PHY 6200
Theoretical Mechanics
Chapter 9
Motion in a non-inertial reference frame
Prof. Claude A Pruneau
Notes compiled by L. Tarini
Introduction
• Sometimes it is simply easier to write
equations in a non-inertial reference frame.
– E.g. Motion near the surface of the Earth
• A non inertial reference frame
Rotating coordinate systems
x3
x3
x2
R
x 2
x1
Fixed system
r
P
r
x1
Rotating system
r r
r  R  r
First consider a fixed point in the rotating frame
x

3
x3
x2
R
x1
x 2
r
P
r
x1
r r
r
dr fixed  d  r
r
r
d r r r
 dr 

r  r
 
dt fixed dt
Next, consider a moving point in the rotating frame
Motion of the point in the rotating frame:
r
 dr 
 
dt rotating
Motion of the point in the fixed frame:
r
r
r r
 dr 
 dr 
 
  r
 
dt fixed  dt  rotating
Consider r  x1ê1  x2 ê2  x3ê3 in rotating system
Assume fixed and rotating system have same origin
r

d
 dr 
   xi êi 
 

dt fixed dt  i


ˆi
  x&i êi  xi e&
i
r&
 xi êi  ri in the rotating system
i
r
r&
 dr 
ˆi
 rr   xi e&
 
dt fixed
i
Example: Consider a d rotation about the 3-axis.
dê1  d 3ê2
3
x2
d 3
3 
dt
dê1 d 3

ê2
dt
dt
ê1
x1
dê1
dê1
  3ê2
dt
Similarly consider a d rotation about the 2-axis.
x3
dê1  d 2 ê3
x2
2
ê1
x1
dê1
d 2
2 
dt
dê1
d 2

ê3
dt
dt
dê1
  2 ê3
dt
dê1
  3ê2
dt
dê1
  2 ê3
dt
dê1
  3ê2   2 ê3
dt
Similarly
dê1
  3ê2   2 ê3
dt
dê2
  3ê1  1ê3
dt
dê3
  2 ê1  1ê2
dt
Conclusion
r
&
eˆi    êi
r
r&
r& r r
 dr 
&
 rr   xi eˆi  rr    r
 
dt fixed
i
The result is valid for any vector.
r
r
 dQ 
 dQ 
r r

 Q
 dt 

 dt  rotating
fixed
Note that  , the angular acceleration, is
the same in both systems.
r
r
r
r r  d 
 d 
 d 

   







dt fixed
dt rot
dt rot
r&

r
r
 dr  
 dR 
 dr 
 
 


dt fixed  dt  fixed  dt  fixed
 dR 
 
 dt  fixed
r
r r
 dr 
 
  r
 dt  rotating
Now define
Conclusion
r
r&  dr  
v f  rf  
 dt  fixed
r
r r&  dR 
v  Rf   
 dt  fixed
r
r r&  dr 
vr  rr   
 dt  rotating
r r
r r
v f  v  vr    r
Acceleration
r
F  ma is valid only in inertial frames
r
 dv f 
r
F  ma f  m 
 dt  fixed
r r
r r
v f  v  vr    r
r
 dv f 
 dt 
fixed
r
r
r
r& r r  dr 
 dv 
 dvr 
 

  r   

 dt  fixed  dt  fixed
 dt  fixed
r
 dv f 
 dt 
fixed
r
r
r
r& r r  dr 
 dv 
 dvr 
 

  r   

 dt  fixed  dt  fixed
 dt  fixed
Rf
r
r r
 dvr 

    vr
dt rot
r
r
ar    vr
The last term
r
r
r  dr 
r
r r
 dr 
  
          r 
 dt  fixed
 dt  rot
r
r
r r
r
   vr      r 
Finally…
r&
r& r
r
r r
r r
r
r
&
F  ma f  mR f  mar  m  r  m  (  r )  2m  vr
For an observer in the rotating coordinate system
r
r
Feff  mar
(Newton's 2nd law)
r
r&
r& r
r
r r
r r
 F  mR f m  r  m  (  r )  2m  vr
Sum of the
forces acting on
the particle as
measured in the
fixed (inertial)
frame
Translation +
angular
acceleration
Centrifugal
Force
Coriolis Force
Centrifugal and Coriolis forces are not forces in the
usual sense of the word.
They are pseudo-forces introduced by our desire
to write
r
Feff  mar
We thus have:
r
Feff  ma f 
(non inertial terms)
The Centrifugal Term
Magnitude:
r
r
r
m  (  r )
m r
Direction:
2

r
 r
r
r
  (  r )
rr r
m  (  r )
r
Directed outward
Example 2
Hockey puck on a large merry-go-round (M-G-R)
with a smooth frictionless horizontal flat 
surface
Assume M-G-R has constant angular velocity
rotating clockwise (seen from above).
a) Find the effective force on the hockey puck
after it has been given a push.
b) Plot path
r
r
r r
r
Solution: Feff  m  (  r )  2m  vr
Neglecting friction; vr measured by observer on
rotating surface.
aeff 
r
Feff
r
r
r r
r
   (  r )  2  vr
m
r
veff   aeff dt
r
reff   veff dt
Assume the puck is initially at (0.5R, 0)
Initial velocity
Motion relative to the Earth
z
z
In the fixed inertial frame
r
r
F  s  mgo
y
x
r
y
x
 External forces
Gravitational
attraction
me
g0  G 2 êr
R
along a radius
Assume the Earth is a perfect sphere and neglect
the fact g 0 varies because of oblateness, density,
nonuniformities, altitude changes, etc.
r& r
r
r r
r r
r
r
&
&
Feff  s  mg0  mRf  m  r  m  (  r )  2m  vr
r
let  be along +z
  7.3  10 rad s
5
Practically constant in time
r
So we neglect   r
 dRr 
&
Rf    Rf  
 dt 
14 2 43
0
r
r
r
    R
r


r
r
r r
r
r
r r
Feff  s  mg0  m    r  R   2m  vr
1 4 4 4 42 4 4 4 43

r
g effective

r
r
r
r r
g  g0      r  R 
1 4 44 2 4 4 43


centrifugal
Since near the Earth r = R, the centrifugal force
r
is dominated by     R

r

The centrifugal force is responsible for the oblate
shape of the Earth.
Earth is deformed
• Equatorial radius is +21.4 km > polar radius
g  0.052 m 2 at the pole relative to the equator
s
• On a calm ocean, the water is perpendicular to
g , not g 0
r
r
r
  (  r )
g0
g
• In simpler terms
r r
r
r
Feff  s  mg  2m  v f
• Effective g
(Pendulum period + direction of equilibrium)
•
 2 R  0.034 m
s
2
about 0.35% of g0
Coriolis Force Term
r
r
2m  vr
Magnitude 2m 2 r
north

Direction
r
  vr
vr
r
r
2m  vr
Deflected path to the right of
the particle motion
Basics Properties of the Jovian Planets
Planet
Distance
(AU)
Period
(years)
Diameter
(km)
Mass
(Earth=1)
Density
(g/cm3)
Rotation
(hours)
Jupiter
5.2
11.9
142800
318
1.3
9.9
Saturn
9.5
29.5
120540
95
0.7
10.7
Uranus
19.2
84.1
51200
14
1.2
17.2
Jet
stream flows eastward
30.1
Jupiter
164.8 ~ 300 km/h
49500
: speed
Saturn : speed ~ 1300 km/h
17
1.6
16.1
Neptun
e
zones – upwelling air,capped by white ammonia
cirrus clouds – top of enormous convection current.
Darker zones – cooler atmosphere – downward motion
– complete convection cycle.
Light
Example: Find horizontal deflection of a plumb lines
caused by Coriolis effect for a particle falling from
height “h”
êz
êy
Solution:

s0
r
Feff  mar
r r
r
r
ar  g  2  vr
êx
 x   cos 
y  0
x0
y& 0
 z   sin 
z& gt
Initially…
êx
êy
êz
r
  v f   cos  o  sin    gt cos êy
o
o
gt
g  (0, 0, g)
r
arx  0
r
ary  2 gt cos 
r
arz  g
1
y(t)   gt 3 cos 
3
1 2
z(t)  z(o)  gt
2
t  2h g
y  0; y& 0 at t  0
1
8h 3
1
8h 3
d  g
cos    cos 
3
3
g
3
g
Ex 5 Pendulum Precession (Foucault pendulum)
z
Small oscillations
l
y
T
Ty
x
Tx
x
Tx  T
l
y
Ty  T
l
Tz  T
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
http://www.astro.louisville.edu/foucault/
r
r r
r T
ar  g   2  vr
m
z
l
 x   cos 
y  0
y
T
 z   sin 
gx  0
gy  0
Tx
Ty
x
gz  g
& (vr )y  y&, (vr )z  z& 0
(vr )x  x,
êx
êy
êz
r
  vr   cos  0  sin 
x&
y&
0
r
  vr  ( y& sin , x& sin , y cos  )
T
ax  
m
T
ay  
m
x
 2 y& sin 
l
y
 2 x& sin 
l
x   2 x  2 z y&
&
y&  2 y  2 z x&
&   (x  iy)  2 z (ix& y&)
(&
x& i&
y)
2
 2 zi( x& iy&)
Let
q  x  iy
2
&
q  2i z q   q  0
That’s a damped oscillator! Except we have an
imaginary term.
The solution is thus:
q(t)  exp(i zt)  A exp(  z2   2 t)  B exp(  z2   2 t)


if
z  0
thus
then
 ? z
q   2 q  0;  z  0
oscillation frequency
q(t)  ei z t (Aei t  Bei t )
q(t)  q(t)e
 i z t
rotation (precession)
x(t)  x cos  z t  y sin  zt
y(t)   x sin  z t  y cos  z t
 x(t)  cos
 y(t)   sin 
sin    x (t)
with



cos   y(t) 
   zt