Rotating noninertial frames

We saw that in a (non-inertial) linearly accelerating reference frame with acceleration $\myv A$ (measured in the IF), that Newton's law becomes.... $$ m\ddot{\myv{r}} = \myv{F}-m\myv{A}$$ where $-m\myv{A}$ is the "inertial force".

Next, we try to come up with Newton's law in terms of derivatives in another non-inertial reference frame: one that's rotating--e.g. Earth.

We shall find two inertial forces: the centrifugal force and the Coriolis force.

A body rotating with one point fixed

What do we mean by rotation? Let's generalize the result that you know from General Physics for circular motion: $$\omega=\frac{v}{r}\Rightarrow v=r\omega.$$ to get a relationship between a rotation vector for a rotating body, and the position vector for any particle within the rotating body...

Consider:

• A body that rotates,
• But (at least) one point is motionless (fixed) in an inertial reference frame.
• Call that point the origin $O$.
• Rotation can be described as happening around an axis running through $O$,
• Only one axis at a time, specified by a unit vector $\myv{u}$, using a right-hand rule...
• Angular speed of rotation is $\omega=d \phi/dt$.
• Combining the magnitude and direction into a vector $$\myv{\omega} = \omega \myv{u}$$
• $\myv{\omega}$ is not necessarily constant. $\myv{\omega}(t)$ might be just an instantaneous angular velocity.

The velocity for a point at some position $\myv{r}$ in the rotating body can be written in terms of the angular speed as $\rho \omega=r\omega \sin \theta $ which we recognize as the cross product: $$\myv{v}= \dot{\myv{r}}= \myv{\omega} \times \myv{r}.$$

This gives the time rate of change of any vector which is fixed relative to the rotating body.

Note that as long as the origin is somewhere--anywhere--along the axis of rotation, this relation will give the same velocity.

Addition of angular velocities

How do plain ol', non-angular velocities of a body in one reference frame relate to those in another? Once again....

In an inertial frame $\mathscr{S}_0$ With these velocities:

$\myv v = \dot{ \myv r}=$ the velocity of a body in the (possibly non-inertial) reference frame $\mathscr{S}$, like the train car.

$\myv{v}_0 = \dot{ \myv r}_0=$ the velocity of a body in the inertial reference frame $\mathscr{S}_0$, like the rail station platform.

$\myv{V}$ = the velocity of reference frame $\mathscr{S}$ as measured in $\mathscr{S}_0$.

According to Galilean relativity (but this gets updated in special relativity!), the velocities are related by $$\myv{v} = \myv{v}_0+\myv{V}.$$

It doesn't matter whether $\mathscr{S}$ is inertial or not.

Now, suppose that $\mathscr{S}$ and $\mathscr{S}_0$ share the same origin, but that frame $\mathscr{S}$ is rotating with angular velocity $\myv{\Omega}$ with respect to frame $\mathscr{S}_0$ .

Think of a a reference frame (like Earth) rotating with angular velocity $\myv \Omega$. Now, imagine a solid body (football) spinning with angular velocity $\myv{\omega}$ and $\myv{\omega}_0$ relative to frames $\mathscr{S}$ and $\mathscr{S}_0$. The velocities for a point in the body at position $\myv{r}$ in these two frames must still relate according to the equation above, which we calculate using our cross product... $$\begineq \myv v=&\myv v_0+\myv V\\ \myv{\omega} \times \myv{r} =& \myv{\omega}_0 \times \myv{r} + \myv{\Omega} \times \myv{r}\\ \myv{\omega}\times \myv{r} =& (\myv{\omega}_0 + \myv{\Omega}) \times \myv{r}.\endeq $$

But since there's nothing special about $\myv{r}$, it could be any position in the body. So this requires

$$\myv{\omega} = \myv{\omega}_{0} +\myv{\Omega}.$$

Time derivatives in a rotating frame

We'd like to find equations of motion for:

• an object viewed in a noninertial frame $\mathscr{S}$,
• that is rotating with angular velocity $\myv{\Omega}$ relative to,
• an inertial frame $\mathscr{S}_0$.

We'll assume that the two frames share a common origin.

[ $\Omega$ is a capital $\omega$ to parallel the previous usage of capitals $A$ and $V$ which designated the linear acceleration and velocity of a non-inertial frame relative to an inertial one.]

Let $\myv{Q}$ be a vector which has three components related to three convenient unit vectors which are not moving in our noninertial frame $\mathscr{S}$: $$\begineq\myv{Q} =& Q_x \uv x+Q_y \uv y+Q_z \uv z\\ \equiv& Q_1\uv{e}_1+Q_2\uv{e}_2+Q_3\uv{e}_3=\sum_{i=1}^3Q_i\uv{e}_i \endeq$$

• This could be any vector: a position vector, a velocity vector, what-evuh.
• This de-composition is equally valid in the inertial frame $\mathscr{S}_0$. It's just more awkward, since the unit vectors are moving around as seen from $\cal{S}_0$.

To differentiate $\myv{Q}$ with respect to time in the non-inertial frame $\mathscr{S}$, (the unit vectors are not moving relative to $\mathscr{S}$): $$\left(\frac{d\myv{Q}}{dt}\right)_{\mathscr{S}} = \sum_i \frac{dQ_i}{dt} \uv{e}_i .$$

In $\mathscr{S}_0$, we have to use the product rule to account for the changing unit vectors of the rotating frame, $$\left(\frac{d\myv{Q}}{dt}\right)_{\mathscr{S}_0} = \sum_i \frac{dQ_i}{dt} \uv{e}_i +\sum_i Q_i \left(\frac{d \uv{e}_i}{dt}\right)_{\mathscr{S}_0}.\label{dQ}$$

Since each of the unit vectors is fixed relative to the rotating frame, the time rate of change of any one is $$\begineq \left(\frac{d\uv{e}_i}{dt}\right)_{\mathscr{S}_0} =& \left(\frac{d\uv{e}_i}{dt}\right)_{\mathscr{S}}+\myv{\Omega} \times \uv{e}_i\\ =& 0 +\myv{\Omega} \times \uv{e}_i\\ =& \myv{\Omega} \times \uv{e}_i. \endeq$$

The second sum in Eq \ref{dQ} becomes $$\begineq \sum_i Q_i \left(\frac{d \uv{e}_i}{dt}\right)_{\mathscr{S}_0}=&\sum_i Q_i \myv{\Omega} \times \uv{e}_i = \myv{\Omega} \times \sum_i Q_i \uv{e}_i \\ =& \myv{\Omega} \times \myv{Q}.\endeq$$

Putting this into the time derivative results in

$$\left(\frac{d\myv{Q}}{dt}\right)_{\mathscr{S}_0} = \left(\frac{d\myv{Q}}{dt}\right)_{\mathscr{S}} + \myv{\Omega}\times \myv{Q}.\label{dQdt}$$

Newton's "apparent" second law in a rotating frame

To evaluate Newton's law, we'll need to carry out some second order derivatives.

Let's specialize to the case where the rotational speed is constant (true to a very good approximation on Earth), so that at least $\myv\Omega$ is constant, $$\left(\frac{d\myv{\Omega}}{dt}\right)_{\mathscr{S}}=0.$$

If we use our relation ($\ref{dQdt}$) above on $\myv{\Omega}$, we find that its time derivative in the inertial frame also vanishes.

In what follows, it will be useful to use the dot notation to indicate time derivatives in the non-inertial frame. That is $$\dot{\myv{Q}} \equiv \left(\frac{d \myv{Q}}{dt}\right)_{\mathscr{S}}.$$

In particular, this means that $\ddot{\myv r}$ is *not* the same as the acceleration which appears in Newton's 2nd law in the inertial frame: $$m\left(\frac{d^2\myv{r}}{dt^2}\right)_{\mathscr{S}_0} = \myv{F}.$$

Now, let's use (\ref{dQdt}) to write Newton's 2nd law in terms of time derivatives in the non-inertial reference frame, $\mathscr{S}$. The second order differential in the inertial frame becomes.... $$\begineq\left(\frac{d^2\myv{r}}{dt^2}\right)_{\mathscr{S}_0} =& \left(\frac{d}{dt}\right)_{\mathscr{S}_0} \left(\frac{d\myv r}{dt}\right)_{\mathscr{S}_0} \\ =& \left(\frac{d}{dt}\right)_{\mathscr{S}_0} \left[ \left(\frac{d\myv{r}}{dt}\right)_{\mathscr{S}} + \myv{\Omega}\times \myv{r}\right]\\ =&\left(\frac{d}{dt}\right)_{\mathscr{S}} \left[\left(\frac{d\myv{r}}{dt}\right)_{\mathscr{S}} + \myv{\Omega}\times\myv{r}\right]\\ &&\ \ +\myv{\Omega} \times \left[\left(\frac{d\myv{r}}{dt}\right)_{\mathscr{S}} + \myv{\Omega}\times\myv{r}\right]\\ =&\ddot{\myv{r}} + \myv{\Omega}\times \dot{\myv{r}} + \myv{\Omega}\times \dot{\myv{r}} +\myv{\Omega}\times \myv{\Omega} \times \myv{r}\\ =&\ddot{\myv{r}} - 2\dot{\myv{r}} \times \myv{\Omega} - \myv{\Omega}\times \myv{r} \times \myv{\Omega} \endeq$$

Multiplying both sides by $m$ and solving for $m\ddot{\myv r}$:

$$\begineq m\ddot{\myv{r}} =& \myv{F}+ 2m\dot{\myv{r}} \times \myv{\Omega} + m\myv{\Omega}\times \myv{r} \times \myv{\Omega} \\ =& \myv{F} + \myv{F}_{co r} + \myv{F}_{cf}.\endeq$$

The terms, including two inertial force terms, are

• $\myv{F}$ is the sum of all the forces identified in the inertial frame,
• $\myv{F}_{co r}= 2m\dot{\myv{r}} \times \myv{\Omega}$ is the Coriolis force.
• $\myv{F}_{cf}=m\myv{\Omega}\times \myv{r} \times \myv{\Omega}$ is the centrifugal force.

For objects with a position on Earth's surface, and with velocities tangent to Earth's surface: Where does the centrifugal force vanish? Where can the Coriolis force vanish?

For an object on the surface of the earth ($r=R_e$) moving with speed $v=|\dot{\myv{r}}|$ the ratio of the two forces is very roughly

$\frac{F_{co r}}{F_{cf}} \approx \frac{v}{R\Omega}$

With $R\approx6400$km and $\Omega=(2\pi {\rm Radians})/(24 {\rm hr})$ we get $R\Omega \approx 1700$ km/hr ~ 1000 mi / hr. So for moderate speeds much less than this, the Coriolis force is small compared to the centrifugal force.

The centrifugal force

On the surface of the earth, the centrifugal force is generally not acting "straight up".

The centrifugal force vanishes at the poles and its maximum value occurs at the equator, where $a_{cf}=F_{cf}/m = \Omega^2R_e \approx 0.034 m/s^2$. So things weigh about 0.3% less at the equator than at the poles.

Coriolis force

$$\myv{F}_{co r}= 2m\dot{\myv{r}} \times \myv{\Omega}.$$

Using the right-hand rule...What direction is the Coriolis force on winds with the initial velocities shown below in the northern hemisphere?

Let's say, that air from N / E / W / S is being pulled to the intersection of that arrow by a massive Low Pressure center, as happens with hurricanes. Which direction do you anticipate that the winds will circle around the LP center?

Prevailing winds

• On average, more sunlight per meter^2 at the equator than at the poles.
• Earth's surface absorbs a portion of the sunlight and heats up: More so near the equator than at either pole.
• Air rises more vigorously above the warmest land. Drops most vigorously above the coldest land.

So, air near the surface should be drawn ($\Rightarrow$ wind) towards the equator and away from either pole.

How many circulating "convection cells" are there between the equator and either pole?

At any latitude where the cells meet, air should be rising (or dropping) in *both* adjacent cells.

This implies that there must be an odd number of cells from equator to pole.

The answer turns out to be....

3 cells.

Can you tell why Columbus landed in the Bahamas and not Newfoundland?

Prevailing surface winds across the continental U.S. are West-> East. ...As we can also see from...

(DOE-NREL)

Circulation in the 3 cells, and the jetstreams at the boundaries are partly driven by sunlight absorption, partly by the prevailing land temperature differences. Climate change has arguably caused the average global temperature increase of 0.8 C since 1885 (with more to come!) But the Arctic has warmed by about twice that amount. Northern hemisphere circulation has weakened, with some consequences...

How the Jet Stream and Climate Change Bring on Cold Snaps [Inside Climate News]

Also: Climate Whiplash [Yale Environment 360].

 

Homework

Chapter 9, problems: 7, 8, 10, 16, 19 (