $\dfrac{d}{dt}_{f}$ turns $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ into $0$, and $\dfrac{d}{dt}_{r}$ turn $\mathbf{i}^{\prime}$, $\mathbf{j}^{\prime}$, and $\mathbf{k}^{\prime}$ into $0$.

Acceleration

Since it was initially $\mathbf{r}=\mathbf{r}^{\prime}$, any vector can be substituted in place of $\mathbf{r}(=\mathbf{r}^{\prime})$ and the equation still holds.

$$ \dfrac{d \mathbf{A}}{dt}_{f} = \left[ \dfrac{d}{dt}_{r} + \boldsymbol{\omega} \times \right] \mathbf{A} $$

Substituting the velocity $\mathbf{v}=\mathbf{v}^{\prime}+\boldsymbol{\omega} \times \mathbf{r}^{\prime}$ gives

$$ \begin{align*} \dfrac{d \mathbf{v}}{dt}_{f} &=\left[ \dfrac{d}{dt}_{r} +\boldsymbol{\omega} \times \right] (\mathbf{v}^{\prime} + \boldsymbol{\omega} \times \mathbf{r}^{\prime}) \\ &= \dfrac{d}{dt}_{r}(\mathbf{v}^{\prime} +\boldsymbol{\omega} \times \mathbf{r}^{\prime}) + \boldsymbol{\omega} \times (\mathbf{v}^{\prime}+\boldsymbol{\omega}\times \mathbf{r}^{\prime} ) \\ &= \dfrac{d \mathbf{v}^{\prime}}{dt}_{r} + \dfrac{d}{dt}_{r} (\boldsymbol{\omega} \times \mathbf{r}^{\prime} )+ \boldsymbol{\omega} \times \mathbf{v}^{\prime} + \boldsymbol{\omega}\times ( \boldsymbol{\omega} \times \mathbf{r}^{\prime} ) \\ &= \dfrac{d\mathbf{v}^{\prime}}{dt} _{r} + \dfrac{d\boldsymbol{\omega}}{dt}_{r} \times \mathbf{r}^{\prime} + \boldsymbol{\omega} \times \dfrac{d \mathbf{r}^{\prime} }{dt}_{r} + \boldsymbol{\omega}\times \mathbf{v}^{\prime} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}^{\prime} ) \end{align*} $$

Simplifying this yields

$$ \begin{align*} \mathbf{a} &= \mathbf{a}^{\prime} + \dot{\boldsymbol{\omega}} \times \mathbf{r}^{\prime}+ \boldsymbol{\omega} \times \mathbf{v}^{\prime}+\boldsymbol{\omega}\times \mathbf{v}^{\prime}+\boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}^{\prime}) \\ &= \mathbf{a}^{\prime} + \dot{\boldsymbol{\omega}} \times \mathbf{r}^{\prime}+ 2\boldsymbol{\omega} \times \mathbf{v}^{\prime}+\boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}^{\prime}) \end{align*} $$

This means that the movement of an object on the disco ride seen from outside can be expressed in terms of displacement, velocity, and acceleration in the rotating coordinate system. If the rotating coordinate system also involves translational motion (straight-line motion, parallel movement) relative to a fixed coordinate system, only the velocity and acceleration of the translational motion need to be added.