Lecture 8: Rolling Constraints II
Generalizations and review
Nonhorizontal surfaces
Coin rolling on a slope
Small sphere rolling on a larger sphere’s surface
Hoop rolling inside a hoop
1
What can we say in general?
vector from the inertial origin to the center of mass





 


2
2
1
Ý
Ý
Ý
Ý
Ýcos
Ý 
L  IXX cos  sin sin  IYY sin  sin cos  IZZ 
2
1
 mxÝ2  yÝ2  zÝ2  mg R
2

2

Ýcos  
Ýsin  
Ý 
Ýsin sin  i  
Ýsin cos j 
Ýcosk
  
v  r


vector from the contact point to the center of mass
2
We can restrict our attention to axisymmetric wheels
and we can choose K to be parallel to the axle
without loss of generality





2
2
1
Ý
Ý
Ý
Ý
Ýcos
Ý 
L  A cos  sin sin  B sin  sin  cos  C 
2
1
 mxÝ2  yÝ2  zÝ2  mg R
2

 
1
Ý2  
Ý2 sin2   C 
Ýcos
Ý 
L  A 
2


2

2
1
 mxÝ2  yÝ2  zÝ2  mg R
2
mgz
3
If we don’t put in any simple holonomic constraint (which we often can do)
x 
 
y 
z 
q   
 
 
 
 

4
We know v and  in terms of q
any difficulty will arise from r
v  r
r  aJ 2

Actually, it’s something of a question as to where the difficulties will arise in general

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
5
d L 
 i  mg x   j C1j
dt xÝ 
d L 
j
 i  m gy   j C2
dt yÝ 

d L 
 i  mg z   j C3j
dt zÝ

d L 
j
 Ýi   j C4
dt  




d L  L
  j C5j
 Ý
dt  
d L 
j
 Ý  j C6
dt 
6
We have the usual Euler-Lagrange equations
d L  L
j
 i  i   j Ci
dt qÝ q
and we can write out the six equations

7
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 plane
and choose i to to be horizontal, so that j points down hill
(This is a rotation of the usual system about i)
8
y  cosy  sinz'
z  cosz siny 

9
so the potential energy in the primed coordinates is
V  mgcosz siny 
the kinetic energy is unchanged

We can go forward from here exactly as before
everything is the same except for gravity
10
This has the same body system as before
but the angle  can vary
(it’s equal to -0.65π here)
r remains equal to –aJ2
but we need the whole 
11
??
Let’s look at a rolling coin on a tilted surface in Mathematica
12
Curved surfaces
Spherical surface: spherical ball on a sphere
Two-d surface: wheel inside a wheel
General surface
13
Spherical ball on a sphere
a
holonomic constraint
x 2  y 2  z 2  R  a
2
R

14
The Lagrangian simplifies because of the spherical symmetry


1 Ý2 Ý2 Ý2
Ý
Ýcos  1 mxÝ2  yÝ2  zÝ2  mgz
L  A 
     2
2
2
We have a constraint, which we can parameterize

x 2  y 2  z 2  R  a
2

x  R  asin cos, y  R  asin sin , z  R  acos

15
which transforms the Lagrangian
L


1 Ý2 Ý2 Ý2
Ý
Ýcos
A       2
2
1
Ý2  
Ý2 1 cos2   mR  agcos 
 mR  a 2
2



We can now assign generalized coordinates

 
 


 

q   
 
 

 

16
We have rolling constraints
 is unchanged, and r is as shown on the figure
and we recalculate v

 

Ýcos  
Ýsin  
Ý 
Ýsin sin  i  
Ýsin cos j 
Ýcosk
  
asin  cos 


rC  asin  sin  
 acos 



Ýcos cos  

Ýsin  sin  
Ý

Ý
v

a

R

cos

sin



cos

sin




 


Ýsin 





17
The rolling constraint appears to have three components
but the normal component has already been satisfied
v   r  0
The normal is parallel to r, so I need two tangential vectors

k  r v    r  0
k  k  r v    r   0

18
We have the usual Euler-Lagrange equations
d L  L
j
 i  i   j Ci
dt qÝ q
and we can write out the five equations

19
d L 
j
 Ý  j C1
dt 

d L  L
  j C2j
 Ý
dt  
d L 
j
 Ý  j C3
dt 


d L  L
  j C4j
 Ý
dt  
d L 
j
   j C5
Ý
dt 


20
The constraint matrix is
 2 2
1 2
a
sin


a sin2 cos    a 2 sin  sin  cos   sin   
0

2

1
1 2
1
 0
 a 2 sin2 sin   
a sin2  sin  cos   
 aR  asin2 

2
2
2

R  asin  
0
The last two Euler Lagrange equations are suitable for eliminating the Lagrange multipliers
21



After some algebra
R a
Ý sin 
Ý
Ý
Ý
2cos
asin 
R  a Ý2
R a
mg
Ý
Ý
2  m
  m

a
acos sin 
acos
1  m


I have three remaining Euler-Lagrange equations
and
 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??
22
Wheel within a wheel
Treat them both as hoops
radii r1 > r2
y 2  r1  r2 sin 
z2  r1  r1  r2 cos

23
We have holonomic constraints
Put us in two dimensions
x1  0  x 2
1 

2
1  
 2

2
 2
realize that z1 = r1

24
We have an interesting connectivity constraint —
define the position of the small wheel in terms of the angle 
y 2  y1  r1  r2 sin , z2  r1  r1  r2 cos
Putting
all this in gives us a Lagrangian

L
1
1
1
1
1
2 2
2
2
2Ý
2Ý
Ý
Ý
ÝÝ
m

m
y

m
r


m
r


m
r

r


m
r

r
 1 2 1
11 1
2 2 2
2 1
2
2 1
2  cos y1
2
2
2
2
2
 gm2 r1  r2 cos  gr1 m1  m2 
25
We define a vector of generalized coordinates
y1 
 
 
q   1 
 2 

 


26
There will be two nonholonomic constraints
Ý1,
yÝ1  r1
Ý2
Ý r2
r1  r2 
The corresponding constraint matrix is

1 r1 0
C  
0 0 r2
0 

r1  r2 

27
The second and third Euler-Lagrange equations are fairly simple
so I will use those to find the two Lagrange multipliers
Ý
Ý1, 2  m2r2
Ý
Ý2
1  m1r1
To solve the problem we use the first and fourth Euler-Lagrange equations
and
 the differentiated constraints
The solution is numerical and we need to go to Mathematica to look at it.
QUESTIONS FIRST??
28
That’s All Folks
29