How Do Rotating Reference Frames Affect Physics Calculations?

AI Thread Summary
Rotating reference frames introduce complexities in physics calculations, particularly when relating them to inertial frames. The acceleration of a particle in a rotating frame includes additional terms: Coriolis acceleration and centripetal acceleration, which account for the effects of rotation. The equations governing these effects illustrate how the velocity and acceleration change when viewed from a rotating frame. Understanding these dynamics is crucial for accurate calculations in systems like the Earth's rotation. This knowledge enhances our grasp of motion in non-inertial frames and is essential for various applications in physics.
Messages
19,773
Reaction score
10,726
Definition/Summary

Often in physics we need to consider frames of reference that are non-inertial (the Earth spinning on its axis for example). We must therefore see how these rotating reference frames relate to an inertial reference frame.

Equations

\frac{d^2\mathbf{r}}{dt^2} = \ddot{\mathbf{r}} + 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) +\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})

Extended explanation

Effect on 1st derivatives:

Consider a rotating frame with an instantaneous angular velocity \mathbf{\Omega}. A unit vector \mathbf{e}_i traces a circle about \mathbf{\Omega} at a rate:

\frac{d\mathbf{e}_i}{dt} = \mathbf{\Omega} \times \mathbf{e}_i

A particle will have a position in the rotating frame given by \mathbf{r} = x_i\mathbf{e}_i (where i is summed from 1 to 3) and thus the velocity in an inertial frame is then:

\frac{d\mathbf{r}}{dt} = \frac{d}{dt}( x_i\mathbf{e}_i) = \frac{dx_i}{dt}\mathbf{e}_i + x_i\frac{d\mathbf{e}_i}{dt} = \frac{dx_i}{dt}\mathbf{e}_i + x_i(\mathbf{\Omega} \times \mathbf{e}_i)

Example: torque equation:

For example, in a fixed frame of reference, the equation relating net torque on a body to its rate of change of angular momentum is:

\mathbf{\tau}_{net}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\Omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\Omega}\right)\ +\ \frac{d\tilde{I}}{dt}\left(\mathbf{\Omega}\right)

but in a frame rotating with the body, it is:

\mathbf{\tau}_{net}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\Omega}\right)\ \ +\ \ \mathbf{\Omega}\,\times \left(\tilde{I}\,\mathbf{\Omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\Omega}\right)\ \ +\ \ \mathbf{\Omega}\,\times \left(\tilde{I}\,\mathbf{\Omega}\right)

Effect on 2nd derivatives:

The acceleration is then (assuming \mathbf{\Omega} is constant):

\frac{d^2\mathbf{r}}{dt^2} = \frac{d^2r}{dt^2}+ 2\frac{dx_i}{dt}(\mathbf{\Omega}\times\mathbf{e}_i) + x_i(\mathbf{\Omega}\times(\mathbf{\Omega}\times \mathbf{e}_i))

Tidying up a bit we have:

\frac{d^2\mathbf{r}}{dt^2} = \ddot{\mathbf{r}} + 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) +\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})

The 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) term is called the Coriolis acceleration and the \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) term is called the centripetal acceleration.

What we have essentially is:

Acceleration seen by inertial observer = Acceleration seen by rotating observer + extra terms

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
Physics news on Phys.org
That's interesting! It's helpful to know how these rotating reference frames relate to an inertial reference frame so we can better understand how to calculate the acceleration of a particle in a rotating frame. Thanks for sharing this information!
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top