2024-03-01 Non-Inertial reference frames

Non-Inertial reference frames

Recall: an inertial reference frame is at rest or moves with a constant velocity & newtons laws are valid

• Non-inertial reference frame:
  • Rotating Reference frame
    • WWII Artillerymen had to account for the spin of the earth to accurately hit their targets

Rotating coordinate systems: Example: Rotating reference frame on earth compared to stationary reference frame

From last time:

$${\vec{v}}_{fixed}=\vec{V}+{\vec{v}}_r+\vec{\omega }\times \vec{r}$$

Where

$$\begin{array}{c}{\vec{v}}_{fixed}=\text{velocity relative to fixed frame}\\ \vec{V}=\text{Linear velocity relative to origin of moving frame}\\ {\vec{v}}_r=\text{ velocity relative to rotating frame}\\ \vec{\omega }\times \vec{r}=\text{angular velocity}\end{array}$$

Using Newtons Second Law:

$$\vec{F}=m\vec{a}=m\frac{d\vec{v}}{dt}{|}_{\text{fixed}}$$ $$\frac{d{\vec{v}}_{fixed}}{dt}{|}_{fixed}=\frac{d\vec{V}}{dt}{|}_{fixed}+\frac{d{\vec{v}}_r}{dt}{|}_{fixed}+\vec{\omega }\times \vec{r}+\vec{\omega }\times \frac{d\vec{r}}{dt}{|}_{fixed}$$

In which the first term on the RHS is ${\ddot{\vec{R}}}_f$ , the acceleration relative to the origin. The second term becomes

$$\frac{d\vec{v}}{dt}{|}_{rot}+\vec{\omega }\times \vec{r}$$

The third term remains the same, angular velocity is the same, and the forth term becomes

$$\vec{\omega }\times \frac{d\vec{r}}{dt}{|}_{rot}+[\vec{\omega }\times \vec{r}]$$

Putting this all together we get

$$=\vec{\omega }\times \frac{d\vec{r}}{dt}{|}_{rot}+\vec{\omega }\times (\vec{\omega }\times \vec{r})$$ $$=\vec{\omega }\times {\vec{v}}_r+\vec{\omega }\times (\vec{\omega }\times \vec{r})$$

Using this we can find force measured in the fixed frame

$$\vec{F}=m{\vec{a}}_f=m{\ddot{\vec{R}}}_f+m{\vec{a}}_r+m\dot{\vec{\omega }}\times \vec{r}+m\vec{\omega }\times (\vec{\omega }\times \vec{r})+2m\vec{\omega }\times {\vec{v}}_r$$

Finding the force measured in the rotating frame

$${\vec{F}}_{eff}=m{\vec{a}}_r=\vec{F}-m{\ddot{\vec{R}}}_f-m\vec{\dot{\omega }}\times \vec{r}-m\vec{\omega }\times (\vec{\omega }\times \vec{r})-2m\vec{\omega }\times {\vec{v}}_r$$

What each term represents in physical quantities

• $m{\ddot{\vec{R}}}_f$ is the translation of the rotating frame relative to the fixed frame
• $m\dot{\vec{\omega }}\times \vec{r}$ is the angular acceleration of the coordinate system
• $m\vec{\omega }\times (\vec{\omega }\times \vec{r})$ is the centrifugal “force”
• $-2m\vec{\omega }\times {\vec{v}}_r$ is the Coriolis force

$$\vec{F}=m{\vec{a}}_f$$ $${\vec{F}}_{eff}=m{\vec{a}}_r=m{\vec{a}}_f+\text{ non inertial terms}$$

Forces in the fixed frame

$${\vec{F}}_f=m{\vec{a}}_r=\vec{S}+m{\vec{g}}_0$$

where

$${\vec{g}}_0=-\frac{G{m}_{\otimes }}{\vec{R}}{\vec{\rho }}_r$$

See HW 8

Gravity is different across the world based on lattitudes

$${\vec{F}}_{eff}=\vec{S}+m{\vec{g}}_0-m{\ddot{\vec{R}}}_f-m{\vec{R}}_f-m\vec{\omega }\times (\vec{\omega }\times \vec{r})-2m\vec{\omega }\times {\vec{v}}_r$$

For the earth, $\vec{\omega }=\omega {\vec{\rho }}_z$ and $\vec{\dot{\omega }}{\approx }_0$