Mechanics Rigid Body Motion 4, Lecture Notes - Physics, Study notes of Mechanics

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Mechanics

Physics 151

Lecture 11

Rigid Body Motion

(Chapter 5)

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Heavy Top „

Top is spinning on a fixed point

„

Lagrangian is

„

and

are cyclic

„

Symmetry

„

p

φ

and

p

ψ

2 are conserved^

2

2

2

3

1

sin

cos

cos

I^2

I

L

Mgl

z

y

x

φ

ψ

θ

Conserved Momenta

„

Solve them for

and

„

We need

t ) to get

t ) and

t

2

2

2

2

3

1

sin

cos

cos

I^2

I

L

Mgl

2

1

3

1

sin

cos

cos

const.

L

I

p

I b

I

φ

3

3

3

1

co

con

s

st.

p

I a

L

I

I

ψ

cos^2

sin

b

a

1

2

3

cos

cos

sin

I a

b

a

I

Got rid of 2 degrees of freedom

1-D Equation of Motion „

Simplify the equation of motion by defining

„

Switch variable from

to

u

= cos

„

Integrate

2

3

3

1

1

and

E

I

Mgl

I

I

2

2

cos

cos

sin

b

a

2

2

2

u

u

u

b

au

( )

2

2

(0)

u t u

du

t

u

u

b

au

∫

Ellipticintegral

EoM becomesEoM

Qualitative Behavior „

Try to extract qualitative behavior

„

Same way as we did with central force problem

„

Consider the

RHS

of the last equation

„

Physical range is

and

„

f (

u

) is a cubic function of

u

with

„

2

2

2

3

2

2

2

u

f u

u

u

b

au

u

a

u

ab

u

b

α

β

β

α

β

α

2

f u

u

u

1 2

MglI

β ≡

2

f

b

au

These conditions constrain

the shape of

f

( u

)

cos

u

Nutation „

Consider the sign of

„

changes sign at „ „

2

2

cos

sin

b

a

b

au u

u

u

b

a

1

2

or

u

u

u

u

′^

1

2

u

u

u

is monotonous

switches direction

locus

Initial Condition „

Suppose the figure axis is initially at rest

„

Spin the top, then release it “quietly”

„ „

0

t

=

0

t

f u

=

0

1

2

or

t u

u

u

=

0

t

=

0

t

b

au

=

0 t u

u

=

t^

= 0

„

Initially, the figure axis falls

„

It then picks up precession in

„

How does it know which way to go?

Uniform Precession „

Can we make a top precess without bobbing?

„

i.e.

„

We need to have a double root for

f

u

const

( ) f u

u

1

1 −

0 u

3 u

2

2

0

0

0

0

f u

u

u

b

au

2

0

0

0

0

0

f

u

u

u

u

a b

au

′^

Combine

2

0

a

u

1

3

3

I a

I

1

Mg

l

I

3

3

1

0

cos

Mgl

I

I

cos^2

sin

b

a

Uniform Precession

„

For any given value of

3

and cos

0

, you must give exactly

the right “push” in

to achieve uniform precession

„

Quadratic equation

Æ

2 solutions

„

Same top can do “fast” or “slow” precession

„

For the solutions to exist

„

Uniform precession is achieved only by a fast top

3

3

1

0

cos

Mgl

I

I

2

2

3

3

1

0

cos

I

MglI

3

1

0

3 2

cos

MglI

I

Magnetic Dipole Moment

„

Using

„

Explicit calculation using polar coordinates

„

Take time average

Æ

Assume rotation is fast

i^

i^

i^

i^

i^

i

q

q

m

m

m

m

×

×

×

N

r

v

B

ω

r

r

B

i^

i

v

ω

× r

ω

B

i r

2

sin

sin

cos

(sin

cos

sin

cos

cos

i^

i^

r Bi

×

⋅^

ω

r

r

B

(

)

2

sin

i^

i

q

q

m

r

m

m

N

ω

× B

L × B

Magnetic Dipole Moment

„

Magnetic dipole

M

in

B

feels the torque

„

Fast spinning charged rigid body has a magneticmoment

„

Equation of motion

„

This makes

L

to precess around

B

„

Angular velocity of precession is

d dt

L

L × B

precess

q m

ω

B

B

q m

N

L × B

N

M × B

M

L

q m

gyromagnetic ratio

B

L

Larmor frequency

Anomalous Magnetic Moment „

μ

of electron and muon known very accurately

„

Not pure Dirac particles, but surrounded by thin cloud ofvirtual particles due to quantum fluctuation

„

Measurement uses spin precession

„

Store particles with known spin orientation in B field

„

Measure spin direction after time

t

precess

g 2

q m

ω

B

Need to know B very accurately

electron

g

muon

g

Muon g–2 Experiment

BNL E-821 muon storage ring

muon

g