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7/27/2019 Mechanics Symon

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MOTION OF A PARTICLE IN ONE DIMENSION

I.MOMENTUM AND ENERGY THEOREMS

The motion of the particle is governed, according to Eqs.

, by the equation ...................................................... (1.1)

Before considering the solution of Eg. (1.1), we shall define some concept which are

useful in discussing mechanical problems and prove some simple general theorems about

one- dimensional motion. The linear momentump, according to eq. P = mv, is defined as

....................................................... (1.2)From Eq. (1.1), using Eq. (1.2) and the fact that m is constant, we obtain

.............................................................................................(1.3)

If we multiply Eq. (1.3) by dt and integrate from t 1 to t2, we obtain an integral form of the

momentum theorem:

......................................................................... (1.4)Equation (1.4) gives the change in momentum due to the action of the force Fbetween

the times t1 and t2. The integral on the right is called the impulse delivered by the force F

during this time; F must be known as a function of t alone in order to evaluate theintegral. IfF is given as F( x, v, t ), then the impulse can be computed for any particular

given motionx(t), v(t).

A quantity which will turn out to be of considerable importance is the kinetic

energy, defined (in classical mechanics) by the equation


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................................................................................. (1.5)If we multiply Eq. (1.1) by v, we obtain

, or ....................................(1.6)Equation (1.6) gives the rate of change of kinetic energy, and may be called the

(differential) energy theorem. If we multiply by dtand integrate from t1 to t2, we obtain

the integrated form of the energy theorem:

..................................................................... (1.7)In general, when F is given as F( x, v, t), the work can only be computed for a

particular specified motion x(t), v(t). Since v = dx/dt, we can rewrite the work integral in a

form which is convenient when Fin known as a function ofx:

...................................................................... (1.8)2. APPLIED FORCE DEPENDING ON THE TIME

If the force Fis given as a function of the time, then the equation of motion can be

solved in the following manner. Multiplying Eq. by dt and integrating

from an initial instant t0to any later (or earlier) instant t, we obtain Eq. (1.4), which in this

case we write in the form

................................................................. (2.1)Since F(t) is a known function oft, the integral on the right can, at least in principle, be

evaluated and the right member is then a function oft(and t0). We solved for v:

........................................................... (2.2)Now multiply by dt and integrate again from t0 to t:


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* + .................................. (2.3)To avoid confusion, we may rewrite the variable of integration as t in the first integral

and tin the second:

.................................(2.4)This gives the required solutionx(t) in terms of two integrals which can be evaluated shen

F(t) is given.

3. DAMPING FORCE DEPENDING ON THE VELOCITY

Another type of force which allows an easy solution of Eq. is the casewhen Fis a function ofvalone:

...........................................................(3.1)To solve, we multiply by dt and integrate from to to t:

...............................................................................(3.2)

The integral on the left can be evaluated, in principle at least, when F(v) is given, and an

equation containing the unknown vresults. If this equation is solved for v (we assume in

general discussions that this can always be done), we will have an equation of the form

........................................................................(3.3)The solution forxis then

..............................................................(3.4)In certain cases and over certain ranges of velocity, the frictional force is proportional to

some fixed power of the velocity:


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.................................................................................(3.5)If n is an odd integer, the negative sign should be chosen in the above equation.

Otherwise the sign must be chosen so that the force has the opposite sign to the velocityv. The frictional force is always opposed to the velocity, and therefore does negative

work, i.e, absorbs energy from the moving body. A velocity-dependent force in the same

direction as the velocity would represent a source of energy; such cases do not often

occur.

As an example, we consider the problem of a boat traveling with initial velocity vo,

which shuts off its engines at to = 0 when it is at the position x0 = 0. We assume the force

of friction given by Eq. (3.5) with n = 1:

.................................................................................(3.6)We solve Eq (3.6), following the steps outline above [Eq. (3.1) through (3.4)]:

Ln

............................................................................(3.7)We see that as , as it should, but that the boat never comes completely torest in any finite time. The solution forxis

..................................................................(3.8)

As approaches the limiting value .....................................................................................(3.9)


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Thus we can specify a definite distance that the boat travels in stopping. Although

according to the above result, Eq. (3.7), the velocity never becomes exactly zero, when t is

sufficiently large the velocity becomes so small that the boat is practically stopped. Let us

choose some small velocity vs such that when v < vs we are willing to regard the boat as

stopped (say, for example, the average random speed given to an anchored boat by the

waves passing by it). Then we can define the time ts required for the boat to stop by

, ...........................................(3.10)Since the logarithm is a slowly changing function, the stopping time ts will not depend to

any great extent on precisely what value of vs we choose so long as it is much smaller

than vo. It is often instructive to expand solutions in a Taylor series in t. If we expand the

right side of Eqs. (3.7) and (3.8) in power series in t, we obtain

t + ..., .................................................................(3.11) + ... ................................................................ (3.12)

4. CONSERVATIVE FORCE DEPENDING ON POSITION. POTENTIAL ENERGY

One of the most important types of motion occurs when the force F is a function

of the coordinate x alone:

.................................................................... (4.1)We have than, by the energy theorem (1.8),

......................................................(4.2)

The integral on the right is the work done by the force when the particle goes from xo to

x. We now define the potential energy V(x) as the work done by the force when the

particle goes fromxto some chosen standard pointxs:


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V(x) = = - ................................................(4.3)The reason for calling this quantity potential energy will appear shortly. In terms ofV(x),

we can write the integral in Eq. (4.2) as follows:

....................................................... (4.4)With the help of Eq. (4.4). Eq. (4.1) can be written

..................................................... (4.5)The quantity on the right depends only on the initial conditions and is therefore constant

during the motion. It is called the total energy E, and we have the law of conservation of

kinetic plus potential energy, which holds, as we can see, only when the force is a

function of position alone;

........................................................ (4.6)

Solving for v, we obtain

......................................................... (4.7)The function x(t) is to be found by solving for x equation

.................................................. (4.8)From the definition (4.3) we can express the force in terms of the potential energy:

.................................................................................... (4.9)

This equation can be taken an expressing the physical meaning of the potential energy.

The potential energy is a function whose negative derivative gives the force. The effect of

changing the coordinate of the standard pointx, is to add a constant to V(x). Since it is the


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derivative of V which enters into the dynamical equations as the force, the choice of

standard point xs is immaterial. A constant can always be added to the potential V(x)

without affecting the physical results. (The same constant must, of course, be added to

E).

As an example, we consider the problem of a particle subject to a linier restoring

force for example, a mass fastened to a spring:

F = -kx .................................................................................(4.10)

The potential energy, if we take xs = 0, is

..........................................................(4.11)Equation (4.8) becomes, for this case, with to = 0

...........................................................(4.12)Now make the substitution

....................................................................................(4.13)

, .....................................................................................(4.14)So that

And, by Eq.(4.12)

We can now solve for x in Eq. (4.13):


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sin = A sin ( , ................................................(4.15)Where

......................................................................................(4.16)Thus the coordinate x oscillates harmonically in time, with amplitude A and frequency

. The initial conditions are here determined by constants A and , which arerelated to E and xo by

E = , ......................................................................................(4.17)

...................................................................................(4.18)Notice that in this example we meet the sign difficulty in taking the square root in Eq.

(4.12) by replacing (1-sin2-1/2 by (cos )-1, a quantity which can be made either positive

or negative as required by choosing in the proper quadrant.5. FALLING BODIES

One of the simplest and most commonly occurring types of one-dimensional

motion is that of falling bodies. We take up this type of motion here as an illustration of

the principles discussed in the preceding sections.

A body falling near the surface of the earth, if we neglect air resistance, is subject

to a constant force

F = -mg, ...............................................................................(5.1)

Where we have taken the positive direction as upward. The equation of motion is

...................................................................................... (5.2)


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The solution may be obtained by any of the three methods discussed in Sections 2, 3, and

4 since a constant force may be considered as a function of either t, v or x. The reader will

find it instructive to solve the problem by all three methods.

In order to include the effect of air resistance, we may assume a frictional force

proportional to v, so that the total force is

F = -mg -bv ..............................................................................(5.3)

The constant b will depend on the size and shape of the falling body, as well as on the

viscosity of the air. The problem must now be treated as a case ofF(v):

............................................................................(5.4)Taking v0= 0 at t= 0, we proceed as in Eq. (3.2):

.................................................................................(5.5)We integrate and solve for v:

......................................................................(5.6)

We may obtain a formula useful for short times of fall by expanding the exponential

function in a power series:

, .................................................................(5.7)Thus for a short time (t > The velocity mg/b is called the terminal velocityof the falling body in question. The body

reaches within 1/e of its terminal velocity in a time t = m/b. We could use the


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experimentally determined terminal velocity to find the constant b. We now integrate

Eq.(5.6), takingxo = 0:

................................................................(5.8)

By expanding the exponential function in a power series, we obtain

................................................................(5.9)If t > m/b

This result is easily interpreted in term of terminal velocity. Why is the positive constant

present?

It is worth noting that we may obtain the series solution (5.7) directly from the

differential equation (5.4) without solving it exactly. Let us first neglect altogether the

term involving b, so that the solution is

Substitute this result in the last term in Eq. (5.4) and integrate again:

This result agrees with the first two terms in Eq. (5.7). If we put v = v

(1)into the last term

in Eq. (5.7) and integrate, we get a better approximation v(2)

, good to order b2, and so on.

This method of successive approximation is often useful in solving an equation containing

a small term which in a zero-order approximation may be neglected. A similar method

can be used to solve by successive approximations an algebraic equation containing one

or more small terms.


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For small heavy bodies with large terminal velocities, a better approximation may

be

F = bv

2

.......................................................................................(5.10)

The reader should be able to show that with the frictional force given by Eq. (5.10), the

result (takingxo = vo = 0 at to= 0) is

tanh

=

.................................................(5.11)

=

{

.............................................(5.12)

In the case of bodies falling from a great height, the variation of the gravitational

force with height should be taken into account. In this case, we neglect air resistance (in

order to be able to use the energy method), and measure x from the center of the earth.

Then if M is the mass of the earth and m the mass of the falling body.

, .......................................................................................(5.13)

And .................................................................(5.14)Where we have taken xs = in order to avoid a constant term in V(x). Equational (4.7)

becomes


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........................................................(5.15)6. THE DAMPED HARMONIC OSCILLATOR

The equation of motion for a particle subject to a linear restoring force and a

frictional force proportional to its velocity is

....................................................................... (6.1)Where the dots stand for time derivatives. Applying the method of Section 5, we make

.........................................................................(6.2)The solution is

[ ]

...............................................................(6.3)

We distinguish three cases: (a) k/m > (b/2m)2, (b) k/m < (b/2m)

2, and (c) k/m = (b/2m)

2.

In case (a), we make substitutions

.............................................................................................(6.4) .............................................................................................(6.5)

.............................................................................(6.6)Where is called the damping coefficient and is the natural frequency of theundamped oscillator. There are now two solutions forp:

......................................................................................(6.7)The general solution of the differential equation is therefore

........................................................(6.8)


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Setting

..........................................................(6.9)We have

...................................................................(6.10)This corresponds to an oscillation of frequency ( with an amplitude whichdecreases exponentially with time (Fig. 6.1). The constants A and depend upon theinitial conditions.

The frequency of oscillation is less than without damping. The solution (6.10) can also be

written

.................................................(6.11)In terms of the constants and , Eq. (6.1) can be written

............................................................................(6.12)

This form of the equations often used in discussing mechanical oscillations.

The total energy (kinetic plus potential) of the oscillator is

-A

A

0

Fig. 6.1


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............................................................................(6.13)It is no longer constant; the friction - b plays the role of F. In the important case of smalldamping,


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We now show that, in this case, another solution is

..........................................................................................(6.20)To prove this, we compute

....................................................................(6.21)

The left side of Eq. (6.12)is, for this x

...............................................(6.22)This is zero if0 = Hence the general solution in case 0 = is

............................................................................(6.23)If we keep either 0 or fixed and let the other vary, we see from Eq. (6.16) that

1 > c > 2 ..........................................................................................(6.24)Where

c is the value when

=

0

MOTION OF A PARTICLE IN

TWO OR THREE DIMENSIONS

1. PLANE AND VECTOR ANGULAR MOMENTUM THEOREMS

The angular momentum L is taken as positive when the particle is moving in a

counterclockwise sense with respect to O. It is expressed most simply in terms of polar

coordinates with O as origin. Let the particle have mass m. Then its momentum is mv, and

the component of momentum perpendicular to the radius vector from O is mvo:

............................................................................(1.1)If we write the force in terms of its polar components,


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........................................................................................(1.2)

Then in plane polar coordinates the equation of motion,

........................................................................(1.3) ....................................................................(1.4)

We now note that

Thus, multiplying Eq. (1.4) by r, we have

( ) ..................................................................(1.5)

The quantity is the torque exerted by the force Fabout the point O. Integrating Eq.(1.5), we obtain the integrated from of the angular momentum theorem for motion in a

plane:

.............................................(1.6)As a final generalization of the concept of angular momentum, we define the

vector angular momentum Lo about a point O as the vector moment of the momentum

vector about O:

Lo = r x p = m(r x v), ..................................................................(1.7)

Where the vector r is taken from the point O as origin to the position of the particle of

mass m. Again we shall omit the subscript o when no confusion can arise. The component

of the vector L in any direction is the moment of the momentum vectorp about an axis in

that direction through O.


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By taking the cross product of r with both members of the vector equation of

motion, we obtain

...............................................................................(1.8)

By the rules of vector algebra and vector calculus,

We substitute this result in Eq.(1.8):

dL/dt = r x F= N .....................................................................(1.9)

The time rate of change of the vector angular momentum of a particle is equal to the

vector torque acting on it. The integral form of the angular momentum theorem is

..............................................................................(1.10)

The theorems for plane angular momentum and for angular momentum about an axis

follow from the vector angular momentum theorems by taking components in the

appropriate direction.

2. PROJECTILES

An important problem in the history of the science of mechanics is that of

determining the motion of a projectile. A projectile moving under the action of gravity


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near the surface of the earth moves, if air resistance is neglected, according to the

equation

, ...................................................................................(2.1)

Where the z-axis is taken in the vertical direction. In component form:

, ........................................................................................(2.2) , .......................................................................................(2.3)

.....................................................................................(2.4)

The solution of these equations are

..................................................................................(2.5) ..................................................................................(2.6) .........................................................................(2.7)

Or, in vector form,

.......................................................................(2.8)We assume the projectile stars from the origin (0,0,0), with its initial velocity in

the xz-plane, so that = 0. This is no limitation on the motion of the projectile, butmerely correspond to a convenient choice of coordinate system. Equation (2.6), (2.7),

(2.8) then become

................................................................................................ (2.9) ......................................................................................................(2.10)


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...................................................................................(2.11)

These equations give a complete description of the motion of the projectile. Solving the

first equation for t and substituting in the third, we have an equation for the path in the

xz-plane:

.............................................................................(2.12)This can be rewritten in the form

..........................................................(2.13)

This is a parabola, concave downward, whose maximum altitude occurs at

.............................................................................................(2.14)And which crosses the horizontal plane z = 0 at the origin and at the point

.....................................................................................(2.15)

If the surface of the earth is horizontal, is the range of the projectile.Let us now take account of air resistance by assuming a frictional force

proportional to the velocity:

.........................................................................(2.16)In component notation, if we assume that the motion takes place in the xz-plane,

..................................................................(2.17)


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It should be pointed out that the actual resistance of the air against a moving projectile is

a complicated function of velocity, so that the solutions we obtain will be only

approximate, although they indicate the general nature of the motion. If the projectile

stars from the origin at t = 0, the solutions of Eg. (2.17) are

................................................................................(2.18) ( ), ....................................................................(2.19) .........................................................(2.20)

( ) ............................................(2.21)

Solving Eq. (2.19) for t and substituting in Eq. (2.21), we obtain an equation for the

trajectory:

..................................................(2.22)For low air resistance, or short distances, when (bx)/(mvxo)


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Which agrees with Eq. (2.15). Let us substitute this solution in the second term of Eq.

(2.24), to obtain a second approximation

...................................................................(2.25)The second term gives the first-order correction to the range due to air resistance, and

the first two terms will give a good approximation when the effect of air resistance is

small. Higher-order terms could be calculated by repeating the process of substituting

approximate solutions back in Eq. (2.24). We thus obtain successive terms for xm as a

power series in b. The extreme opposite case, when air resistance is predominant in

determining range, occurs when the vertical drop at x = (mvxo

)/b begins above the

horizontal plane z = 0. The range is then, approximately,

................................................................(2.26)We can treat (approximately) the problem of the effect of wind on the projectile

by assuming the force of air resistance to be proportional to the relative velocity of the

projectile with respect to the air:

.........................................................(2.27)Where is the wind velocity. If is constant, the term b in Eq. (2.27) behaves as aconstant force added to , and the problem is easily solved by the method above,the only difference being that there may be constant forces in all three directions x, y,z.

The air resistance to a projectile decreases with altitude, so that a better form for

the equation of motion of a projectile which rises to appreciable altitudes would be

.........................................................(2.28)Where h is the height (say about five miles) at which the air resistance falls to 1/e of its

value at the surface of the earth. In component form


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We may note that the vector angular momentum of a particle under the action of a

central force is constant, since the torque is

............................................................(2.29)Therefore, by Eq. (1.9),

...................................................................................................(2.30)As a consequence, the angular momentum about any axis through the center of force is

constant. It is because many physical forces are central forces that the concept of angular

momentum is of importance.

In solving for the motion of a particle acted on by a central force, we first show

that the path of the particle lies an a single plane containing the center of force. To show

this, let the position ro and velocity vo be given at any initial time to, and choose the z-axis

through the initial position ro of the particle, and the z-axis perpendicular to the initial

velocity vo. Then we have initially:

Xo = |ro|, yo = zo = 0 .........................................(2.31)

....................................(2.32)The equation of motion in rectangular coordinates are,

.................................(2.33)

A solution of the zequation which satisfies the initial conditions on z0 and vz0 is

Z(t) = 0


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We have now reduced the problem to one of motion in a plane with two

differential equations and four initial condition remaining to be satisfied. If we choose

polar coordinates r, in the plane of the motion, the equations of motion in the r and direction are,

..........................................................................(2.43) ............................................................................(2.44)

Multiplying Eq. (2.44) by r, as in the derivation of the (plane) angular momentum

theorem, we have

.............................................................................(2.45)

This equation expresses the conservation of angular momentum about the origin and is a

consequence also of Eq.(2.30) above. It may be integrated to give the angular momentum

integral of the equations of motion:

...................................................................(2.46)

The constant L is to be evaluated from the initial conditions. Another integral of Eq.

(2.43) and (2.44), since the force is conservative, is

............................................(2.47)Where V(r) is given by Eq. V(r) = and Eis energy constant, to be evaluatedfrom the initial conditions. If we substitute for

from Eq. (2.46), the energy becomes

...................................................................(2.48)

We can solve for :


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..........................................................(2.49)Therefore`

..................................................................(2.50)

The integral is to be evaluated and the resulting equation solved for r(t). We then obtain

(t) from Eq.(2.46): .........................................................................(2.51)

It will be noted that our treatment based on Eq. (2.48) this analogous to our

treatment of the one-dimensional problem based on the energy integral. The coordinate r

here plays to the role ofx, and the term in the kinetic energy, when is eliminated byEq. (2.46), plays the role of an addition to the potential energy. We may bring out this

analogy further by substituting from Eq.(2.46) into Eq. (2.43):

..............................................................................(2.52)If we transpose the termL

2/mr

3to the right side, we obtain

..............................................................................(2.53)


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