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$\renewcommand{\vec}{\boldsymbol}$ $\newcommand{\highlight}{\boxed}$

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.

Mark H
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Xiang Li
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2 Answers2

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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

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

Farcher
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Subtle point but worth noting: These equations do not calculate the velocity or acceleration "in the moving frame." Trying to understand what is going on by placing a miniature version of yourself in the frame may lead to additional confusion -- how do you even "know" about the rotation if you stand at the center of the turntable? What the equations are expressing is the TOTAL velocity and acceleration but EXPRESSED in a moving coordinate system.

Instead of fixed unit vectors, you have to account for the fact that they change direction. So the moving unit vectors are differentiable. To take the derivative of a moving unit vector, you take the cross product with its angular velocity as is done above.

Mariano G
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