Transcript Testing

Class #23 Centrifugal and Coriolis forces
Accelerated reference frames
Rotating reference frames



Vector angular velocity
Newton’s laws on rotating frame
Two “fictitious forces”
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Linearly Accelerated Reference Frames
OK to use
 F  ma but
F  F
inertial
 Ffictitious
Ffictitious  maexternal
so...
F
inertial
 Ffictitious  ma( relative to accelerated frame )
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Vector angular velocity
v

r
Theta=90-Latitude = “CoLatitude”
Earth obeys right-hand rule
Omega points North Earth spins
clockwise Sun rises in east
dr
|S 0  v    r |S
dt
A fixed vector “r” in rotating
frame “S” appears to have a velocity “V”
when seen from an inertial (nonrotating) frame S0.
The velocity may be calculated
from cross product of Omega as seen
from S0 and “r” as seen in “S”.
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Vector velocity – 1 minute problems
v

r
v   r
earth  7.3 105 rad / s
rearth  6400 km
Latitudes –
Socorro – 34 N
Nome, Alaska – 64 N
Hilo, Hawaii – 19 N
Hobart, Tazmania – 43 S
What is circumferential velocity of
each of these four cities?
Spaceflight –
A rocket needs a velocity of 8 km/s for
orbit. What fraction is fuel use reduced if
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launch from equator instead of poles?
Vectors expressed
in rotating frames

r
Imagine same axes
ˆ y,
ˆ zˆ
eˆ1 ,eˆ 2 ,eˆ 3 = x,
expressed in two
frames S0 stationary)
and S (fixed to earth).
r
S
  ri eˆi eˆi are fixed in S
i
dri
r   eˆi
S
i dt
dri
deˆi
r   eˆi   ri
S0
dt
i dt
i
Because eˆi is any arbitrary FIXED
vector in frame S , it transforms
deˆi
in S 0 as
   eˆi
dt
dr
r   i eˆi   ri   eˆi
S0
i dt
i
 r   eˆ
i
i
r
i
    ri eˆi    r 
i
S0
dri
  eˆi    r  r    r
S
i dt
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Derive Newton in rotating frame
We know
r
S0
 r   r  v S   r
S
S
We differentiate again with
same procedure.
d
d
r 
v S  v
S
0
dt
dt
d
r  a S    v S  (  v )
S0
dt
r  a S    v S  (  v )    (  v )


S0
r
S0
 a S  2  v S    (  r )
S
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Derive Newton in rotating frame II
We know
r
S0
 a S  2 v S   ( r )
S
So
m a S  m a S 0  2m r  m ( r )
mr
S
 Fexternal  2mr    m( r )  
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Newton in a rotating frame
mr  Fexternal  2mr   m( r ) 
Newtonian Term
Coriolis Term
Centrifugal Term
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Coriolis force for a rocket going up

r
r 
r 
r
is West

mr  Fexternal  2mr   m( r ) 
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Problem L23-1
Albuquerque latitude  35 N ; earth  7.3105 rad / s
A 100 kg weather rocket is launched straight up
from Langmuir Lab. At t=10 sec, it’s velocity is
100 m/s. At t=100 sec, it’s velocity is 1000 m/s.
What is the magnitude and direction of the
Coriolis force at t=10 and 100 sec?
An airplane flies due North from Albuquerque.
Because of Coriolis force, the airplane will be
deflected from its true North path
A) Which direction will the deflection be in?
B) How many miles off course will the airplane
be after one hour? Assume v=300 m/s.
10 :30