Datasets:
vnpthack
/
vnpt_data

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
0
6.73k
distance đh to the distance đs along the path. In other words,
tỳ = —mmgsin 8 = —rmg —,
so that
m đs đhÀ ( ds dh
—— —= — T\|, —— —— =—= —†TT\( —
° “Áás J (ái “án”
since the đs°s cancel. Thus we get —rng(dh/đ£#), which is equal to the rate of
change of —rngh, as before.
Tn order to understand exactly how the conservation of energy works in general
in mechanics, we shall now discuss a number of concepts which will help us to
analyze it.
First, we discuss the rate of change of kinetic energy in general in three
dimensions. 'Phe kinetic energy in three dimensions is
T= ÿm(w + b2 +?).
'When we differentiate this with respect to time, we get three terrifying terms:
dđT duy đuy dù;
—= „—— —= +0; —_—- ]. 18.4
dt mẮn Ai tờ TP x) 034)
But m(doz/đÐ) is the force F„ acting on the object in the z-direction. Thus the
right side of Eq. (13.4) is Fxu„ + Fyuy + Fxo„. We recall our vector analysis and
recognize this as #'- 0; therefore
đT /dt = F- 0. (13.5)
This result can be derived more quickly as follows: if œ and b are two vectOrs,
both of which may depend upon the time, the derivative of a - b is, in general,
d(œ - b)/dt = a- (db/df) + (da/di) - b. (13.6)
--- Trang 254 ---
We then use this in the form œ = b = 0:
d($mœ2 d(3m®-® du S
"“ “_--=. _- .Ắ. (13.7)
Because the concepts of kinetic energy, and energy in general, are so important,
various names have been given to the important terms in equations such as these.
smu2 is, as we know, called kớứnetic energu. F`-0 is called pouer: the force acting
on an object times the velocity of the object (vector “dot” produet) is the power
beïng delivered to the obJect by that force. We thus have a marvelous theorem:
the rate 0ƒ change oƒ kinetic energ oƒ an object is cqual to the potuer ezpended
bụ the forces acting on tt.
However, to study the conservation oŸ energy, we want to analyze this still
more closely. Let us evaluate the change in kinetic energy in a very short tỉme đi.
If we multiply both sides of Bq. (18.7) by đý, we ñnd that the diferential change
in the kinetic energy is the force “dot” the diferential distance moved:
đi = F':- d3. (13.8)
TỶ we now integrate, we get
AT= II F'- ds. (13.9)
What does this mean? lt means that if an object is moving 7n am wøœw under
the infuence of a force, moving in some kind of curved path, then the change
in K.E. when it goes from one poïnt to another along the curve is equal to the
integral of the component of the force along the curve times the diferential
displacement đs, the integral being carried out from one point to the other. 'This
integral also has a name; it is called the tuork done bụ the ƒorce ơn the object. VWe
see Immediately that pouer equals tuork done per second. W©e also see that 1t 1s
only a component oŸ force #n the direclfion oƒ motion that contributes to the work
done. In our simple example the forces were only vertical, and had only a single
component, say #;, equal to —mng. No matter how the obJect moves in those
circumstances, falling in a parabola for example, È' : s, which can be written
as F„ dz + Eụ dụ + F> dz, has nothing left of it but F; dz = —?rng đz, because the
other components of force are zero. Therefore, in our simple case,
2 Z2
J F`-ds—= J —ng đz = —rng(za — Z1), (13.10)
1 Z1
--- Trang 255 ---
so again we fñnd that it is only the 0ertical height from which the object falls
that counts toward the potential energy.
A word about units. Since forces are measured in newtons, and we multiply
by a distanece in order to obtain work, work is measured in øeufon - meters (Ñ-m),
but people do not like to say newton-meters, they prefer to say jøuwes (J). A
newton-meter is called a joule; work is measured in joules. Power, then, is joules
per second, and that is also called a øø## (W). IÝ we multiply watts by time, the
result is the work done. 'Phe work done by the electrical company in our houses,
technically, is equal to the watts times the time. That is where we get things like
kilowatt hours, 1000 watts times 3600 seconds, or 3.6 x 108 joules.
Now we take another example of the law of conservation of energy. Consider
an object which initially has kinetic energy and is moving very fast, and which
slides against the Hoor with friction. It stops. At the start the kinetic energy
1s mo‡ zero, but at the end it 2s zero; there is work done by the forces, because
whenever there is friction there is always a component of force in a direction
opposite to that of the motion, and so energy is steadily lost. But now let us
take a mass on the end of a pivot swinging in a vertical plane in a gravitational
feld with no friction. What happens here is diferent, because when the mass is
goïing up the force is downward, and when it is coming down, the force is also
downward. Thus #'- đs has one sign going up and another sign coming down. At
each corresponding point of the downward and upward paths the values of F' - đs
are exactly equal in size but of opposite sign, so the net result of the integral
will be zero for this case. Thus the kinetic energy with which the mass comes
back to the bottom is the same as it had when it left; that is the principle of the
conservation of energy. (Note that when there are friction forces the conservation
of energy seems at first sipght to be invalid. We have to fnd another ƒorm of
energy. Ït turns out, in fact, that heaf 1s generated in an object when ï§ rubs
another with friction, but at the moment we supposedly do not know that.)
13-2 Work done by gravity
The next problem to be discussed 1s mụch more difficult than the above; it
has to do with the case when the forces are not constant, or simply vertical, as
they were in the cases we have worked out. We want to consider a planet, for
example, moving around the sun, or a satellite in the space around the earth.
W© shall first consider the motion of an object which starts at some point 1
and falls, say, đirecfu toward the sun or toward the earth (Fig. 13-2). WilI there
--- Trang 256 ---
s c——=———s
Fig. 13-2. A small mass mm falls under the influence of gravity toward
a large mass Mĩ.
be a law oŸ conservation of energy in these circumstances? The only difference is