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2.12. Specification of a particle motion and the information compression principle 31
2.12. Specification of a particle motion and the information
compression principle
Even though kinematics can not be considered as an ideal theory, it possesses the system approach techniques.
In a fixed system of coordinates the position, velocity and acceleration of a
particle are defined by:
r = xi + yj + zk,
2
d
a =
dt
dr
V =
2
d
r
2
x
=
2
dt
i +
dt
=
2
y
d
j +
2
dt
dy
dx
i +
dt
2
z
d
k. (2.12)
2
dt
dt
j +
dz
dt
k,
In a moving system of coordinates the compression of information permits
to obtain more compact expressions:
s = s(t),V =˙sτ,
where ρ
2
d
r
a =
−1
=¨s ·τ +˙s ·
2
dt
is the curvature of curve at given point. Only the mathematical
dτ
=¨s ·τ +˙s
dt
dτ
2
=¨s ·τ +
·
ds
2
˙s
n, (2.13)
ρ
component of the information operator compresses an information, whereas the
physical component doesn’t, since unit vectort (tangent to the trajectory) and
unit vector n (normal to the osculating plane) should be controlled.
2.13. Differentiation of a vector of unit length and the analogy
principle
In a vector from the differentiation of a vector of unit length can written as
db
= w ×b, (2.14)
dt
where w is the angular velocity of the rotation of vectorb. Let’s replace vectorb
by position vectort which locates the point on a rigid body rotating about a fixed
axis. Then we should replace vectorb in formula (2.13) by position vector r
given with respect to reference point O on the axis of the rigid – body rotation.
After that we obtain Euler’s formula
v = w ×R. (2.15)
32 Chapter 2
Formula (2.14) is applied also for the rigid – body rotation about a fixed
point. But now w denotes an instantaneous axis. One can see the technique of an
ideal theory, since the mathematical “shell” of the formula holds but the physical
feeling changes. This is properly obvious in definition of an acceleration of a
point on the rigid body rotating about a fixed axis, or rotating about a fixed
point (see Figs. 2.7, 2.8). In the first case the angular acceleration vector
coincides in direction with the angular velocity vector, but in the second case
the instantaneous angular acceleration doesn’t coincide with the instantaneous
angular velocity. The analogy holds also for Poisson’s formulas:
O
1
di
= w ×i,
dt
dj
= w ×j,
dt
dk
= w ×k. (2.16)
dt
w
r
w
r
e
O
Figure 2.7. Rigid-body rotation
body about a fixed axis
Figure 2.8. Rigid-body about a fixed point
2.14. The information operator and velocity and acceleration
diagrams for a body moving with a general plane motion
When the velocity and acceleration diagrams are constructed for a body
moving with general plane motion, the mathematical component of the information operator is most evident in the vector form:
V
=VA+VBA, (2.17)
B
= aA+ aBA. (2.18)
a
B
2.14. The information operator and velocity and acceleration diagrams 33
The mathematical analogy is obvious in expressions (2.16) and (2.17). The
physical component of information operator is presented here by the theorem
about projections of points velocities onto the line connecting these points. The
velocity of a body along line AB is constant (see Fig. 2.9), since the body is a
rigid and can not be deformed.
r
e
V
A
C
r
V
A
Figure 2.9. General plane motion of a rod
O
r
V
B
r
V
C
B
There is an analogy between the constancy of a linear velocity along the
line connecting two points on a rigid body, and the moving of a force along
it’s line of action in statical loading. The simplest method of constructing the
velocity polygon for the determining of the velocity of point B on the road AB
involves the following operations. The component of the velocity vector of
point A along the line AB is found, then this component is moved to point B,
and finally the perpendicular from the tip of the latter vector is constructed to
intersect the direction of velocity of point B (see Fig. 2.10).
The system component of the information operator is
represented by two techniques: the replacement of a compound motion by two simple motions (the motion of reference
point A, and the motion about A so a simple rotation; here
velocity of point A is the velocity of transport, and velocity of
point B with respect to A is the relative velocity v
;there-
BA
placement of a compound motion by one simple motion. The
r
V
AB
C
r
V
B
r
V
C
r
V
A
latter technique (in a combination with the technique involving a known information) is used for finding fixed point O
(see Fig. 2.9). We know that a body will rotate about this
point, hence the linear velocity of any point on the body will
Figure 2.10. Velocity diagram of
arod
be perpendicular to the line connecting this point with the
fixed reference point O. This reference is an instantaneous one, and also it is a
projection of the instantaneous axis of a rigid body rotation onto the plane.
P
34 Chapter 2
The similar way is used if an instantaneous center of accelerations of a
coplanar body is to be determined. Since the instantaneous angular acceleration
doesn’t coincide with the instantaneous angular velocity always (see Fig. 2.8),
the angle α is not equal in general to 90
0
as it is for the location of the instanta-
neous center of velocities (see Fig. 2.11).
e
A
B
r
a
r
a
A
B
P
Figure 2.11. Instantaneous center of acceleration of a rod
2.15. Graphical method of successive analysis of velocity and
acceleration in a rigid body plane motion
In kinematics it is very important to determine velocities and accelerations of body points in a successive way. We’ll consider a simple problem of
finding velocities and accelerations of points situated between extremities of
link AB (Fig. 2.12).
Before finding velocity of point C the instant center of velocity P
be located. To do this we need only lines of action of V
and Va, not their abso-
b
lute values (the method of informational mechanics).In classical mechanics the
component of a velocity acting along a link is not specially specified. We designate this velocity by˜Vab, and determine its absolute value as the projection of
velocity at point A onto link AB. This velocity has constant line of action (along
link AB), constant sense and constant absolute value. We slide this velocity to
point C, where the line of action of a velocity is known (V
to CP
). Then we drop perpendicular intersecting it with the line of velocity
v
is perpendicular
c
at point C (on the contrary with that done at point A). The most of lecturers
concerned in classical mechanics knows this method but it is not presented in
should
v
2.15. Graphical method of successive analysis 35
V
B
V
B
C
P
V
C
V
AB
A
V
A
O
Figure 2.12. Finding velocities
textbooks. The method of location of point with zero acceleration (instant center
of acceleration) instant center of acceleration) without determining acceleration
at point B of the link AB, is discussed in this book.
In 1927 A. P. Kotelnikov introduced the notion of reduced vectors and used
these vectors to find accelerations of body points in general plane motion.
Propagation of this method is started by Y. M. Zingerman — “About one
poorly known way of finding accelerations in rigid body plane motion” [6].
If the accelerations of points A and B are known, then this method allows
to find the instant center if acceleration at the point if crossing of the circles
of diameters equal modules of acceleration vectors for points A and B.This
approach contradicts a task of kinematic. For location of the instant center of
acceleration P
we lay of the reduced acceleration of point A
a
˜a
= aa· w−2, (2.19)
a
determine
tan α = A
· AB−1, (2.20)
1B1
angle α
α =arctanA
1B1
· AB−1.
We draw a straight line making angle α which acceleration vector ˜a
a perpendicular to it from the end of vector AA
acceleration Q
for link AB is found (Fig. 2.13).
a
. The instant center of reduced
1
, and drop
a
36 Chapter 2
B
C
α
A
a
A
C
1
a
B
1
α
A
O
Q
Figure 2.13. Finding accelerations
Let’s consider the second example for location of the instant center of
acceleration (Fig. 2.14). Here we are provided with velocity and acceleration of
point A as well as distance AB. The instant center of velocity for link AB is
point B itself. It is easy to find
w = v
· AB−1.
a
~
a
A
a
~
a
C
a
Figure 2.14
2.15. Graphical method of successive analysis 37
In our example vector A1B1coincides with vector ˜aa.Then
tan α =˜a
· AB−1= aa· (va· ω)−1. (2.21)
a
We come back to common acceleration for point C as follows
a
=˜ac· w
c
2
.
ab
The instant center of acceleration for the given can be found in two more
ways: angle α is determined from the diagram directly, therefore it is possible
to draw at once angle α at point A; if angle BAA
is right one then the instant
1
center of acceleration is at the crossing of a perpendicular from the head of a
reduced vector ˜a
on a line segment connecting the tail of this vector and the
o
instant center of velocity, with this line segment.
K
1
a
K
a
α
o
O
K
M
α
A
Q
Figure 2.15. Graphic way of finding accelerations for a wheel rolling without sliding on
aplane
Determining of accelerations for a wheel rolling without sliding on a plane,
causes difficulties in students. In Fig. 2.15 the simple graphic way of finding
acceleration for any point on a wheel is shown. Given quantities: radius R of
a wheel, velocity V
is easy to find angular speed ω of the wheel and reduced acceleration ˜a
and acceleration aoof the center of the wheel. Point M
o
o
.We
drop a perpendicular on line segment MA, and the instant center of acceleration
for reduced vectors of our problem is received. Have graphically received the
angle α necessary to us, we connect the instant center of acceleration Q with
38 Chapter 2
point K, and make angle α. The direction of making angle is should rotate about
the instant center of acceleration as well as vector ˜α (in the same direction). We
draw a perpendicular to line segment QK and receive vector KK
.
1
2.16. A system way to derive Coriolis acceleration
Let‘s consider the elementary example (Fig. 2.16).
The disk rotates about an axis with angular
speed ω. At a fixed disk motion on it with veloc-
(VAof motion of point A) is given. The initial
ityV
r
information is given not for system as a whole, but for
two independent subsystems. The first subsystem —
disk only rotates, and the point A is fixes. The second
subsystem — a disk does not rotate, and the point A
V
A
with velocity V goes only.
Motion in the first subsystem is a motion of
A
ω
Figure 2.16
tion; relative motion in motion of transport. It is due to our choice only what
motion we should take as a relative one or motion of transport. Therefore these
both motion should be equal and consist of the basic information. The Coriolis
acceleration determined by these two additional motion is
transport. Motion in the second subsystem is a relative motion. We have the basic information for Coriolis acceleration: ω
— angular speed in the motion of
e
transport (the basic information for the first subsystem),V
— velocity of a point in the relative motion
r
(the basic information for the second subsystem). In
Fig. 2.17 we shall present these two mutually opposite
problems:
To the two problem their interconnected prob-
lems are added: motion of transport in relative mo-
=2ωe×Vr. (2.22)
a
c
Now it is necessary to choose what acceleration we shall consider for relative motion (subsystem 1) are accepted fixed then the acceleration of point
should be considered as a total relative acceleration.
Let‘s introduce relative acceleration by
r
dV
a
=
r
, (2.23)
dt
2.16. A system way to derive Coriolis acceleration 39
where sign ∼ designates local derivative (there is no rotation of axes). the
absolute acceleration in compound motion we shall present as
= ak+ ar+ ae, (2.24)
a
A
Figure 2.17
where aeis the acceleration of transport for compound motion of a particle.
It means from the point if view of mathematics that we allocated two simple
expressions from a complex expression for acceleration of point A, and all other
members are incorporated in a
.
e
To derive absolute acceleration in compound motion of a particle we need:
Poisson‘s formulas
di
= ω ×i ;
dt
Bur‘s formula
We derive Bur‘s formula as
R = R
dR
dt
dR
=
dt
dR
x
i +
dj
= ω ×j ;
dt
dR
dR
=
dt
dt
·i + Ry·j + Rz·k,
x
dR
y
j +
dt
dt
dk
= ω ×k, (2.25)
dt
+ ω ×R, (2.26)
z
k + R
di
x
dt
+ R
dj
y
dt
+ R
dk
,
z
dt
40 Chapter 2
mean
dR
dt
dR
=
dt
dR
y
x
i +
dt
j +
dR
z
k. (2.27)
dt
From (1.27) the concept of a local derivative is clear. It is necessary to know
features of the mixed product of vectors:
a ·(b ×c)=b · (c ×a)
and transform
di
R
x
dt
= ω ×iR
dj
+ R
y
dt
+ ω ×jRy+ ω ×kRz= ω ×(R
x
+ R
dk
z
dt
(ω ×i)+Ry(ω ×j)+Rz(ω ×k)=
= R
x
x
i + R
j + R
y
k)=ω ×R.
z
The displacement of point M (Fig. 1.16) will consist of two displacement: one
of transport — r
, and relative displacement — ρ.
A
= rA+ ρ
r
M
(2.28)
ρ = x ·i + y ·j + z ·k.
The absolute velocity of point M is
dρ
dt
dr
=
dt
dr
V
M
A
dt
+
=
dρ
A
+
+ ω ×ρ. (2.29)
dt
We assume
dρ
=V
, (2.30)
dt
V
=VA+ ω ×ρ +V
M
r
r
(2.31)
It we keep our division of motion onto relative and transport, then relative
velocity in (2.29)has already taken its place by definition (1.30). Therefore the
compound expression for velocity of transport is
e
V
=VA+ ω ×ρ. (2.32)
Two components of absolute acceleration (a
,ac) are known to us, it is necessary
r
to determine the third component. We derive absolute acceleration of point M
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