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Lecture 8: Rolling Constraints II

Coin rolling on a slope

Hoop rolling inside a hoop

Generalizations and review

Nonhorizontal surfaces

Small sphere rolling on a larger sphere’s surface

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What can we say in general?

€

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

€

L =12

IXX˙ θ cosψ + ˙ φ sinθ sinψ( )

2+ IYY − ˙ θ sinψ + ˙ φ sinθ cosψ( )

2+ IZZ

˙ ψ + ˙ φ cosθ( )2

( )

+12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

€

v =ω × r

vector from the inertial origin to the center of mass

vector from the contact point to the center of mass

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We can restrict our attention to axisymmetric wheelsand we can choose K to be parallel to the axle

without loss of generality

€

L =12

A ˙ θ cosψ + ˙ φ sinθ sinψ( )2

+ B − ˙ θ sinψ + ˙ φ sinθ cosψ( )2

+ C ˙ ψ + ˙ φ cosθ( )2

( )

+12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

€

L =12

A ˙ θ 2 + ˙ φ 2 sin2 θ( ) + C ˙ ψ + ˙ φ cosθ( )2

( ) +12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

mgz

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If we don’t put in any simple holonomic constraint (which we often can do)

€

q =

xyzφθψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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We know v and ω in terms of qany difficulty will arise from r

This will depend on the surface

flat, horizontal surface — we’ve been doing this

flat surface — we can do this today

general surface: z = f(x, y) — this can be done for a rolling sphere

€

v =ω × r

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r = −aJ2

Actually, it’s something of a question as to where the difficulties will arise in general

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€

ddt

∂L∂˙ x i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgx = λ jC1

j

€

ddt

∂L∂ ˙ φ i ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC4

j

€

ddt

∂L∂ ˙ θ ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂L∂θ

= λ jC5j

€

ddt

∂L∂ ˙ ψ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC6

j

€

ddt

∂L∂˙ y i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgy = λ jC2

j

€

ddt

∂L∂˙ z i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgz = λ jC3

j

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We have the usual Euler-Lagrange equations

€

ddt

∂L∂˙ q i ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂L∂qi = λ jCi

j

and we can write out the six equations

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The key to the problem lies in the constraint matrix

The analysis is pretty simple for flat surfaces, whether horizontal or tilted

Let’s play with the tilted surface

Choose a Cartesian inertial system such that i and j lie in the tilted planeand choose i to to be horizontal, so that j points down hill

(This is a rotation of the usual system about i)

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€

y = cosα ′ y + sinαz'z = cosα ′ z − sinα ′ y

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so the potential energy in the primed coordinates is

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V = mg cosα ′ z − sinα ′ y ( )

the kinetic energy is unchanged

We can go forward from here exactly as beforeeverything is the same except for gravity

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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 ω

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??Let’s look at a rolling coin on a tilted surface in Mathematica

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Curved surfaces

Spherical surface: spherical ball on a sphere

Two-d surface: wheel inside a wheel

General surface

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Spherical ball on a sphere

R

a

€

x 2 + y 2 + z2 = R + a( )2

holonomic constraint

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€

L = 12

A ˙ θ 2 + ˙ φ 2 + ˙ ψ 2 + 2 ˙ φ ˙ ψ cosθ( ) + 12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mgz

The Lagrangian simplifies because of the spherical symmetry

We have a constraint, which we can parameterize

€

x 2 + y 2 + z2 = R + a( )2

€

x = R + a( )sinξ cosχ , y = R + a( )sinξ sin χ , z = R + a( )cosξ

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which transforms the Lagrangian

€

L = 12

A ˙ θ 2 + ˙ φ 2 + ˙ ψ 2 + 2 ˙ φ ˙ ψ cosθ( )

+ 12

m R + a( ) 2 ˙ ξ 2 + ˙ χ 2 1− cos 2ξ( )( )( ) − m R + a( )gcos ξ( )

We can now assign generalized coordinates

€

q =

φθψξχ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

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⎭

⎪ ⎪ ⎪

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We have rolling constraintsω is unchanged, and r is as shown on the figure

and we recalculate v

€

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

€

rC =asinξ cosχasinξ sin χ

acosξ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

v = a + R( )

˙ ξ cosξ cosχ − ˙ χ sinξ sinχ˙ ξ cosξ sinχ + ˙ χ cosξ sin χ

− ˙ ξ sinξ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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The rolling constraint appears to have three componentsbut the normal component has already been satisfied

€

v −ω × r = 0

€

k × r( ) ⋅ v −ω × r( ) = 0

k × k × r( )( ) ⋅ v −ω × r( ) = 0

The normal is parallel to r, so I need two tangential vectors

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We have the usual Euler-Lagrange equations

€

ddt

∂L∂˙ q i ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂L∂qi = λ jCi

j

and we can write out the five equations

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€

ddt

∂L∂ ˙ φ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC1

j

€

ddt

∂L∂ ˙ ξ ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂ξ= λ jC4

j

€

ddt

∂L∂ ˙ χ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC5

j

€

ddt

∂L∂ ˙ θ ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂θ= λ jC2

j

€

ddt

∂L∂ ˙ ψ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC3

j

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The constraint matrix is

€

a2 sin2 χ − 12

a2 sin2χ cos φ −ξ( ) a2 sin χ sinθ cos χ −θ( )sin φ −ξ( ) 0 R + a( )sinχ

0 − 12

a2 sin2χ sin φ −ξ( )12

a2 sin2χ sinθ cos φ −ξ( ) − 12

a R + a( )sin2χ 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

The last two Euler Lagrange equations are suitable for eliminating the Lagrange multipliers

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After some algebra

€

λ1 = m R + aasinχ

2cosχ ˙ χ ˙ ξ + sin χ ˙ ̇ χ ( )

λ 2 = m R + aa

˙ ξ 2 − m R + aacosχ sinχ

˙ ̇ χ + mgacosχ

I have three remaining Euler-Lagrange equationsand two constraint equations that I need to differentiate

to give me five equations for the generalized coordinates

We need to go to Mathematica to see how this goes.

QUESTIONS FIRST??

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Wheel within a wheel

Treat them both as hoopsradii r1 > r2

c

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y2 = r1 − r2( )sinχ

z2 = r1 − r1 − r2( )cosχ

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We have holonomic constraints

Put us in two dimensions

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x1 = 0 = x2

φ1 = π2

= φ2

θ1 = − π2

= θ2

realize that z1 = r1

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We have an interesting connectivity constraint —define the position of the small wheel in terms of the angle c

€

y2 = y1 + r1 − r2( )sinχ , z2 = r1 + r1 − r2( )cosχ

Putting all this in gives us a Lagrangian

€

L = 12

m1 + m2( ) ˙ y 12 + 1

2m1r1

2 ˙ ψ 1 + 12

m2r22 ˙ ψ 2 + 1

2m2 r1 − r2( )

2 ˙ χ 2 + 12

m2 r1 − r2( )2 cosχ ˙ χ ̇ y 1

+ gm2 r1 − r2( )cosχ − gr1 m1 + m2( )

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We define a vector of generalized coordinates

€

q =

y1

ψ1

ψ 2

χ

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

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There will be two nonholonomic constraints

€

˙ y 1 = r1 ˙ ψ 1, r1 − r2( ) ˙ χ = r2˙ ψ 2

The corresponding constraint matrix is

€

C =1 r1 0 00 0 −r2 r1 − r2

⎧ ⎨ ⎩

⎫ ⎬ ⎭

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The second and third Euler-Lagrange equations are fairly simpleso I will use those to find the two Lagrange multipliers

€

λ1 = −m1r1˙ ̇ ψ 1, λ 2 = −m2r2˙ ̇ ψ 2

To solve the problem we use the first and fourth Euler-Lagrange equationsand the differentiated constraints

The solution is numerical and we need to go to Mathematica to look at it.

QUESTIONS FIRST??

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That’s All Folks