Acceleration in a rotating frame

29 Oct, 2024

We consider two frames of reference:

Inertial Frame ( $ℐ$ ): A frame that is not accelerating, where Newton's laws hold in their simplest form.
Rotating Frame ( $ℛ$ ): A frame that rotates with an angular velocity $\vec{ω}$ relative to the inertial frame.

For any vector $\vec{A}$ , the relationship between the time derivative in the inertial and rotating frames is given by:

{(\frac{d \vec{A}}{d t})}_{ℐ} = {(\frac{d \vec{A}}{d t})}_{ℛ} + \vec{ω} \times \vec{A}

where:

${(\frac{d \vec{A}}{d t})}_{ℐ}$ is the time derivative in the inertial frame,
${(\frac{d \vec{A}}{d t})}_{ℛ}$ is the time derivative in the rotating frame,
$\vec{ω}$ is the angular velocity vector of the rotating frame with respect to the inertial frame.

Let $\vec{r}$ be the position vector of the particle in both frames (assuming the origins coincide).

{\vec{v}}_{ℐ} = {(\frac{d \vec{r}}{d t})}_{ℐ}

{\vec{v}}_{ℛ} = {(\frac{d \vec{r}}{d t})}_{ℛ}

Using the derivative relationship:

{\vec{v}}_{ℐ} = {\vec{v}}_{ℛ} + \vec{ω} \times \vec{r}

The acceleration in the inertial frame is:

{\vec{a}}_{ℐ} = {(\frac{d {\vec{v}}_{ℐ}}{d t})}_{ℐ}

To express ${\vec{a}}_{ℐ}$ in terms of quantities in the rotating frame, we differentiate ${\vec{v}}_{ℐ}$ with respect to time:

{\vec{a}}_{ℐ} = {(\frac{d}{d t} ({\vec{v}}_{ℛ} + \vec{ω} \times \vec{r}))}_{ℐ}

Applying the time derivative to both terms:

{\vec{a}}_{ℐ} = {(\frac{d {\vec{v}}_{ℛ}}{d t})}_{ℐ} + {(\frac{d}{d t} (\vec{ω} \times \vec{r}))}_{ℐ}

First Term:

{(\frac{d {\vec{v}}_{ℛ}}{d t})}_{ℐ} = {(\frac{d {\vec{v}}_{ℛ}}{d t})}_{ℛ} + \vec{ω} \times {\vec{v}}_{ℛ}

Second Term:

{(\frac{d}{d t} (\vec{ω} \times \vec{r}))}_{ℐ} = \frac{d \vec{ω}}{d t} \times \vec{r} + \vec{ω} \times {(\frac{d \vec{r}}{d t})}_{ℐ}

Since $\frac{d \vec{r}}{d t} = {\vec{v}}_{ℐ} = {\vec{v}}_{ℛ} + \vec{ω} \times \vec{r}$ :

\vec{ω} \times {\vec{v}}_{ℐ} = \vec{ω} \times {\vec{v}}_{ℛ} + \vec{ω} \times (\vec{ω} \times \vec{r})

Combining the terms, we obtain:

{\vec{a}}_{ℐ} = {\vec{a}}_{ℛ} + 2 \vec{ω} \times {\vec{v}}_{ℛ} + \frac{d \vec{ω}}{d t} \times \vec{r} + \vec{ω} \times (\vec{ω} \times \vec{r})

To find the acceleration as observed in the rotating frame, we solve for ${\vec{a}}_{ℛ}$ :

{\vec{a}}_{ℛ} = {\vec{a}}_{ℐ} - \vec{ω} \times (\vec{ω} \times \vec{r}) - 2 \vec{ω} \times {\vec{v}}_{ℛ} - \frac{d \vec{ω}}{d t} \times \vec{r}

Centrifugal Acceleration ( $- \vec{ω} \times (\vec{ω} \times \vec{r})$ ): Acts outward from the axis of rotation.
Coriolis Acceleration ( $- 2 \vec{ω} \times {\vec{v}}_{ℛ}$ ): Depends on the velocity of the particle in the rotating frame.
Euler Acceleration ( $- \frac{d \vec{ω}}{d t} \times \vec{r}$ ): Occurs when the angular velocity $\vec{ω}$ is changing over time.