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Файл:A General Course of Physics. Mechanics. Textbook
.pdf
41
So, the total mechanical energy increment in a system of bodies,
p
W
acted upon by conservative forces only, equals the work of external
forces, applied to the system bodies.
If the system is closed, i.e. it has no influence of external forces,
then А = 0. Consequently, W2–W1 = 0; W = 0; W = const.
This equation expresses the law of energy conservation in mechanics, which is formulated as follows:
The total mechanical energy of a closed system, having only conservative forces acting among its objects, remains constant.
In various processes, occurring in a closed conservative system,
when the total mechanical energy is conserved, the potential energy can
be converted into the kinetic energy and vice versa, in equal proportions.
For this reason, this law is sometimes called the law of energy conservation and transformation.
If in a closed system, apart from only conservative forces, nonconservative forces act as well, such as friction forces, then the total me-
chanical energy isn’t conserved. The presence of friction forces in a
closed system leads to the diminishing of its total mechanical energy and
to transforming it into other (non-mechanical) kinds of energy. In this
case, a more general law of energy conservation comes into force: in a
closed system, the sum of all kinds of energy, including non-mechanical,
remains constant.
3.5. Relationship between Potential Energy and Force
Let us determine the functional relationship between the potential
energy and force. To this end, let us consider an elementary displacement
in the field of conservative (potential) forces. The displacement work is
performed due to the decrease of the stored potential energy of the body,
and so elementary displacement work А and an elementary change in
the potential energy
are opposite in sign:
Wp = –A. (30)
42
In (30), partial differentials are used because a particular (partial)
rFrFA
r
cos
r
FF cos
r
r
W
F
p
r
x
zyxW
F
p
х
),,(
y
zyxW
F
p
y
),,(
z
zyxW
F
p
z
),,(
k
z
W
j
y
W
i
x
W
kFjFiFF
ppp
zyx
kji
,,
displacement in a particular direction is under consideration.
On the other hand, the displacement work is
. (31)
viz. the quantity
direction
equals the partial derivative of the potential energy in this direction, with the
sign reversed. Formula (32) allows the getting of a general equation for the
force vector. To this end, let us choose displacement along the axis of abscises in the Cartesian reference frame as a particular direction. In this case,
coordinates y and z will be constant, while it is necessary to write correlation
(32) in the form:
specifically, Fx is the force projection on the axis of abscises x; Wp(x, y,
z) / (x) is a partial derivative of the potential energy with respect to
coordinate «x».
Similarly, for the projections on the other axes we will get:
. After comparing (30) and (31), we get:
Therefore, the numerical force projection on a particular direction
is the force projection on the displacement
. (32)
,
;
.
Now we will express the force vector in terms of its projections:
,
viz.
tem. After that, the equation can be written in a shorter form, using the
are unit vectors of the coordinate axes in the Cartesian sys-
43
gradient operator (grad), which is a reduced form of the expression in
WF grad
brackets:
.
This correlation is a general expression of the relationship between the force acting on a body and the potential energy of this body in
conservative systems.
3.6. Central Collision of Spheres
By the example of collisions of spheres we will consider how the
law of momentum conservation and the law of energy conservation are
used in practice.
There are two limiting forms of collision: a perfectly elastic collision and perfectly inelastic collision.
A perfectly elastic collision is defined as one in which mechanical energy is not converted into non-mechanical kinds of energy.
After such a collision, the kinetic energy of the collided objects is
firstly converted into resilience. Then, the objects return to their initial
state, repelling each other. Ultimately, resilience is again converted into
kinetic energy, and the objects fly apart at the velocities, whose magnitude and direction are controlled by two laws: the law of energy conservation and the law of momentum conservation.
A perfectly inelastic collision is characterized by the fact that the
kinetic energy totally or partially turns into the internal energy of the objects, leading to a temperature rise. After the collision, the objects either
move at a uniform velocity or stay at rest. After a perfectly inelastic collision, only the law of momentum conservation is relevant.
Let us suffice with the central collision of two spheres only.
44
A collision is called
2010
vvand
v
vvv
)(
21202101
mmmm
21
202101
mmmm
vv
v
v
10
v
20
v
v
101
vm
201
vm
10
v
21
202101
mmmm
vv
v
m
1
m
2
v
10
m
1
m
2
v
10
v
20
v
20
а
b
Fig. 15
central if the balls until the
collision are moving along a
straight line, crossing their
centers (Fig. 15).
In cases of central impact, mutual collision may
happen, if
the spheres are moving towards each other (Fig. 15, a),
one of the spheres “is following” the other (Fig. 15, b).
Let us assume that the spheres comprise a closed system or external forces, applied to the spheres, balance each other (a quasi-closed
system).
Perfectly inelastic collision. Let us introduce the following symbols: m1, m2 denote the mass of the spheres;
ities of the balls before the collision;
is the velocity of both balls after
mean the veloc-
the collision.
We will write the law of momentum conservation in the form:
;
.
rection
In Fig. 15, b the velocity
depends on the correlation of impulses
is directed as
and
and
, and the di-
(Fig.
15, a).
We will consider that the direction of velocity
is positive.
Then, turning to the scalar equality, we will receive:
, (33)
45
where the positive sign and the negative sign are for the cases shown in
101
vm
202
vm
2
)(
22
2
21
2
202
2
101
vmm
vmvm
Q
21
101
mmm
vv
2
2
101
vm
Q
10
21
101
v
v
v
mm
m
2010
vvand
1
v
2
v
Fig. 15, a and b, respectively. If
<
, v < 0 and is directed as
v20.
The amount of mechanical energy, changed into the internal
energy (heat), equals the energy differences before and after the collision:
. (34)
Let us consider a particular case when a collided object (m2) stays
at rest (v
= 0), then from formula (33) it follows that
20
. (35)
We will assume that the collided object is heavy (m2 >>m1), then
v 0 and from (34) it follows that
It means that in this case almost the entire kinetic energy turns into heat (in a blacksmith’s forge an anvil has a heavy mass).
If m2<< m1 (while hammering a nail with a mass of m2 with
hammer m1 into a plank), from formula (36) we get:
i.e. the hammer velocity is almost entirely communicated to the nail.
Then, from formula (34) we have Q 0. In other words, the kinetic
energy from the hammer turns into the kinetic energy of the nailhammer system (which is afterwards consumed by overcoming the
plant resistance).
Perfectly elastic collision. Let us introduce the symbols:
We will write the equations in conformity with the law of momentum and energy conservation:
,
are the velocities of spheres before the collision;
are the velocities of the spheres after the collision.
and
46
221202101
vvvv
1
mmmm
;
2222
2
22
2
11
2
202
2
101
vmvm
vmm
v
)()(
20221101
vvvv
mm
))(
2
20
2
22
2
1
2
101
vv(vv
mm
)()(
202110
vvvv
1
v
2
v
21
1021202
1
)(2
mm
mmm
vvv
21
2012101
2
)(2
mm
mmm
vvv
21
1021202
1
)(2
mm
mmm
vvv
21
2012101
2
)(2
mm
mmm
vvv
.
We will transpose these equations:
; (36)
. (37)
We will divide (36) by (37):
. (38)
Formulas (36) and (38) make up a system of two equations, from
which we can find the velocities
and
.
Having multiplied (38) by m2, subtracted (38) from (36) and having
made a few transformations afterwards, we will get the velocity of the first
sphere after the collision:
. (39)
Having multiplied (38) by m1 and having added (38) and (36) afterwards, we will get the velocity of the second sphere after the collision:
. (40)
We will consider the velocity direction v
positive. Then, moving
10
to scalar quantities,
; (41)
. (40)
In equations (41) and (42) the positive sign corresponds to the
case in Fig. 15, b (the spheres are moving in the same direction), while
47
the negative sign – to the case in Fig. 15, a (the spheres are moving to-
201
vv
102
vv
10201
2 vvv
202010
2
1
2
2
vvvv
m
m
101
vv
F
rd
wards each other).
Now we will examine a few particular cases.
1. Collision of similar spheres: m1 = m2.
It follows from formulas (39) and (40) that
;
,
i.e. the spheres communicate their velocities to each other.
If one of the spheres stays at rest, for example, v20 = 0, then after
the collision it will move at the velocity equal to the velocity of the first
sphere (and in the same direction), whereas the first ball will stop.
2. Collision between a sphere and a massive wall: m2>>m1.
From formulas (39) and (40) we will get:
;
.
The wall velocity stays unchanged. If the wall stays at rest,
(v20 = 0), then
, i.e. the collided sphere will bounce back almost
at the same velocity.
Test Questions
1. Express the law of energy conservation in mechanics.
2. Express the law of energy conservation and transformation.
3. Express the law of momentum conservation.
4. Express the law of angular momentum conservation.
5. What is the relation between elementary work and the force
vector
vector
, on the one hand, and the elementary displacement
, on the other hand? Which units of measurement are
used for measuring work in the SI system?
48
6. What physical quantity is called power? Which units of measurement are used for measuring power in the SI system?
7. What is a conservative physical system? What features does a
conservative system possess?
8. A bomb with a mass of 20 kg flies from a gun barrel with a mass
of 2000 kg. The kinetic energy of the bomb at the start equals
107 J. What kinetic energy will the gun barrel produce as a result
of the recoil?
9. A body with a mass of 3 kg is moving at a velocity of 4 m/s
and collides with an immovable object of equal mass. On condition that the collision is central and inelastic, find the amount
of heat released on impact.
10. A bullet, flying horizontally, collides with a ball supported by a
very light but rigid rod and penetrates it. The bullet mass is 100
times less than the ball mass. The distance between the rod sus-
pension point and the ball’s center is 1 m. Find the velocity of
the bullet, if it is known that the rod with the ball deviated by
60° as a result of the bullet impact.
11. Coal is being offloaded from a barge onto a quay which is 2.5
m high with the help of a conveyor belt which consumes a
power of 10 kW. Provided that the coefficient of efficiency
equals 75%, determine how many tones of coal it is possible to
offload in a 20-minute time.
12. A nuclear reactor, working on a continuous basis, develops a
power of 1000 MW. Assuming that nuclear fuel replenishment
does not take place annually, determine how much the nuclear
fuel mass decreased after a year of continuous reactor work.
13. Find the kinetic energy change in an isolated system comprised
of two balls with a mass of m1 = 1 kg and m2 = 2 kg on front
49
(central) impact. Before the collision they had been moving at
oppositely directed velocities v1 = 1 m/s and v2 = 0.5 m/s. What
velocity will the balls be travelling at after the collision? What
energy is produced in the form of heat upon impact?
14. Construct a graph of time as a function of the kinetic, potential
and total energy of a stone with a mass of 1 kg, thrown vertically with an initial velocity of 9,8 m/s for the time interval 0-1
s for every 0.2 s.
15. A mineshaft cage (which is a kind of lift cabin) with a mass of 3
tonnes is being lifted at a constant velocity to a height of 200 m.
The hoisting mechanism efficiency is 80%. The lifting time is 3
min. Find the work done and the power consumed by the hoisting
mechanism.
16. A wagon with a mass of 1.5 tonnes is being towed horizontally
at a constant velocity of 5 m/s. The rope tension is 19,5 kN.
Find the work done during the wagon displacement for a distance of 1.5 km. Determine the friction coefficient and the
power developed during the towage.
17. A conveyor belt is used for loading coal into wagons for transportation up a slope to a height of 5 m. The conveyor belt delivers 12
tonnes of coal per minute to the wagons. What work will the conveyor belt perform in a 5-minute period?
18. A wagon loaded with coal with a mass of 40 tonnes is moving at a
speed of 2 m/s towards another wagon with the same mass standing still and collides with it. Determine the maximum car buffer
spring compression, considering the spring deformation to be elastic, if it is known that under the action of force F = 200 kN the
spring compression is 1 sm.
50
4. GYRODYNAMICS
22
22
iiii
ki
mm
W
rv
2
The action of forces together with the fact that these forces determine the mode of motion, they also cause deformation of bodies, i.e. their
amount and form of change. It is common practice that deformations are so
negligible that it is possible to disregard them when describing motion.
An object, whose deformations can be overlooked in the context of
the problem considered, is called a perfectly rigid body.
Any motion of a rigid body can be broken down into two basic
categories – translational and rotational.
During the translational motion, all body particles at any moment of time have the same velocity and acceleration; therefore, all
displacements of body particles are equal.
During the rotational motion, all particles of a rigid body
move along the circumference, whose centers are located along a
straight line, called the axis of rotation. Displacements of the particles
vary. If a rigid body is travelling in space and rotating at the same
time, this complex motion can be described as the sum of translational
and rotational motions, occurring simultaneously.
4.1. Rotational Kinetic Energy.
Moment of Rigid Body Inertia
Let us consider a rigid body, rotating about an axis OO, fixed
in space.
Then, we will break down the body volume in mind into small
elements and number them from 1 to N. After that, we will select a
body element with a mass of mi, being at a distance of ri away from
the axis of rotation (Fig. 16). The linear velocity of the element is vi.
The kinetic energy of the small selected body element is
, (43)
viz. is the rotational velocity of the body (which is equal for all
elementary masses).
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