Mechanics
Physics 151
Lecture 10
Rigid Body Motion
(Chapter 5)
What We Did Last Time
!
Found the velocity due to rotation
!
!
!
Connected ω with the Euler angles
Lagrangian " translational and rotational parts
!
!
Used it to find Coriolis effect
d d
  =   + ω×
 dt  s  dt r
Often possible if body axes are defined from the CoM
Defined the inertia tensor
!
Calculated angular momentum
and kinetic energy
I jk = mi ( ri 2δ jk − xij xik )
ω ⋅ Iω 1 2
L = Iω T =
= Iω
2
2
Diagonalizing Inertia Tensor
 mi (ri 2 − xi2 )
− mi xi yi
−mi xi zi 


2
2
−mi yi zi  = mi ( ri 2δ jk − xij xik )
mi (ri − yi )
I =  −mi yi xi
2
2 
 −mi zi xi
−
−
m
z
y
m
(
r
z
i i i
i i
i )

!
Inertia tensor I can be made diagonal
 I1
! = 0
I D = RIR

 0
0
I2
0
0
0 
I 3 
R is a rotation matrix
Ii > 0
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 I D = RIR
Rotate x-y-z by R " New body axes x’-y’-z’
!
In x’-y’-z’ coordinates
ω′ = Rω L′ = RL
= RIω
!
= RIRRω
! =1
RR
= I D ω′
How do I
find them?
One can choose a set of
body axes that make the
inertia tensor diagonal
" Principal Axes
Finding Principal Axes
!
0
0 
I 3 
ni is an eigenvector of I with eigenvalue Ii
To find the principal axes and principal moments:
!
!
!
!
!
0
I2
0
Consider unit vectors n1, n2, n3 along principal axes
Ini = I i ni
!
 I1
! = 0
I D = RIR

 0
Express I in in any body coordinates
Solve eigenvalue equation
λ = I1 , I 2 , I 3
I−λ = 0
(I − λ )r = 0
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
Rotational Equation of Motion
!
Concentrate on the rotational motion
!
Newtonian equation of motion gives
 dL 
 dL 
N
=
+ ω×L = N




 dt  s
 dt b
“space” axes
!
d d
  =   + ω×
 dt  s  dt b
“body” axes
Take the principal axes as the body axes
 I1
L = Iω =  0
 0
0
I2
0
0  ω1   I1ω1 
0  ω 2  =  I 2ω 2 
I 3  ω3   I 3ω3 
Euler’s Equation of Motion
 I1ω1  ω1   I1ω1   N1 
d 
 + ω  ×  I ω  =  N 
I
ω
2 2
 2  2 2  2
dt 
 I 3ω3  ω3   I 3ω3   N 3 
I1ω"1 − ω 2ω3 ( I 2 − I 3 ) = N1
I 2ω" 2 − ω3ω1 ( I 3 − I1 ) = N 2
I 3ω" 3 − ω1ω 2 ( I1 − I 2 ) = N 3
Euler’s equation of motion for rigid body
with one point fixed
!
Special cases:
! ω 2 = ω3 = 0
I1ω"1 = N1
I 2 = I3
I1ω"1 = N1
!
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
I1ω"1 − ω 2ω3 ( I 2 − I 3 ) = 0
I 2ω" 2 − ω3ω1 ( I 3 − I1 ) = 0
I 3ω" 3 − ω1ω 2 ( I1 − I 2 ) = 0
!
Integrating these equation will give us
energy and angular momentum
conservation
We will do something more intuitive (hopefully)
!
Geometrical trick by L. Poinsot
Inertia Ellipsoid
!
For any direction n, I = n ⋅ In = ni I ij n j
!
Moment of inertia
about axis n
If we express n using principal axes x’-y’-z’
I = I i ni2 = I1n12 + I 2 n22 + I 3 n32
! Consider a vector ρ =
n
I
1 = I i ρi2 = I1 ρ12 + I 2 ρ 22 + I 3 ρ32
Inertia ellipsoid
Inertia Ellipsoid
n
ρ=
I
1
I3
1 = I1 ρ12 + I 2 ρ 22 + I 3 ρ32
Inertia ellipsoid represents
the moment of inertia of a
rigid body in all directions
z′
ρ
y′
x′
1
1
I1
I2
Usefulness of this
definition will become
apparent soon …
n
ρ=
I
Inertia Ellipsoid
!
Inertia along axis n is I = n ⋅ In
F (ρ) ≡ ρ ⋅ Iρ = I1 ρ12 + I 2 ρ 22 + I 3 ρ32 = 1
!
F is a function (like potential) defined in the ρ space
!
!
!
F(ρ) = 1 " Inertia ellipsoid
Normal of the ellipsoid
given by the gradient
∇ρ F = 2Iρ
Using ρ = n
2
∇ρ F =
L
T
I =ω
2T
Surface of the
inertia ellipsoid is
perpendicular to L
∇ρ F
ρ
F=1
Invariable Plane
2
∇ρ F =
L
T
!
!
L is conserved!
Surface of inertia ellipsoid at ρ is
perpendicular to the angular momentum L
As ρ moves, inertia ellipsoid must rotate to satisfy this
condition continuously
Consider the projection of ρ on L
ρ⋅L ω⋅L
2T
=
=
L
L
L 2T
!
constant
The ellipsoid is touching
a fixed plane perpendicular to L
Invariable plane
ρ
2T
L
L
Invariable Plane
!
Inertia ellipsoid touches the invariable plane
!
Distance between the center and the plane is
2T L
Determined by the initial conditions
!
Touching point = instantaneous axis of rotation
!
!
Tip of the ρ vector is momentarily at rest in space
dρ
ω ×ω
= ω×ρ =
=0
dt
2T
i.e. it’s not sliding, but
ρ
rolling without slipping
on the invariable plane
Invariable Plane
!
Inertia ellipsoid rolls on the invariable plane
!
!
!
Touching point draws curves
!
!
!
Rotation of ellipsoid gives the rotation of the body
Direction of ρ gives the direction of ω in space
Curve drawn on the ellipsoid = polhode
Curve drawn on the invariable plane
= herpolhode
Let’s examine a few
simple 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
I1ω"1 − ω 2ω3 ( I 2 − I 3 ) = 0
I 2ω" 2 − ω3ω1 ( I 3 − I1 ) = 0
I 3ω" 3 − ω1ω 2 ( I1 − I 2 ) = 0
!
Small deviation leads to instability
1
1
I3
I1
Simple Cases
!
!
Since I1 > I2 > I3, distance 1 I 2 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
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
I1ω"1 = ω 2ω3 ( I1 − I 3 )
I1ω" 2 = ω3ω1 ( I 3 − I1 )
I 3ω" 3 = 0
ω3 is constant
Consider it as a given initial condition
ω"1 = −Ωω 2 , ω" 2 = Ωω1
Ω≡
I 3 − I1
ω3
I1
ω1 = A cos Ωt , ω 2 = A sin Ωt
!
ω precesses around the I3 axis
!
Draws the body cone
A
ω3
ω
Rotation Under Torque
!
!
!
I1ω"1 − ω 2ω3 ( I 2 − I 3 ) = N1
I 2ω" 2 − ω3ω1 ( I 3 − I1 ) = N 2
We introduce torque
Things get messy
I 3ω" 3 − ω1ω 2 ( I1 − I 2 ) = N 3
Consider a spinning top
!
Define Euler angles
z
θ
y
φ
ψ
x
Lagrangian
!
Assume I1 = I2 ≠ I3
!
!
1
1
2
2
Kinetic energy given by T = I1 (ω1 + ω 2 ) + I 3ω32
2
2
Use Euler angles
φ" sin ψ sin θ + θ" cosψ 


ω = φ" cosψ sin θ − θ" sin ψ 
" cos θ + ψ"


φ


I3 "
I1 " 2 "2 2
T = (θ + φ sin θ ) + (φ cos θ + ψ" ) 2
2
2
Lagrangian
!
!
Potential energy is given by the height of the CoM
V = Mgl cos θ
Lagrangian is
θ
I3 "
I1 " 2 "2 2
L = (θ + φ sin θ ) + (φ cos θ + ψ" ) 2 − Mgl cos θ
2
2
!
!
Finally we are in real business!
How we solve this?
! Note φ and ψ are cyclic
!
!
Can define conjugate momenta that conserve
To be continued …
l
Summary
!
Discussed rotational motion of rigid bodies
!
!
Analyzed torque-free rotation
!
!
!
!
Euler’s equation of motion
Introduced the inertia ellipsoid
It rolls on the invariant plane
Dealt with simple cases
1
I3
z′
ρ
y′
x′
1
1
Started discussing heavy top
!
Found Lagrangian " Will solve this next time
I3 "
I1 " 2 "2 2
L = (θ + φ sin θ ) + (φ cos θ + ψ" ) 2 − Mgl cos θ
2
2
I1
I2