Dynamics of a nonhomogeneous rolling ball and its impulsive control

Proiect
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Domeniu: Fizică
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Pagini : 36 în total
Cuvinte : 2274
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Publicat de: Isabela Duma
Puncte necesare: 7
Profesor îndrumător / Prezentat Profesorului: Self

Cuprins

  1. Kinematics of the rolling ball
  2. 2 Dynamics of a symmetric rolling ball
  3. 3 Optimal control
  4. 4 Impulsive approximation to optimal control

Extras din proiect

1 Kinematics of the rolling ball

Description of the model: kinematics

Consider a ball of unit radius rolling on a

at surface (plate,

ground).

A con guration is an element (x;A) 2 R2 ș SO(3) (position and

orientation).

If the canonical basis fe1; e2; e3g is attached to the ball at its

original position, then these vectors get rotated to fAe1; Ae2;Ae3g

Recall the usual isomorphism^: R3 ! so(3) given by

v 7!

2

64

0 ????v3 v2

v3 0 ????v1

????v2 v1 0

3

75

The spatial angular velocity ! 2 R3 is de ned through !^ = A_ A????1.

Nonholonomic constraints:

No slipping: ! ș e3 = _ x (velocity of the contact point on the

ball is zero)

No spinning: !3 = h!; e3i = 0 (Veselova's constraint)

These can be combined into

! = e3 ș _ x

One of the main consequences of including the no spinning"

condition is that if you make the ball roll along a given path

x(t) on the ground, then its orientation A(t) is uniquely

determined.

e1

e2

e3

A(t1)

x(t)

Principal bundle and principal connection

Con guration space R2 ș SO(3)

Action of the group SO(3) on the con guration space:

g  (x; A) = (x; Ag????1)

Quotient space: R2

The rolling constraint is SO(3)-invariant and complementary

to the vertical distribution

Principal connection, with connection 1-form

A: T(R2 ș SO(3)) ! so(3)  R3 given by

A(x;A; _ x; _A) = A????1(e3 ș x_ )A ???? A????1_A =

(body angular velocity consistent with _ x)

????

(actual body angular velocity)

Allowed motions: A = 0 (horizontal", in principal bundle

terms)

Given a path x(t), the resulting reorientation is computed by

taking the horizontal lift of x(t).

Curvature two-form

Let X; Y 2 TxR2 ,! TR3. Consider their horizontal lifts

Xh

(x;A) =



x; A; X; (e3ș X)A



Y h

(x;A) =



x; A; Y ; (e3ș Y )A



:

The curvature two-form B = dA ???? [A;A] of the principal

connection A for these two vectors turns out to be

B(x;A)



Xh

(x;A); Y h

(x;A)



= ????A????1 (X ș Y ) 2 R3

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