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3.4. Compact of dynamics problems (resonance) 71
Antiresonance
The problem of resonance is illustrated in Fig. 3.13.
We add an analogous system (see Fig. 3.14) such that w = p
1
The vibration of the second system is in an anti — phase with the loading
vibration, then weight m
becomes motionless.
1
The original system is significantly changed (see Fig. 3.22):
+ c1x1− c2(x2− x1)= F sin wt;
m
1¨x1
m
+ c2(x2− x1)=0.
2¨x2
The solution:
= a1sin wt, x2= a2sin wt.
x
1
−m
w2a1+ a1(c1− c2)+a2(−c2)=F,
1
(−c2+(−m2w2+ c2)a2=0.
a
1
= p2.
a
=
1
−m
a
=
2
−m
F (−m
w2+ c1− c
1
(−c2) −m2w2+ c
w2+ c1− c
1
(−c2) −m2w2+ c
w2+ c2)
2
2
Fc
2
2
−c
−c
, (3.38)
2
2
.
2
2
If we substitute p for w in the denominator, then we’ll obtain a frequency
equation of the free vibration:
−m
w2+ c1− c
1
4
m1m2+ p2(−m1c2− m2c1+ m2c2)+c1c2− 2c
p
(−c2) −m2w2+ c
2
−c
2
=0, (3.39)
2
2
=0. (3.40)
2
We have adjusted root
c
p
= w, p2=
2
−m
w2+ c2=0,a1=0,a2=0.
2
m
2
;
2
72 Chapter 3
3.5. Energy mechanics
Kinetic and potential energy
According to properties of an object two mechanics are distinguished: a
rigid-body mechanics, and mechanics of a deformable body.
According to the basic information there are three mechanics:
• force mechanics Newton’s mechanics;
• energetic mechanics;
• forceless mechanics Hertz’s mechanics.
According to mathematical form of the basic equations we have two me-
chanics:
• vector mechanics (Newton’s mechanics);
• analytical mechanics (Lagrange’s mechanics).
There exist also other principles of the mechanics creation besides men-
tioned ones. We see elements of a new approach in this course of lectures. In
paragraph 3.3 we used two levels of a physical representation of problem: an
energetic level; equations of motion.
A force level is one where equations of motion are represented in the form
of equations of forces.The energetic level is necessary as a special method for
the problem analysis (equations of energy) and as an intermediate base for Lagrange’s method. In the latter method specially introduced coordinates and additional principles (virtual displacement principle) allow to come from the second
level to the first one and to remove the constraints reactions.
Let’s now consider the energetic mechanics. Earlier, we selected the indi-
visible information of m, r in Newton’s mechanics. We also formed the infor-
mation matrix:
−→ d/dt
r
mrm˙rm
0 r ×m˙rr × m¨r
←−
˙
r
¨
r
¨
r
(3.41)
3.5. Energy mechanics 73
and introduced the system operator for Newton’s mechanics:
d
{
dt
;
;0;=0; •;
; }, (3.42)
where
d
denotes differential laws,
dt
— integrational laws,
0 — laws of conservation,
=0— laws of change,
• — laws for participle,
— laws for system of particles,
— laws for center of mass.
Our problem is to find the place of kinetic and potential energies in New-
ton’s mechanics.
Let’s multiply the expression form the second line of the information matrix
by r:
m¨r · dr =F · dr = dA, (3.43)
F = F
i + F
x
y
k, (3.44)
j + F
z
dr = dxi + dyj + dzk, (3.45)
dA= F
dx+ Fydy+ Fzdz. (3.46)
x
We denoteF · dr by dA, the elementary work. We transform the left side
of equation (1.43):
2
˙
dA= m¨r · dr = m
d˙r
· dr = md˙r ·˙r = d (m
dt
r
)= d (T ). (3.47)
2
We introduce the notion of elementary work dA and elementary kinetic energy dT . In the basic information we used term dr which is not a new one, but
obtained as a derivative of r.
As a result we obtain the principle of work and kinetic energy:
dT = dA. (3.48)
The integrating form of this principle is:
T
− T0=
1
where T
— initial and final kinetic energies of a system, respectively;
0,T1
e
A
+
k
i
A
, (3.49)
k
74 Chapter 3
e
A
k
,A
i
— works of external forces and internal forces, respectively.
k
The work of the internal forces in a rigid body equals zero.
The principle states that change in kinetic energy of the system is equal to
the total work done by forces acting on the system of particles or on the body.
Conservative force field. Scalar function
Let’s chose among all forces one which is mathematically convenient, i. e.
which gives complete differential in equation (3.46). We introduce an arbitrary
function U having the complete differential
dU =
∂U
∂x
dx+
∂U
∂y
dy+
∂U
dz. (3.50)
∂z
Comparing (3.47) and (3.49) we can write our proposal:
∂U
∂x
+ F
=
F
x
∂U
∂y
+ F
=
y
∂U
∂z
.
=
z
Such forces are called the potential ones. They can move a body from point
B to point A in different ways. Only the difference between values of forces
at B and A is of importance. In our problems these forces are the elastic forces
and weight forces.
Let’s introduce the gradient of scalar function U :
grad U =
∂U
∂x
i +
∂U
y
j +
∂U
∂z
k,
then grad U =F .
We also consider the rectangular Cartesian representation of a vector field
called the curl
rotF =i (
∂F
∂y
∂F
z
x
−
∂z
)+j (
∂F
∂z
∂F
x
z
−
∂x
)+k (
∂F
∂x
∂F
y
x
−
∂y
).
Scalar function U contains null inside itself:
2
∂
U
∂y∂z
−
∂z∂y
2
∂
U
=0,
2
∂
U
∂z∂x
−
∂x∂z
2
∂
U
=0,
2
∂
U
∂x∂y
∂
−
∂y∂x
2
U
=0,
because of scalar function U has a complete differential.
2
∂
U
∂y∂z
−
∂z∂y
2
∂
U
=0,
3.5. Energy mechanics 75
whence
∂F
∂y
∂F
z
y
−
∂z
=0.
The necessary and sufficient condition for forceF to be conservative is that curl
of a vector – rotF =0. Potential forces don’t cause rotation of a body moving
along a line.
The negative of scalar function is called a potential energy
V (r)=−U(r),
whence
dA= −dV.
Writing the expression of the work and kinetic energy principle for a parti-
cle moving in a conservative force field
mv
2
2
2
−
mv
2
2
1
= −(V
− V1)=V1− V2,
2
and manipulating it, we obtain the principle of conservation of total mechanical
energy
Here v
2
mv
2
+ V
=
2
are v2the velocities of a particle at positions 1 and 2.
1
2
mv
2
2
1
+ V
.
1
Thus, the information compact for energetic mechanics can take the follow-
ing form:
{grad U, rotF }.
If vector curl rotF =0,thenF =gradU andF is a potential force.
Compact for Lagrange’s dynamics. Principle of total energy conservation
Let’s consider an information compact for Lagrange’s dynamics. What is
the limitation of the compact for Newton’s dynamics? The linear displacement
and corresponding moments — statical moment mr,momentumm˙r, inertia
force m¨r — are considered separately. The generalized coordinates (displace-
ment, angle) are introduced in Lagrange’s mechanics, hence the second and
third lines of the compact for Newton’s dynamics must be determined. There
are no kinetic and potential energies in Newton’s mechanics, therefore we need
a special energy mechanics.
76 Chapter 3
Let’s introduce Lagrange’s function
L = T + U = T − V.
For a conservative system Lagrange’s equations can be written in the form:
where q
∂T
d
(
dt
∂ ˙q
— generalized displacement, ˙qi— generalized velocity, Qi— general-
i
i
) −
∂T
∂q
= Qi,
i
ized force.
In case of
∂T
T = T(q
, ˙qi),
i
But the most frequently case is when T = T(˙q
∂q
i
i
=0.
).
For conservative systems (without the exchange of energy with the envi-
ronment) Lagrange’s equations can be written in the form
d
dt
(
∂ ˙q
∂L
i
) −
∂L
∂q
=0,
i
and for systems with active forces:
∂L
d
dt
(
∂ ˙q
i
) −
∂L
∂q
= Qi.
i
We represent the compact for Lagrange’s dynamics in the following form
)
i
− L)
⎫
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎭
.
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
˙q
i
L
∂L
∂q
∂L
∂ ˙q
i
i
− L
d
dt
dt
(˙q
dL
dt
d
∂L
(
∂ ˙q
∂L
i
∂ ˙q
i
dL
Here
is a power (in case of a stationary force field).
dt
We show upon correspondence between the second lines of Newton’s and
Lagrange’s compacts:
L
m˙r ⇒
∂L
∂q
,m¨r ⇒
i
d
dt
(
∂ ˙q
).
i
3.5. Energy mechanics 77
The element
∂L
i
∂ ˙q
− L)
i
(˙q
is called Hamilton’s function and it coincides, in particular case, with the total
mechanical energy of a system. Indeed,
˙q
i
∂T
∂ ˙q
=2T,
i
=
i
)=mv, v·mv=2(
mv
2
2
)=2T.]
[T =
mv
∂L
˙q
i
∂ ˙q
2
∂
,
2
∂v
∂L
˙q
i
∂ ˙q
2
mv
(
2
− L =2T − (T −V )=T + V = E.
i
Principle of the mechanical energy conservation:
d
dt
d
dt
(˙q
∂L
− L)= 0,
i
∂ ˙q
i
(E)=0,E=const.
Energy mechanics and kinematics. Analogies
In kinematics a body is moved from one point to another without rotation
(this corresponds to the case of a conservative force field). The pure rotation
occurs at a point (this corresponds to a solenoidal field). Both these cases have
the analogy with indivisible information of a force and a couple in statics. In
energetic mechanics (and theory of field) a vector field whose curl is zero, i. e.
is no need in rotation.
The solenoidal field corresponds to
there is no need in displacement.
Here
∇ =
∂
∂x
i +
∂
∂y
∇×F =0, (3.51)
∇·
F =0, (3.52)
∂
j +
k,F = F
∂z
i + F
x
i + F
x
i.
x
78 Chapter 3
The force system cannot rotate (it is rotationally balanced) if
∂F
−
dF
dx
∂x
y
y
=0,
,d F
· dx= dFydy.
x
x
x
=
then
∂F
∂y
∂F
∂y
∂F
dF
x
y
=
∂x
,
dy
If there is no need in a linear displacement,then
∂F
x
∂x
∂F
∂y
∂F
∂z
=0,F
y
=0,F
z
=0,F
=const;
x
=const;
y
=const.
z
The analogy with kinematics: in the solenoidal field the projections of forces
onto axes are constant, this allows to translate a body as well as to rotate it.
The kinetic energy of a translating body is
T =
, (3.53)
2
2
Mv
one of a rigid body rotating about a fixed axis is
2
T = J
w
, (3.54)
z
2
where J
— moment of inertia about axis OZ.
z
The analogy between (1.52) and (1.53) is used in Lagrange’s equation.
If we consider motion of the mass center of a body being in a general plane
motion, then we have Konig’s theorem:
T =
Mv
2
2
c
+ J
2
w
, (3.55)
cz
2
where M — mass of a body,
— velocity of the mass center,
v
c
J
— moment of inertia about axis through the mass center,
cz
w — angular velocity of a body.
3.6. Elements of Lagrange’s analytical mechanics 79
For the round disk of radius R
J
=
cz
.
2
2
mR
If the disk rotates about an axis (see Fig. 2.16), then
J
cz
=
·
2
2
w
mR
2
2
2
w
·
2
=
mv
4
2
c
.
Maxwell’s pendulum
The uniform cylinder of weight P falls down without an initial speed and
reels out the rope. Determine the speed of the axle of cylinder as a function of
the distance travelled
A
=
e
= Ph, T=
k
v
P
=
g
2
v
c
P
g
2
P
3
v
g
4
T − T
2
c
2
+
2
= Ph, v
=
0
Mv
+
R
P
g
2
2
2
e
A
, (3.56)
k
2
c
mR
2
w
·
2
=
2
+ J
2
w
·
2
2
=
2
4
3
·
cz
2
=
34P
gh.
2
w
=
2
2
v
;
g
Did we need the value of R? No! Why? Did we need the value of m?No!
Why?
The mass is in both left and right sides of equation (3.55). But we need
mass in implicit from since it is connected with free – fall acceleration g!
3.6. Elements of Lagrange’s analytical mechanics
Virtual and actual displacements
The basic purpose of analytical mechanics is to solve static and dynamic
problems for constrained systems. Three types of infinitesimal displacements
are distinguished in analytical mechanics (see Fig. 3.17).
F (t) — denotes active forces,
R(t) — denotes reactions of constraints.
80 Chapter 3
If the constraints reactions don’t depend on time, then these constraints are
referred to the stationary ones.
The infinitesimal displacements are classified as follows:
r — virtual (“possible”) displacements;
this is an abstraction, there exists the infinite number of them;
dr — real displacements; each system has unique real displacements;
δr — virtual displacements in proper sense, the virtual displacements in
case of the stationary constraints (time is “frozen”).
Virtual displacement principle
The mechanical system constrained by ideal, stationary and “fixed-guide”
constraints, is in equilibrium if, and only if the sum of works of all the active
forces is zero for any arbitrary virtual displacement from the initial configuration
(see Fig. 3.17), i. e.
N
F
· δrk=0, (3.57)
k
k=1
whereF is active force acting on k −th particle in the system;
r is position vector of k − th particle.
If the mechanical system is in equilibrium, then for each particle active
forceF and reaction forceR satisfy the static equilibrium condition
F +R =0,k =1, ..., N.
Let’s multiply both sides of equation by virtual displacement δr and find sum of
such expressions for all particles of the system:
N
k=1
F
k
· δrk+
N
k=1
R
· δrk=0.
k
For the ideal constraints the condition
N
R
· δrk=0 (3.58)
k
k=1
is satisfied, and we get equation (3.56).
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