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Particle Dynamics inOne Dimension

R

2.1 INTRODUCTION

Suppose a particle of mass m is acted on by several forces F1; F2, . . ., Fn. The net force F act-ing on the particle is given by the superposition principle as

(2.1)

(2.2a)

and the motion of the particle is described by Newton's second law as

F =dt

where p is the linear momentum of the particle. Only when the mass m remains constant maywe write

d\F = m —y = ma

dt(2.2b)

If we describe the motion in rectangular coordinates, Eq. (2.2b) may be written in the form ofthree components as

(2.3)= mx = max

with similar expressions for Fy and Fz. If the acceleration a or its components ax, ay, az areknown, then Eq. (2.2b) may be used to solve for the force F. In general, the situation in particledynamics is just the reverse; that is, we know the net force F acting on a particle and we want

28

Sec. 2.1 Introduction 29

:o solve Eq. (2.2b) to find the position of the particle as a function of time t. In this chapter, since•ve are limiting our motion of the particle to one dimension, the only equation of interest isEq. (2.3), which after dropping the subscript x may be written as

dh

To be more explicit, we may write this equation as

F(x, = md2

A

dt2

(2.4)

(2.5)

•vhere x = dxldt = v is the velocity of the particle and Eq. (2.5) states that the force acting on:he particle is a function of position, velocity, and time. Such a problem in which the appliedforce is a function of all three variables simultaneously is difficult to solve. On the other hand,if the applied force is a function of only one variable, the problem is much simplified. Hence weihall divide our discussion into the following four cases:

1. The applied force is constant; that is, F = constant, such as freely falling bodies and every-day motion.

2. The applied force is time dependent; that is, F = (Ft), such as in the case of electromag-netic waves.

3. The applied force is velocity dependent; that is, F = F(v), such as air resistance to fallingor rising objects.

4. The applied force is position dependent; that is, F = F(x), such as restoring force to vi-brating springs.

Before we start solving Eq. (2.4) for these different cases, we may remind ourselves that since

d x

dt1

dv—dt

dv dx

dx dt

dv—dx

Eq. (2.4) may be written in the following different forms:

d2xF = m

F = m

dt2

dv

dt

dvF = mv —

dx

(2.6)

(2.7a)

(2.7b)

(2.7c)

Also, since momentum p is defined as p = mv = m{dxldt), we may write Eq. (2.7a) as [or di-rectly from Eq. (2.2a) as applied to the one-dimensional case]

F =dp

dt(2.8)

30 Particle Dynamics in One Dimension Chap. 2

That is, the applied force is equal to the rate of change of momentum. If the applied force actsbetween the time interval tx and t2, then, by integrating Eq. (2.8), we get

Pi - Pi = Fdt (2.9)

which is the integral form of Newton's second law, while Eqs. (2.7) are the differential forms.The integral on the right side of Eq. (2.9) is the impulse delivered by a force F during a shorttime interval (t2 — t{); that is, the change in the linear momentum is equal to the impulse deliv-ered. Thus Eq. (2.9) is a statement of the impulse-momentum theorem.

2.2 CONSTANT APPLIED FORCE: F= CONSTANT

We are interested in studying the motion of a particle when the applied force acting on the par-ticle is constant in time. Since F is constant, so will be the acceleration a, and we may writeNewton's second law as

dt2

dv F— = -- = a = constantdt m

(2.10)

The equation may be solved by direct integration provided we know the initial conditions. Solv-ing Eq. (2.10) gives us the familiar results obtained in elementary mechanics, as we will shownow. Let us assume that at t = 0, the initial velocity is v0, and at time / the velocity is v. Thus,from Eq. (2.10),

dv = dt

which on integration yields

v = vQ + at (2.11)

Substituting v = dx/dt in Eq. (2.11) and again assuming the initial condition that x = x0 at t0, we get by direct integration

x = vQt wBy eliminating t between Eqs. (2.11) and (2.12), we get

v2 = VQ + 2a(x - x0)

(2.12)

(2.13)

Equations (2.11), (2.12), and (2.13) are the familiar equations that describe the translational mo-tion of a particle in one dimension.

One of the most familiar examples of motion with constant force, hence constant acceler-ation, is the motion of freely falling bodies. In this case, a is replaced by g, the acceleration dueto gravity, having the value g = 9.8 m/s2 = 32.2 ft/s2. The magnitude of the force of gravity act-ing downward is mg.

Sec. 2.3 Time-Dependent Force: F = F(t)

2.3 TIME-DEPENDENT FORCE: F= F(fl

31

In this case, the force being given by F = F(t) implies that it is an explicit function of time;hence Newton's second law may be written as

(2.14)

(2.15)

which on integration gives, assuming that v = v0 at t = t0,

1 f'v = v0 + — F(t) dt

Since v - v(t) = dx{t)ldt, Eq. (2.15) takes the form

dt mdt

or, integrating again,

x = x0 + vo(t ~ t0) + — F{t)dt\dt (2.16)

Since there are two integrations, we may use two variables t' and t" and write Eq. (2.16) as

x = xnvo(t -to)+ - f dt' \ F(f) df (2.17)

We will illustrate this discussion by applying it to the interaction of radio waves with elec-trons in the ionosphere, resulting in the reflection of radio waves from the ionosphere. Theionosphere is a region that surrounds Earth at a height of approximately 200 km (about125 miles) from the surface of Earth. The ionosphere consists of positively charged ions andnegatively charged electrons forming a neutral gas. When a radio wave, which is an electro-magnetic wave, passes through the ionosphere, it interacts with the charged particles and ac-celerates them. We are interested in the motion of an electron of mass m and charge —e initiallyat rest when it interacts with the incoming electromagnetic wave of electric field intensity E,given by

E = Eo sin(a)t (2.18)

where w is the oscillation frequency in radians per second of the incident electromagnetic waveand (f) is the initial phase. The interaction results in a force F on the electron given by

F = —eE= — eE0 (2.19)


while the acceleration of the electron is given by

Fa = — =

m

eEn

m+

Let a0 = eEJm be the maximum acceleration so that Eq. (2.20) becomes

a = —a0 sin(tot + <j>)

Since a = dvldt, the equation of motion of the electron may be written as

dv eE(l— = sm(cot + <p)dt m

(2.20)

(2.21)

(2.22)

Assuming initially the electron to be at rest, that is, t = t0 = 0, u0 = 0, the integration ofEq. (2.22) yields

eEn eEnv = cos (j> + cos(cot + <p) (2.23)

mco mco

Since i; = dxldt, and assuming that* = xQ at t0 = 0, the integration of Eq. (2.23) yields

x = — sin (/> —mco

eE

mojc o s </> }t +

eEn

mcosin(tot + 4") (2.24)

The first two terms indicate that the electron is drifting with a uniform velocity and this veloc-ity is a function of the initial conditions only. Superimposed on this drifting motion of the elec-tron is an oscillating motion represented by the last term. The oscillating frequency to of theelectron is independent of the initial conditions and is the same as the frequency of the incidentelectromagnetic waves. In the following, we want to investigate how such coherent oscillationsof free electrons can modify the propagation characteristics of incident electromagnetic waves.

Comparing Eq. (2.19) for F with Eq. (2.24) for x, it becomes quite clear that the oscillat-ing part of the displacement x is 180° out of phase with the applied force that results from theelectric field of incident electromagnetic waves. Ordinarily, in a dielectric at low frequencies,the charges are displaced in the direction of the applied force, resulting in the polarization of thecharges in phase with the applied force. In such situations, the resulting dielectric coefficient ofthe material is greater than 1. In the case of the ionosphere, it can be shown that the resultingpolarization is 180° out of phase with the electric field; hence the dielectric coefficient of theionosphere is less than 1. This result has two consequences.

1. The phase velocity v of electromagnetic waves in the ionosphere is greater than the speedof light c.

2. The refractive index of the ionosphere for incoming electromagnetic waves is less than therefractive index of the free space from where the waves are coming (the incident medium,which is a vacuum in this case).

Sec. 2.3 Time-Dependent Force: F= F(t) 33

Ionosphere

Empty spaceFigure 2.1 Reflection of radiowavesfrom the ionosphere. The total internalreflection of electromagnetic wavesfrom the ionosphere.

This leads to the possibility of total internal reflection, that is, the reflection of incident electro-magnetic waves from the ionosphere back to Earth, as illustrated in Fig. 2.1.

y Example 2.1

A block of mass m is initially at rest on a frictionless surface at the origin. At timet = 0, a decreasing force given by F = F0exp(-A,t), where % = 0.5 is positive and less than1, is applied. Calculate x(t) and v(t). Graph x, v, and F versus t.

SolutionFrom Newton's second law, afterrearranging and integrating, we get thevelocity vl at time tl to be

dt

vl

0

rtl1 dv=

F0V^dt

Once again rearranging andintegrating, we get the displacement xlto be

xl

ldx=

(km)

•tl

X -m

1)


Now we can write expressions forFj, vj, and xj, keeping in mind thatat t = 0, F = Fo. These calculations aremade for N = 100 values even thoughonly 15 values are shown in the graph.The values of F, x, and v at fourdifferent times are given below.

N:=100

m:=2

R: = Fo-(e

Fo

-X-t.

V =2

m-X

i : = 0 . . N t. :=i

F o : = l X\=.5

Fo , ,v. : = -\1 - e

m-X

-x-t.

i) Fo-X,1

2

m-X

(a) Look at the variation in F, v, and xversus t and explain the conclusionsyou draw from such variations.

(b) What does the leveling of thevalues of F and v for high t mean?

(c) What changes in (b) will beobserved if X is 0.01, 1.0, and 5?(You can explain by regraphing fordifferent values of A,.)

= 0.607

F5= 0.082

F25 =3.727-10

F50 = 1.389-10

x, =0.213

x5= 3.164

x25=23

X 5 0 = 4 8

= 0.393

. =0.918

4

X.1

-X-V.

F.

/

J

(1

I

_H—t—1—t—

1 n o p n

• t t t t

10

t.

Time

15

Motion due to a decreasing force

Exercise 2.1: A particle of mass m is at rest at the origin of the coordinate system. At t = 0, a force

F = - te-kt)

is applied to the particle. Find the acceleration, velocity, and position of the particle as a function of time.Graph these values and answer (a), (b), and (c) in the example.

Sec. 2.4 Velocity-Dependent Force: F= F(v)

Z4 VELOCITY-DEPENDENT FORCE: F=F{v)

35

There are many situations of common everyday occurrence where, in addition to constant ap-r ad forces, forces are present that are a function of velocity. For example, when a body is in arra\ itational field, in addition to the gravitational force, there exists a force of air resistance on-j-.e falling or rising body, and this resisting force is some complicated function of velocity. The-dme is true of objects moving through fluids (gases and liquids). Such opposing forces to the-.otion of objects through fluids are called viscous forces or viscous resistance. In these cases,Newton's second law may be written in the following form:

F(v) = mdv

dt

dv dx dvF(v) = m — — = mv —

dx dt dx

(2.25)

(2.26)

Knowing the form of the force F(v), either of these two equations may be solved to analyze the~otion, that is, to calculate x as a function of t. Starting with Eq. (2.25), we may write

si,,dt = m

F{v)•- hich on integration yields

t = t(v) = m\ — (2.27)J F(v)

Solving this gives v as a function of t; that is v = v(t). Thus, knowing v(t), we can solve for x.

dxV = ~ = V(t)

: r dx - rit) dt

- hich on integration gives

x = x{t) = I v{t)dt

Thus the problem is solved. Similarly, if we start with Eq. (2.26), we get

vdv

(2.28)

dx = m

- hich on integration yieldsF(v)

x = x(t) = mv dv

(2.29)

(2.30)

(2.31)

Equations (2.29) and (2.31), which describe the displacement x as a function of t, may appear:o be quite different, but when evaluated, they yield the same relationships as can be demon--•j-ated. We shall divide our discussion into two parts. First, we shall discuss those cases in which


there is no externally applied force besides the viscous resistance opposing the motion of thebody. Later, we shall investigate more practical situations in which both types of forces, fric-tional as well as applied, are present.

Special Case

Suppose an automobile is moving with velocity v0 on a smooth frictionless surface when itsengine is suddenly shut off. Let us assume that the air resistance is proportional to velocity;that is,

Fr = Fr(v) = ~kv (2.32)

Assuming that at t = 0, v = % calculate v and x as functions of t. We may write the differen-tial equation of motion as

Fr(v) = —kv = m

m r" dvThat is, dt = —

k \ v

which on integration gives (J dv/v = In v)

dv

dt(2.33)

m vt= - - i n -

k \v0

After rearranging

v = vne-(k/m)t

(2.34)

(2.35)

That is, the velocity decreases exponentially with time.Substituting v = dxldt in Eq. (2.35) and rearranging, we get

= voe-(k/m)'dt

which on integration, setting the limits as t = 0 when x = 0, and allowing the displacement tobe v at time t, gives

mvnx = [1 (2.36)

It is clear from Eqs. (2.35) and (2.36) that when t = 0, v = % x = 0, as it should be. We notefrom Eq. (2.35) that v = 0 only when t = o° , and then from Eq. (2.36), x = mvjk = xb whereA. is the limiting distance. The body never goes beyond this distance. (But it takes infinite timeto reach there! We shall discuss this shortly.)

Reconsidering Eqs. (2.35) and (2.36), we know that the motion cannot continue foreverand the automobile must come to rest long before the infinite time as calculated earlier. Let usassume that when the automobile reaches a certain minimum velocity ve it has almost reachedthe f.nal distance xe. This will be true as long as v is less than the certain minimum value ve.

Sec. 2.4 Velocity-Dependent Force: F= F(v) 37

By substituting v = v( in Eq. (2.35), we can calculate the time t( it takes to reach velocity ve;that is,

_ p-{k/m)tc

or

u( - vOi

tt = - l n M

(2.37)

(2.38)

Another interesting fact is revealed if we consider the motion in a short time interval whenthe retarding or the resistive force just begins acting on the moving body. To discuss this, let usexpand the right sides of Eqs. (2.35) and (2.36) by using a Taylor series (ex = 1 + x + x2/2\ +xV3! + --O-Thatis,

kvnv = v0 1 + ••• = v0 +

= vo arOt

1 rO

mt +

(2.39)

where a^ = F^m is the acceleration at t = 0. Similarly,

1 jtw(

X = ^ " 2 ma .2 1 Fo

2 m

(2.40)

If we ignore the higher terms, Eqs. (2.39) and (2.40) reveal that they describe the motionof a particle acted on by a constant force provided t is very small. Note that — kv0 = Fr0 = mar0,which is the force acting on the particle initially when t = 0; that is, these equations are simplythe equations of motion of a particle under a constant force.

General Case

The preceding situation was limited to a simple case in which the retarding force Fr(v) was pro-portional to velocity. In actual situations, Fr is a complicated function of velocity and the solu-tions cannot be obtained by simply using the integration tables. Instead it becomes necessary todo numerical integrations. But, in many cases over a wide range of velocities, it is possible, inpractice, to use the following approximation in which the retarding force or frictional force isproportional to some power of velocity. That is,

Fr = (+)kvn (2.41)

where k is the positive constant of proportionality for the strength of the retarding force and nis a positive integer. If n is an odd integer, the negative sign in Eq. (2.41) must be used. If n isan even integer, if will be positive, and the sign + or — in Eq. (2.41) is chosen in such a waythat it gives the direction of Fr to be opposite to that of v (Fr in the direction of v will be addingenergy to the system instead of retarding it!). For small objects moving in air with velocities lessthan 25 m/s, it is found that, if we take n — 1, we get good agreement with experimental results,


while for velocities greater than 25 m/s but less than — 32 m/s, the use of n = 2 gives goodagreement with experimental values.

Let us apply these ideas to the case of a freely falling body, that is, to the vertical motionof an object in a resisting medium, the medium being air in this case. Let us assume that the airresistance is proportional to v, which can be written as — kv, independent of the sign of v. Thusthe net force acting on the body is

F = Fg + Fr = —mg — kv

and the differential equation describing the motion of the freely falling body is

dvm — = — mg — kv

at

(2.42)

(2.43)

Taking v = v0 at t = 0, we may write

/dt= -

m dvr

Vo mg + kv

mrk

kv)

orm mg + kv

t = — - I nk mg + kv0

Solving this equation for v, we get

mg mi

kvo\e

,-(k/m)t

Note that if the initial velocity is zero, that is, if at t = 0, vQ = 0, we get

_ e-(k/m)t.

(2.44)

(2.45)

(2.46)

In Eq. (2.45), we may substitute v = v(t) = dxldt and the limits x = x0 when t = 0 and x = xwhen t — t. After integrating, we get the following result:

mik

nfik2

mv, [1 _ .-(k/m)t (2.47)

Terminal Velocity

Let us consider Eq. (2.45) once again. As t increases, the exponential term decreases and dropsto zero when t is very large as compared to mlk; that is, for t > mlk,

e-(k/m)t _ e-[ = 0

And for such large values of t, the velocity v reaches a limiting value, which from Eq. (2.45) isvt= — (mg/k). This limiting velocity v, is called the terminal velocity of a falling body. The ter-minal velocity of a body is defined as the velocity of the body when the retarding force (the forceof air resistance) is equal to the weight of the body, and the net force acting on the body is zero(hence there will be no acceleration of the body); that is,

or

Fr + Fg = 0 or — kv, — mg = 0

mg(2.48)


The magnitude of the terminal velocity, which is equal to mglk, is called the terminal speed. Asan example, the terminal speed of raindrops varies anywhere from 3 to 7 m/s. Different bodiesstarting with different initial velocities will approach terminal velocities in different times.There are three different possibilities for initial velocities: |vo| = 0, |v0 < vj , and v0 > v,|. Fig-ure 2.2 illustrates the time taken by the bodies to reach terminal velocity for the three cases.

Characteristic Time

It should be clear from the previous discussion that the dimensions of mlk are time. We definethe quantity mlk as the characteristic time T; that is, from Eq. (2.48) (ignoring the sign),

mT =

Using the definitions of v, and T, we rewrite Eqs. (2.45) and (2.47) as

v = -v, + (vt + vQ)e~'/T

vtt + (gT2 + VOT)(1x = x0,-t/r

(2.49)

(2.50)

(2.51)

Let us consider a special case in which the body starts from rest. Substituting v0 = 0 inEq. (2.50), we obtain

v= -vt(l - e~'/T) (2.52)

and, if t = T, Eq. (2.52) takes the form

v = -vt(l - \/e) = -0.63u(

That is, in one characteristic time the body reaches 0.63 of its terminal speed. This allows us todefine the characteristic time as that in which the body reaches 0.63 of its terminal speed. Ift = 2T, then v = -0.87ur, while for t = 10T, V = -0.99995u(.

Let us go back to Eqs. (2.45) and (2.47) again. Using a series expansion, we can show thatthese equations reduce to the familiar equations of motion for the case of constant force. Fort < T(=m/k = v/g), we get

and

v = v0 - gt

x = x0 + vot - \gt2

(2.53a)

(2.53b)

That is, for small t, the effect of air resistance is negligible. On the other hand, for t > r (= mlk),Eqs. (2.45) and (2.47) reduce to

v = -v,

x = x0 VOTm m

= Xo+ VQT (r2g - grt)

(2.54a)

(2.54b)


Figure 2.2

The graph illustrates the time it takes the bodies to reach terminal velocity fordifferent initial velocities for given values of m, g, and k. vt is the terminal velocity,vOl, vO2, and vO3 are three different initial velocities

k:=5

vO = 0v02 < vtv03 > vt

(For absolute values)

Note that the terminal velocity isvt = -4.9 m/secand, depending on the initialcondition, the terminal velocity isreached between 20 and 30seconds. (In the graphs, t is dividedby 10.)

vt=-4.9

vl5 =-3.097 v25 =-3.833 v35=-6.04

vl10 =-4.237 v2,0 =-4.508 v31Q=-5.32

n:=40

m:=2.5

vol :=0

t :=0..n

g: = 9.8

vt=-4.9

"•B-(T

vo2: = -2

vo3 :=-8

v 2 . = _

v 3 : = .

l 1 0 2,01Q

vl20

vl30

vl40

= -4.81

= -4.888

= -4.898

v220

v230

v240

= -4

= -4

= -4

.847 v

.893 v

.899 v

3 2 0 =-4.957

330 =-4.908

3.0 =-4.901

v2.

v3

-101

0.6 1.2 1.8

(a) What are the different factors that affect theterminal velocity?

(b) If the initial velocity is positive and upward,how will it affect the terminal velocity? How willthis graph differ from the others before terminalvelocity is reached? Graph this.

t

Totime

10

k\ t

2.4

Time to reach terminal velocity

Sec. 2.4 Velocity-Dependent Force: F= F{v)

A Better Approximation

41

For small, compact, heavy bodies, a better approximation is that in which the retarding or vis-cous force is proportional to v2. Thus the equation

takes the form

-r ;,,,2kv — mg = mdv

dt(2.55)

where - kv2 is used for rising (or ascending) bodies, while +kv2 is used for falling bodies. Theterminal speed is given by

F+F=0 or kit = mg

or

and the characteristic time T is given by

(2.56)

(2.57)

(2.58)

Following the procedure outlined before, Eq. (2.55) can be solved for v and x. The results ob-tained are

f m dvt = \ —; , for rising objects

J -mg - kxr

f m dvt = ~ c , for falling objects

J -me + kxr

which give

(2.59)

(2.60)

(2.61)

(2.62)

where C\ and C2 are constants to be determined from initial conditions. Solving these equationsfor v, we get

t = — r tan '( — for rising objects

, v \t = -T tanh [—] + C2, for falling objects

C, - tv - v.

v = — v, tanh

T

t-C,

for rising objects

for falling objects

(2.63)

(2.64)

As compared to a retarding force ±kv, the terminal speed for the case of a retarding force±kv2 is reached much faster, as illustrated in Fig. 2.3. In this case, when t = 5T, V =—0.99991 vt; that is, it reaches the same speed in half the time.


Figure 2.3

The graph shows the motion of a vertically falling object under a linearforce and a quadratic force with initial conditions to = 0, xo = 0 and vo = 0.

Linear retarding force: F=-kx Quadratic retarding force: F=-kv2 or kv2

The terminal velocity v2 = -0.808 for i N :=50quadratic retarding force (± kv2) is reachedmuch faster (in about 0.2 seconds) than m :=.1the terminal velocity vl = —0.653 for alinear retarding force (-kv) is reached in mabout 0.3 seconds. To calculate vl and kv2 use Eqs. (2.46) and (2.64).

Tl =0.067

:=0..N

;:=9.8

_ I

' 50

k:=1.5

x2 := -m_

x2 = 0.082

(a) In a given time interval, which object vl :=BJwill travel a larger distance and why?

vl =0.653

(b) For the vertically falling objectswhich of the two situations is moredesirable and why? v l . : = _ ( v l ) - l l - e

v2 :=

v 2 = 0.808

v2. :=-v2-tanh

(c) What do the values of vl and v2 attime t = 20 and 30 indicate?

vl5 =-0.508

t 3 ( )=0.6

=-0.677

vl3 Q =-0.653 v23 Q=-0.808

t 1 5 = 0 . 3

v l | Q =-0.621 v 2 m =-0.796 g •

v l | 5 =-0.646 v215 =-0.807 > '

vl =-0.652 v2,n =-0.808

0.1

1

U.It)

0.5

0.25

n:

i a a • • [ l a a a

' X X X-

-B—( 1—B-

f—X-

• •

X X

-B-H

-X—>

0.2 0.3

t.l

time

0.5

Speed for quadratic and linear force

Note that the terminal velocity v2 is reached much faster (in about 0.3 second) than theterminal velocity vl (in about 0.6 second). Explain why.


Example 2.2

A ball of mass m is thrown with velocity vo on a horizontal surface where theretarding force is proportional to the square root of the instantaneous velocity. Calculatethe velocity and the position of the ball as a function of time and graph the results.

SolutionThe retarding force is given byRearrange this equation and integrateassuming that initial velocity at t = 0 isvo and vl at time t l .

Solving, we get the value of vlat time tl as

Integrating vl we find thedisplacement xl at time t l .

Now we may graph x and v as function oft using Eqs. (i) and (ii), rewritten as (iii)and (iv)

dv 2F=m-—=-kv

vl

vO

xl

dt

-dv=

k tl

2 m

rti

l dx=

-Idtm simplifies to

2-A/vl-2-A/v0=-k-—m

(i)

- - — d t

2 m

xl - xO=tl-vO + —-k2-—12 m2

i :=0..20 t. :=i xo:=0 vo:=20 k := 1.1 m:=.5

k 2 k / Nv. :=vo t.-vo -( (t.

m ' 4-m2 V V(iii)

x. : = xo+- vo-t.--'---ft.)1 2 m V V T2~^

(iv)


Displacement and velocity versus time

1500

.1000

500

i1/

10

Time

20

200

150

Displacement and velocity versus time

1 \100

- K -

50

/

2.5

Time

7.5 10

Explain the decrease in v and then the increase in v as a function of time t. How does itaffect the value of x? (Refer to the zoomed graph on the right).

EXERCISE 2.2 Repeat the example for a retarding force that is proportional to thecube root of the instantaneous velocity.

2.5 POSITION-DEPENDENT FORCES: F= F(x), CONSERVATIVEFORCES, AND POTENTIAL ENERGY

This is one of the most important cases considered so far. There are many situations in whichmotion depends on the position of the object. Examples of position-dependent forces are grav-itational force, Coulomb force, and elastic (tension and compression) forces. The differentialequation that describes the rectilinear motion of an object under the influence of a position-dependent force is

d2xm —y = F(x)

at(2.65)

which may also be written in such a manner that v is a function of position; that is,

dvmv— = F(x)

dx(2.66)

or -{-my* )=F(x) (2.67)

Sec. 2.5 Position-Dependent Forces 45

Since the kinetic energy of the particle is K = \m^, we may write Eq. (2.67) as

dK

dx= F(x)

which on integration gives

or

K- Ko= F(x) dx

F(x)dxmtfo= f

(2.68)

(2.69a)

(2.69b)

The right side is equal to the work done when the particle is displayed from position x0 to x.It is convenient at this point to introduce potential energy or a potential energy function

• or simply & potential function) V(x) such that

dV(x)

dx= F(x) (2.70)

We define V(x) as the work done by the force when the particle is displaced from x to some ar-bitrary chosen standard point xs; that is,

V(x) = F(x) dx = - F(x) dx•'x 'x,

which is consistent with Eq. (2.70). Thus the work done is going from x0 to x is

(2.71)

= -j'dV(x)- j dV(x)

= +V(x0) - V(x) = -V(x) + V(x0)

Combining Eqs. (2.69) and (2.72), we get

K + V(x) = Ko + V(x0) = constant = E

or

(2.72)

(2.73)

(2.74)

This equation states that if a particle is moving under the action of a position-dependent force,then the sum of its kinetic energy and potential energy remains constant throughout its motion.Such forces are called conservative forces. For nonconservative forces, K + V =£ constant, anda potential energy function does not exist for such forces. An example of a nonconservative force


is frictional force. [It may be pointed out that if V(x) is replaced by V(x) + constant, the pre-ceding discussion still holds true. In other words, the sum of the kinetic and potential energywill still remain constant and will be equal to E.] E is the total energy, and Eq. (2.74) states thelaw of conservation of energy, which holds only if F = F(x). A description of the motion of aparticle may be obtained by solving the energy equation, Eq. (2.74); that is,

dx

dt

which on integration yields

-,-f

± A - [E - V(x)]I m

±dx

V(2/m)[£ - V(xJ]

(2.75)

(2.76)

and gives t as a function of x. [We shall not discuss the significance of the negative sign, whichdeals with time reversal.]

In considering the solution of Eq. (2.76), it is essential to note that only those values of xare possible for which the quantity E — V(x) is positive. Negative values lead to imaginary solu-tions and hence are unacceptable. Also, the motion is limited to those values of x for whichE — V(x) 3= 0; that is, the roots of this equation give the region or regions to which the motionis confined. This is demonstrated in Fig. 2.4. The function \mx2 + V(x) is called the energy in-tegral of the equation of motion m{dvldt) = F{x), and such an integral is called a constant ofmotion. (This is the first integral of a second-order differential equation.)

Before we give specific examples of solving the equation of motion for x(t), we shall showthat much can be learned about the motion by simply plotting V(x) versus x. Figure 2.5 showsthe plot of a potential energy function for one-dimensional motion. As mentioned earlier, themotion of the particle is confined to those regions for which E - V(x) > 0 or V(x) < E. Let uskeep Eq. (2.75) in mind and discuss different cases.

E - V(x) I Allowed limits of motion of a particle

Figure 2.4 Allowed regions of motion for a particle in a position-dependentforce field.

Sec. 2.5 Position-Dependent Forces

V(x)

47

Figure 2.5 The solid curve corresponds to a potential function V(x), and EO,E l , . . . , are different energies of a particle moving in such a potential.

If E = Eo, as shown in Fig. 2.5, then Eo - V(x) = 0 and, according to Eq. (2.75), x = 0;that is, the particle stays at rest in equilibrium at x = x0. Let us consider the case in which theparticle energy is slightly greater than Eo, say E1. For x < xx and x > x[, v will be imaginary;hence the particle cannot exist in these regions. Thus a particle with energy Ex is constrained tomove in the potential well (or valley) between xx and x[. A particle moving to the right is re-flected back at x[; and when traveling to the left, it is reflected back at xx. The points xx and x\are called the turning points and are obtained by solving Ex - V(x) = 0. The velocity of the par-ticle at these points is zero. Between xx and x[, the velocity of the particle will change as V(x)changes. We briefly explain the motion of a particle corresponding to different energies andmoving in a potential V(x), as shown in Fig. 2.5.

Eo: The particle is in stable equilibrium.Ex: The particle moves between the turning points xx and x[.E2: The particle moves between the turning points x2 and x'2 with changing velocity. While

moving between the turning points x"2 and x"2, the particle has constant velocity and henceis in the region of neutral equilibrium. The particle can also exist in the region for x > x"2".

E3: When a particle with this energy is at x3, it is at a position of unstable equilibrium. It canalso move in the valley on the left with a motion similar to that of a particle with energyE2. Once it starts moving to the right, it keeps on moving, first with increasing velocity tox"2 and then with constant velocity up to x"2.

E4: A particle with this energy can move anywhere. When passing over the hills, it slowsdown, while over the valleys, it speeds up, as it should.


Continuing our discussion of position-dependent forces, we shall examine two specialcases of interest in the next sections:

1. Motion under a linear restoring force2. Variation of g in a gravitational field

Example 2.3

A particle of mass m is subjected to a force F = a -2bx, where a and b are constants.Find the potential energy V = V(x). Then graph F(x) and V(x). Discuss the motionof the particle for different values of energy.

SolutionSubstitute for F in the expression for V and integrate to get the value of V.

The motion is limited tothe region x = 0 and x = a/b p_a _ 2.t,-x

For E < 0, the motion is limited tothe left of V(x) and cannot crossthe barrier. V=-

V=- I F dx

For V(x)=-a-x+b-xzss0 x=-b

For — V(x)=-a+2-b-x=0 x = —dx 2-b

At x= Vmin=2-

F=0 at

Differentbelow

- a - = 52-b

vo = o

FQ = 20

(min(V))

max(F) =

b

x=0

4-b

and2-b

values are calculated

1 = 10b

V5 =-50

F 5 =0

= -50

20

2

- - ^ = - 5 04 b

V10=°

F]0=-20

max(V) =150

min(F) =-40

a-2-b-xdx simplifies to V=-a-x-i-b-x2 (i)

: = 0.. 15

V. :=-

>

ial

Ene

S l

B —oft. V.•§ '

200

150

100

50

-100

a:=20

F. :=a- 2-b-x:

5 10x.

l

Distance x

b:=2

(ii)

Sec. 2.6 Motion under a Linear Restoring Force 49

(a) Looking at the variation in the values of V and F versus x, what do youconclude from this variation?

(b) What are the values of F and V where these graphs intersect? What is thesignificance of this? Explain.

(c) What is the significance of x = 5 where F = 0 and V is minimum? Explain.

EXERCISE 2.3 Repeat the example for F=a-2bx2

2.6 MOTION UNDER A LINEAR RESTORING FORCE

Let the motion of a particle subject to a linear force be given by

F(x) = -kx (2.77)

This equation is a statement of Hooke's law. A typical example of such a motion is that of a massfastened to a spring. The resulting motion is simple harmonic, as we shall discuss in detail inChapter 3. For the time being, we shall use the energy method discussed in the previous sectionto obtain the solution. Taking the standard point to be at the origin (also the equilibrium point),that is xs = 0, we may write the potential energy to be

V(x) = - [ F{x)dx= - f (-kx) dx

or V(x) = {kx2 (2.78)

Once again, the total energy is a constant of motion and may be arrived at in the same manner;that is,

which on integration gives

or

dvmv— = F(x) = -kx

dx

mv dv = —kx dx

\mv2 = - \kx2 + constant

\mv2 + {kx2 = E = total energy

(2.79)

(2.80)


We can now use this equation or Eq. (2.76) with V(x) given by Eq. (2.78) to solve for the dis-placement x. For the conditions t0 = 0 at x = x0, Eq. (2.80) or (2.76) takes the form (keepingonly the positive sign)

J Vf2dx

m)(E - \kx2) -I dx

Substituting

we get

k k—; x = sin 0 and A| — dx = cos 0 d0

t = m

As usual, let the angular velocity be denned as

Therefore,

or

Combining Eq. (2.84) and (2.82), we get

or

t=\ie- 0O)

8= cot +

sin ' l - v / — x | =

x = A sin(«f + 0O)

where A is the amplitude given by

(2.81)

(2.82)

(2.83)

(2.84)

(2.85)

(2.86)

Thus Eq. (2.85) states that the motion of the particle is simple harmonic with the coordinate xoscillating harmonically in time with amplitude A and frequency w. A and 0 can be determinedfrom the initial conditions; that is, if E and x0 are given, from Eqs. (2.86) and (2.85)

and x0 = A sin 0O

(2.87)

(2.88)

Sec. 2.7 Variations of g in a Gravitational Field

2.7 VARIATION OF g IN A GRAVITATIONAL FIELD

51

For small heights just above the surface of Earth, the value of g is almost constant and is equalio 9.8 m/s2 = 32.2 ft/s2. But at large distances above the surface of Earth, the value of g varieswith distance and may be calculated in a simple manner. According to Newton's law of gravita-tion, the force between an object of mass m at a distance x from the center of Earth of mass Mis

F(x) = -GMm

(2.89)

If we neglect air resistance, the differential equation of motion of an object in a gravitional fieldmay be written as

dv~-dx

Mm

Since v = x, we may write

•vhich on integration gives

m I xdx = - GMm idx

x~2

. , MmL - G = constant = E

?r \ mx2 + V(x) = E

.vhere we have defined the gravitational potential energy V(x) to be

MmV(x) = ~G

(2.90a)

(2.90b)

(2.91)

We can derive the same expression for the gravitational potential energy by the direct defini-::on of the potential function. But it is convenient to take the initial conditions to be xs = °o, in->tead of xs = 0, as we did previously. Thus

V(x) == -J F{x)dx=-\ - GMm GMm

. defined previously.]We may rewrite Eq. (2.90a) as

(2.92)

/-here the positive sign corresponds to ascending motion, while the negative sign corresponds: .•> descending motion. Equation (2.92) may be solved for x(t) by integrating; that is,

dx

ro [£• + (GMm/x)] 1/2 (2.93)


V(x)i

E?

•V(x)=-G-^

Figure 2.6 Particle of energy E in agravitational potential V(x) versus x.

The integration of this equation is not as simple as in the case of a linear restoring force. Weshall not pursue this any further at the present time, but shall give a graphical interpretation anddiscuss some simple cases.

The plot of V(x) versus x is shown in Fig. 2.6. It is quite clear that for negative values ofE the motion is bound with a turning point at xT. That is, when E — Ex the body will go a heightx = xT, come to a stop, and turn around. For values of E greater than zero, there is no turningpoint and the body will never return to Earth. Thus, for a minimum energy E = 0, the velocitycorresponds to v = ve, the escape velocity, which we shall calculate below. We can calculate theturning point by substituting velocity v = x = 0 and x = xT in Eq. (2.92) (by definition v is zeroat the turning point), and we get

Mm Mm— G = E or xT = — G ——-

x t(2.94)

Since E is negative, xT will be positive.Let us consider Eq. (2.90a) once again:

1 . , Mm-mx — G = constant2 x

(2.95)

Let a body be dropped from a height x = x0 with zero initial velocity. That is, substituting v =x = 0 at x = x0 in Eq. (2.95),

Mm— G = constant

Using this value for the constant in Eq. (2.95) and rearranging, we get

x2 = 2GM(~ - -\x x0

(2.96)

Sec. 2.7 Variations of g in a Gravitational Field 53

Let g0 be the value of the acceleration due to gravity on the surface of Earth, where x =?o that

or ~, Mm

mg0

is.

go = ~ or G = go~ (2.97)

. Y-ially the distance is measured from the surface of Earth; hence we may write

x = R + r, x = r and x0 = R + h

-.ere h = x0 — R equals the height (as measured from the surface of Earth) from which the>:dy is dropped. Using this notation and Eqs. (2.96) and (2.97), we may write

1 1

\R + r R + h)

hus, for r = 0, that is, when the body reaches the surface, v = v0; hence

_- d may be written asR R + h

R

(2.98)

(2.99)

(2.100)

•• -jch reduces to v0 — 2g$h ifh<R, as it should. Equation (2.98) also applies to the case when?ody is projected upward with a velocity % and it will reach a height h when v = 0.

We can arrive at an expression for the escape velocity by substituting h = <* in Eq. (2.99),r-ulting in

= 28oR =2GM

R

2GM

R- 11 km/s = 7 miles/s (2.101)

we could say that at the surface of the Earth E = 0; hence potential energy must be equal toe kinetic energy. That is,

GMm

v. = x =


PROBLEMS2.1. Force F acting on a particle of mass m has the following dependencies:

(a) F(x, t) = f(x)g(t)(b) F(x,t)=f(x)g(t)(c) F{x,x)=f(x)g{x)

Write the differential equations describing these situations. Which of these differential equationscan be solved to describe the motion of the particle? Explain.

2.2. A particle of mass m is acted on by the force (a), (b), (c), (d), or (e) as given below. Solve theseequations to describe the motion of the particle.(a) F(x, t) = k{x + t2), for t = 0, x = x0, and v = v0 = 0(b) F(JC, t) = kx2x, for t = 0, x = x0, and v = D0 = 0(c) F(x, t) = k(ax + t), for t = 0, v = v0

(d) F{x, x) = axVx(e) Fix, x,t) = k(x + xt)

2.3. A block of mass m is initially at rest on a frictionless surface. At time t = 0, an increasing forcegiven by by F = kt2 is applied to the block. Find the velocity and the displacement of the block asa function of time and graph x and v versus t.

2.4. A block of mass m is initially at rest on a frictionless surface at the origin. At time t = 0, a forcegiven by F = Fote

Ar is applied. Calculate x(f) and vit) and graph them. What are these values when(a) t is very small, and (b) t is very large?

2.5. A particle of mass m is at rest at t = 0 when it is subjected to a force F = Fo sin (otf + <£). (a) Cal-culate the values of :c(f) and vit) (b) Make plots of x(0 and vit) versus t. What are the maximumand minimum values of x and vl

2.6. A particle of mass m is subjected to a force given by

F = F0<rA'sin(crf + (f>)

Calculate the values of v(t) and x(t) and graph them. What is the magnitude of the terminal veloc-ity in this case?

2.7. A particle of mass m is at rest at t = 0 when it is subjected to a force F = Fo cos2 cot.

(a) Calculate the values of x(t) and vit)(b) Make plots of jc(f) and vit) versus t.(c) Describe the outstanding characteristic of these graphs.

2.8. A ball of mass m is thrown with velocity t>0 on a horizontal surface where the retarding force is pro-portional to the square root of the instantaneous velocity. Calculate the velocity and the position ofthe ball as a function of time. Discuss any limitations.

2.9. An object of mass m is thrown up an inclined plane of an angle 6 with an initial velocity v0. If themotion is resisted by the retarding force Fr = —kv, how far will the mass travel before coming torest? Assuming the same retarding force, how long will it take the object to travel back to the ini-tial position?

2.10. Repeat Problem 2.9 for the retarding force Fr = ±kvL.

2.11. A boat is slowed down by a frictional force Fiv). Its velocity decreases according to therelation D = kit - ts)

2, where ts is the time it takes to stop the boat and k is the constant. Calcu-late Fiv).

Problems 55

2.12. The motor of a speed boat is shut off when it has attained a speed of v0. Now the boat is sloweddown by a retarding force Fr = Ce ~kv. Calculate v(i) and x(i). How long will it take for the boat tostop, and how much distance will it travel before stopping?

2.13. A particle of mass m starting with an initial velocity u0 is acted on by a force F = m(kv + cv2),where k and c are constants. Calculate the displacement as a function of time.

2.14. For the situation in Problem 2.12 graph F and x versus t.

2.15. For the situation in Problem 2.13 graph F and x versus t. Compare the results with those in Prob-lem 2.14.

2.16. A body of mass m is dropped from a height h. Calculate the speed when it hits the ground if (a) thereis no air resistance, (b) air resistance is proportional to the instantaneous velocity, and (c) air re-sistance is proportional to the square of the instantaneous velocity. Graph velocity versus time ineach case and compare the results.

2.17. A projectile is thrown vertically with a velocity J;0. Calculate and compare times and maximumheights reached when air resistance is (a) zero, (b) proportional to the instantaneous velocity, and(c) proportional to the square of the instantaneous velocity. Graph distance versus time in all casesand compare the results.

2.18. A ball is thrown vertially upward with an initial velocity v0. The air resistance is proportional tothe square of the velocity. Show that the velocity with which the ball returns to the original posi-tion is

VnV.

where vt is the terminal velocity.2.19. Derive Eq. (2.47); that is,

2.20. Using Eqs. (2.45) and (2.47), show that for t < T, we get the familiar equations of motion,Eqs. (2.53a) and (2.53b).

2.21. Show that for the case t < T, Eqs. (2.45) and (2.47) reduce to Eqs. (2.54a) and (2.54b).2.22. Starting with Eq. (2.55), derive Eqs. (2.61) and (2.62).

2.23. Starting with Eq. (2.60) and initial conditions that v0 = 0, x0 = 0 at t = 0, find the expressions forx and v.

2.24. Using the law of conservation of energy, derive the general equations of motion for freely fallingbodies.

2.25. A particle of mass m is attracted toward the origin by a force that is inversely proportional to thesquare of the distance; that is, F = —Kir2. If this mass is released from a distance L, show that itwill take time t to reach the origin, where t is given by

ImL'V'2i


2.26. The velocity of a particle of mass m subjected to a certain force varies with the distance accordingto the relation v = K/x", where A" is a constant. Assuming at t = 0, x = x0, calculate the force act-ing on the particle as a function of (a) distance x, and (b) time t. (c) Calculate the position of theparticle as a function of time /. (d) Graph in (a), (b) and (c) and discuss any outstandingfeatures.

2.27. A particle of mass m is subjected to a force given by F = — ax + bx2, where a and b are constants.(a) Find the potential energy V(x).(b) Make plots of F(x) and V(x).(c) Discuss the motion of the particle for different values of energy and also point out the regions

in which the motion is forbidden.

2.28. A particle of mass m is subjected to a force represented by a potential function V(x) = —ax2 + bx4,where a and b constants.(a) Calculate F(x).(b) Make plots of F{x) and V(x).(c) Discuss the motion of the particle for different values of energy. Also discuss the restrictions

on the motion.

2.29. A particle at time t = 0 is at rest at a distance x0from the origin and is subjected to a force that isinversely proportional to the distance from the origin. Solve the equation of motion for this parti-cle; that is, find v(i) and x(i).

2.30. A particle of mass m is subjected to a force

KF(x) = - Cx + -r

x

where C and K are constants.(a) Find the potential function V(JC).(b) Make plots of F(x) and V(x).(c) Discuss the nature of the motion for different values of E and also find the regions where the

motion is not possible.

2.31. The force between two particles in a diatomic molecule is such that it may be represented by a po-tential function of the form

x' x

where Cx and C2 are the positive constants and x is the distance between the two atoms.(a) FindF(x).(a) Make plots of F(x) and V(x).(c) Assume that one of the atoms in the molecules is very heavy and remains at rest at the origin.

Discuss the possible motions of the other atom in the molecule.2.32. An alpha particle when inside the nucleus is bound by the potential shown for —R < x < R, while

outside the nucleus the interaction between the alpha particle and the nucleus is represented by thecoulomb potential, as shown in Fig. P2.32.(a) Write a potential function for the different regions shown in the figure.(b) Calculate F(x) and make a plot of F(x) versus x.(c) Discuss the motion of the alpha particle for different values of energy and different regions.

Suggestions for Further Reading 57

-VnFigure P2.32

2.33. Show that for h<R, Eq. (2.100) reduces to v0 = 2gji.

2.34. Starting with Eq. (2.90a) and substituting

/ E \1/2

GMm

show that x = xT cos2 6, where

and x = x0 = xT = turning point.

2GM

SUGGESTIONS FOR FURTHER READINGARTHUR, W., and FENSTER, S. K., Mechanics, Chapter 3, New York: Holt, Rinehart and Winston, Inc., 1969.BARGER, V., and OLSSON, M., Classical Mechanics, Chapter 1. New York: McGraw-Hill Book Co., 1973.BECKER, R. A., Introduction to Theoretical Mechanics, Chapter 6, New York: McGraw-Hill Book Co., 1954.DAVIS, A. DOUGLAS, Classical Mechanics, Chapter 2, New York: Academic Press, Inc., 1986.FOWLES, G. R., Analytical Mechanics, Chapter 3, New York: Holt, Rinehart and Winston, Inc., 1962.FRENCH, A. P., Newtonian Mechanics, Chapter 7, New York: W. W. Norton and Co., Inc., 1971.HAUSER, W., Introduction to the Principles of Mechanics, Chapter 4. Reading, Mass.: Addison-Wesley

Publishing Co., 1965.

KLEPPNER, D., and KOLENKOW, R. J., An Introduction to Mechanics, Chapter 4. New York: McGraw-HillBook Co., 1973.

MARION, J. B., Classical Dynamics, 2nd ed., Chapter 2. New York: Academic Press, Inc., 1970.

ROSSBERG, K., Analytical Mechanics, Chapter 4. New York: John Wiley & Sons, Inc., 1983.

STEPHENSON, R. J., Mechanics and Properties of Matter, Chapter 2. New York: John Wiley & Sons, Inc., 1962.

SYMON, K. R., Mechanics, 3rd. ed., Chapter 2. Reading, Mass.: Addison-Wesley Publishing Co., 1971.

TAYLOR, E. F., Introductory Mechanics, Chapter 5. New York: John Wiley & Sons, Inc., 1963.

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