Non-Inertial Reference Frames
Vern Lindberg
June 15, 2010
In the first 4 chapters of the text we have considered various types of motion all observed
from an inertial reference frame. Sometimes we are in a non-inertial reference frame, such
as the rotating earth, and want to describe motion relative to that non-inertial frame.
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Translating Reference Frame
~ 0 relative to
We first consider a reference frame O0 xyz that has a constant acceleration A
an inertial frame Oxyz. This is shown in Figure 5.1.1 in Fowles and Cassiday.
~ 0 can
For convenience we require the axes xyz and x0 y 0 z 0 to be parallel. The acceleration A
be in an arbitrary direction. Since we are non-relativistic, we assume the time is measured
identically in the two systems.
~ 0 with velocity V
~0 . Observers in the
At some time t0 , the non-inertial origin is located at R
two frames observe an object. According to the inertial observer the object has location
~r, velocity ~v , and acceleration ~a that arise because of forces F~ . The non-inertial frame
measures the object and finds a location ~r 0 , velocity ~v 0 , and acceleration ~a 0 .
From geometry we see a relation between location measurements, and by taking derivatives
find
~ 0 + ~r 0
~r = R
~0 + ~v 0
~v = V
~ 0 + ~a
~a = A
0
(1)
(2)
(3)
Newton’s Second Law becomes
~ 0 + m~a 0
F~ = m~a = mA
1
(4)
It is more common to write this for the non-inertial frame as
~ 0 = m~a 0
F~ − mA
(5)
since the acceleration is what is measured in the non-inertial frame. Thus in addition to
the real forces (those acting between two objects) there is a fictitious force, or inertial force
that arises because of the acceleration of the frame. To the non-inertial observer this acts
just like a regular force, with the only issue being that there is no second object for a
Newton’s Third Law to apply.
We shall look at a several examples of this.
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Pure Rotation
Here we have a common origin for the two coordinate systems, but the non-inertial system
will rotate with an angular speed ω about an arbitrary direction n̂, i.e. an angular velocity
ω
~ = ωn̂. Although the text does its best to draw the situation, it is best to actually get
some sticks and build a 3D model. The angular speed may be changing in time.
The location vectors are identical, ~r = ~r 0 since the origins coincide. This does not mean
that the components or directions are the same in the two frames. So
~r = ~r 0
0
(6)
0
0
0
0 0
îx + ĵy + k̂z = î x + ĵ y + k̂ z
(7)
For the inertial frame the unit vectors are fixed and their time derivatives are zero. In
the rotating system the unit vectors change direction, and we must evaluate the time
derivatives. Thus taking a time derivative
î
dx0
dy 0
dz 0
dî 0
dĵ 0
dk̂ 0
+ ĵ 0
+ k̂ 0
+ x0
+ y0
+ z0
dt
dt
dt
dt
dt
dt
0
0
0
d
î
d
ĵ
d
k̂
~v = ~v 0 + x0
+ y0
+ z0
dt
dt
dt
dx
dy
dz
+ ĵ
+ k̂
dt
dt
dt
= î 0
(8)
(9)
Figure 1 is my version of Figure 5.2.3 in the text that describes how to evaluate the
derivatives of the unit vectors. The angular velocity vector, ω
~ , of the rotating reference
0
frame is shown as well as the initial unit vector î that is inclined from ω
~ by an angle φ. In
a small time ∆t the unit vector is rotated to a new position, and we can draw ∆î 0 .
Two “moment arms” extend from the the tip and tail of ∆î 0 to ω
~ , and are perpendicular to ω. The two moment arms are separated by an angle ∆θ and have length 1 sin φ.
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[ht]
Figure 1: Computing time rate of change of unit vector in a rotating reference frame,
Hence
|∆î 0 | ≈ sin φ∆θ
∆î 0 ∆θ
= ω sin φ
≈ sin φ
∆t ∆t
(10)
(11)
Since the direction of ∆î 0 is perpendicular to both ω
~ and î 0 , we can write
dî 0
=ω
~ × î 0
dt
(12)
with similar relations for the other unit vectors. Thus the last terms in Eq. 9 can be
written
x0
dî 0
dĵ 0
dk̂ 0
+ y0
+ z0
dt
dt
dt
= x0 ω
~ × î 0 + y 0 ω
~ × ĵ 0 + z 0 ω
~ × k̂ 0
(13)
= ω
~ × ~r 0
(14)
so that Eq. 9 becomes (using ~r = ~r 0 )
~v = ~v 0 + ω
~ × ~r
" #
d~r
d~r
d
=
+ω
~ × ~r =
+ω
~ × ~r
dt f ixed
dt rotating
dt rotating
(15)
(16)
What does this mean? The left hand side is the rate of change evaluated in the fixed
inertial system, using the coordinates from the fixed system. The derivative on the right
is the rate of change of the vector in the rotating system, using the coordinates from the
non-inertial frame.
Fowles and Cassiday now state “A little reflection shows that the same applies to any vector
~ I am at a point that I believe it but don’t think I can make it clear to others.
Q.”
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Taking it as fact, and using ~v = ~v 0 + ω
~ × ~r 0 , we can write
d~v
d~v
=
+ω
~ × ~v
dt f ixed
dt rotating
d(~
ω × ~r 0
0
~a = ~a +
+ω
~ × ~v 0 + ω
~ × (~
ω × ~r 0 )
dt
rotating
d~
ω
× ~r 0 + ω
~ × ~v 0 + ω
~ × ~v 0 + ω
~ × (~
ω × ~r 0 )
~a = ~a 0 +
dt rotating
~a = ~a 0 + ω
~˙ × ~r 0 + 2~
ω × ~v 0 + ω
~ × (~
ω × ~r 0 )
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(17)
Rotation and Translation
Combining results from the two previous sections we get
~0
~a = ~a 0 + ω
~˙ × ~r 0 + 2~
ω × ~v 0 + ω
~ × (~
ω × ~r 0 ) + A
(18)
Let’s look at each term in Equation 18.
The two accelerations, a and a0 , are the values measured in the appropriate reference
frames. For example if you hold a ball in your hand while riding a roller coaster, the
acceleration in the inertial frame will be non-zero and have both radial and tangential
components, while for you in the non-inertial frame the ball has zero acceleration.
The term ω
~˙ × ~r 0 , called the transverse acceleration, deals with effects that arise because
a reference frame has a non-constant angular velocity. A centrifuge during the speed up
and slow down parts of its motion is a good example of this. For planets this term is
negligible.
The 2~
ω × ~v 0 is called the Coriolis acceleration and is present whenever a particle moves in
a rotating reference frame, unless the motion is parallel to the rotation axis.
The ω
~ ×(~
ω ×~r 0 ) is our old friend the centripetal acceleration. You should be able to see that
this acceleration points toward the center of the circle in which the particle travels.
~ 0 is the acceleration of the origin of the non-inertial reference system relative to
Finally, A
the inertial system.
This only makes sense if we do some examples.
E.g. 5.2.1 A wheel of radius b rolls without slipping with a constant forward velocity of
magnitude V0 . Find the acceleration of any point on the rim.
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E.g. 5.2.2 A wheel of radius b rolls without slipping around a circular track of radius ρ.
The speed of the wheel (its center) is V0 . Find the acceleration at the highest point
on the wheel.
Extending E.g. 5.2.2 For the same wheel, find the acceleration at the lowest point, the
leading point, and the trailing point.
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Putting it Together: Dynamics
We have the relation between ~a and ~a 0 , now we want to connect it to Newton’s Second
Law and get equations of motion.
In the inertial frame we have F~ = m~a, and would like to apply this to the non-inertial
frame. Using Equation 18 we get
~ 0 = m~a 0
F~ − mω
~˙ × ~r 0 − 2m~
ω × ~v 0 − m~
ω × (~
ω × ~r 0 ) − mA
(19)
Defining fictitious (inertial) forces
0
trans
~
F 0Cor
Transverse force F~
Coriolis force
Centrifugal force F~
0
centrif
= −mω
~˙ × ~r 0
= −2m~
ω × ~v
(20)
0
(21)
0
= −m~
ω × (~
ω × ~r )
(22)
and using F~physical for the real forces having Newton’s Third Law partners, we can define
the net force in the non-inertial frame as
F~ 0 = F~physical + F~
0
trans
+ F~
0
Cor
+ F~
0
centrif
~0
− mA
(23)
and hence write
F~ 0 = m~a 0
(24)
The remainder of this chapter is devoted to application of this equation to a variety of
systems.
E.g. 5.3.1 A horizontal bicycle wheel rotates uniformly and a bug crawls outwards with
a constant speed along a spoke. Find the fictitious forces.
E.g. 5.3.2 The above bug starts at the center of the wheel, and there is a coefficient of
static friction µs . At what distance from the center does the bug slip?
E.g. 5.3.3 A rod of length ` rotates at constant angular speed about a fixed axis located
at one end of the rod. A bead of mass m initially is at the axis but is given a very
small initial speed ω down the rod. At what time does the bead leave the rod?
Assume no friction.
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Sect. 5.4 Plumb Line A plumb bob is hung near the earth at latitude λ. In what
direction does the plumb bob point, and how is this related to the location of the
center of the earth?
Sect 5.4 Projectile Motion on a Rotating Earth Ignoring air resistance, and making appropriate approximations, find equations for the position components and velocity components when the projectile is launched from the origin with some initial
velocity at a latitude λ.
E.g. 5.4.1 A ball is dropped from rest at a height h above the surface of the earth. Find
the deflection of the ball.
E.g. 5.4.2 A bullet is shot “horizontally” with a high initial speed. Find the deflection of
the bullet when it has travelled a horizontal distance H that is of moderate size.
Sect 5.6 The Foucault Pendulum How is it analyzed, and what is the period of precession?
Sect 5.5 Rendezvous With Rama Motion of a projectile in a uniformly rotating cylinder.
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