Physics 430: Lecture 21
Rotating Non-Inertial Frames
Dale E. Gary
NJIT Physics Department
9.4 Time Derivatives in a Rotating
Frame
Let’s explicitly take the Earth as the rotating frame. The angular frequency
of rotation of the Earth is
2 rad
W
 7.3 10 5 rad/s.
24  3600 s
 We will assume that the inertial frame So and rotating frame S share the
same origin, so the only motion of S relative to So is a rotation with angular
velocity W. For example, the common origin could be the center of the
Earth.
 Now consider an arbitrary vector Q. The time rate of change of Q in frame
So is related to its time rate of change in S by
 dQ 
 dQ 

 
  Ω  Q.
 dt  So  dt  S


This says that even if Q is not changing length or direction in frame S, it is
still changing in frame So by virtue of its rotation.
November 17, 2009
9.5 Newton’s Second Law in a
Rotating Frame

Let’s now see what Newton’s second law looks like. Obviously, in frame So,
it is just
 d 2r 
m 2   F,
 dt  So
where, as usual, F is the net force on the particle as identified in the inertial
frame.
 We can now use our result from the previous slide, identifying Q and the
position vector r, i.e.
 dr 
 dr 
      Ω  r.
 dt  So  dt  S
 Differentiating a second time

 d 2r 
 d   dr 
 2        Ω  r .
 dt  So  dt  So  dt  S

 Now the vector Q is the entire quantity in square brackets, hence:

 dr 

 d 2r 
 d   dr 
 2        Ω  r   Ω     Ω  r .
 dt  So  dt  S  dt  S

 dt  S

November 17, 2009
Newton’s 2nd Law in a Rotating
Frame-2
We will use the dot notation for vectors in frame S, hence:

 dr 

 d 2r 
 d   dr 
 2        Ω  r   Ω     Ω  r 
 dt  So  dt  S  dt  S

 dt  S

 r  2Ω  r  Ω  Ω  r 
 d 2r 
 Thus, Newton’s second law, which was m 2   F,
 dt  So

is now mr  F  2mr  Ω  m(Ω  r )  Ω.
 Once again, we just have Newton’s second law, but this time with two
inertial force terms on the right:
Fcor  2mr  Ω. coriolis force
Fcf  m(Ω  r)  Ω.

centrifuga l force
We will look at each of these in more detail.
November 17, 2009
Centrifugal Force
You are already quite familiar with the centrifugal force (centripetal force in
the inertial frame). You know it as mv2/r, but you may protest that this
expression: Fcf  m(Ω  r)  Ω looks nothing like that.
 But if you recall that v = wr, and that we are using capital letters for
quantities related to motion of one frame relative to another, then v = Wr.
So you see that mv2/r = mW2r, which does have the right variables.
 But what do we make of these complicated looking
W
cross-products? First, let’s look at the magnitude:
W×r
Ω r  Wr sin 
r
and since the cross-product of two vectors is
 r
perpendicular to both of the two vectors, the
angle between W × r and W is /2, so
Fcf  m(Ω  r)  Ω
2
2
(Ω  r)  Ω  W r sin   W r ,
where r is as shown in the figure.
 Now look carefully at the directions. Fcf points
radially outward from its circular path. Fcf  mW 2 rρˆ .

November 17, 2009
Non-Vertical Gravity
Due to the centrifugal force, a plumb bob does not actually point in the
direction to the center of the Earth except at the pole or equator.
 Because a plumb bob will respond to both the force of gravity Fgrav downward,
and Fcf outward, the effective force is
Feff  Fgrav  Fcf  mg o  mΩ 2 R sin  ρˆ

from which we can identify an effective gravity
g  g o  Ω 2 R sin  ρˆ .
The centrifugal term is zero at the pole, and largest
at the equator, where it is W2R = 0.034 m/s2. The
tangential component of this effective gravity is
gtang = W2R sin cos and is maximum at  = 45o.
 The maximum angular deviation is
 max  Ω 2 R / 2 g o  0.1o.
W
Fcf

R

Fgrav = mgo
November 17, 2009
Coriolis Force

The Coriolis force may seem mysterious at first, but its origin is almost trivial.
Before examining the equation, let me start by considering a missile at the
equator. It’s speed in the inertial frame is just that of the Earth,
v  ΩREarth  (7.3 10 5 rad/s )(6.38 106 m)  466 m/s  1000 mi/h .
If we fire the missile straight north, heading for latitude l = /2   ( is called
the colatitude), it will, of course, maintain its (the equator’s) sideways velocity,
but it will be traveling over land for which the sideways
W
velocity of the Earth is less:
v( )  Ωr  ΩREarth sin   1000sin   mi/h .
 Because of that difference in speed, the rocket
actual path
seems to drift to the east in the frame of Earth.
 This appears as an inertial force, the Coriolis
attempted path
force Fcor  2mr  Ω. Note that it depends on the
velocity of the moving object (the missile), and
if you are used to finding the direction of
Sideways distance traveled
v × B in E&M, note Fcf ~ v × W.
by launch point at equator during flight

November 17, 2009
Direction of the Coriolis Force
The Coriolis force also acts in a flat geometry, such as movement on a rotating
disk.
v
 The force is always perpendicular to the velocity, so the force
and velocity have the relationships shown in the figure.
W
Fcor
Check the direction using the right-hand rule.
Fcor v
 You can see how the Coriolis force works in the flat geometry
by imagining a puck on a frictionless surface, kicked radially
from the center of the rotating platform. From an inertial frame
the puck will follow a straight line. However, from the rotating frame the path
is a curve, as shown in the figures below.
C
C

View from
inertial frame
B
B
W
A
A
W
View from
rotating frame
November 17, 2009
Effect of Coriolis Force on Free-Fall
Let’s do a simple example that exercises the calculation of Coriolis force and
introduces the useful technique of successive approximations.
 Consider an object at the surface of the Earth in free-fall with no other forces
acting (i.e. no air resistance). We have to include both Fcor and Fcf.
mr  mg o  Fcf  Fcor .

Since we are at the surface of the Earth, we can let r = R. Let’s also combine
the first two terms as mg, the effective gravity force. Then
r  g  2r  Ω.
W
 Since this does not depend on position r, only on its
y (north)
derivatives, we can move the origin to the surface of
z (up)
the Earth. Let’s now write the components of the
O r
vectors in east-north-up coordinates.
 R
x (east)
r  ( x , y , z ),
Ω  (0, W sin  , W cos  ).
 Thus
r  Ω  ( yW cos   zW sin  , xW cos  , xW sin  ),

November 17, 2009
Successive Approximations





The equation of motion resolves to the three equations:
x  2W( y cos   z sin  )
y  2Wx cos 
z   g  2Wx sin  .
Let’s solve these using the method of successive approximations. First, write
the equations for small W (i.e. set W to zero). This is the zeroeth-order
approximation. This gives the equations you have solved in freshman Physics:
x  0,
y  0, z   g.
These lead to the solutions
x  0,
y  0,
z  h  12 gt 2 ,
where we have used initial conditions appropriate to dropping the object from
rest from high h. We then plug these back into our original equations to get
y  0, z   g.
the first-order approximation: x  2Wgt sin  ,
The last two equations have not changed, but the x equation has. Integrating
twice, we get x  13 Wgt 3 sin  , which shows that there will be an eastward
deviation. We could repeat to get second-order approx., etc.
An object dropped down a 100-m mine shaft at the equator will move east by:
x  13 Wg (2h / g ) 3 / 2  2.2 cm.  uses t  (2h / g )1/ 2
November 17, 2009
Coriolis Force and Weather





You may think that, since Coriolis force depends on the velocity, the force
would be miniscule for something moving as slowly as a cloud in the
atmosphere—and you would be right. However, a small force does not
necessarily mean a small effect. The force on a slowly moving body can act
over a long period of time (weeks in the case of a weather pattern), and so the
effect can be important.
It is the Coriolis force that causes the cyclonic (counter-clockwise) weather
patterns seen in the northern hemisphere. Consider a low pressure system that
draws air from the North and South. The air moving south will be turned to the
west, while the air moving north will be turned to the east.
Although not directly due to Coriolis force, the flows from east and west also
turn due to the turning of the north and south flows.
To properly understand this, you have to consider the
motions on a sphere, i.e. the north and south flows
are not parallel to W.
L
What is the flow pattern like in the southern hemisphere?
November 17, 2009