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Lecture 18 of a physics course, covering the topics of symmetric top and oscillations. The lecture begins with an example of a symmetric top, discussing its characteristic motions and the euler equations of motion. The lagrangian approach is then introduced as a simpler solution for the symmetrical top problem. The document also covers the concept of integrals of motion and their significance in the context of symmetric top. The lecture concludes with a brief discussion on oscillations, their formalism, and potential and kinetic energy expansions.
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Symmetric Top example (Section 5.7)
Oscillations, theory of (Section 6.1/6.2)
last time... gave us Euler’s equations of motion:
1
˙ω 1 − ω 2 ω 3 ( I 2 − I 3
1
2
˙ω 2 − ω 3 ω 1 ( I 3 − I 1
2
3
˙ω 3 − ω 1 ω 2 ( I 1 − I 2
3
Continuing the dumbbell example, if
ω
then
given
ω
and remembering
~ω
, ω
sin
(^) θ, ω
cos
(^) θ
]
Then only one Torque component is non-zero N x = − ω y ω z
2
−
3 ) =
− ( m 1 + m 2 ) b 2 ω 2
sin
θ
cos
(^) θ
ie spins the dumbbell around in the
x
-direction
For gyroscope and spinning top
ψ
θ
φ
now
3
(since symmetry along
z )
from Euler’s eqns, torque with
1 , N
2
won’t change
ω
3
ω 3 − ω 1 ω 2 ( I 1 − I 2
3
˙ω 3
initial torque
3
=
2
= 0
only
1
6 = 0
and initial rotation
ω
1
=
ω
2
= 0
only
ω
3
6 = 0 =
constant
But time evolution has
ω
1
and
ω
2
changing
ie complicated motion in terms of
ψ,
φ,
θ˙
A (somewhat) easier solution from the Lagrangian...
2 1 (^) I 1 ( ω
(^12)
ω
(^22) (^) ) +
2 1 (^) I 3 ω
(^32)
Now, Goldstein 4.87 has exactly these 3
ω
’s in terms of
~ω
ω
x ′
ω
y ′
ω
z ′
φ
(^) sin
(^) θ
sin
(^) ψ
θ˙
cos
ψ
φ
(^) sin
(^) θ
cos
(^) ψ
θ˙
sin
(^) ψ
φ
cos
(^) θ
ψ
and in terms of Euler angle’s
mgl
cos
(^) θ
and
1
θ˙ 2
φ
2
sin
2 θ ) + I 3
ψ
φ
(^) cos
(^) θ
)
2
giving a Lagrangian with two cylic coords (
φ, ψ
ie the
p
φ
, p
ψ
components are constant in time
Given that the
ω
3
is constant in time
this means that the fourth constant of motion really is
3 ω
(^32)
1
θ 2
1
b
−
a
cos
(^) θ
) 2
sin
2
θ
mgl
(^) cos
(^) θ
1
˙
θ 2
′ ( θ )
this can be solved for
θ ( t )
, but it’s complicated
darkness (ie. elliptical integrals).we won’t solve it here since it just leads down into
Goldstein 5.7There are limited regimes of soln see remainder
mounted such that centre of mass is fixed...Gyroscopes pg 222,223... are rapidly rotating devices
interested in motion about equilibrium
system is in equilibrium when generalised forces
i
=
∂q ∂V
i )
0
= 0
ie extremum at equilibrium
q 01
, q
02
,... , q
on
As per usual equilibrium classified stable/unstable
equilibrium:in terms of the deviations of system energy from
In the distant past
~v i
=
d~ r i
dt
j
∂~ r i
∂q
j
˙
q j
∂~ r i
∂t
(see Goldstein equations 1.71 and 1.72 on page 25)
our generalised coords don’t explicitly depend on time:
i
m
i v i 2
=
i
m
i (
j
r i
∂q
j
˙q j )
2
=
i,j
ij
q i ˙
q j
The coeffs
ij
are functions of
q k ,
ij (^) ( q 1 , q
2 , .., q
n ) =
m
ij (^) ( q 01
, q
02
, .., q
on
∂m
ij
∂q
k ) 0 η k +
only include constants since
has quadratic terms in
˙q i
so we have
ij
m
ij (^) ( q 01
, q
02
, .., q
on
ij
and
m
ij
η i ˙
η j
=
ij
η i ˙
η j