MechanicsPhysics 151

Lecture 10Rigid Body Motion

(Chapter 5)

What We Did Last Time

! Found the velocity due to rotation! Used it to find Coriolis effect

! Connected ω with the Euler angles! Lagrangian " translational and rotational parts

! Often possible if body axes are defined from the CoM

! Defined the inertia tensor! Calculated angular momentum

and kinetic energy

s r

d ddt dt = + ×

ω

( )2jk i i jk ij ikI m r x xδ= −

=L Iω 212 2

T Iω⋅= =ω Iω

Diagonalizing Inertia Tensor

! Inertia tensor I can be made diagonal

( )2 2

2 2 2

2 2

( )( )

( )

i i i i i i i i i

i i i i i i i i i i i jk ij ik

i i i i i i i i i

m r x m x y m x zm y x m r y m y z m r x xm z x m z y m r z

δ − − − = − − − = − − − −

I

1

2

3

0 00 00 0

D

II

I

= =

I RIR! R is a rotation matrix0iI >

Kinematical properties of a rigid body are fully described by its mass, principal axes, and moments of inertia

Principal Axes

! Consider a rigid body with body axes x-y-z! Inertia tensor I is (in general) not diagonal! But it can be made diagonal by

! Rotate x-y-z by R " New body axes x’-y’-z’! In x’-y’-z’ coordinates

D =I RIR!

′ =ω Rω

D

′ ==

=′=

L RLRIωRIRRωI ω

!=RR 1!

One can choose a set of body axes that make the inertia tensor diagonal" Principal AxesHow do I

find them?

Finding Principal Axes

! Consider unit vectors n1, n2, n3 along principal axes

! To find the principal axes and principal moments:! Express I in in any body coordinates! Solve eigenvalue equation

! Eigenvectors point the principal axes! Use them to re-define the body coordinates to simplify I

! You can often find the principal axes by just looking at the object

i i iI=In n

1

2

3

0 00 00 0

D

II

I

= =

I RIR!

ni is an eigenvector of I with eigenvalue Ii

( ) 0λ− =I r 0λ− =I 1 2 3, ,I I Iλ =

Rotational Equation of Motion

! Concentrate on the rotational motion! Newtonian equation of motion gives

! Take the principal axes as the body axes

s

ddt

=

L N

“space” axes

b

ddt

+ =

L ω×L N

“body” axes

1 1 1 1

2 2 2 2

3 3 3 3

0 00 00 0

I II I

I I

ω ωω ωω ω

= =

L = Iω

s b

d ddt dt = + ×

ω

Euler’s Equation of Motion

! Special cases:!

!

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

I I Nd I I Ndt

I I N

ω ω ωω ω ωω ω ω

+ × =

1 1 2 3 2 3 1

2 2 3 1 3 1 2

3 3 1 2 1 2 3

( )( )( )

I I I NI I I NI I I N

ω ω ωω ω ωω ω ω

− − =− − =− − =

"""

Euler’s equation of motion for rigid body with one point fixed

2 3 0ω ω= = 1 1 1I Nω ="

2 3I I= 1 1 1I Nω ="

Torque-Free Motion

! No linear force " Conservation of linear momentum! No torque " Conservation of angular momentum

! Try N = 0 in Euler’s equation of motion

! We will do something more intuitive (hopefully)! Geometrical trick by L. Poinsot

1 1 2 3 2 3

2 2 3 1 3 1

3 3 1 2 1 2

( ) 0( ) 0( ) 0

I I II I II I I

ω ω ωω ω ωω ω ω

− − =− − =− − =

"""

Integrating these equation will give us energy and angular momentum conservation

Inertia Ellipsoid

! For any direction n,! If we express n using principal axes x’-y’-z’

! Consider a vector

i ij jI n I n= ⋅ =n In

2 2 2 21 1 2 2 3 3i iI I n I n I n I n= = + +

I= nρ

Inertia ellipsoid2 2 2 2

1 1 2 2 3 31 i iI I I Iρ ρ ρ ρ= = + +

Moment of inertia about axis n

Inertia Ellipsoid

I= nρ

x′

y′

z′

11 I21 I

31 I

ρ

Inertia ellipsoid represents the moment of inertia of a rigid body in all directions

2 2 21 1 2 2 3 31 I I Iρ ρ ρ= + +

Usefulness of this definition will become

apparent soon …

Inertia Ellipsoid

! Inertia along axis n is

! F is a function (like potential) defined in the ρ space! F(ρ) = 1 " Inertia ellipsoid! Normal of the ellipsoid

given by the gradient

! Using

I= nρ

I = ⋅n In2 2 2

1 1 2 2 3 3( ) 1F I I Iρ ρ ρ≡ ⋅ = + + =ρ ρ Iρ

F = 1

ρ

F∇ ρ

2F∇ =ρ Iρ

2I T= =ρ n ω

2FT

∇ =ρ LSurface of the

inertia ellipsoid is perpendicular to L

Invariable Plane

! As ρ moves, inertia ellipsoid must rotate to satisfy this condition continuously

! Consider the projection of ρ on L

! The ellipsoid is touchinga fixed plane perpendicular to L

2FT

∇ =ρ L Surface of inertia ellipsoid at ρ is perpendicular to the angular momentum L

22

TL LL T⋅ ⋅= =ρ L ω L

constant

ρ

L

2TL

Invariable plane

L is conserved!

Invariable Plane

! Inertia ellipsoid touches the invariable plane! Distance between the center and the plane is

! Touching point = instantaneous axis of rotation! Tip of the ρ vector is momentarily at rest in space

! i.e. it’s not sliding, butrolling without slippingon the invariable plane

2T LDetermined by the initial conditions

ρ

02

ddt T

= = =ρ ω×ωω×ρ

Invariable Plane

! Inertia ellipsoid rolls on the invariable plane! Rotation of ellipsoid gives the rotation of the body! Direction of ρ gives the direction of ω in space

! Touching point draws curves! Curve drawn on the ellipsoid = polhode! Curve drawn on the invariable plane

= herpolhode

! Let’s examine a fewsimple cases

Simple Cases

! Inertia ellipsoid is a sphere (I1 = I2 = I3)! ρ is constant and parallel to L! Stable rotation

ρ

L

Simple Cases

! Initial axis is close to one of the principal axes! Assume I1 > I2 > I3

! Stable rotation around I1 and I3

! Not so obvious around I2

! If ω1 = ω3 = 0, ω2 is constant

! Small deviation leads to instability

11 I

31 I1 1 2 3 2 3

2 2 3 1 3 1

3 3 1 2 1 2

( ) 0( ) 0( ) 0

I I II I II I I

ω ω ωω ω ωω ω ω

− − =− − =− − =

"""

Simple Cases

! Since I1 > I2 > I3, distance allows a polhode that wraps around the inertia ellipsoid

! Rotation around a principal axis is stable except for the one with the intermediate moment of inertia

21 I

Simple Cases

! Inertia ellipsoid is symmetric around one axis! I1 = I2 ≠ I3

! ρ draws a cone (space cone) on the invariable plane! ρ draws a cone (body cone) in the inertia ellipsoid! Body cone rolls on the space cone

ρ

L

Precession

! I1 = I2 turns the Euler’s equation of motion to

! ω precesses around the I3 axis! Draws the body cone

1 1 2 3 1 3

1 2 3 1 3 1

3 3

( )( )

0

I I II I II

ω ω ωω ω ωω

= −= −=

"""

ω3 is constantConsider it as a given initial condition

1 2 2 1,ω ω ω ω= −Ω = Ω" " 3 13

1

I II

ω−Ω ≡

1 2cos , sinA t A tω ω= Ω = Ω

3ω ω

A

Rotation Under Torque

! We introduce torque! Things get messy

! Consider a spinning top! Define Euler angles

1 1 2 3 2 3 1

2 2 3 1 3 1 2

3 3 1 2 1 2 3

( )( )( )

I I I NI I I NI I I N

ω ω ωω ω ωω ω ω

− − =− − =− − =

"""

z

y

xφ

ψ

θ

Lagrangian

! Assume I1 = I2 ≠ I3! Kinetic energy given by! Use Euler angles

2 2 21 1 2 3 3

1 1( )2 2

T I Iω ω ω= + +

sin sin coscos sin sin

cos

φ ψ θ θ ψφ ψ θ θ ψ

φ θ ψ

+ = − +

ω

" "" "

" "

2 2 2 231 ( sin ) ( cos )2 2

IIT θ φ θ φ θ ψ= + + +" " " "

Lagrangian

! Potential energy is given by the height of the CoM

! Lagrangian is

! Finally we are in real business!! How we solve this?

! Note φand ψ are cyclic! Can define conjugate momenta that conserve

! To be continued …

l

θcosV Mgl θ=

2 2 2 231 ( sin ) ( cos ) cos2 2

IIL Mglθ φ θ φ θ ψ θ= + + + −" " " "

Summary

! Discussed rotational motion of rigid bodies! Euler’s equation of motion

! Analyzed torque-free rotation! Introduced the inertia ellipsoid! It rolls on the invariant plane! Dealt with simple cases

! Started discussing heavy top! Found Lagrangian " Will solve this next time

x′

y′

z′

11 I21 I

31 I

ρ

2 2 2 231 ( sin ) ( cos ) cos2 2

IIL Mglθ φ θ φ θ ψ θ= + + + −" " " "