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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 =1

2IXX

˙ θ cosψ + ˙ φ sinθ sinψ( )2

+ IYY − ˙ θ sinψ + ˙ φ sinθ cosψ( )2

+ IZZ˙ ψ + ˙ φ cosθ( )

2

( )

+1

2m ˙ 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 =1

2A ˙ θ cosψ + ˙ φ sinθ sinψ( )

2+ B − ˙ θ sinψ + ˙ φ sinθ cosψ( )

2+ C ˙ ψ + ˙ φ cosθ( )

2

( )

+1

2m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

€

L =1

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

2

( ) +1

2m ˙ 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 =

x

y

z

φ

θ

ψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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We know v and w 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

€

r = −aJ2

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

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€

d

dt

∂L

∂˙ x i ⎛

⎝ ⎜

⎞

⎠ ⎟+ mgx = λ jC1

j

€

d

dt

∂L

∂ ˙ φ i ⎛

⎝ ⎜

⎞

⎠ ⎟= λ jC4

j

€

d

dt

∂L

∂ ˙ θ

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂θ= λ jC5

j

€

d

dt

∂L

∂ ˙ ψ

⎛

⎝ ⎜

⎞

⎠ ⎟= λ jC6

j

€

d

dt

∂L

∂˙ y i ⎛

⎝ ⎜

⎞

⎠ ⎟+ mgy = λ jC2

j

€

d

dt

∂L

∂˙ z i ⎛

⎝ ⎜

⎞

⎠ ⎟+ mgz = λ jC3

j

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We have the usual Euler-Lagrange equations

€

d

dt

∂L

∂˙ q i ⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂qi = λ jCij

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

€

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 w

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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 =1

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

1

2m ˙ 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 =1

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

+1

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

We can now assign generalized coordinates

€

q =

φ

θ

ψ

ξ

χ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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We have rolling constraintsw 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 cosχ

asinξ sin χ

acosξ

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

€

v = a + R( )

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

€

d

dt

∂L

∂˙ q i ⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂qi = λ jCij

and we can write out the five equations

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€

d

dt

∂L

∂ ˙ φ

⎛

⎝ ⎜

⎞

⎠ ⎟= λ jC1

j

€

d

dt

∂L

∂ ˙ ξ

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂ξ= λ jC4

j

€

d

dt

∂L

∂ ˙ χ

⎛

⎝ ⎜

⎞

⎠ ⎟= λ jC5

j

€

d

dt

∂L

∂ ˙ θ

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂θ= λ jC2

j

€

d

dt

∂L

∂ ˙ ψ

⎛

⎝ ⎜

⎞

⎠ ⎟= λ jC3

j

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The constraint matrix is

€

a2 sin2 χ −1

2a2 sin2χ cos φ −ξ( ) a2 sin χ sinθ cos χ −θ( )sin φ −ξ( ) 0 R + a( )sinχ

0 −1

2a2 sin2χ sin φ −ξ( )

1

2a2 sin2χ sinθ cos φ −ξ( ) −

1

2a 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 = mR + a

asinχ2cosχ ˙ χ ˙ ξ + sin χ ˙ ̇ χ ( )

λ 2 = mR + a

a˙ ξ 2 − m

R + a

acosχ sin χ˙ ̇ χ +

mg

acosχ

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

€

y2 = r1 − r2( )sin χ

z2 = r1 − r1 − r2( )cosχ

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We have holonomic constraints

Put us in two dimensions

€

x1 = 0 = x2

φ1 =π

2= φ2

θ1 = −π

2= 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 =1

2m1 + m2( ) ˙ y 1

2 +1

2m1r1

2 ˙ ψ 1 +1

2m2r2

2 ˙ ψ 2 +1

2m2 r1 − r2( )

2 ˙ χ 2 +1

2m2 r1 − r2( )

2cosχ ˙ 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 0

0 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