lecture 5: constraints i

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Lecture 5: Constraints I

holonomic constraints

external constraints (bead on a wire)

internal constraints (connectivity constraints)

orientation constraints

Constraints reduce the number of degrees of freedom of a mechanism

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Holonomic constraints can be written in terms of the coordinates

Simple holonomic constraints are linear in the coordinates

Nonsimple holonomic constraints are nonlinear in the coordinates

(I will address nonholonomic constraints in Lecture 7.)

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Bead on a wire

I call it a bead, and I’ll treat it as a point — three degrees of freedom

mg

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A little digression on the properties of curves in spacethe tangent vector

€

λ =dpds

This is the unit tangent vectorwe’ll do a lot with nonunit tangent vectors

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Bead on a wire

Curve is defined by a parameterization

€

λ =

dxdsdydsdzds

⎧

⎨

⎪ ⎪

⎩

⎪ ⎪

⎫

⎬

⎪ ⎪

⎭

⎪ ⎪

=

dxdχdydχdzdχ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

dχds

€

x = x s( ), y = y s( ), z = z s( )

arclength

€

x = x χ( ), y = y χ( ), z = z χ( )

general parameter

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Bead on a wire

Kinetic energy

€

T =12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) =12

m ′ x 2 + ′ y 2 + ′ z 2( )dχdt ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

add in the potential energy and write the Lagrangian

€

L =12

m ′ x χ( )2 + ′ y χ( )2 + ′ z χ( )2( )

dχdt ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

− mgz χ( )

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Bead on a wire

The only variable here is c (a proxy for s), so we build one EL equation

€

∂L∂˙ χ

= m ′ x χ( )2 + ′ y χ( )2 + ′ z χ( )2( ) ˙ χ

€

ddt

∂L∂˙ χ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= m ′ x χ( )2 + ′ y χ( )2 + ′ z χ( )2

( )˙ ̇ χ + mddt ′ x χ( )2 + ′ y χ( )2 + ′ z χ( )2

( ) ˙ χ

€

∂L∂χ

= m ′ ′ x χ( ) + ′ ′ y χ( ) + ′ ′ z χ( )( ) ˙ χ 2 − mg ′ z χ( )

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Bead on a wire

The Euler-Lagrange equation

€

m ′ x χ( )2 + ′ y χ( )

2 + ′ z χ( )2

( ) ˙ ̇ χ + m ′ ′ x + ′ ′ y + ′ ′ z ( ) ˙ χ 2 − mg ′ z = 0

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Bead on a wire

We can solve this analytically in the simple case where

€

′ x χ( )2 + ′ y χ( )

2 + ′ z χ( )2

is a constant.

The helix is such a case

€

x = acos χ( ), y = asinχ , z = −hχ

€

′ x χ( )2 + ′ y χ( )

2 + ′ z χ( )2 = a2 sin2 χ + a2 cos2 χ + h2 = a2 + h2

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Bead on a wire

€

∂L∂ ˙ χ

= m ′ x χ( )2 + ′ y χ( )

2 + ′ z χ( )2

( ) ˙ χ = m a2 + h2( ) ˙ χ

€

m 1+ h2( ) ˙ ̇ χ − mgh = 0

so the Euler-Lagrange equation is

which is easily integrated to give

€

c =c0 + ˙ χ 0t + g ht 2

2 a2 + h2( )

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Bead on a wire

From which we obtain

€

x = acos χ 0 + ˙ χ 0t + g ht 2

2 a2 + h2( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

y = asin χ 0 + ˙ χ 0t + g ht 2

2 a2 + h2( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

z = hχ 0 + h ˙ χ 0t + g h2t 2

2 a2 + h2( )

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Bead on a wire

The force on the bead from the wire

€

fx = m˙ ̇ x = −amcos Φ( ) ˙ χ 0 + g hta2 + h2( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

− amsin Φ( )gh

a2 + h2( )

fx = m˙ ̇ y = −amsin Φ( ) ˙ χ 0 + g hta2 + h2( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

+ amcos Φ( )gh

a2 + h2( )

fz = m˙ ̇ z − mg = − mga2

a2 + h2( )

€

Φ=c0 + ˙ χ 0t + g ht 2

2 a2 + h2( )

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Bead on a (complicated) wire

wire profile

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This one has to be done numerically

€

T = 12

m ˙ x 2 + ˙ y 2 + ˙ z 2( )

€

V = mgz

Constraints are from the previous slide

x y z

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The constrained Lagrangian

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L = 14

m 25cos 20χ( ) + 8cos 6χ( ) − 35( ) ˙ χ 2 − 12

mg cos2 5χ( )( )


€

25cos 20χ( ) + 8cos 6χ( ) − 35( ) ˙ ̇ χ + 12

500sin 20χ( ) + 48sin 6χ( )( ) ˙ χ 2 − 20gcos 5χ( )sin 5χ( ) = 0

Make two first order equations out of this

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€

˙ u =2 125sin 20χ( ) +12sin 6χ( )( )u2 − 5gcos 5χ( )sin 5χ( )( )

25cos 20χ( ) + 8cos 6χ( ) − 35€

˙ χ = u

The maximum height of the track is unity at c = 0

Starting from rest below unity leads to an oscillation

Starting at c = 0 with no motion gives (unstable) equilibriumAdd motion and we get a roller coaster-like motion

Let’s look at the Mathematica code

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The Lagrangian, the Euler-Lagrange equations and the partition into two equations


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The odes, the initial conditions and a space for the answer

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The solution procedure and the answers

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Start from c = 0 and u =0.1 and let 0 ≤ t ≤ 8πThis gives several runs around the track

We can look at speed vs. timeheight vs. time

energy vs. time (conserved)

RESULTS

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speed

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height

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energy

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speed vs.height

initial speed is 0.1

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Link on a wire

I will eventually specialize to an axisymmetric link on the helical wire,but we can say some general things first

I will align the K axis of the link with the local curve

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Link on a wire

I suppose the link to be small enough that I can neglect the curvature of the wire locally, so I can still impose the position constraints

€

x = x χ( ), y = y χ( ), z = z χ( )

But now I have an orientation constraint

€

K = λ ⇒ sinθ sinφi − sinθ cosφj+ cosθk∝ ′ x i + ′ y j+ ′ z k

€

tanφ = −′ x ′ y ⇒ sinθ

′ x ′ x 2 + ′ y 2

i − sinθ′ y

′ x 2 + ′ y 2j+ cosθk∝ ′ x i + ′ y j+ ′ z k

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Link on a wire

divide lhs by sinq

€

′ x ′ x 2 + ′ y 2

i − ′ y ′ x 2 + ′ y 2

j+ cotθk∝ ′ x i + ′ y j+ ′ z k

€

cotθ =′ z

′ x 2 + ′ y 2

We have determined two Euler angles: f and qWe are left with y, and the system has two degrees of freedom

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Link on a wire

Doing this in general is really difficultand it will obscure what is going on in a sea of algebra

so let’s stick to the helix

€

x = acos χ( ), y = asin χ , z = −hχ

€

′ x = −asin χ( ), ′ y = acosχ , ′ z = −h

€

K = sinθ sinφi − sinθ cosφj+ cosθkλ ∝−asin χ i + acosχ j− hk

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Link on a wire

Start by letting f = c, which is almost intuitive

€

K = sinθ sin χ i − sinθ cosχ j+ cosθkλ ∝−asin χ i + acosχ j− hk

sinq needs to be proportional to –acosq needs to be proportional to -h

€

sinθ = − aa2 + h2

, cosθ = − ha2 + h2

€

q =−sin−1 ha2 + h2

⎛ ⎝ ⎜

⎞ ⎠ ⎟− π

2

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Link on a wire

Two degrees of freedom — c and y — as we noted before

If the link is axisymmetric this problem has a closed from solution!

I’m going to leave this as a problem for you

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We’ve been looking at external constraints on a single link

I’d like to start on internal constraints connecting links

I will split these into orientation constraints and connectivity constraints

I will also split them into simple and nonsimple constraints

Orientation constraints are often simpleConnectivity constraints are usually nonsimple

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Some notation

I’m going to work, for now, with fairly symmetric bodiestheir centers of mass are also their “actual center”

rectangular solidscylinders

ellipsoids. . . .

I will use body coordinates parallel to the principal moments

I will usually (at least for now) connect bodies at their endsand suppose the ends to be in the K direction

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I suppose the semiaxes to be a, b and c in the I, J and K directions

Here b = a.

The connectivity of these two links relates the center of mass of link twoto the center of mass of link one

€

r2 = r1 + c1K1 + c2K 2

This relates x2, y2 and z2 to the coordinates of link 1 and all six Euler angles

It’s a connectivity constraint and it is not simpleit is nonsimple — by which I mean not linear

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€

r2 = r1 + c1 sinθ1 sinφ1i − sinθ1 cosφ1j+ cosθ1k( ) + c2 sinθ2 sinφ2i − sinθ2 cosφ2j+ cosθ2k( )

€

x2 = x1 + c1 sinθ1 sinφ1 + c2 sinθ2 sinφ2

y2 = y1 − c1 sinθ1 cosφ1 − c2 sinθ2 cosφ2

z2 = z1 + c1 cosθ1 + c2 cosθ2

You can see that substituting this into the Lagrangian will lead to quite a mess

It gets worse as we add links

We’re going to learn cool ways to deal with this, but not just yet

The system has nine degrees of freedomIt started with twelve and we took three out

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We can further constrain this system by attaching the first link to the ground at the origin of the inertial system

€

r1 = c1K1, r2 = 2c1K1 + c2K 2

€

x1 = c1 sinθ1 sinφ1, x2 = 2c1 sinθ1 sinφ1 + c2 sinθ2 sinφ2

y1 = −c1 sinθ1 cosφ1, y2 = −2c1 sinθ1 cosφ1 − c2 sinθ2 cosφ2

z1 = c1 cosθ1, z2 = 2c1 cosθ1 + c2 cosθ2

and this system now has six degrees of freedom

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Suppose the two links we looked at to be connected by a hingeinstead of a magic spherical joint

Here I denotes the common Ivector for both links

the direction of the hinge pin

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€

I1 = cosφ1 cosψ1 − sinφ1 cosθ1 sinψ1( )i + sinφ1 cosψ1 + cosφ1 cosθ1 sinψ1( )j+ sinθ1 sinψ1k

I2 = cosφ2 cosψ 2 − sinφ2 cosθ2 sinψ 2( )i + sinφ2 cosψ 2 + cosφ2 cosθ2 sinψ 2( )j+ sinθ2 sinψ 2k

and these will be equal if

€

φ2 = φ1, ψ 2 = 0 =ψ1

(It is possible to have a more complicated hinge relationand we may get to that near the end of the course or we may not.)

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Think about this operationally

We rotate both of these through the same f angle

Then we rotate each about a different q angle

We do not rotate either about its y angle

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initial position

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first rotation

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second rotations

lecture 5: constraints i

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