lecture 7 rolling constraints

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Lecture 7 Rolling Constraints

The most common, and most important nonholonomic constraints

They cannot be written in terms of the variables aloneyou must include some derivatives

The resulting differential is not integrable

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The major assumption/simplification is point contact

This is true of a rigid hoop on a rigid surface

or a rigid sphere on a rigid surface

I will also generally assume that the surface is horizontalso that gravity is normal to the surface

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rigid hoop on a rigid surface

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rigid sphere on a rigid surface

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It’s not a bad approximation for a well-inflated bicycle or motorcycle tireon hard pavement

It’s not so good for an automobile tire, which has a pretty big patch on the ground

Cylindrical wheels, like those on grocery carts and children’s toysmust slip over parts of their surfaces when turning

We aren’t going to care about any of thisWe’re going to suppose that all the wheels we consider have point contact

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Two fundamental ideas:

The speed of a point rotating about a fixed point is

€

v = ω × r

The contact point between a wheel and the ground is not moving

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For a sphere of radius a r = ak

€

v = aω × k = aωy

−ωx

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

This is inertial , so we have

€

v = aωy

−ωx

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= a

˙ θ sinφ − ˙ ψ cosφ sinθ− ˙ θ cosφ − ˙ ψ sinφsinθ

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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€

˙ x = a ˙ θ sinφ − ˙ ψ cosφsinθ( ), ˙ y = −a ˙ θ cosφ + ˙ ψ sinφsinθ( )

We can write the constraints as

Of course, we also have the simple constraint for this problem: z = a.

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An erect wheel has the same r

If we choose to let the axle be the K body coordinatethen q = ±π/2, fixed, and the constraint becomes

€

v = aωy

−ωx

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= a

˙ θ sinφ − ˙ ψ cosφsinθ− ˙ θ cosφ − ˙ ψ sinφsinθ

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

v = a

˙ ψ cosφ− ˙ ψ sinφ

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪⇒ ˙ x = −a ˙ ψ cosφ, ˙ y = −a ˙ ψ sinφ

The general wheel is fancier; I’ll get to that shortly

BUT FIRST

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How do we impose these?

The classical method is through Lagrange multipliers,and that’s what we’ll do today

(There’s an alternate method that we will see later in the semester)

To get at this we need to go back to first principlesand it helps to note that we can write the constraints in matrix form

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€

˙ x = a ˙ θ sinφ + ˙ ψ cosφsinθ( ), ˙ y = −a ˙ θ cosφ + ˙ ψ sinφsinθ( )

€

q =

xyzφθψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

€

˙ q 1 − a ˙ q 5 sinq4 + ˙ q 6 cosq4 sinq5( ) = 0 = ˙ q 2 + a ˙ q 5 cosq4 + ˙ q 6 sinq4 sinq5( )

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€

1 0 0 0 −asinq4 −acosq4 sinq5

0 1 0 0 acosq4 asinq4 sinq5

⎧ ⎨ ⎩

⎫ ⎬ ⎭

˙ q 1

˙ q 2

˙ q 3

˙ q 4

˙ q 5

˙ q 6

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

= 0

€

˙ q 1 − a ˙ q 5 sinq4 + ˙ q 6 cosq4 sinq5( ) = 0 = ˙ q 2 + a ˙ q 5 cosq4 + ˙ q 6 sinq4 sinq5( )

€

C ˙ q = 0 ⇔ C ji ˙ q j

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We can view the time derivative of q as a surrogate for dq

€

C jiδq j = 0

€

dI η( )dη

= 0 = ∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟∂qk

∂ηdt

t1

t2∫ = ∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟δqkdt

t1

t2∫

Now let’s go back to lecture two and the action integral it is stationary if

Consider the final version

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€

∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟δqkdt

t1

t2∫ = 0

and let’s add multiples of the constraints to thisThey are zero, so there’s no problem

€

∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟δqkdt

t1

t2∫ = 0

€

λiC jiδq j = 0

€

λiCkiδqk = 0

multiply each row by its own constant

change the dummy index j to k

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€

∂L *∂qk − d

dt∂L *∂˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ λ iCk

i ⎛

⎝ ⎜

⎞

⎠ ⎟δqkdt

t1

t2∫ = 0

drop the asterisk and incorporate external forces directlythe constrained Euler Lagrange equations

€

ddt

∂L∂ ˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qk = λ iCki + Qk

I have added as many variables as there are nonholonomic constraints

I need as many extra equations, which are the constraints themselves

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€

ddt

∂L∂ ˙ q k ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qk = λ iCki + Qk

We have a physical interpretation of the Lagrange multipliersThey are the forces (and torques) of constraint —

what the world does to make the system conform

How do we actually apply this?

The explanation is on pp 30-32 of chapter three, perhaps a bit telegraphically

I’d like to go through the erect coin, example 3.5, pp. 32-35

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First let’s see how we can set up a wheel

I like the axle to be the K axis; let’s see how we can erect the wheel

Starting with all Euler angles equal to zero

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Pick a direction by adjusting f

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erect the wheel with q = -π/2(the text example uses + π/2)

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€

r = −aJ2, ω = ˙ ψ K

€

v = ω × r = − ˙ ψ K × aJ2 = a ˙ ψ I2 = a ˙ ψ cosφsinφ

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

˙ x − a ˙ ψ cosφ = 0 = ˙ y − a ˙ ψ sinφ

Variables are x, y, f and y — z and q are fixed

w has a k component, but it’sperpendicular to r

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The Lagrangian is simple

€

L = 12

m ˙ x 2 + ˙ y 2( ) + 12

A ˙ φ 2 + 12

C ˙ ψ 2

I’ve left out the gravity term because the center of mass cannot move vertically

The constraint matrix is

€

C =1 0 0 acosφ0 1 0 asinφ ⎧ ⎨ ⎩

⎫ ⎬ ⎭

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The product

€

λC = λ1 λ 2{ }1 0 0 acosφ0 1 0 asinφ ⎧ ⎨ ⎩

⎫ ⎬ ⎭= λ1 λ 2 0 a λ1 cosφ + λ 2 sinφ( ){ }

The Lagrange equations are then

€

m˙ ̇ x = λ1, m˙ ̇ y = λ 2, ˙ ̇ φ = 0, ˙ ̇ ψ = aC

λ1 cosφ + λ 2 sinφ( )

and the constraints are

€

˙ x − a ˙ ψ cosφ = 0 = ˙ y − a ˙ ψ sinφ

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So I have six equations in six unknowns: the variables and the multipliers

Solve the first two dynamical equations for the multipliers

€

λ1 = m˙ ̇ x , λ 2 = m˙ ̇ y

The remaining dynamical equations become

€

˙ ̇ φ = 0, ˙ ̇ ψ = amC

˙ ̇ x cosφ + ˙ ̇ y sinφ( )

and I can use the constraints to eliminate the acceleration terms

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€

˙ x − a ˙ ψ cosφ = 0 = ˙ y − a ˙ ψ sinφ⇓

˙ ̇ x = a ˙ ̇ ψ cosφ − a ˙ ψ ˙ φ sinφ, ˙ ̇ y = a ˙ ̇ ψ sinφ + a ˙ ψ ˙ φ cosφ

substitute all this back in

€

˙ ̇ ψ = amC

˙ ̇ ψ cosφ − ˙ ψ ˙ φ sinφ( )cosφ + ˙ ̇ ψ sinφ + ˙ ψ ˙ φ cosφ( )sinφ( ) = amC

˙ ̇ ψ ⇒ ˙ ̇ ψ = 0

€

φ=φ0 + ˙ φ 0t, ψ =ψ 0 + ˙ ψ 0t

so, we have an exact solution for the two angles

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and we can substitute this into the constraints and integrate to get the positions

€

˙ x = a ˙ ψ 0 cos φ0 + ˙ φ 0t( ), ˙ y = a ˙ ψ 0 sin φ0 + ˙ φ 0t( )

x = a˙ ψ 0˙ φ 0

sin φ0 + ˙ φ 0t( ), y = −a˙ ψ 0˙ φ 0

cos φ0 + ˙ φ 0t( )

The coin rolls around in a circle of constant radius,which can be infinite, in which case

€

x = a ˙ ψ 0t sinφ0, y = a ˙ ψ 0t cosφ0

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Suppose we consider a general rolling coin,one not held vertical

The book doesn’t deal with this until chapter five,but we can certainly set it up here.

It’s a hard problem and does not have a closed form solution

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This has the same body system as beforebut the angle q can vary

(it’s equal to -0.65π here)

r remains equal to –aJ2

but we need the whole w

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€

J2 =−cosθ sinφcosθ cosφ

sinθ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪, ω =

cosφ ˙ θ + sinφ sinθ ˙ ψ sinφ ˙ θ − cosφsinθ ˙ ψ

˙ φ + cosθ ˙ ψ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

and, after some simplification

€

v =˙ x ˙ y ˙ z

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= −ω × aJ2 = a

cosφcosθ ˙ φ − sinφsinθ ˙ θ + cosφ ˙ ψ sinφcosθ ˙ φ + cosφsinθ ˙ θ + sinφ ˙ ψ

−cosθ ˙ θ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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€

˙ x − a cosφcosθ ˙ φ − sinφ sinθ ˙ θ + cosφ ˙ ψ ( ) = 0

˙ y − a sinφcosθ ˙ φ + cosφsinθ ˙ θ + sinφ ˙ ψ ( ) = 0

˙ z + acosθ ˙ θ = 0

The last one is actually integrable, but I won’t do that

We have a constraint matrix that appears to disagree with the one on p. 15 of chapter five.

The difference stems from the assumption in the book that q ≥ 0whereas in the picture here, q ≤ 0.

Choose the one that fits your idea of the initial conditions.

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I will use the book’s picture, and I will go to Mathematicato analyze this system