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PC2132 CLASSICAL MECHANICS
AY2005/2006 Semester 2
Part II
x = a cos ωt + b sin θ
y = a sin ωt − b cos θ
x˙ = −aω sin ωt + b
θ cos θ
y ˙ = aω cos ωt + b
θ sin θ
v
2 = x˙
2
2
= a
2
ω
2
sin
2
ωt − 2 abω
θ sin ωt cos θ + b
2 ˙ θ
2
cos
2
θ + a
2
ω
2
cos
2
ωt + 2abω
θ cos ωt sin θ + b
2 ˙ θ
2
sin
2
θ
= a
2 ω
2
θ
2
θ sin(θ − ωt).
T =
mv
2
=
m[a
2
ω
2
2 ˙ θ
2
θ sin(θ − ωt)].
V = mgy = mg(a sin ωt − b cos θ).
L = T − V = m
[
a
2
ω
2
b
2 ˙ θ
2
θ sin(θ − ωt) − ga sin ωt − gb cos θ
]
∂L
∂θ
= m
[
abω
theta cos(θ − ωt) + gb sin θ
]
∂L
θ
= m
[
b
2 ˙ θ + abω sin(θ − ωt)
]
Using the Euler-Lagrange equation,
d
dt
(
∂L
θ
)
∂L
∂θ
b
2 ¨ θ + abω(
θ − ω) cos(θ − ωt) − abω
θ cos(θ − ωt) + gb sin θ = 0,
θ =
b
2
[
abω
2
cos(θ − ωt) − gb sin θ
]
2a. Since potential energy can be arbitrarily defined, we can set K = 0.
mgz −
m
ω
2
r
2
= 0 ,
ω =
√
2 gz
r
2
= 15 .3 rad s
− 1 .
2b. Defining i: east, j: north and k: up, we have ˙x = 0, ˙y = −800 cos 37
◦ and ˙z = 800 sin 37
◦ − gt.
Using the equation of motion for Coriolis force,
mx¨ = 2 mω( ˙y cos θ − z˙ sin θ)
= 2 mω(−800 cos 3
◦ − gt sin 40
◦ ).
x ˙ = − 1600 ωt cos 3
◦
− gωt
2
sin 40
◦
.
x = − 800 ωt
2 cos 3
◦ −
gωt
3 sin 40
◦
. (1)
The constants of integration for ˙x and x can be dropped because ˙x = 0 initially and by re-defining
coordinate origin respectively. To find t, we solve for z = 0.
z˙ = 800 sin 37
◦
− gt
z = 800t sin 37
◦ −
gt
2 = 0
t = 0 or t =
1600 sin 37
◦
g
Substituting this into (1) and solving, one will find a deflection of 706 m to the west.
3a.
T
i
+ V
i
= T
f
+ V
f
For minimum escape speed, T f
= V
f
Ti = −Vi,
mv
2
esc
Gm E
m
r
v esc
√
2 Gm E
r
= 1. 07 × 10
4 m s
− 1 .
3b. By substituting θ = 0
◦ , r = (300 + 6372) km and θ = 180
◦ , r = (3500 + 6372) km into the equation
l
r
= 1 + cos θ, we get
l
l
Solving (2) and (3) for , we get
6672
9872
6672
9872
(i) Solving (2) and (3) for l, we have l = 7962 km.
l
r
= 1 + cos θ,
r + 6372
= 1 + 0.193 cos 90
◦
,
r = 1590 km.
(ii)
a =
(300 + 6372 + 3500 + 6372) = 8272 km.
