MATH10222, Chapter 4: Frames of Reference 1 Motion relative to a

MATH10222, Chapter 4: Frames of Reference
Dr. R.E. Hewitt, http://hewitt.gotdns.org/courses.html
Last modified: 2015-04-29
These notes are provided as a revision/overview of the lectures. Any expressions/formulae
that I expect you to have memorised for the examination are highlighted with a surrounding box.
Definition: An “inertial frame of reference” is a coordinate system in which Newton’s
laws hold.
1
Motion relative to a translating origin
Suppose that we have an inertial frame of reference centred about an origin O. By the
definition of the inertial frame, we know that a particle P of constant mass m and position
vector r(t) (relative to O) satisfies Newton’s second law:
m¨r = F ,
where F is the resultant force acting upon P .
O′
r¯(t)
P
b(t)
r(t)
O
Figure 1: O is an origin in an inertial frame of reference, whereas O ′ is moving relative to
O.
Now suppose there is another coordinate system that is defined relative to a moving
origin O ′ as in figure 1. Let’s further suppose that the position of O ′ relative to O is b(t)
and that the position vector of P relative to O ′ is r¯(t). We therefore have the relationship
r(t) = b(t) + r¯(t) ,
and therefore Newton’s second law reduces to
m
d2 r¯
= F − m¨b .
dt2
1
Thus, relative to the moving frame of reference centred at O ′ , the particle feels an effective
force of
F − m¨b .
Newton’s second law clearly only holds in the moving coordinate system if
¨b = 0 ,
that is, if O ′ is moving at constant velocity relative to O. We refer to the moving coordinate
system as a “non-inertial frame of reference” whenever ¨b 6= 0.
1.1
Example: ‘zero-g’ motion in a non-inertial frame of reference
Quite often a particle will be referred to as being ‘weightless’ when in fact it is still being
acted upon by a gravitational acceleration. This is commonly the case in non-inertial
frames of reference where the ‘observer’ and the particle are both in free fall.
For example, consider a particle P of mass m in a uniform gravitational field. The
force acting on P comes from its weight F = −mgk in the obvious notation. Clearly if our
frame of reference (i.e., O ′ ) is also moving in the k direction with the same acceleration
−g (m/s/s), then we have
¨b = −gk .
As above let’s denote the position vector of P relative to O ′ as r¯. Thus, in the noninertial frame of reference the appropriate equation of motion is
d2 r¯
m 2 = F − m¨b = −mgk − m(−gk) = 0 ,
dt
and we may view the particle in this frame as (effectively) being free from any external
force.
2
Two-dimensional rotating frames of reference
Instead of a coordinate system that is translating, we now consider the more complicated
case of a rotating coordinate system. Consider two coordinate systems as shown in figure
2:
• A Cartesian coordinate system {i, j, k} centred at an origin O in an inertial frame
of reference.
• A second Cartesian coordinate system {e1 , e2 , e3 } that is also centred at the origin
O, but which is rotating relative to {i, j, k}.
If we assume that the coordinate system {e1 , e2 , e3 } is rotating about the k axis, with
e3 = k and a rotation angle of θ(t), then we have (from figure 2):
e1 = cos θi + sin θj ,
e2 = − sin θi + cos θj ,
e3 = k .
2
j
e2
e1
θ
θ
i
Figure 2: A Cartesian coordinate system i, j, k (k out of the plane) relative to an origin
O, together with a second coordinate system {e1 , e2 , e3 } that is rotated by an angle θ(t)
about the axis e3 = k.
We’re interested in the rate of change of the basis vectors {e1 , e2 , e3 }, which is easy to
determine via
de
de
˙ sin θ i + cos θ j) ,
e˙1 = 1 = 1 θ˙ = θ(−
dt
dθ
de
de
˙ cos θ i − sin θ j) ,
e˙2 = 2 = 2 θ˙ = θ(−
dt
dθ
e˙3 = 0 .
We note that the above expressions are all equivalent to
˙ ∧e ,
e˙1 = θk
1
˙
e˙2 = θk ∧ e2 ,
˙ ∧e = 0.
e˙3 = θk
3
(1)
(2)
(3)
This is in fact a special case of a general result that we state next.
3
The angular frequency vector
The results of (1)–(3) generalise to (we do not prove it here)
e˙i = ω ∧ ei ,
(4)
for i = 1, 2, 3, where ω is the ‘angular frequency vector’. The magnitude ω = |ω| is then
the rotation rate (or just ‘angular frequency’) of the rotating coordinate system, whist
ω/ω is a unit vector that defines the axis of rotation. In the simpler case of (1)–(3) we
˙ because the rotation rate was θ˙ and the axis was k.
simply had ω = θk
4
Velocity relative to a rotating frame
Suppose that we have an inertial frame of reference (labelled S). Further we suppose
that we wish to use an alternative frame of reference S ′ that rotates relative to S with an
angular frequency vector of ω.
3
Relative to the rotating frame S ′ , we know the position of the particle:
r=
3
X
xi ei ,
i=1
that is, in terms of three coordinates x1,2,3 in the directions of the three basis vectors e1,2,3 .
The velocity relative to the inertial frame of reference is then the rate of change of the
position vector, so
!
3
X
d
dr
r˙ =
xi ei .
=
dt
dt i=1
S
However, we have to be careful (as in Chapter 1, section 5), because in the frame S the
three basis vectors e1,2,3 change with time as the coordinate system rotates. Therefore
3
X
(x˙ i ei + xi e˙ i ) ,
r˙ =
S
i=1
but using (4) we can write this as
r˙ =
S
3
X
(x˙ i ei + xi ω ∧ ei ) ,
i=1
3
X
dr +ω∧
=
xi ei ,
dt S ′
i=1
= r˙ + ω ∧ r .
′
S
The simply says that the velocity relative to the frame S is equal to the velocity relative
to the rotating frame S ′ plus an extra contribution due to the rotation of S ′ relative to S.
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A particle in a rotating frame of reference
As in the preceding section we suppose that S ′ is rotating relative to S with (constant)
angular frequency vector ω, where S is an inertial frame. We know that (by definition of
an inertial frame) the equation of motion for a particle P of mass m in the frame S is
d2 r m 2 = F .
(5)
dt S
Suppose we prefer to describe the problem relative to the rotating frame S ′ , then
what is the equation of motion of P ? To determine the correct equation we need the
acceleration in the rotating frame.
We know from the previous section that the velocity is related by
dr dr = +ω∧r.
dt S
dt S ′
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However we need the acceleration, so differentiating this again (as before) we find that
d2 r d2 r + ω ∧ (ω ∧ r) .
=
+
2ω
r
˙
∧
′
dt2 S
dt2 S ′
S
Substituting this into (5) we obtain
d2 r m 2 = F − 2mω ∧ r˙ − mω ∧ (ω ∧ r) ,
dt S ′
S′
where the additional acceleration terms that arise from the rotation of S ′ are moved to
the right-hand side of the equation.
So in a rotating frame S ′ we simply apply Newton’s second law as usual, but include
two additional “fictitious forces”. We give these ‘fictitious forces’ some names:
• −2mω ∧ r˙ is the “Coriolis force”,
S′
• −mω ∧ (ω ∧ r) is the “centrifugal force”,
but they are purely a consequence of the rotating frame of reference.
5.1
Example: Plane polars and a rotating frame
Suppose we consider a particle P of mass m that moves in a rotating frame of reference.
The angular frequency vector of the rotating frame is ω = ωk, for some constant ω.
You are given that, in the rotating frame of reference P moves in a plane with position
vector r relative to an origin O on the axis of rotation where
r = rˆ
r,
and rˆ is the usual unit vector that points radially outwards from the axis of rotation.
(Aside: from here we will drop the cumbersome notation of O ′, r|
˙ S ′ and just revert to
our previous notation, recognising that this is a non-inertial frame that leads to additional
(fictitious) forces.)
Question: Simplify the vector equation of motion (in the non-inertial frame)
m¨
r = F − 2mω ∧ r˙ − mω ∧ (ω ∧ r) ,
to give two scalar equations for the polar coordinates r and θ in the case F = 0.
Answer: To simplify things we need to determine the components of the Coriolis and
r and we know
centrifugal forces. This is easy to do as we are told that ω = ωk, r = rˆ
from Chapter 1, section 5 that
r˙ = rˆ
˙ r + r θ˙θˆ .
The Coriolis force is then
ˆ = −2mω r˙ θˆ + 2mωr θˆ
˙r ,
−2mω ∧ r˙ = −2mωk ∧ (rˆ
˙ r + r θ˙ θ)
whilst the centrifugal force is
−mω ∧ (ω ∧ r) = −mrω 2 k ∧ (k ∧ rˆ) = −mrω 2 k ∧ θˆ = mrω 2 rˆ .
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Using Chapter 1, section 5 for the acceleration in polar basis vectors, leads to an
equation of motion in the form
¨ θˆ = −2mrω
˙r + mrω 2 rˆ, ,
˙ θˆ + 2mωr θˆ
m (¨
r − r θ˙ 2 )ˆ
r + (2r˙ θ˙ + r θ)
after setting F = 0 (as there is no force acting).
The vector equation therefore simplifies to two scalar equations:
r¨ − r θ˙2 = 2ωr θ˙ + rω 2 ,
or equivalently
r¨ = rω 2
θ˙
1+
ω
!2
,
(6a)
and
2r˙ θ˙ + r θ¨ = −2rω
˙ .
(6b)
Note: It is possible to simplify these further (left as an exercise).
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Two final examples
6.1
A long pendulum in an inertial frame
Question: A planar pendulum consists of a particle P of mass m attached to a string of
length a in a uniform gravitational field of strength −gk in the usual notation. Simplify
Newton’s second law, as applied to P , under the assumption that the deflection of P from
the vertical is small.
Answer: We define the origin of a Cartesian coordinate system to be at the point of
attachment of the string, such that the position vector of P is r = x(t)i + z(t)k, with k
pointing vertically upwards. For deflections of P that are small compared to the length
of the string we will write
x(t) = ǫx1 (t) + · · · ,
z(t) = −a + O(ǫ2 ) ,
(7a)
(7b)
where ǫ ≪ 1. Note that the ǫ2 correction in the vertical position of P is because it moves
on a circle if the string remains tight.
If the angle of the string to the vertical is θ then the forces acting on P are
F = −T sin θi + T cos θk − mgk ,
(8)
where
x1
x
= ǫ +··· ,
a
a
−z
cos θ =
= 1 + O(ǫ2 ) .
a
sin θ =
6
(9)
(10)
Newton’s second law provides
m
d2
(x(t)i + z(t)k) = −T sin θi + T cos θk − mgk .
dt2
(11)
A substitution of the above series expansions shows that, at leading order,
T = mg ,
(12)
d2 x1 g
+ x1 = 0 ,
dt2
a
(13)
from the k component at O(ǫ0 ) and
from the i component at O(ǫ1 ).
This is the harmonic equation for the horizontal displacement of the pendulum mass,
with a general solution of
x1 (t) = A cos(Ωt) + B sin(Ωt) ,
p
where Ω = g/a is the angular frequency of oscillation of the linear pendulum and A, B
are constants. If we started the motion by releasing the pendulum from rest, x(t
˙ = 0) = 0,
at a distance of x(t = 0) = L, then B = 0 and A = L, leaving
x1 (t) = L cos(Ωt) .
(14)
The above reduces to (in effect) setting z(t) = a (constant), thereby ignoring the
vertical acceleration in Newton’s second law and resulting in T = mg. We take this
simpler approach in a more complicated problem below.
Exercise:
Extra
By introducing a similar perturbation expansion for the tension in the string, deduce
the correction to the leading order result of T = mg.
6.2
A long pendulum in a rotating frame
The results of the previous example only hold in an inertial frame of reference. If the
pendulum motion persists on a time scale that is comparable to the rotation of the Earth,
then we must instead use the equations appropriate for the non-inertial rotating frame of
reference.
In 1851 Jean Bernard L´eon Foucault hung a 28kg mass from a 67m string in the
Panth´eon in Paris. Even though the pendulum’s motion was started in such a way that
it moved in a plane, the plane of oscillation was seen to rotate slowly at a rate of 11o per
hour about the vertical axis.
Foucault’s pendulum provided the first terrestrial demonstration of the Earth’s rotation. This slow precession of the plane of oscillation is entirely a consequence of the
(fictitious) Coriolis force that arises in the non-inertial frame of reference (i.e., Paris).
We can solve a slightly simplified version of this problem by assuming that the angular
frequency vector ω is vertical; see figure 3.
From the schematic diagram of figure 3, we can write the tension force in the string
as
T = (−T sin θ cos φ)i + (−T sin θ sin φ)j + T cos θk, .
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z, k
replacemen
ω = ωk
y, j
a
y, j
θ
T sin θ
T sin θ sin φ
r
φ
T sin θ cos φ
x, i
x, i
Figure 3: A schematic view of the Foucault pendulum. The string is of length a and the
magnitude of the tensionpin the string is T . The radial distance of the mass/bob from
the downard axis is r = x2 + y 2. For convenience we’ll choose the z = 0 plane to be at
the point of support of the pendulum string.
On using sin θ = r/a, cos φ = x/r, sin φ = y/r, we can write the tension force as
T = −(T x/a)i − (T y/a)j + (T cos θ)k .
If we assume that the deflection angles are sufficiently small, then cos θ ≈ 1 and, to
leading order, the magnitude of the string tension, T , balances the weight of the pendulum,
mg, where m is the mass of the bob. The horizontal components of the tension force (i.e.,
the restoring forces of the pendulum) are then −(mgx/a)i − (mgy/a)j.
The equation of motion in the non-inertial (Paris) frame of reference is
m¨
r = F − 2mω ∧ r˙ − mω ∧ (ω ∧ r) ,
(15)
which we can now simplify. To do this we write the position vector of the particle as
r = (x, y, −a) (for small deflections the vertical position of the particle does not differ
significantly from z = −a, as seen in the previous example), ω = ωk (the axis of rotation
is assumed to be vertical) and F = T − mgk (the only forces acting are tension in the
string and the weight).
The Coriolis force is then proportional to
ω ∧ r˙ = −ω yi
˙ + ω xj
˙ ,
whilst the centrifugal force is proportional to
ω ∧ (ω ∧ r) = −ω 2 xi − ω 2 yj .
Hence (15) simplifies to the two scalar ODEs:
g
2
x¨ − 2ω y˙ +
−ω x = 0,
a
8
and
y¨ + 2ω x˙ +
g
− ω2 y = 0 ,
a
where {x(t), y(t)} is the coordinate of the pendulum in the plane.
These two equations are coupled together through the Coriolis force. However we can
reduce them to a single equation for the position (represented in the complex plane):
g
2
¨
˙
−ω ψ = 0,
ψ + 2iω ψ +
a
where ψ = x(t) + iy(t). This is simple to solve using the methods of the first half of this
course to give a general solution:
ψ(t) = e−iωt AeiΩt + Be−iΩt ,
p
where Ω = g/a is the usual planar pendulum oscillation frequency.
If we start the pendulum from rest in the y = 0 plane, then initial conditions are
x(0) = L, y(0) = 0, x(0)
˙
= y(0)
˙
= 0, where L is the initial displacement. These lead to
˙
ψ(0) = L, ψ(0)
= 0. The two constants A and B are therefore determined from
A + B = L,
A−B =
Lω
.
Ω
The period of oscillation of the pendulum in the Panth´eon is approximately 16 seconds
which is much shorter than the time taken for the Earth to rotate, so ω/Ω ≪ 1 and these
equations show that (at leading order)
A=B=
L
,
2
so that
x(t) + iy(t) = eiωt L cos Ωt .
This solution only differs from (14) through the eiωt term, and it is that term which will
cause the precession of the plane in which the pendulum swings. Clearly for ωt ≪ 1 the
plane of oscillation will remain close to y = 0 as the right-hand side is real. However,
when ωt = π/2 the plane of oscillation will be x = 0 (ie. it will have rotated though π/2
radians), which is easily seen as in this case the right-hand side is purely imaginary.
9
The room the pendulum is located in is not really rotating through 2π radians in Extra
one day however. The angular frequency vector we need is the component of the Earth’s
rotation in the vertical direction. So, in fact,
ω = ωE sin Θ ,
where Θ is the latitude of Paris (approx. 48o north) and ωE = 2π radians per day. The
angular frequency required is
ω = 360 sin Θ , degrees per day
= 15 sin Θ , degrees per hour
≈ 11 , degrees per hour ,
which is indeed the observed rate of precession.
Additional comments
Anyone interested in the history/people/topics discussed in this
course may also be interested in the ‘popular’ science books:
‘Isaac Newton’ by James Gleick and ‘Pendulum: Leon Foucault
and the Triumph of Science’ by Amir D. Aczel.
This material is covered by a Creative Commons, noncommerical, attribution, share-alike license.
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