Download presentation

Presentation is loading. Please wait.

Published byBraxton Hill Modified about 1 year ago

1
Lecture 7 Rolling Constraints The most common, and most important nonholonomic constraints They cannot be written in terms of the variables alone you must include some derivatives The resulting differential is not integrable 1

2
The major assumption/simplification is point contact This is true of a rigid hoop on a rigid surface or a rigid sphere on a rigid surface 2 I will also generally assume that the surface is horizontal so that gravity is normal to the surface

3
3 rigid hoop on a rigid surface

4
4 rigid sphere on a rigid surface

5
5 It’s not a bad approximation for a well-inflated bicycle or motorcycle tire on hard pavement It’s not so good for an automobile tire, which has a pretty big patch on the ground Cylindrical wheels, like those on grocery carts and children’s toys must slip over parts of their surfaces when turning We aren’t going to care about any of this We’re going to suppose that all the wheels we consider have point contact

6
6 Two fundamental ideas: The speed of a point rotating about a fixed point is The contact point between a wheel and the ground is not moving

7
7 For a sphere of radius a r = ak This is inertial, so we have

8
8 We can write the constraints as Of course, we also have the simple constraint for this problem: z = a.

9
9 An erect wheel has the same r If we choose to let the axle be the K body coordinate then = ±π/2, fixed, and the constraint becomes The general wheel is fancier; I’ll get to that shortly BUT FIRST

10
10 How do we impose these? The classical method is through Lagrange multipliers, and that’s what we’ll do today (There’s an alternate method that we will see later in the semester) To get at this we need to go back to first principles and it helps to note that we can write the constraints in matrix form

11
11

12
12

13
13 We can view the time derivative of q as a surrogate for q Now let’s go back to lecture two and the action integralit is stationary if Consider the final version

14
14 and let’s add multiples of the constraints to this They are zero, so there’s no problem multiply each row by its own constant change the dummy index j to k

15
15 drop the asterisk and incorporate external forces directly the constrained Euler Lagrange equations I have added as many variables as there are nonholonomic constraints I need as many extra equations, which are the constraints themselves

16
16 We have a physical interpretation of the Lagrange multipliers They are the forces (and torques) of constraint — what the world does to make the system conform How do we actually apply this? The explanation is on pp of chapter three, perhaps a bit telegraphically I’d like to go through the erect coin, example 3.5, pp

17
17 First let’s see how we can set up a wheel I like the axle to be the K axis; let’s see how we can erect the wheel Starting with all Euler angles equal to zero

18
18 Pick a direction by adjusting

19
19 erect the wheel with = -π/2 (the text example uses + π/2)

20
20 Variables are x, y, and — z and are fixed has a k component, but it’s perpendicular to r

21
21 The Lagrangian is simple I’ve left out the gravity term because the center of mass cannot move vertically The constraint matrix is

22
22 The product The Lagrange equations are then and the constraints are

23
23 So I have six equations in six unknowns: the variables and the multipliers Solve the first two dynamical equations for the multipliers The remaining dynamical equations become and I can use the constraints to eliminate the acceleration terms

24
24 substitute all this back in so, we have an exact solution for the two angles

25
25 and we can substitute this into the constraints and integrate to get the positions The coin rolls around in a circle of constant radius, which can be infinite, in which case

26
26 Suppose we consider a general rolling coin, one not held vertical The book doesn’t deal with this until chapter five, but we can certainly set it up here. It’s a hard problem and does not have a closed form solution

27
27 This has the same body system as before but the angle can vary (it’s equal to -0.65π here) r remains equal to –aJ 2 but we need the whole

28
28 and, after some simplification

29
29 The last one is actually integrable, but I won’t do that We have a constraint matrix that appears to disagree with the one on p. 15 of chapter five. The difference stems from the assumption in the book that ≥ 0 whereas in the picture here, ≤ 0. Choose the one that fits your idea of the initial conditions.

30
30 I will use the book’s picture, and I will go to Mathematica to analyze this system

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google