Rotating Reference Frames
Rotating Reference Frames
Rotating Reference Frames
Suppose that a given object has position vector in some non-rotating inertial reference
frame. Let us observe the motion of this object in a non-inertial reference frame which
rotates with constant angular velocity
about an axis passing through the origin of the
inertial frame. Suppose, first of all, that our object appears stationary in the rotating reference
frame. Hence, in the non-rotating frame, the object's position vector will appear to precess
about the origin with angular velocity
. It follows, from Equation (A.1309), that in the
non-rotating reference frame
(406)
Suppose, now, that our object appears to move in the rotating reference frame with
instantaneous velocity . It is fairly obvious that the appropriate generalization of the above
equation is simply
(407)
Let and
and
denote apparent time derivatives in the non-rotating and rotating
frames of reference, respectively. Since an object which is stationary in the rotating reference
frame appears to move in the non-rotating frame, it is clear that
. Writing
, the above
(408)
or
(409)
since is a general position vector. Equation (409) expresses the relationship between
apparent time derivatives in the non-rotating and rotating reference frames.
, in the rotating
(411)
or
(412)
Applying Newton's second law of motion in the inertial (i.e., non-rotating) reference frame,
we obtain
(413)
Here,
is the mass of our object, and is the (non-fictitious) force acting on it. Note that
these quantities are the same in both reference frames. Making use of Equation (412), the
apparent equation of motion of our object in the rotating reference frame takes the form
(414)
The last two terms in the above equation are so-called ``fictitious forces''. Such forces are
always needed to account for motion observed in non-inertial reference frames. Note that
Centrifugal Acceleration
Let our non-rotating inertial frame be one whose origin lies at the center of the Earth, and let
our rotating frame be one whose origin is fixed with respect to some point, of latitude , on
the Earth's surface--see Figure 24. The latter reference frame thus rotates with respect to the
former (about an axis passing through the Earth's center) with an angular velocity vector,
, which points from the center of the Earth toward its north pole, and is of magnitude
(415)
Let the non-fictitious force acting on our object be the force of gravity,
. Here, the
local gravitational acceleration, , points directly toward the center of the Earth. It follows,
from the above, that the apparent gravitational acceleration in the rotating frame is written
(417)
where is the displacement vector of the origin of the rotating frame (which lies on the
Earth's surface) with respect to the center of the Earth. Here, we are assuming that our object
is situated relatively close to the Earth's surface (i.e.,
).
It can be seen, from Equation (417), that the apparent gravitational acceleration of a
stationary object close to the Earth's surface has two components. First, the true gravitational
acceleration,
, of magnitude
, where
is the
It is convenient to define Cartesian axes in the rotating reference frame such that the
points vertically upward, and
- and
-axis
-axis pointing
and
are written
(419)
(420)
According to the above equation, the centrifugal acceleration causes the magnitude of the
apparent gravitational acceleration on the Earth's surface to vary by about
, being
largest at the poles, and smallest at the equator. This variation in apparent gravitational
acceleration, due (ultimately) to the Earth's rotation, causes the Earth itself to bulge slightly at
the equator (see Section 12.6), which has the effect of further intensifying the variation, since
a point on the surface of the Earth at the equator is slightly further away from the Earth's
center than a similar point at one of the poles (and, hence, the true gravitational acceleration
is slightly weaker in the former case).
Another consequence of centrifugal acceleration is that the apparent gravitational acceleration
on the Earth's surface has a horizontal component aligned in the north/south direction. This
horizontal component ensures that the apparent gravitational acceleration does not point
directly toward the center of the Earth. In other words, a plumb-line on the surface of the
Earth does not point vertically downward, but is deflected slightly away from a true vertical
in the north/south direction. The angular deviation from true vertical can easily be calculated
from Equation (421):
(423)
Here, a positive angle denotes a northward deflection, and vice versa. Thus, the deflection is
southward in the northern hemisphere (i.e.,
hemisphere (i.e.,
). The deflection is zero at the poles and at the equator, and reaches
its maximum magnitude (which is very small) at middle latitudes.
Coriolis Force
We have now accounted for the first fictitious force,
, in
Equation (414). Let us now investigate the second, which takes the form
,
and is called the Coriolis force. Obviously, this force only affects objects which are moving
in the rotating reference frame.
Consider a particle of mass
free-falling under gravity in our rotating reference frame. As
before, we define Cartesian axes in the rotating frame such that the
-axis points vertically
upward, and the
- and
Here,
, and is the local acceleration due to gravity. In the above, we have
neglected the centrifugal acceleration, for the sake of simplicity. This is reasonable, since the
only effect of the centrifugal acceleration is to slightly modify the magnitude and direction of
the local gravitational acceleration. We have also neglected air resistance, which is less
reasonable.
Consider a particle which is dropped (at
) from rest a height above the Earth's
surface. The following solution method exploits the fact that the Coriolis force is much
smaller in magnitude that the force of gravity: hence,
can be treated as a small parameter.
To lowest order (i.e., neglecting
), the particle's vertical motion satisfies
which can be solved, subject to the initial conditions, to give
(427)
Substituting this expression into Equations (424) and (425), neglecting terms involving
and solving subject to the initial conditions, we obtain
, and
(428)
-direction). Now,
(429)
Note that this deflection is in the same direction as the Earth's rotation (i.e., west to east), and
is greatest at the equator, and zero at the poles. A particle dropped from a height of 100m at
the equator is deflected by about
, east is
, etc.).
(431)
(432)
To lowest order in
If follows that the Coriolis force causes the compass bearing of the particle's velocity vector
to rotate steadily as time progresses. The rotation rate is
(437)
Hence, the rotation is clockwise (looking from above) in the northern hemisphere, and
counter-clockwise in the southern hemisphere. The rotation rate is zero at the equator, and
greatest at the poles.
The Coriolis force has a significant effect on terrestrial weather patterns. Near equatorial
regions, the intense heating of the Earth's surface due to the Sun results in hot air rising. In
the northern hemisphere, this causes cooler air to move in a southerly direction toward the
equator. The Coriolis force deflects this moving air in a clockwise sense (looking from
above), resulting in the trade winds, which blow toward the southwest. In the southern
hemisphere, the cooler air moves northward, and is deflected by the Coriolis force in a
counter-clockwise sense, resulting in trade winds which blow toward the northwest.
Furthermore, as air flows from high to low pressure regions, the Coriolis force deflects the air
in a clockwise/counter-clockwise manner in the northern/southern hemisphere, producing
cyclonic rotation--see Figure 26. It follows that cyclonic rotation is counter-clockwise in the
northern hemisphere, and clockwise in the southern hemisphere. Thus, this is the direction of
rotation of tropical storms (e.g., hurricanes, typhoons) in each hemisphere.