Lecture 5: Constraints I
Constraints reduce the number of degrees of freedom of a mechanism
holonomic constraints
external constraints (bead on a wire)
internal constraints (connectivity constraints)
orientation constraints
1
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.)
2
Bead on a wire
I call it a bead, and I’ll treat it as a point
— three degrees of freedom
mg
3
A little digression on the properties of curves in space
the tangent vector

dp
ds

This is the unit tangent vector
we’ll do a lot with nonunit tangent vectors
4
Bead on a wire
Curve is defined by a parameterization
x  xs, y  ys, z  zs
x  x, y  y, z  z
arclength
general parameter


dx  dx 
ds  d 
dy  
dy 
d
     
ds  d ds
dz  dz 
ds  
d 


5
Bead on a wire
Kinetic energy
2


1
1
d

T  mxÝ2  yÝ2  zÝ2   mx2  y2  z2  
dt 
2
2

add in the potential energy and write the Lagrangian

1
2
2
2
L  m x  y  z
2
d2
   m gz
dt 


6
Bead on a wire
The only variable here is  (a proxy for s), so we build one EL equation


L
2
2
2
Ý
 m x  y  z 
Ý

d L 
d
2
2
2
2
2
2
Ý
Ý
Ý

m
x


y


z



m
x  y  z 
 
 
 
 
Ý
dt 
dt




L
Ý2  mg z
 mx  y  z



7
Bead on a wire
The Euler-Lagrange equation


Ý
Ý mx  y  z
Ý2  m gz 0
m x    y    z  
2
2
2

8
Bead on a wire
We can solve this analytically in the simple case where
x  y   z
2
2
2
is a constant.

The helix is such a case
x  acos , y  asin , z  h
x  y  z  a2 sin2   a2 cos2   h 2  a2  h 2

2
2
2

9
Bead on a wire


L
2
2
2
Ý ma 2  h 2 
Ý
 m x    y    z  
Ý


so the Euler-Lagrange equation is
Ý
Ý mgh 0
m1 h 2 
which is easily integrated to give

ht2
Ý0t  g
   0  
2a2  h 2 

10
Bead on a wire
From which we obtain


ht 2

Ý
x  acos
 0  0 t  g 2 a 2  h 2 





ht 2

Ý
y  asin
 0  0 t  g 2 a 2  h 2 



h 2t 2
Ý0 t  g
z  h 0  h
2a 2  h 2 

11
Bead on a wire
The force on the bead from the wire

2
ht
gh

Ý amcos
Ý0  g 2
f x  mxÝ

amsin

  2 2
2 

a

h


a  h 

2


ht
gh

Ý amsin
Ý0  g 2
f x  myÝ

amcos

  2 2
2 

a

h


a  h 

m ga2
f z  mÝ
zÝ m g  2
a  h 2 

ht2
Ý0 t  g
   0  
2a2  h 2 
12
Bead on a (complicated) wire
wire profile
13
Bead on a (complicated) wire
This one has to be done numerically
1
T  mxÝ2  yÝ2  zÝ2 
2
V  mgz
 the previous slide
Constraints are from

x
y
z
14
Bead on a (complicated) wire
The constrained Lagrangian
L
1
1
Ý2  mgcos2 5
m25cos20  8cos6   35
4
2
The Euler-Lagrange equation

Ý
Ý
25cos20  8cos6  35
1
Ý2  20gcos5sin5   0
500sin20  48sin6

2
Make two first order equations out of this
15
Bead on a (complicated) wire
Ý u

uÝ


2 125 sin20    12 sin6  u 2  5gcos 5  sin5  
25 cos 20    8cos 6    35
 The maximum height of the track is unity at  = 0
Starting from rest below unity leads to an oscillation
Starting at  = 0 with no motion gives (unstable) equilibrium
Add motion and we get a roller coaster-like motion
Let’s look at the Mathematica code
16
Bead on a (complicated) wire
The Lagrangian, the Euler-Lagrange equations and the partition into two equations
The Euler-Lagrange equation
17
Bead on a (complicated) wire
The odes, the initial conditions and a space for the answer
18
Bead on a (complicated) wire
The solution procedure and the answers
19
Bead on a (complicated) wire
RESULTS
Start from  = 0 and u =0.1 and let 0 ≤ t ≤ 8π
This gives several runs around the track
We can look at speed vs. time
height vs. time
energy vs. time (conserved)
20
Bead on a (complicated) wire
speed
21
Bead on a (complicated) wire
height
22
Bead on a (complicated) wire
energy
23
Bead on a (complicated) wire
speed vs.height
initial speed is 0.1
24
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
25
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, y  y, z  z
But now I have an orientation constraint

K    sin sin i  sin  cos j  cos k  x i  y j  zk

tan  
x 
x 
y 
 sin
i

sin

j  cosk  x i  y j  zk
2
2
2
2
y 
x   y 
x   y 

26
Link on a wire
divide lhs by sin
x 
x   y 
2

2
i
y 
x   y 
2
2
cot 
j  cotk  x i  y j  zk
z
x 2  y 2
We have determined two Euler angles:  and 
 with y, and the system has two degrees of freedom
We are left
27
Link on a wire
Doing this in general is really difficult
and it will obscure what is going on in a sea of algebra
so let’s stick to the helix
x  acos , y  asin , z  h
x  asin , y  acos, z h


K  sin  sin i  sin  cosj  cosk
  asin i  acosj  hk

28
Link on a wire
Start by letting  = , which is almost intuitive
K  sin  sin i  sin  cosj  cosk
  asin i  acosj  hk
sin needs to be proportional to –a
 cos needs to be proportional to -h
sin  

a
a2  h
, cos  
2
h
a2  h 2
 h
 
  sin  2

2
 a  h  2
1
29
Link on a wire
Two degrees of freedom —  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
30
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 simple
Connectivity constraints are usually nonsimple
31
Some notation
I’m going to work, for now, with fairly symmetric bodies
their centers of mass are also their “actual center”
rectangular solids
cylinders
ellipsoids
....
I will use body coordinates parallel to the principal moments
I will usually (at least for now) connect bodies at their ends
and suppose the ends to be in the K direction
32
33
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 two
to the center of mass of link one
r2  r1  c1K1  c 2K 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 simple
it is nonsimple — by which I mean not linear
34
r2  r1  c1sin1 sin 1i  sin1 cos1j  cos1k  c 2 sin2 sin 2i  sin2 cos2 j  cos2k
x 2  x1  c1 sin 1 sin 1  c 2 sin 2 sin  2
y 2  y1  c1 sin1 cos1  c 2 sin 2 cos 2
z2  z1  c1 cos1  c 2 cos 2
The system has nine degrees of freedom
It started
 with twelve and we took three out
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
35
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  c 2K 2
x1  c1 sin 1 sin 1, x 2  2c1 sin 1 sin 1  c 2 sin  2 sin  2
y1  c1 sin 1 cos1, y 2  2c1 sin1 cos1  c 2 sin 2 cos 2
z1  c1 cos1, z2  2c1 cos1  c 2 cos 2
and this system now has six degrees of freedom

36
Suppose the two links we looked at to be connected by a hinge
instead of a magic spherical joint
Here I denotes the common I
vector for both links
the direction of the hinge pin
37
I1  cos1 cosy1  sin 1 cos1 siny1i  sin 1 cosy1  cos1 cos1 siny1j  sin 1 siny1k
I2  cos2 cosy2  sin 2 cos2 siny2 i  sin 2 cosy2  cos2 cos2 siny2 j  sin2 siny2k
and these will be equal if
 2  1, y 2  0  y1
(It is possible to have a more complicated hinge relation
 may get to that near the end of the course
and we
or we may not.)
38
Think about this operationally
We rotate both of these through the same  angle
Then we rotate each about a different  angle
We do not rotate either about its y angle
39
initial position
40
first rotation
41
second rotations
42