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Файл:A General Course of Physics. Mechanics. Textbook
.pdf
21
dt
pd
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dt
pd
dt
pd
dt
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dt
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.
Let us summarize the ratios. In the left-hand part of the equation,
we will get the sum of all internal and external forces. Keeping in mind
that according to Newton’s third law, internal forces are reduced when
added because each action has an equal action and an oppositely directed
counteraction:
.
The sum of all external forces in the left-hand part of the equation, represent the net force
, acting upon the whole physical system.
The sum of derivatives in the right-hand part can be replaced with the
derivative of the sum. Then,
. (13)
The impulse of the system, consisting of N objects, is called the
vector sum of impulses of all objects, which make up the system
In this case, equation (13) can be written as follows:
.
Assume that the system is closed (no external forces are applied)
or quasi-closed (the sum of all external forces is zero). Then, we have
.
Therefore,
.
22
This formula expresses the law of momentum conservation: the
NNNN
mmmmmm vvvvvv
22112211
1
v
N
vv
2
1
v
N
vv
2
v
1
v
u
vv
u
1
vv
1u
momentum of a closed system (isolated or quasi-isolated) remains constant.
In other words, if objects inside a closed system interact with
each other, then the aggregate momentum of the system before interaction equals the aggregate momentum of the system after the interaction.
For example, if as a result of the objects interaction, only the objects velocities change without any alterations in the objects masses, then
this law can be written in the following way:
,
viz.
,
are the objects velocities before interaction,
,
are the objects velocities after interaction.
Examples of the law’s effect are shock power, reactive motion.
2.4. Motion of a Body with Variable Mass.
Reactive Motion
Variable body mass occurs when some mass leaves the body at a
certain velocity (addition of mass to the body on the move is also possible). The separated mass can be, for example, the jet exhaust of a rocket
engine. To begin with, let us consider a rocket motion in space, where
except for the jet stream force, there are no other forces acting upon the
rocket. In this case, it is necessary to regard the mass of the jet stream as
a body acting upon the second object – the rocket. The jet exhaust and the
rocket make up a closed (isolated) system which complies with the law of
momentum conservation.
Let us introduce the following symbols:
and
which are the
rocket and jet exhaust velocities, respectively, in relation to an inertial
reference frame;
is the jet stream velocity related to the rocket.
According to the law of velocities addition,
or
(14)
23
At a certain point in time, the rocket with the mass m+dm is
v
1
v
vvd
1
)()( vvv
dmdmdmm
1
vvvvv
dmmdmdmm
vvv
mddm )(
1
v
mddmu
)()( dtdmdtdmu v
)( dtdm v
u
dt
dm
uF
reactive
ext
F
dt
d
m
dt
dm
uF
v
ext
moving with the velocity
. After the time interval dt, the mass dm (the
mass of the jet exhaust) leaves the rocket and moves with the velocity
, while the rocket itself with the mass m continues motion with the ve-
locity
.
According to the law of momentum conservation,
;
;
.
Taking into account equations (14), we will rewrite
(15)
and divide by dt:
,
viz.
is the rocket mass and the rocket acceleration,
is the
velocity of the jet outflow in relation to the rocket body. The reactive
force is expressed in the following way:
If apart from the reactive force, the rocket body is also acted
upon by an external force
(e.g. the gravity force), then the rocket mo-
tion equation will be modified:
.
24
This equation is called Meshchersky's equation (1897). Its solu-
v
u
udmmd v
m
dm
ud v
u
vv
0
0
m
m
dmud
m
m
u0lnv
tion is quite complicated. For this reason, we will specify the law, according to which the rocket velocity accelerates if the rocket engine works
continuously, for the particular case when external forces have no effect
on the rocket.
Let us return to equation (15) and focus the vector equation on
the direction of the rocket motion. Meanwhile, we will take into consid-
eration that the two kinds of velocity
and
are oppositely directed.
Moreover, when the rocket mass is decreasing (dm < 0), its velocity is
increasing (dv > 0). Then we will get
.
Therefore,
.
Let us take the integral of the equation, keeping in mind that if
the rocket engine is under permanent operation, the jet outflow velocity
is a constant quantity in relation to the rocket body:
.
Having accomplished the velocity integration from 0 to v and the
mass integration from m0 to m, we will get the Tsiolkovsky formula
(1903):
,
in particular, m0 is the initial rocket mass (including the rocket fuel on
board); m is the rocket mass with the velocity v; v is the terminal (maximum) rocket velocity.
The formula implies that the higher the jet outflow velocity u, the
higher the terminal rocket velocity and, probably, the terminal mass (payload
mass) for the given initial rocket mass.
25
2.5. The Center of Mass and the Law of Motion
i
F
ij
f
0
2
2
j
i
iiij
dt
rd
mFf
0j
ij
f
)(
22
dtrdm
ii
i
i
i
dt
rd
mF
2
2
net
i
iic
rm
М
r
1
Let us consider a rigid body of an arbitrary geometric form
with a free distribution of mass. Then, we will divide the body in our
mind into smaller (not necessarily equal) parts and number them. As-
sume that an external force
is applied to a part of the body with the
number i and mass mi and, in addition, internal forces
of the oth-
er parts of the body also act upon the part under consideration. As a
result of the net force application to the body part, the motion will
comply with Newton’s second law:
,
in particular,
part i from the other body parts;
is the sum total of all internal forces which act upon the
is the part’s mass and acce-
leration.
Let us fulfill the aggregation of all the body parts. After the addition,
the external forces become mathematically reduced, as any force which acts
upon the i element from the j element, in compliance with Newton’s third
law, has an equal force with the opposite direction which acts upon the j
element from the i element. The sum of all external forces (the net force):
. (16)
Now let us consider a point particle with the radius-vector:
,
specifically, M is the body mass. Let us call the point particle the center
of mass and elaborate on the meaning of the term.
We will differentiate the latter equation twice with respect to
time:
26
i
i
i
ñ
dt
rd
m
Ì
dt
rd
2
2
2
2
1
. (17)
2
2
net
dt
rd
МF
с
22
dtrd
c
0
net
F
Having compared (16) and (17), we will have that
. (18)
As
is the acceleration of the body’s center-of-mass,
equation (18) implies that the center of mass is moving in compliance
with Newton’s second law, and also the motion is performed as if the
mass was concentrated in the center of mass. Equation (18) represents the
law of motion for the center of mass.
Now let us examine a closed (isolated) system of bodies where
(the sum of all external forces equals zero), acceleration is zero
and the center-of-mass velocity remains constant. Consequently, the center of mass of a closed system will either move rectilinearly or stay at rest
irrespective of the way in which separate constituent parts of the system
are moving.
This statement is an alternative wording to the law of momentum
conservation. Internal forces existing among individual elements in the
system might not be equal to zero. On the one hand, they cannot affect
the motion of center-of-mass; on the other hand, they influence the motion of individual objects comprised within the system.
It is worth mentioning that instead of the term center-of-mass the
terms center-of-inertia and center of gravity are used.
In some particular cases the problem of the center-of-gravity position determination can be simplified. For instance, if the motion of a solid
body with a particular symmetry type is being studied, then it is possible to
determine the position of the center of mass without in-depth calculations.
There are a few examples: 1) the center of mass of a uniform-density ball
coincides with its center; 2) the center of mass of a thin uniform-density
bar is located in its center; 3) three point particles with identical masses
situated at the vertexes of a hard equilateral triangle have the center of mass
at the intersection point of bisecting lines, which divide the internal angles
of the triangle into two equal angles; etc.
27
Solving problems aimed at
c
r
21
21
)(mmlamam
Х
с
21
2
mm
lm
Х
с
x
y
X
c
m
1
m
2
0
l
Fig. 7
determining the center of gravity
for a particular object can be simplified in cases when all three radius-vector projections
are
worked out separately.
As an example, let us ex-
amine two spheres with the masses
m1 and m2, placed at a fixed distance l from each other, and find the center of
mass of this system (Fig. 7). Each sphere can be considered as a point particle with the mass located in its center.
Then, we will draw the axis of abscises along the straight line
crossing the center of the spheres, and the central point of coordinates – at a
distance a away from the left sphere. So, the center-of-mass coordinate X
can be written as follows:
.
If the central point of the coordinates is placed in the center of the
first sphere (а = 0), then the equation for Xc will have a simpler form:
.
c
In the specific case, when m1 = m2, Хс = l/2, i.e. the spheres are
absolutely identical, the center of mass will be located in the point directly in the middle between the two balls.
Test Questions
1. What is an inertial reference frame? What is the difference between inertial and non-inertial reference frames?
2. How is Newton’s first law stated?
3. Express Newton’s second law and write its formula.
28
4. Express Newton’s third law and write its formula.
5. What physical quantity is called the momentum? Write the formula, which determines the momentum. Which unit is used for
measuring the momentum in the International System of Units
(SI)?
6. What physical quantity is called the impulse of force? Write the
formula, determining the momentum. Which unit is used for
measuring the momentum in the International System of Units
(SI)?
7. An object with a mass of 1 kg is moving along a horizontal surface. The friction ratio is = 0.1. What is the friction force acting upon the object if gravity acceleration is 10 m/s2?
8. A railroad train with the mass М = 3000 Т. is moving at a constant speed v = 72 km/h. The friction ratio is = 0.01. What is
the locomotive haulage capacity? What power can it develop?
9. A chain with the mass m = 2 kg and the length l = 2 m hangs
by a thread touching a table surface with its lower end. After
burning out the thread, the chain falls onto the table. Determine
the total momentum passed to the table.
10. Imagine a boat away from the lake shore. A person with a mass
of m = 70 kg changes his position from the bow to the stern of
the boat. The boat length is l = 5 m. The boat mass is 210 kg.
What is the distance of displacement of the boat?
11. Determine the position of the center of mass of a system consisting of three small-sized objects. Two of them have the mass
m = 2 kg and are situated at a distance of 60 cm from each other. The third body with the mass m = 4 kg is at a distance of 50
cm from the other two objects.
29
12. A small body is at rest on a horizontally placed flat disk. The
disk is forced to rotate about its symmetry axis. The friction ratio of the body on the disk face is μ = 0.1. The distance between the body and the axis of rotation is l = 20 cm. What rotational velocity of the disk will make it impossible for the body
to maintain its position?
13. A plane is performing Nesterov’s loop. The loop radius is
R = 150 m. The plane velocity at the top of the loop is
v = 320 km/h. What is the pressure on the pilot at this point (in
terms of gravitational acceleration)?
14. Imagine a highway with a curve of a 10° slope, and a radius of
100 m. What speed is the curve designed for?
15. What is the pressure upon a passenger on a plane in his seat during
takeoff if the acceleration of the plane is аi = 15 m/s and the
weight of the passenger is 70 kg?
16. A rocket is being launched from the surface of the Earth. The
rocket mass is m = 2000 kg. The rocket engine is releasing jet ex-
haust at a rate of 3 km/s, consuming 50 kg/s of rocket fuel (including the oxidizer). What lifting force does the rocket engine possess? What rocket acceleration does the rocket engine provide at
the start?
17. A rocket in outer space away from any planets, is gaining speed
due to the work of the rocket engine. By how much will the
rocket speed increase if its mass is М0 = 3000 kg before starting the engine, and М = 1000 kg after switching off the engine,
the jet outflow rate is v = 3 km/s in relation to the rocket, and
the engine will be working for 1.5 min.? What overpressure did
the rocket crew on board endure at the start of the rocket engine?
30
3. WORK AND ENERGY
Fl
FdldlFFdldlFdA cos)(
cosFF
l
cosdldl
F
dlFA
l
l
cosFF
l
constlF
cosFllFA
l
Fig. 8
F
dl
l
Fig. 9
A
l
Fl
3.1. Work and Power
Let us assume that an object acted upon by force F, travels over a
short distance dl, moving along a trajectory (Fig. 8). The force action F
over the distance dl is characterized by a quantity called work.
Work is a scalar quantity equal to the scalar product of force by
displacement dl:
viz. is the angle between the direction of force and displacement;
is the projection of force onto the direction of displacement;
is the projection of displacement on the force direction.
The total work along the entire distance in a general case of variable force equals
Represented graphically, the work of force numerically equals
the area (the crosshatched area in Fig. 9), specifically, the force projection is given along the axis of ordinates:
ment – along the axis of abscesses. In a special case, when
; while displace-
,
.
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