Lecture 8: Rolling Constraints II
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Transcript of Lecture 8: Rolling Constraints II
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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
€
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
€
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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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
€
y2 = r1 − r2( )sinχ
z2 = r1 − r1 − r2( )cosχ
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We have holonomic constraints
Put us in two dimensions
€
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