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Lecture 4 Rotations II: Angular Dynamics

angular velocity

angular momentum

“angular energy”

We’ll get all of this from our vision of constrained particles/points

2

angular momentum of a collection of points

€

l = ri × mivii=1

K

∑

€

ri = rCM + ′ r i, vi = vCM + v

€

l = rCM + ′ r i( ) × mi vCM + v( )i=1

K

∑ =

rCM × mivCMi=1

K

∑ + mi ′ r ii=1

K

∑ ⎛

⎝ ⎜

⎞

⎠ ⎟× vCM + rCM × mi ′ v i

i=1

K

∑ + ′ r i × mi ′ v ii=1

K

∑

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€

l = MrCM × vCM + ′ r i × mi ′ v ii=1

K

∑CM wrt reference point

wrt center of mass

I care about the second one

The particles are glued together, so their only possible motion is rotation

€

′ v i = Ω × ′ r i

and, of course, it’s the same W for every particle: they rotate in the same body

€

l2 = ′ r i × mi ′ v ii=1

K

∑ = mi ′ r i × Ω × ri( )i=1

K

∑

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Find a nice vector identity

5

€

l2 = ′ r i × mi ′ v ii=1

K

∑ = mi ′ r i × Ω × ri( )i=1

K

∑

becomes

€

′ r i × Ω × ′ r i( ) = ′ r i ⋅ ′ r i( )Ω − ′ r i ⋅Ω( ) ′ r i

€

l2 = Ω mi ′ r i ⋅ ′ r i( )( )i=1

K

∑ − mi ′ r i ⋅Ω( ) ′ r i( )i=1

K

∑

W factors out

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Now I’d like to pass to the limit and replace the sums by volume integrals

€

l2 = Ω ρ ′ r ⋅ ′ r ( )dVvol∫ − ρ ′ r ⋅Ω( ) ′ r dV

vol∫

I haven’t yet said what coordinate system I am using

Since I am integrating over the volume, it makes sense to use the body systemand let

€

′ r =XYZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪, Ω =

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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What does this do to the integrals?

€

l2 = ρ X 2 + Y 2 + Z 2( )dVvol∫ ⎛ ⎝ ⎜

⎞ ⎠ ⎟ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪− ρ XΩX + YΩY + ZΩZ( )

XYZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪dV

vol∫

vectors

scalars

We can combine all of this into three component equations

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€

l2 = ρ X 2 + Y 2 + Z 2( )dVvol∫ ⎛ ⎝ ⎜

⎞ ⎠ ⎟ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪− ρ XΩX + YΩY + ZΩZ( )

XYZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪dV

vol∫

€

l2 = ρ X 2ΩX + Y 2ΩX + Z 2ΩX − X 2ΩX − XYΩY − XZΩZ( )IdVvol∫

+ ρ X 2ΩY + Y 2ΩY + Z 2ΩY −YXΩX −Y 2ΩY −YZΩZ( )JdVvol∫

+ ρ X 2ΩZ + Y 2ΩZ + Z 2ΩZ − ZXΩX − ZYΩY − Z 2ΩZ( )KdVvol∫

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€

l2 = ρ Y 2 + Z 2( )ΩX − XYΩY − XZΩZ( )IdVvol∫

+ ρ X 2 + Z 2( )ΩY −YXΩX −YZΩZ( )JdVvol∫

+ ρ X 2 + Y 2( )ΩZ − ZXΩX − ZYΩY( )KdVvol∫

Now we can recognize the moments and products of inertia

The products of inertia vanish if the body axes are aligned with the principal moments

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€

l2 =IXX IXY IXZ

IYX IYY IYZ

IZX IZY IZZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

That’s in body coordinatesWe can put it in inertial coordinates using the rotation matrices

€

l2 = RzT φ( )Rx

T θ( )RzT ψ( )

IXX IXY IXZ

IYX IYY IYZ

IZX IZY IZZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

We’ll look with more specificity using Mathematica later

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If we are in principal coordinates this is simply

€

l2 = RzT φ( )Rx

T θ( )RzT ψ( )

IXX 0 00 IYY 00 0 IZZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

which becomes

€

l2 = RzT φ( )Rx

T θ( )RzT ψ( )

IXXΩX

IYYΩY

IZZΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

and we need an expression for the rotation in terms of body coordinates

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What is the angular velocity?Can we express it in terms of the Euler angles?

Change in f corresponds to rotation about k

Change in q corresponds to rotation about I1

Change in y corresponds to rotation about K2

€

ω = ˙ φ k + ˙ θ I1 + ˙ ψ K 2The vector rotation rate will be

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This is not an orthogonal basis, and, indeedit may not even be a basis

We have a choice of bases: inertial or body

We want to use body coordinatesto allow us to go back to the earlier slides and get l

can be put into inertial coordinatesWe have expressions for the three vectors wrt an inertial frame

So let’s go take a look at that and see what we have

€

ω = ˙ φ k + ˙ θ I1 + ˙ ψ K 2

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€

ω = ˙ φ k + ˙ θ I1 + ˙ ψ K 2

€

K 2 = −sinθJ1 + cosθK1

€

I1 = cosφI0 + sinφJ0, J1 = −sinφI0 + cosφJ0, K1 = K 0

€

K 2 = −sinθ −sinφI0 + cosφJ0( ) + cosθK 0

€

ω = ˙ φ k + ˙ θ cosφI0 + sinφJ0( ) + ˙ ψ −sinθ −sinφI0 + cosφJ0( ) + cosθK 0 ( )

and I can clean this up

€

ω = ˙ θ cosφ + ˙ ψ sinθ sinφ( )i + ˙ θ sinφ − ˙ ψ sinθ cosφ( )j+ ˙ φ + ˙ ψ cosθ( )k

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€

ω = ˙ θ cosφ + ˙ ψ sinθ sinφ( )i + ˙ θ sinφ − ˙ ψ sinθ cosφ( )j+ ˙ φ + ˙ ψ cosθ( )k

goes into body coordinates using the rotation matrices

€

W=Rz ψ( )Rx θ( )Rz φ( )ω

= ˙ θ cosψ + ˙ φ sinθ sinψ( )I3 + − ˙ θ sinψ + ˙ φ sinθ cosψ( )J3 + ˙ ψ + ˙ φ cosθ( )K 3

and the angular momentum in body coordinates is simply

€

l = ˙ θ cosψ + ˙ φ sinθ sinψ( )IXXI3 + − ˙ θ sinψ + ˙ φ sinθ cosψ( )IYY J3 + ˙ ψ + ˙ φ cosθ( )IZZK 3

If the body coordinates are principle

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We’ll look at the angular momentum in inertial coordinateswhen we go to Mathematica

It’s a big expression!

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??OK, let’s do the same thing for kinetic energy

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kinetic energy of a collection of points

€

T = 12

mivi ⋅vii=1

K

∑

€

vi = vCM + ′ v i

€

T = 12

mi vCM + ′ v i( ) ⋅ vCM + ′ v i( )i=1

K

∑

= 12

mivCM ⋅vCMi=1

K

∑ + 12

vCM ⋅ mi ⋅ ′ v ii=1

K

∑ + 12

mi ′ v ii=1

K

∑ ⎛

⎝ ⎜

⎞

⎠ ⎟⋅vCM + 1

2mi ′ v i ⋅ ′ v i

i=1

K

∑

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€

T = 12

MvCM ⋅vCM + 12

mi ′ v i ⋅ ′ v ii=1

K

∑

center of mass motion

motion wrt center of mass

€

′ v i = Ω × ′ r i

€

T = 12

MvCM ⋅vCM + 12

mi Ω × ′ r i( ) ⋅ Ω × ′ r i( )i=1

K

∑

and now I need another vector identity

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Apply that

€

W× ′ r i( ) ⋅ Ω × ′ r i( ) = Ω ⋅Ω( ) ′ r i ⋅ ′ r i( ) − Ω ⋅ ′ r i( ) ′ r i ⋅Ω( )

€

T = 12

MvCM ⋅vCM + 12

mi Ω ⋅Ω( ) ′ r i ⋅ ′ r i( ) − Ω ⋅ ′ r i( ) ′ r i ⋅Ω( )( )i=1

K

∑

I want to pass to the limit and rearrange the second part

€

TROT = 12

ρ Ω ⋅Ω( ) ′ r ⋅ ′ r ( ) − Ω ⋅ ′ r ( ) ′ r ⋅Ω( )( )dVvol∫

€

TROT = 12

ρ ΩX2 + ΩY

2 + ΩZ2( ) X 2 + Y 2 + Z 2( ) − ΩX X + ΩYY + ΩZ Z( )

2( )dV

vol∫

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Again we have a lot of cancellation

€

TROT = 12

ρ ΩX2 Y 2 + Z 2( ) + ΩY

2 X 2 + Z 2( ) + ΩZ2 X 2 + Y 2( )( )dV

vol∫

− 12

2ρ ΩX XΩYY + ΩX XΩZ Z + ΩYYΩZ Z + ΩZ Z( )dVvol∫

We can recognize the integrals as moments and products of inertia as before

€

TROT = 12

ΩX ΩY ΩX{ }IXX IXY IXZ

IYX IYY IYZ

IZX IZY IZZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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If we have chosen our body axes to be principal this simplifies

€

TROT = 12

ΩX ΩY ΩX{ }IXX 0 00 IYY 00 0 IZZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

ΩX

ΩY

ΩZ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

TROT = 12

IXXΩX2 + IYYΩY

2 + IZZΩZ2( )

€

W= ˙ θ cosψ + ˙ φ sinθ sinψ( )I3 + − ˙ θ sinψ + ˙ φ sinθ cosψ( )J3 + ˙ ψ + ˙ φ cosθ( )K 3

€

TROT = 12

IXX˙ θ cosψ + ˙ φ sinθ sinψ( )

2+ IYY − ˙ θ sinψ + ˙ φ sinθ cosψ( )

2+ IZZ

˙ ψ + ˙ φ cosθ( )2 ⎛

⎝ ⎜ ⎞ ⎠ ⎟

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??

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Now we are in a position to write equations of motion for a link

€

L = T −V = TTRANS + TROT − mg

I have the kinetic energy, and I can add a simple potential — gravity

I will use m to denote the mass of a linkwe aren’t doing any more points

We are going to do physics, so we need to use the inertial coordinate system.

There are six degrees of freedom, and we have six variables

€

x, y,z,φ,θ,ψ

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Generalized coordinates

€

q =

xyzφθψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

This is the fundamental assignment.We’ll see reductions when we look at constraints.

Let’s apply this to the motion of a block with no external forces.

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The Euler-Lagrange process

1. Find T and V as easily as you can

2. Apply geometric constraints to get to N coordinates

3. Assign generalized coordinates

4. Define the Lagrangian

DONE

NONE

PREVIOUS SLIDE

DONE

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5. Differentiate the Lagrangian with respect to the derivative of the first generalized coordinate

6. Differentiate that result with respect to time

7. Differentiate the Lagrangian with respect to the same generalized coordinate

8. Subtract that and set the result equal to Q1

€

∂L∂˙ q 1

€

ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

∂L∂q1

€

ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q1 = Q1

Repeat until you have done all the coordinates

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€

ddt

∂L∂˙ q 1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q1 = 0 = ddt

∂L∂˙ x ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂x

€

ddt

∂L∂˙ q 5 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q5 = 0 = ddt

∂L∂ ˙ θ ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂θ

€

ddt

∂L∂˙ q 2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q2 = 0 = ddt

∂L∂˙ y ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂y

€

ddt

∂L∂˙ q 3 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q3 = 0 = ddt

∂L∂˙ z ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂z

€

ddt

∂L∂ ˙ q 4 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q4 = 0 = ddt

∂L∂ ˙ φ ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂φ

€

ddt

∂L∂˙ q 6 ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂q6 = 0 = ddt

∂L∂ ˙ ψ ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂ψ

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It happens that we can write the Lagrangian in a suggestive form

€

L = 12

˙ x ˙ y ˙ z ˙ φ ˙ θ ˙ ψ { }

m 0 0 0 0 00 m 0 0 0 00 0 m 0 0 00 0 0 • • •0 0 0 • • •0 0 0 • • •

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

˙ x ˙ y ˙ z ˙ φ ˙ θ ˙ ψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

−V q( )

This is actually general — for bigger problems we’ll have

€

L = 12

˙ q TM q( ) ˙ q −V q( )

vector-matrix notation

€

L = 12

˙ q iM ij qk( ) ˙ q j −V qk( )

indicial notation

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This allows us to reduce the eight step process to a simple formula

€

ddt

M ij qk( ) ˙ q j( ) − 12

˙ q m∂Mmj qk( )

∂qi ˙ q j +∂V qk( )

∂qi = Qi

€

M ij qk( )˙ ̇ q j +∂M ij qk( )

∂qn ˙ q j ˙ q n − 12

˙ q m∂Mmj qk( )

∂qi ˙ q j +∂V qk( )

∂qi = Qi

(It’s not really that simple, and we’ll need Mathematica to do anything real.)

€

M ij qk( ) is symmetric, positive definite, invertible

32

These are six second order equationsI want to convert to twelve first order equations

€

M ij qk( ) ˙ u j +∂M ij qk( )

∂qn u jun − 12

um ∂Mmj qk( )∂qi u j +

∂V qk( )∂qi = Qi

€

˙ q j = u j

These equations are amenable to numerical solution.

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So, what have we done so far?

angular velocityangular momentum

kinetic energythe Lagrangian

the Euler-Lagrange equationsconversion to a first order system

34

??

35

OK. I want to switch to Mathematica and repeat most of this.

Then I want to look at the fall of a single link.

It’s generally known that a block spinning about its long or short axis is stablebut spin about the intermediate axis is unstable

The easiest way to do this is to define the K axisset it horizontal and spin y about K

€

φ =0, θ =π2

, ˙ ψ = λ , ˙ φ = 0.01

perturbation

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long

37

short

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intermediate