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9.5.3 Velocity and Acceleration in a Rotating Coordinate System

Applying Eq. (9.8) to the position vector $\vec{r}$, we have $$\left(\frac{d\vec{r}}{dt}\right)_{in} = \left(\frac{d\vec{r}}{dt}\right)_{rot} + \vec\Omega\times\vec{r},$$ or $$\vec{v}_{in} = \vec{v}_{rot} + \vec\Omega\times\vec{r}$$ We can apply Eq. (9.8) once again to find the acceleration $\vec{a}_{rot}$ in the rotating coordinate system. We have

\begin{align} \left(\frac{d\vec{v}_{in}}{dt}\right)_{in} &= \highlight{\left(\frac{d\vec{v}_{in}}{dt}\right)_{rot}} + \vec\Omega\times\vec{v}_{in} \\ &= \left[\frac{d}{dt}(\vec{v}_{rot} + \vec\Omega\times\vec{r}\right]_{rot} + \vec\Omega\times(\vec{v}_{rot}+\vec\Omega\times\vec{r}) \\ &= \left(\frac{d\vec{v}_{rot}}{dt}\right)_{rot} + \vec\Omega\times\frac{d\vec{r}}{dt}_{rot} + \vec\Omega\times\vec{r}_{rot} + \vec\Omega\times(\vec{v}_{rot}+\vec\Omega\times\vec{r}) \\ &= \left(\frac{d\vec{v}_{rot}}{dt}\right)_{rot} + 2\vec\Omega\times\vec{v}_{rot} + \vec\Omega\times(\vec\Omega\times\vec{r}). \end{align} Expressing this in terms of the acceleration $\vec{a}_{in}$ and $\vec{a}_{rot}$, we have $$\vec{a}_{in} = \vec{a}_{rot} + 2\vec\Omega\times\vec{v}_{rot} + \vec\Omega\times(\vec\Omega\times\vec{r}).\tag{9.9}$$ The acceleration viewed in the rotating system is $$\vec{a}_{rot} = \vec{a}_{int} - 2\vec\Omega\times\vec{v}_{rot} - \vec\Omega\times(\vec\Omega\times\vec{r}).\tag{9.10}$$

I’m confused by the subscript of $rot$ and $in$, short for rotating frame and inertial frame respectively. When we calculate the acceleration in the inertial frame, on the left side of the equation, we need derivative the velocity vector which is seen by an observer in the rotating frame, why is the velocity vector equal to velocity vector in rotating frame plus $\vec\Omega$ cross position vector? I think the velocity vector in rotating frame is just equal to the velocity vector in inertial frame minus $\vec\Omega$ cross position vector.


1 Answer 1


Imagine that you are standing on the ground (inertial frame) and looking at a horizontal turntable rotating (rotational frame) at an angular velocity $\vec \Omega$ which is at right angles to the plane of the turntable.

At position vector $\vec r$ relative to the axis of rotation of the turntable, the turntable is moving with a linear velocity of $\vec \Omega \times \vec r$ relative to the ground, $(\vec v_\text{turntable relative to ground})$.

Suppose that at that position an object is moving with a velocity $\vec v_\text{object relative to turntable} = \left(\dfrac{d\vec r}{dt}\right )_{\rm rot}$.

If $\vec v_\text{object relative to ground} = \left(\dfrac{d\vec r}{dt}\right )_{\rm in}$is the velocity of the object relative to the ground then

$\vec v_\text{object relative to ground} =\vec v_{\text{object relative to }{\rm\color{red} {turntable}}} + \vec v_{\rm \color{red}{turntable} \text{ relative to ground}}\Rightarrow \left(\dfrac{d\vec r}{dt}\right )_{\rm in}=\left(\dfrac{d\vec r}{dt}\right )_{\rm rot}+\,\vec \Omega \times \vec r$

  • $\begingroup$ Thank for reply. I understand how to get the velocity vector. What confuses me is the object labeled by blue, does it mean the derivation of the velocity vector seen in the rotating frame? If so, I think the velocity vector just is v with subscript rot, why the component w cross product position vector exists. $\endgroup$
    – Xiang Li
    Apr 12, 2023 at 8:43
  • $\begingroup$ The subscript gives you the frame relative to the measurement is made. So the subscript "rot" indicates that the measurement is made relative to the rotating frame. $\endgroup$
    – Farcher
    Apr 12, 2023 at 9:31

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