     # Rigid Body Mechanics

• View
214

0

Embed Size (px)

### Text of Rigid Body Mechanics

Rigid Body Mechanics 2

In which we find that a rigid body has six degrees of freedom, learn how to describe the orientation of

a rigid body in terms of Euler angles, define inertial and body coordinates and find the Euler-Lagrange

equations for a single rigid body. . .

Kinematics

I will develop the idea of a rigid body using what we have already from systems of particles. A rigid

body can be viewed as a system of constrained particles. I will need to discuss degrees of freedom and

constraints as part of this process. This will be useful as an introduction to a fuller discussion of

constraints in Chap. 3.

The number of degrees of freedom of a system is the minimum number of variables needed to

specify the system. A particle has three degrees of freedom: translation in three independent

directions (typically with respect to a Cartesian axis, but this is not necessary). Two particles have

six degrees of freedom three each but were they rigidly attached, the total number of degrees of

freedom would be reduced to five. The argument is simple. Denote the positions of the two particles

by r1 and r2 respectively. If the particles are rigidly attached the distance between them is fixed. We

say that there is a constraint enforcing this:

r1 r2 r1 r2 d2 (2.1)

where d denotes the distance between the two particles. I have written the constraint in terms of the

coordinates of the two masses. This is the definition of a holonomic constraint. (I will address

nonholonomic constraints, which cannot be so written, in Chap. 3.) In principle, we can use the

constraint to eliminate one of the six coordinates, but this is an unwieldy task (try it!), and it is easier

to parameterize the constraint1. One useful way to do this is to write

x2 x1 d cos sin ; y2 y1 d sin sin ; z2 z1 d cos (2.2)

1 I will address another alternative in Chap. 3.

R.F. Gans, Engineering Dynamics: From the Lagrangian to Simulation,DOI 10.1007/978-1-4614-3930-1_2, # Springer Science+Business Media New York 2013

31

http://dx.doi.org/10.1007/978-1-4614-3930-1_3http://dx.doi.org/10.1007/978-1-4614-3930-1_3http://dx.doi.org/10.1007/978-1-4614-3930-1_3

(The reader can verify that this satisfies Eq. 2.1). The five degrees of freedom are then described by

the three components of r1 and the two angles. (The angles will look familiar: corresponds to theusual spherical coordinate angle and to . I use and because I want to reserve and [and ]for Euler angles). The parameterization says that r2 lies on a sphere surrounding r1.

Adding a third point leads us directly to the rigid body. The thought process is as follows. Suppose

the third point to be at r3. Denote the fixed distances between the points by d12, d13, d23, respectively.The third point must lie on a sphere of radius d13 from r1 and a sphere of radius d23 from r2. Two

spheres intersect in a circle, so the third point can be anywhere on that circle. This means that it has

but one degree of freedom, so that the system of three rigidly attached particles has six degrees of

freedom, the five associated with the first two rigidly-connected particles, plus the one additional

degree of freedom, the location of the third point on the circle of intersection. Figure 2.1 shows the

original two points and the two spheres.

Each additional particle must lie at the intersection of at least three spheres. Three spheres intersect

at a point (if at all), so additional particles add no additional degrees of freedom. The number of

degrees of freedom of a single rigid body is six.

A Geometrical Digression onto a Plane

It is interesting to show how this last degree of freedom can be written. The vector from r1 to r2 is

perpendicular to the plane defined by the circle of intersection. Figure 2.2 shows a snapshot of the

Fig. 2.1 The intersection of the spheres surrounding the points {0, 0, 0} and {1.56, 0.48, 1.15}, which points areseparated by 2 units

Fig. 2.2 Cross section of the plane defined by r1, r2, r3

32 2 Rigid Body Mechanics

configuration in the plane defined by the three points. This plane passes through the line in Fig. 2.1

and some point on the circle of intersection. r denotes the radius of the circle. a and b define the

location of the center of the circle. Applying the Pythagorean theorem to the two right triangles in the

figure and subtracting gives

a2 b2 a b a b d213 d223 (2.3)

Since a + b d12, one can find expressions for a and b in terms of the three known distances, viz.

a d212 d213 d223

2d12; b d

212 d213 d223

2d12(2.4)

The determination of r is straightforward from the two Pythagorean equations given a and b. Theresult is

r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2d212 d

213 d223

d412 d213 d223 2q

2d12(2.5)

The general vector perpendicular to r12 r2r1 can be formed from two mutually perpendicularvectors. These can be any two of the three Cartesian unit vectors crossed into r12. (At least two of

them will not be parallel to r12.) Given that

r12 d12 cos sin sin sin cos f gT (2.6)

we can write two independent unit vectors (except in the case that i or j is parallel to r12) as

v1 0 cos sin sin f gTffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

cos2 sin2sin2p ; v2 cos 0 cos sin f gTffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

cos2 cos2sin2p (2.7)

The location of r3 can then be written as

r3 r1 ar12 r cos v1 sin v2 (2.8)

This parameterizes the six degrees of freedom by the three coordinates of r1 and the three angles , and . This is an ad hoc formalization of the fact established above that a general rigid body has sixdegrees of freedom, three associated with position and three associated with rotation. I will now move

on to the standard formalization(s) of this fact through the Euler angles.

We need a general way to describe the position and orientation of any rigid body. We know that this

will require six generalized coordinates. Imagine a set of body coordinates coordinates referred to a

Cartesian frame fixed in the body. Let the origin of the body coordinate system be at the center of

mass, which is given by the logical extension to the continuum of the center of mass of a collection of

particles

Kinematics 33

rCM

volume

rdV

m(2.9)

The first three degrees of freedom are the inertial coordinates of the center of mass; the other three

come from the orientation of the body, the angular relations between the body axes and the inertial

axes. These six variables uniquely define the position of a body. I will sometimes refer to this as the

posture of the body. I will replace the ad hoc description above with the standard description in terms

of the Euler angles, which I will now develop.

Denote the body Cartesian coordinates by (X, Y, Z) and denote the corresponding unit vectors by

I, J,K. Denote the unit vectors for the inertial (x, y, z) system by i, j, k. Any vector can be defined with

respect to either coordinate system. Denote the inertial description by v and the body description byV. The connection between these two can be defined using the basis vectors. Well have

v vxi vyj vzk V VXI VYJ VZK

The inertial components in terms of the body components are

vx VXi I VY i J VZi Kvy VXj I VYj J VZj Kvz VXk I VYk J VZk K

which can be written in matrix form as

vx

vy

vz

8>:

9>=>;

i I i J i Kj I j J j Kk I k J k K

8>:

9>=>;

VX

VY

VZ

8>:

9>=>; (2.10)

The body components can be written in terms of the inertial coordinates by a similar process. The

matrix form is

VX

VY

VZ

8>:

9>=>;

i I j I k Ii J j J k Ji K j K k K

8>:

9>=>;

vx

vy

vz

8>:

9>=>; (2.11)

The two matrices are both mutual transposes and mutual inverses. All that we need to complete the

process is to find the dot products. We can do this formally using the Euler angles, but it will be

helpful to go through the process in a less formal, more ad hoc process. The manipulation of vectors

and angles is a remarkably important part of three dimensional rigid body mechanics, so more than

one exposition is not amiss. We need a relationship between the inertial basis vectors and the body

basis vectors, and we can develop that by supposing that the two systems start out coincident and that

we then rotate the body basis vectors with respect to the inertial frame. There are many ways to do

this. I will use the common Euler angles in an intuitive method, which I will formalize shortly. The

idea is to rotate the body frame about three of its own axes successively by the angles , , and . Thedifference between this and the formalism is that I take an inertial point

Recommended ##### Rigid body mechanics in Galilean andrew/papers/pdf/2002f.pdf · Rigid body mechanics in Galilean spacetimes
Documents ##### Geometry, Kinematics, and Rigid Body Mechanics in page.math.tu- gunn/Documents/Papers/ThesisMPKCh1-2.pdf ·
Documents ##### ME 204 Engineering Mechanics: Body Mechanics ... Plane Kinematics of Rigid Bodies 16/23 ... ME 204 Engineering Mechanics: Dynamics – Plane Kinematics of Rigid Bodies
Documents ##### Rigid body dynamics - vilmart/  · Title: Rigid body dynamics Name: Gilles Vilmart
Documents ##### CHAPTER 8 MECHANICS OF RIGID BODIES: PLANAR MOTION Mechanics...آ  1 CHAPTER 8 MECHANICS OF RIGID BODIES:
Documents ##### Chapter Two Laith Batarseh. Definition Engineering mechanics Deformable body mechanics Rigid body mechanics DynamicsStatics Fluid mechanics Constant Velocity
Documents ##### Analytical Mechanics: Rigid Body Rotation - 立命館大学 hirai/edu/2016/analyticalmechanics/handout/...Analytical Mechanics: Rigid Body Rotation Shinichi Hirai Dept. Robotics, Ritsumeikan
Documents Documents