1
1
1.1
Mechanics / ODE application
1.5
Polar basis vectors
Given:
r̂(θ) = cos(θ)i + sin(θ)j
dr̂
dr̂
dθ
dt = dθ = dt = θ̇ θ̂
dr̂
θ̂(θ) = dθ = − sin(θ)i + cos(θ)j
r = rr̂
r̂ = v(t) = ṙr̂ + r dr̂
dt = ṙr̂ + r θ̇ θ̂
repeat for a(t) = (r̈ − rθ̇2 )r̂ + (2ṙθ̇ + rr̈)θ̂
1.2
MECHANICS / ODE APPLICATION
Motion in a plane
At time t, a plane that contains both the particle, the
origin and is also tangential to the path of the particle
is spanned by the vectors r and ṙ. n = r × ṙ.
d
(r × mṙ) (angular momentum equation, obLO = dt
tained from taking the cross product of both sides of
F = ma) with r.
Angular momentum = H O = r × mṙ
H
d
∴ dtO = dt
(r × mṙ) = LO , so angular momentum is
conserved when LO = 0
Newton’s Laws
N1. When all external influences on a particle are
removed, the particle moves with constant velocity
(which may be zero).
N2. When a force F acts on a particle of constant mass
m, the particle moves with an acceleration a where
F = ma = mr̈. If mass is not constant use F = ṗ.
N3. When two particles exert forces upon eachother,
these forces are (i) equal in magni- tude, (ii) opposite
in direction and (iii) parallel to the straight line joining
the two particles.
dot product of each side of F = mr̈ with ṙ
d
=⇒ mṙ · r̈ = ṙ · F =⇒ m
2 dt (ṙ · ṙ) = ṙ · F .
1.6
Conservation of energy
kinetic energy = T =
1
m|ṙ|2
2
Rt
Conservation of angular momentum =⇒ the
work done = t12 F · ṙdt
motion is confined to a plane
conservation of energy: E = T + V , where
V is the
R
potential energy of the particle. V = − F dx
1.3 Motion in a plane: central fields of the turning points of the particle are when ẋ = 0.
force
The constant E is minimum when F(x) = 0, to find the
corresponding energy sub it into V(x).
Angular momentum always conserved. Particle of mass
m, a central field of force is of the form: F = mf (r)r̂.
For a force in this form, the moment is zero
LO = r × F = r × mf (r)r̂ = mf (r)r × r̂ = 0 ∴ angular
momentum = a constant vector = conserved. H O =
mr × ṙ = mrr̂ × (ṙr̂ + rθ̇θ̂) = mr2 θ̇k where k = r̂ × θ̂
we can write angular momentum as H O = mHO k
HO = r2 θ̇.
2
H2
E = m|2ẋ| + V + 2rO2
1.4
Central fields of force:
equations
the path
Often we are interested in the path of the particle
r(θ) rather than the parameterisation of the two coordinates with time r(t) and θ(t). Solve for the path
directly, transform the variables from (r, t) → (u, θ)
where u = 1r .
Hence:
−1
d(u−1 )
du
−2 du
ṙ = dr
= d(udθ ) dθ
dt =
dt
dt = −u
dθ = −HO dθ
repeat the process to find r̈:
d(−HO du
d(−HO du
2 2 d2 u
dθ )
dθ )
r̈ = ddtṙ =
=
θ̇ = −HO
u dθ2 ,
dt
dθ
substituting into newton’s 2nd law gives:
2
r̈ − rθ˙2 = f (r) =⇒ ddθu2 + u = − fH(1/u)
2 2
Ou
Using initial conditions
Typically we know r and ṙ at t = 0 but our equation is
1
for u(θ). Choose θ = 0 =⇒ t = 0 ∴ u(θ = 0) = r(t=0)
Using ṙ = −HO du
dθ we can evaluate it at t = 0, θ = 0
−ṙ(t=0)
to show du
|
=
dθ θ=0
Ho
1
1.7
Stability of equilibrium solutions
(Motion along a line)
An equilibrium solution is x = xe , F (xe ) = 0, V̇ (x) =
0.. Use a perturbation, x = xe + ϵx̃(t) + ...
If x̃(t) grows without bound as t increases then xe is
unstable, if it doesn’t then it is stable. Substituting
the perturbation into newton’s 2nd law:
d2
˜
m dt
2 (xe + ϵx̃(t) + ...) = F (xe + ϵ(t) + ...), perform taylor series
expansion
for
F
(x
+
...)
and simplify to get:
e
d2 x̃
dt2
′′
(xe )
+ V m
x̃ = 0 This is a 2nd order ODE that
when solved for λ gives the frequency of oscillation.
2
2
2.1
MECHANICS / ODE APPLICATION
Mechanics / ODE application
CFoF: planetary motion
If we consider the Sun and Earth with masses m1 = M
and m2 = m respectively. M ¿¿ m so we can approximate m1 to be stationary ∴ gravity due to the Sun
is a central field of force about a fixed origin, with
G
F = − mM
|r|2 r̂ where r is the position of the Earth relative to the stationary ”Sun”.
Introduce a scalar force f (r) s.t F = mf (r)r̂
with f (r) = − rγ2 , |r| = r and γ = M G
Kepler I: elliptical paths subbing f (r) above into the
path equation for a central field of force you get
γ
d2 u
2 which is a linear, inhomogeneous, condθ 2 + u = HO
stant coefficient ODE for u(θ).
Resulting in u(θ) = A cos(θ) + B sin(θ) + Hγ2
relative to O.) None inertial frame of reference when
b̈ ̸= 0.
2.4
2D rotating frames of reference
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 an origin O, but which is rotating
relative to {i, j, k}. Assuming 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:
e1 = cos(θ)i + sin(θ)j
e2 = − sin(θ)i + cos(θ)j
e3 = k.
O
βγ
=⇒ u1 = Hγ2 (1 − β cos(θ − α)) where H
We’re interested in the rate of change of the basis
2 = C is a conO
O
stant. Which is an ellipse in plane polar coordinates. vectors {e1 , e2 , e3 }, which is easy to determine via:
Showing the orbit of earth around the sun is elliptical. e˙1 = de1 = e1 θ̇ = θ̇(− sin(θ)i + cos(θ)j)
dt
dθ
de
2.2
CFoF: practical application
A particle P in an elliptical orbit of semi-major axis
a (or equivalently major axis 2a) has an energy of
G
E = −M
(from Kepler’s 3rd law) ∴ we can apply
2a
the conservation of energy equation:
(U (r))2
−M G
− MrG
2a = E =
2
where U (r) denotes the speed (ṙ2 + r2 θ̇2 )1/2 of P at
radius r.
2.3
e
e˙2 = dt2 = dθ2 θ̇ = θ̇(− cos(θ)i − sin(θ)j)
e˙3 = 0.
We not the above equations are equivalent to θ̇k × e1 ,
θ̇k × e2 , θ̇k × e3 respectively. Generally e˙i = ω × ei .
Where ω is the ’angular frequency vector’. The magnitude |ω| is then the rotation rate.
2.5
Velocity
frame
relative
to
a
rotating
Suppose S = inertial frame of reference. Further we
suppose that we wish to use an alternative frame of
reference S’ that rotates relative to S with an angular
frequency of ω. Relative to the rotating frame S’, we
3
X
know the particle’s position: r =
xi ei , that is in
Frames of Reference
Definition: An ”inertial frame of reference” is a coordinate system in which Newton’s laws hold.
Motion relative to translating origin
i=1
Suppose that we have an inertial frame of reference
centered about origin O. By the definition above we terms of three coordinates x1,2,3 in the directions of
know particle P, mass m and position vector r(t) (rel- e1,2,3 .
The velocity relative to the inertial frame of reference
ative to O) satisfies F = mr̈.
is then the rate of change
! of the position vector:
3
X
d
ṙ|S = dr
xi ei however the three basis vecdt = dt
i=1
tors e1,2,3 change with time as the coordinate system
rotates. Therefore
3
3
X
X
dr ′
ṙ|S =
(ẋi ei + xi e˙i ) =
|S + ω ×
xi ei =
dt
i=1
i=1
O is an origin in ṙ| ′ + ω × r. The velocity relative to the frame S is
S
the inertial frame of reference, O′ is moving relative equal to the velocity relative to the rotating frame S’.
to O.
Suppose there is another coordinate system that is defined relative to moving origin O′ . Further suppose 2.6 ODEs
that the position O′ relative to O is b(t) and that the
dy
Existence and uniqueness theorem: dx
= f (x, y). Theposition vector of P relative to O′ is r̄(t)
∂f (x,y)
∴ r(t) = b(t) = r̄(t). Therefore Newton’s second law orem: If f (x, y) and ∂y are continuous functions of
2
reduces to m ddt2r̄ = F − mb̈. Relative to the moving x and y in a region |x − X| < a and |y − Y | < b where
frame of reference centred at O′ , the particle feels an a, b > 0, then ∃ exactly one solution to the initial
effective force of F − mb̈, Newton’s second law only value problem in an interval |x − X| < h ≤ a, where
holds if b̈ = 0 (if O′ is moving at constant velocity h > 0.
2