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Файл:A General Course of Physics. Mechanics. Textbook
.pdf
71
The resultant of the gravity force
gm
f
F
с
a
с f
amFgm
0
с f
amFgm
gm
f
F
c.i.
F
0
c.i.f
FFgm
cc.i.
amF
rra
22
с
v
rmF
2
c.i.
communicates the centripetal acceleration
and the funicular force
to the ball (Fig. 31). Ac-
cording to Newton’s second law
or
. (54)
Non-inertial reference frame.
Within the frame of reference, involving the rotating disk, the ball is
motionless. Consequently, the resultant of all forces must equal zero, i.e.
apart from the forces
and
, the ball is acted upon by an additional
force, which is called the centrifugal force of inertia
(Fig. 32).
Then, according to Newton’s second law
. (55)
Having compared equations (54) and (55), we will derive the expression of the centrifugal force of inertia:
Because the centripetal acceleration is
, then
. (51)
The centrifugal force of inertia acts upon passengers when a wagon is cornering; it also acts upon pilots during stunt-flying. The centrifugal force is used in certain mechanisms – centrifugal pumps and separators.
The Earth rotates about its axis and, as a result, the reference
frame which involves the Earth is non-inertial. In particular, due to the
centrifugal force of inertia, gravitational acceleration is less at the equator
than at the poles.
However, as it can be seen from formula (56), at a low rotation velocity the magnitude of the centrifugal force is not too strong. That is
why the Earth’s rotation about its axis (let alone its revolution around the
72
Sun) does not have any effect on the majority of phenomena on the Earth,
umF 2
C
u
u
Let us examine the following
test (Fig. 33). We will place a vertical
frame onto a rotating horizontal disk
and fix a pendulum to the frame. We
will force the pendulum to swing in the
plane of the frame and start turning the
disk slowly about the vertical axis. The
pendulum’s oscillation plane keeps
holding its position in relation to the
room walls, whereas the frame plane
continues rotation.
i.e. the reference frame involving the Earth can be regarded as inertial.
3. Inertial forces acting upon a body moving within a rotating ref-
erence frame.
If a body is travelling within the rotating reference frame, then apart
from the centrifugal force of inertia, it will be acted upon by one more inertial force – the Coriolis force:
where
frame. The Coriolis force is directed perpendicularly to the vectors
and
erence frame, the pendulum does not play a part in the disk’s spatial rotation. From the point of view of an observer being in a moving reference
frame involving the disk, the pendulum’s oscillation plane rotates and, as
a result, the pendulum is acted upon by the Coriolis force.
ment (Foucault’s pendulum, 1850), which appeared to be the evidence of
the Earth’s rotation.
Earth, for example, for satellite and rocket trajectory computations. The
effect of the Coriolis force is exposed in wind distributions on the Earth,
the right-hand riversides undermining, flowing in the Northern hemisphere (or the left riversides – in the Southern hemisphere).
is the velocity of the body in relation to the rotating reference
.
From the point of view of an observer being in a motionless ref-
The illustrated example is a model of Foucault’s famous experi-
The Coriolis force should be taken into consideration on the
73
Test Questions
4. Does the moment of
inertia of a solid body depend on: a) the moment of
forces applied to the body;
b) the choice of the rotational axis; c) the shape of
the body; d) the mass of
the body; e) angular acceleration?
1. In Fig. 34, there are objects comprised of identical thin triangular sheets, uniform in density. Identify the objects with minimum and maximum moments of inertia in relation to the axis
ОО
.
1
2. Write the formula, specifying the moment of inertia of a solid
body in relation to the given axis of rotation.
3. What is the measurement unit of the moment of inertia in the SI
system?
5. A disk-shaped flywheel with a radius R = 1 m and a mass
m = 2 t is rotating about its symmetry axis at a frequency of 3
Hz. Find the kinetic energy stored as a result of the rotation of
the flywheel.
6. A cylinder with a radius R = 10 cm and a mass m = 15 kg is
rolling down an inclined plane of 70 cm high (h = 70 cm)
without any sliding motion. Find the cylinder velocity after it
passed the inclined plane. Find the kinetic energy of the translational and rotational cylinder’s motions after it passes the inclined plane.
7. What torque was applied to a cylinder fixed to an axis coinciding with its symmetry axis, if the cylinder radius is R = 10 cm,
its mass is m = 20 kg, and the torque was able to spin the cylinder so that it gained the frequency = 40 Hz for the time interval t = 10 s?
74
8. Find the moment of inertia of a cylinder with a radius
R = 10 cm and a mass m = 20 kg in relation to an axis, tangential to its surface.
9. A shallow beam with a mass m = 2 t and a length l0 = 10 m is
supported horizontally by two props. One of them is at a distance of l1 = 3 m away from one of the beam’s ends; the second
prop supports the other beam’s end. What forces do the props,
bearing the beam, withstand?
10. What force should be applied to one of a post’s ends, lying horizontally on the ground, in order to start raising, if the post’s
mass is m = 600 kg and its length is l = 6 m. The lifting force is
applied perpendicularly to the ground.
11. A wooden post with a height h = 6 m and a mass m = 300 kg
standing vertically is sawn at its foot, which causes it to fall.
What velocity will the top of the post develop while falling to
the ground?
12. Water is flowing through a curved pipe with a radius R = 2 m
lying in a horizontal plane. Find the lateral water pressure
caused by centrifugal force.
13. A train with a mass of 3000 t is moving from the north to the
south at a speed of 144 km/h. The motion takes place at the 60th
parallel north. What is the direction of the Coriolis force acting
upon the train and what is its value?
14. Imagine a thin uniform sheet (m = 2,1 kg) having a shape of a
45-degree right-angled triangle. Find its moment of inertia in
relation to the axis coinciding with one of its sides. The lengths
of the sides are 620 mm.
75
5. UNIVERSAL GRAVITATION
3
1
3
2
2
1
2
2
R
R
T
T
5.1. Kepler’s Laws
The three laws of planetary motion disclosed by Kepler (1571-
1630), together with the laws of dynamics, became the basis of the law of
universal gravitation formulated by Newton:
1. The orbits of all planets are ellipses, and the Sun is located in
one of the foci.
2. The radius-vector, joining a planet and the Sun, creates equal
areas over identical time intervals.
3. The squares of planetary circulation periods are proportional to
the cubes of the major semi-axes of the orbits’ ellipses.
Kepler’s third law can be written in the following form:
,
viz. T1 and T2 are the circulation periods of two particular planets; R1 and
R2 are the major semi-axes of the corresponding ellipses.
5.2. The Law of Universal Gravitation
Let us explain the law of universal gravitation theoretically, on
the basis of Kepler’s laws and Newton’s laws of motion. It should be
mentioned from the start that a circumference is a particular kind of the
ellipse; what is more, a circle’s radius equals the major semi-axis of the
ellipse. Having this in mind, we will consider a hypothetical planetary
system, i.e. a system where all planets move in circular orbits and where
the Sun is at the center (therefore, Kepler’s first law is in operation).
According to Kepler’s second law, the radius-vector of a particu-
lar planet covers equal areas during identical time intervals. It is true if
the velocity of a particular planet’s motion along a circular orbit is a constant value (so, Kepler’s second law is used).
If a planet moves in a circular orbit at a constant velocity, it must
be acted upon by the centripetal force from the Sun. Let us consider two
76
planets moving in circular orbits with the radii R1 and R2. The centripetal
1
2
1
11
R
m
v
F
2
2
2
22
R
m
v
F
122
22121
2
1
R
R
m
mvv
F
F
1
1
1
2TR
v
2
2
2
2TR
v
221
12221
2
1
R
R
m
mTT
F
F
3
1
3
2
2
1
2
2
R
R
T
T
2
1
2
221
2
1
R
F
F R
m
m
forces which act upon them are:
;
,
where т1; т2; v1; v2 are the masses and velocities of the first and second
planets, respectively.
The equations, given above for the centripetal forces, express
Newton’s second law.
Now let us identify the relation of forces:
. (57)
It is worth mentioning that the planetary velocities are deter-
mined by their orbital periods round the Sun and the orbital radii:
;
.
Having inserted the latter equation into (57), we will get the fol-
lowing, after a few reductions:
.
According to Kepler’s third law:
. Using this formula
we have the following after insertion and reduction:
.
After obtaining all the quantities, necessary for the first planet in
the left-hand part of the equation and for the second planet in the righthand side part, we will attain:
77
const
2
2
22
1
2
11
mm
RFRF
)const(2RmF
2
R
mM
GF
2
r
21
MM
GF
gmF
G
or,
for any planet.
Assuming that the constant is proportional to the Sun’s mass, we
will derive for the gravitational pull between the planet and the Sun:
,
specifically, М is the mass of Sun.
It follows that between any two bodies with masses there must be
the force of mutual pull.
The law of universal gravitation can be expressed in the following way: the force of mutual pull acts between any two material bodies
(point particles) which are proportional to the product of the bodies’
masses (М1 and М2 ) and inversely proportional to the squared distance
between them (r2):
,
where G is the gravitational constant; G = 6.6720 10
–11
Nm2/kg2.
5.3. Gravity Force and the Weight of a Body. Weightlessness
A body with a mass m near the Earth surface is acted upon by the
force:
,
which is called the gravity force (g is gravitational acceleration).
In this area of the Earth, gravitational acceleration is the same for
all objects. Near the Earth’s surface, gravitational acceleration varies depending on the latitude: from 9.780 m/s2 at the equator to 9.832 m/s2 at
the poles, – due to the shape and the rotation of the Earth (the equatorial
and polar radii are 6378 and 6357 km, respectively).
78
The force of gravity is caused by the force of gravitational pull of
2
E
G
R
mM
GmgF
2
E
R
M
Gg
2
E
)h
M
Gg
(R
ga
gm
N
y
a
Fig. 35
the Earth which acts upon bodies and for this reason (without considering
the Earth’s rotation) they are equal:
,
viz. M is the Earth’s mass; RE is the Earth’s radius.
Whence, it is clear that gravitational acceleration near the Earth’s
surface is
.
If a body is at a height of h from the Earth’s surface, then
,
i.e. gravitational acceleration decreases as the distance from the Earth’s
surface increases.
The weight of a body is the force with which the body acts upon
the support (or suspension), holding the body from free fall.
The state of a body when it moves only under the action of the
Earth’s gravity is called weightlessness.
Using an example, we will analyze the motion of a body in the area of
the Earth’s gravity with an acceleration
which is dissimilar to gravitational acceleration:
with a mass m is being inside an eleva-
. Assume that a body
tor moving upwards with an acceleration a (Fig. 35).
According to Newton’s second
law:
79
amgmN
After projecting it onto axis y, we get:
mamgN
)( agmNP
a
)( agmNP
)0( ag
0)( agmP
r/
2
1
v
gm
N
y
a
Fig. 36
law, the reaction of the support N numerically equals the pressure of the body on
the support, i.e. the weight of the body P.
Consequently,
ration
, directed downwards, then the weight of the body is (Fig. 36):
If an elevator is free-falling, i.e. its acceleration, directed downwards, equals gravitational acceleration
.
.
According to Newton’s third
.
If a body moves with an accele-
.
, the body inside this
elevator, will be in the state of zero gravity:
.
5.4. Cosmic Velocities
For launching rockets into outer space, it is necessary to communicate to them certain escape velocities, depending on the goals, which
are called cosmic (or space) velocities.
The first cosmic (or escape) velocity v1 is such a velocity which is
necessary to communicate to a body so that it could move along a circular
orbit, i.e. turn into a man-made Earth’s satellite.
A satellite with a mass of m, moving in a circular orbit with a radius r, is acted upon by the Earth’s gravity which communicates the nor-
mal (centripetal) acceleration
to the satellite.
80
If a satellite moves not far from the Earth’s surface, then
Rr
E
R
mmg
2
1
v
gR
1
v
2
v
E
R
R
mM
Gdr
r
mM
G
m
2
2
2
2
v
)(
2
E
RMGg
E22gRv
3
v
(RE is the Earth’s radius). In such a case, the gravity force performs the
function of the centripetal force:
.
Whence, for the first cosmic velocity we obtain:
7.9 km/s.
E
The second cosmic (or parabolic) velocity
is the minimum ve-
locity which is necessary to communicate to the body so that it can es-
cape from the Earth’s gravitational pull and turn into a satellite of the
Sun. In order that the body could overcome the Earth’s gravity and es-
cape into outer space, its kinetic energy has to be equal to the work done
against the Earth’s gravity:
,
specifically, М is the Earth’s mass, RE is the Earth’s radius.
As
, the second cosmic velocity is
11.2 km /s.
The third cosmic (or solar escape) velocity
is such a velocity
which is necessary to communicate to a body at the Earth’s equator in the
direction of the Earth’s orbital motion so that the body could escape from
the Solar system and overcome both the Earth’s and the Sun’s gravity. In
such a case, the Earth itself is a launch site which moves along the orbit
around the Sun and for this reason the velocity of the Earth’s orbital motion is used as a launch velocity for rockets. Moreover, if a launch is executed in the equatorial latitudes, it is also possible to use the tangential
velocity of the Earth’s surface motion.
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