Lecture 18 Outline: Symmetric Top and Oscillations - Prof. M. W. Bromley, Study notes of Physics

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.

Typology: Study notes

Pre 2010

Uploaded on 03/28/2010

koofers-user-24v
koofers-user-24v 🇺🇸

10 documents

1 / 12

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Lecture 18 Outline - transit 5 to 6
Symmetric Top example (Section 5.7)
Oscillations, theory of (Section 6.1/6.2)
pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

Download Lecture 18 Outline: Symmetric Top and Oscillations - Prof. M. W. Bromley and more Study notes Physics in PDF only on Docsity!

• Lecture 18 Outline - transit 5 to 6

Symmetric Top example (Section 5.7)

Oscillations, theory of (Section 6.1/6.2)

Euler Equations of Motion contd

last time... gave us Euler’s equations of motion:

I

1

˙ω 1 − ω 2 ω 3 ( I 2 − I 3

N

1

I

2

˙ω 2 − ω 3 ω 1 ( I 3 − I 1

N

2

I

3

˙ω 3 − ω 1 ω 2 ( I 1 − I 2

N

3

Continuing the dumbbell example, if

ω

then

given

ω

and remembering

= [

, ω

sin

(^) θ, ω

cos

(^) θ

]

Then only one Torque component is non-zero N x = − ω y ω z

I

2

I

3 ) =

− ( m 1 + m 2 ) b 2 ω 2

sin

θ

cos

(^) θ

ie spins the dumbbell around in the

x

-direction

Symmetrical Top - Eulerian Way

For gyroscope and spinning top

ψ

θ

φ

now

I 1 = I 2 6

I

3

(since symmetry along

z )

from Euler’s eqns, torque with

N

1 , N

2

won’t change

ω

3

N 3 = I 3 ˙

ω 3 − ω 1 ω 2 ( I 1 − I 2

I

3

˙ω 3

initial torque

N

3

=

N

2

= 0

only

N

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˙

Symmetrical Top - Lagrangian Way

A (somewhat) easier solution from the Lagrangian...

T = T 1 + T 2 + T 3 =

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

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

V

mgl

cos

(^) θ

and

T

I

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

Applications

Given that the

ω

3

is constant in time

this means that the fourth constant of motion really is

E

E

I

3 ω

(^32)

I

1

θ 2

I

1

b

a

cos

(^) θ

) 2

sin

2

θ

mgl

(^) cos

(^) θ

I

1

˙

θ 2

V

′ ( 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

Oscillations formalism - intro

interested in motion about equilibrium

system is in equilibrium when generalised forces

Q

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

eg. 2-D isotropic SHO

Oscillations formalism - kinetic

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:

T

i

m

i v i 2

=

i

m

i (

j

r i

∂q

j

˙q j )

2

=

i,j

M

ij

q i ˙

q j

The coeffs

M

ij

are functions of

q k ,

M

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

T

has quadratic terms in

˙q i

so we have

M

ij

m

ij (^) ( q 01

, q

02

, .., q

on

T

ij

and

T

m

ij

η i ˙

η j

=

T

ij

η i ˙

η j