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Lecture 19. The Method of Zs When problems get complicated numerical complexity makes computation SLOW The method of Zs speeds the computation up We will eventually look at the third problem of problem set 5 which is computationally impractical (at least in Mathematica) without Zs 1

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We set this in the context of Hamilton’s equations with nonholonomic constraints I generally replace connectivity constraints with pseudononholonomic constraints When I do this we recall that we can write The momentum can be written in terms of u j. where I am borrowing Mathematica’s substitution operator 2

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3 The game is to isolate the us, writing the derivatives as coefficients times us I will denote the coefficients in the momentum (Hamilton and reduced Hamilton) equations by Zs This is going to seem a wee bit weird Bear with me. It’ll make sense eventually

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We note that p i is linear in u j, which we can write as 4 We can obtain the Zs from the usual momentum expression or The latter is often easier

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5 It is clear from the expression that Z does not depend on u

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6 We already have the equations for the evolution of q: Hamilton’s equations become where any explicit We need to replace

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7 So that we have Hamilton’s equations in the form Remember that these are not actually correct because I have not written the Lagrange multipliers is the correct form

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8 We have learned to multiply by S to obtain the correct form the reduced Hamilton’s equations We need to solve these simultaneously with the q equations

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9 is, formally, a terribly complicated expression However, if we relegate the connectivity constraints to the pseudononholonomic world most of the derivatives of M are zero, and it is not at all bad We will see this in context

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10 The full formal expression for the reduced Hamilton’s equations This is a set of nonlinear first order ordinary differential equations in u the coefficients of which are functions of q, often quite complicated functions This would be a snap to integrate numerically were the coefficients to be constant The method of Zs pretends that they are

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11 I have adapted the method of Zs from Kane and Levinson’s 1983 paper, cited in the text Their work used Kane’s method, which I explore in the text, but not in class My method of Zs is not as complete as it might be, because I allow some of the coefficients to remain as explicit functions of q This relies onactually being pretty simple functions of q (and also the generalized forces)

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12 Write Hamilton’s equations as follows We replace the Z variables by constants This choice of substitutions aligns with Hamilton’s equations, not the reduced equations

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13 Of course, they are not constants, so we have to add algebraic equations to our system to allow us to calculate them as we go forward That has been a lot to digest, and we’ll go through an example shortly Let me try to summarize this first

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14 the simple holonomic constraints generalized coordinates Lagrangian nonholonomic and pseudononholonomic constraints Hamilton’s equations Start from square one introduce the Zs reduced Hamilton’s equations with Zs generalized forces

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15 Consider the little sphere on the big sphere — both free to roll

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16 How do we set this up? Set the K axis of the big sphere horizontal as before for rolling I can choose the inertial axis such that the small sphere is on the i axis without loss of generality If all I care about is the ball rolling down starting from rest then the simplest thing to do is but its K axis horizontal as well

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18 It might be fun to start the ball rolling on a line of latitude The easiest way to do that is to put the K axis on a line of longitude which we can do by the proper choice of and . We want K to be Pick

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20 Go to Mathematica to look at this

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SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2. SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10.

SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2. SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10.

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