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Mechanics
Physics 151
Lecture 9
Rigid Body Motion
(Chapter 4, 5)
What We Did Last Time
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Discussed 3-dimensional rotation
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Looked for ways to describe 3-d rotation
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Preparation for rigid body motion
„ Movement in 3-d + Rotation in 3-d = 6 coordinates
Euler angles one of the many possibilities
Euler’s theorem
Defined infinitesimal rotation dΩ
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dΩ = nd Φ
dr = r × dΩ
Commutative (unlike finite rotation)
Behaves as an axial vector (like angular momentum)
Goals For Today
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Time derivatives in rotating coordinate system
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Try to write down Lagrangian for rigid body
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To calculate velocities and accelerations
Coriolis effect
Express angular velocity using Euler angles
Separate rotation from movement of CoM
Define inertia tensor
Will almost get to the equation of motion
Body Coordinates
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Consider a rotating rigid body
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z
z′
Define body coordinates (x’, y’, z’)
y′
Between t and t + dt, the body
coordinates rotate by dΩ = nd Φ
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The direction is CCW because we
are talking about the coordinate axes
Observed in the space coordinates,
any point r on the body moves by
dr = d Φn × r = dΩ × r
Sign opposite from last lecture
because the rotation is CCW
y
x
n
x′
dΦ
dr
r
General Vectors
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Now consider a general vector G
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How does it move in space/body coordinates?
i.e. what’s the time derivative dG/dt ?
Movement dG differs in space and body coordinates
because of the rotation of the latter
( dG )space = ( dG )body + ( dG )rot
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Difference is due to rotation
If G is fixed to the body
( dG )body = 0
and
( dG )space = dΩ × G
( dG )rot = dΩ × G
Generally true
Angular Velocity
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For any vector G ( dG )space = ( dG )body + dΩ × G
⎛ dG ⎞
⎛ dG ⎞
=⎜
⎜
⎟
⎟ + ω×G
⎝ dt ⎠space ⎝ dt ⎠ body
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ωdt = dΩ = d Φn
ω = instantaneous angular velocity
„ Direction = n = instantaneous axis of rotation
„ Magnitude = dΦ/dt = instantaneous rate of rotation
Since this works for any vector, we can say
⎛d⎞ ⎛d⎞
⎜ ⎟ = ⎜ ⎟ + ω×
⎝ dt ⎠ s ⎝ dt ⎠ r
space coordinates
rotating coordinates
Coriolis Effect
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⎛d⎞ ⎛d⎞
⎜ ⎟ = ⎜ ⎟ + ω×
⎝ dt ⎠ s ⎝ dt ⎠ r
Imagine a particle observed in a rotating system
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e.g. watching an object’s motion on Earth
Velocity: v s = v r + ω × r
⎛ dv s ⎞ ⎛ dv s ⎞
Acceleration: a s = ⎜
⎟ =⎜
⎟ + ω × vs
⎝ dt ⎠ s ⎝ dt ⎠ r
= a r + 2ω × v r + ω × (ω × r )
Newton’s equation works in the space (inertial) system,
i.e. F = ma s
ma r = Feff = F − 2m(ω × v r ) − mω × (ω × r )
Object appears to move according to this force
Coriolis Effect
ma r = Feff = F − 2m(ω × v r ) − mω × (ω × r )
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ω
Last term is centrifugal force
vr
Middle term is Coriolis effect
„ Occurs when object is moving in the
rotating frame
−ω × v r
Moving objects on Earth appears to deflect toward right in
the northern hemisphere
„ Hurricane wind pattern
L
„ Foucault pendulum
Coriolis Effect
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v
Free-falling particle
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Drop a ball and ignore air resistance
Coriolis effect is −2m(ω × v )
„ Pointing east (out of screen) Æ x axis
Ignoring the centrifugal force
⎧ mx = −2mω vz sin θ
⎨
⎩ mz = − mg
x=
x=
„
ω gt 3 sin θ
3
ω sin θ
3
gt 2
,z=−
2
(−2 z )3
g
For sinθ = 1 and z = 100 m Æ x = 2.2 cm
ω
z
θ
Euler Angles
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Use angular velocity ω to calculate particles’ velocities
Use Euler angles to describe the rotation of rigid bodies
How are they connected?
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Infinitesimal rotations can be added like vectors
ω = n φ + n θ + n ′ψ
ξ
z
zζ
nz
x φ ξ
ζ′
η
η′
z
θ
nξ
y
x
z
ξ′
z y′
z′
y
n z′
x
ψ
x′
y
Euler Angles
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Let’s express ω in x’-y’-z’
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D
Doing this in x-y-z equally easy (or difficult)
Must express nz, nξ, nz’ in x’-y’-z’
„ nz Æ Anz, nξ Æ Bnξ (nz’ is OK)
zζ
nz
x φ ξ
C
ζ′
η
η′
z
θ
nξ
y
x
B
ξ′
z y′
z′
y
n z′
x
ψ
x′
y
Euler Angles
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From the last lecture
⎡ cos φ
D = ⎢⎢ − sin φ
⎢⎣ 0
sin φ
cos φ
0
0⎤
0
⎡1
0 ⎥⎥ C = ⎢⎢0 cos θ
⎢⎣0 − sin θ
1 ⎥⎦
⎡ cosψ cos φ − cos θ sin φ sinψ
A = ⎢⎢ − sin ψ cos φ − cos θ sin φ cosψ
⎢⎣
sin θ sin φ
0 ⎤
⎡ cosψ
sin θ ⎥⎥ B = ⎢⎢ − sin ψ
⎢⎣ 0
cos θ ⎥⎦
cosψ sin φ + cos θ cos φ sinψ
− sin ψ sin φ + cosθ cos φ cosψ
− sin θ cos φ
sin ψ
cosψ
0
0⎤
0 ⎥⎥
1 ⎥⎦
sinψ sin θ ⎤
cosψ sin θ ⎥⎥
cosθ ⎥⎦
⎡0 ⎤ ⎡ sin ψ sin θ ⎤
⎡1 ⎤ ⎡ cosψ ⎤
⎡0⎤
n z = A ⎢⎢0 ⎥⎥ = ⎢⎢cosψ sin θ ⎥⎥ nξ = B ⎢⎢0 ⎥⎥ = ⎢⎢ − sin ψ ⎥⎥ n z′ = ⎢⎢0 ⎥⎥
⎢⎣1 ⎥⎦ ⎢⎣ cos θ ⎥⎦
⎢⎣0 ⎥⎦ ⎢⎣ 0 ⎥⎦
⎢⎣1 ⎥⎦
Euler Angles
⎡φ sin ψ sin θ + θ cosψ ⎤
⎢
⎥
ω = n zφ + nξθ + n z′ψ = ⎢φ cosψ sin θ − θ sin ψ ⎥
cos θ + ψ
⎢
⎥
φ
⎣
⎦
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Remember: this is in the x’-y’-z’ system
Now we know how to express velocities in terms of
time-derivatives of Euler angles
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We can write down the Lagrangian
Kinetic Energy
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Kinetic energy of multi-particle system is
1
1
2
T = Mv + mi vi′2
2
2
Motion of CoM
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Remember Einstein convention
Motion around CoM
If we define the body axis from the center of mass
1
T = M ( x 2 + y 2 + z 2 ) + T ′(φ,θ,ψ )
2
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T’ depends only on the angular velocity
Must be a 2nd order homogeneous function
Potential Energy
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Potential energy can often be separated as well
V = V1 ( x, y, z ) + V2 (φ ,θ ,ψ )
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Uniform gravity g
V1 = −g ⋅ r
Uniform magnetic field B and magnetic dipole moment M
V2 = −M ⋅ B
Lagrangian can then be written as
L = Lt ( x, y, z , x , y , z ) + Lr (φ ,θ ,ψ , φ,θ,ψ )
Translational
Rotational
It is often possible to separate the translational and rotational motions by
taking the center of mass as the origin of the body coordinate axes
Rotational Motion
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We concentrate on the rotational part
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Translational part same as a single particle Æ Easy
Consider total angular momentum L = mi ri × vi
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vi is given by the rotation ω as v i = ω × ri
L = mi ri × (ω × ri )
⎡ mi (ri 2 − xi2 )
− mi xi yi
− mi xi zi ⎤
⎢
⎥
2
2
2
⎡
⎤
mi (ri − yi )
= mi ⎣ωri − ri (ri ⋅ ω) ⎦ = ⎢ − mi yi xi
− mi yi zi ⎥ ω
2
2 ⎥
⎢ −mi zi xi
m
z
y
m
(
r
z
−
−
i i i
i i
i )⎦
⎣
Inertia tensor I
Inertia Tensor
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Diagonal components are familiar
moment of inertia
I xx = mi (ri − x ) = mi ri sin Θ
2
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2
i
2
2
r
Θ
x
What are the off-diagonal components?
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Iyx produces Ly when the object is turned around x axis
Imagine turning something like:
Unbalanced one has non-zero
off-diagonal components,
which represents “wobbliness”
of rotation
Balanced
Unbalanced
Inertia Tensor
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Using ( xi , yi , zi ) → ( xi1 , xi 2 , xi 3 )
⎡ mi (ri 2 − xi2 )
− mi xi yi
− mi xi zi ⎤
⎢
⎥
2
2
mi (ri − yi )
− mi yi zi ⎥
I = ⎢ − mi yi xi
2
2 ⎥
⎢ −mi zi xi
m
z
y
m
(
r
z
−
−
i i i
i i
i )⎦
⎣
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I jk = mi ( ri 2δ jk − xij xik )
We can also deal with continuous mass distribution ρ(r)
I jk = ∫ ρ (r ) ( r 2δ jk − x j xk )dr
Kinetic Energy
I jk = mi ( ri 2δ jk − xij xik )
1
„ Kinetic energy due to rotation is T = mi v i ⋅ v i
2
„ Using v i = ω × ri
1
ω
ω
mi v i ⋅ (ω × ri ) = ⋅ mi (ri × v i ) = ⋅ L
2
2
2
ω ⋅ Iω
Using L = Iω
T=
2
Defining n as the unit vector in the direction of ω ω = ω n
T=
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n ⋅ In 2 1 2
ω = Iω where I = n ⋅ In = mi ⎡⎣ ri 2 − (ri ⋅ n) 2 ⎤⎦
T=
2
2
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Moment of inertia about axis n
NB: n moves with time
I = I(t) must change accordingly with time
Shifting Origin
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Origin of body axes does not have to be at the CoM
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It’s convenient – Separates translational/rotational motion
If it isn’t, I can be easily translated
from origin
ri = R + ri′
from CoM
I = mi (ri × n) 2 = mi [(R + ri′) × n]2
= M (R × n) 2 + mi (ri′ × n) 2 + 2mi (R × n) ⋅ (ri′ × n)
I of CoM
I from CoM
Summary
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Found the velocity due to rotation
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Connected ω with the Euler angles
Lagrangian Æ translational and rotational parts
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Often possible if body axes are defined from the CoM
I = mi ( ri 2δ jk − xij xik )
Defined the inertia tensor
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Used it to find Coriolis effect
⎛d⎞ ⎛d⎞
⎜ ⎟ = ⎜ ⎟ + ω×
⎝ dt ⎠ s ⎝ dt ⎠ r
Calculated angular momentum
and kinetic energy
L = Iω T =
Next: Equation of motion (finally!)
nIn 2 1 2
ω = Iω
2
2