1. 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

2. What can we say in general?
vector from the inertial origin to the center of mass
Lagrangian
rotation vector
constraint
vector from the contact point to the center of mass

3. mgz We can restrict our attention to axisymmetric wheels
and we can choose K to be parallel to the axle
without loss of generality
g dot R is just gz if z denotes the oriented inertial coordinate
mgz

4. If we don’t put in any simple holonomic constraint (which we often can do)

5. We know v and w in terms of q any difficulty will arise from r
Actually, it’s something of a question as to where the difficulties will arise in general
This will depend on the surface
I don’t know if we’ll get to the general 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

7. We have the usual Euler-Lagrange equations
and we can write out the six equations

8. 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)

10. so the potential energy in the primed coordinates is
the kinetic energy is unchanged
We can go forward from here exactly as before
everything is the same except for gravity

11. This has the same body system as before
but the angle q can vary
(it’s equal to -0.65π here)
r remains equal to –aJ2
but we need the whole w

12. ??
Let’s look at a rolling coin on a tilted surface in Mathematica

13. Curved surfaces
Spherical surface: spherical ball on a sphere
Two-d surface: wheel inside a wheel
General surface

14. Spherical ball on a sphere
holonomic constraint R

15. The Lagrangian simplifies because of the spherical symmetry
We have a constraint, which we can parameterize

16. which transforms the Lagrangian
We can now assign generalized coordinates

17. We have rolling constraints
w is unchanged, and r is as shown on the figure
and we recalculate v

18. The rolling constraint appears to have three components
but the normal component has already been satisfied
The normal is parallel to r, so I need two tangential vectors

19. We have the usual Euler-Lagrange equations
and we can write out the five equations

21. The constraint matrix is
The last two Euler Lagrange equations are suitable for eliminating the Lagrange multipliers

22. QUESTIONS FIRST?? After some algebra
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??

23. Wheel within a wheel
Treat them both as hoops
radii r1 > r2 c

24. We have holonomic constraints
Put us in two dimensions
realize that z1 = r1

25. We have an interesting connectivity constraint —
define the position of the small wheel in terms of the angle c
Putting all this in gives us a Lagrangian

26. We define a vector of generalized coordinates

27. There will be two nonholonomic constraints
The corresponding constraint matrix is

28. The second and third Euler-Lagrange equations are fairly simple
so I will use those to find the two Lagrange multipliers
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??

29. That’s All Folks