Basic Definitions What is Mechanics? Mechanics is the science of motion, which is a change in the position of a body with respect to time. Mechanics studies the interactions between a moving body and the forces acting on it. Hence: Classical Mechanics: Is the science Deals with the motion of objects through absolute space & time in the Newtonian sense. The Coordinate System: In order to define the position of a body in space, it is necessary to have a reference system. In mechanics we use a coordinate system. The basic type of coordinate system is the Cartesian or rectangular coordinate system. The position of a point in such system is specified by three numbers or coordinates, x, y, and z. In case of a moving point, the coordinates change with time; that is, x, y, and z are functions of the quantity t. The particle (or mass point): The particle is a thing that has mass but does not have spatial extension. It is an approximation of a very small body, or one whose size is relatively unimportant in a particular discussion. The earth, for example, might be treated as a particle in Celestial mechanics.

Physical Quantities: The observational data of physics are expressed in terms of certain fundamental entities called physical quantities, for example, energy, time, force, and so on. It has been found that it is possible to define all of the physical quantities of mechanics in terms of three basic ones, namely length, time and mass. Basic Units in (SI System): 1- The Unit of Length (meter): The meter is the distance that light travels during a time interval of exactly 1/299,792,458 second 2- The Unit of Time (second): It is the time required for exactly 9,192,631,770 oscillations of a particular atomic transition of the isotope cesium 133. 3- The Unit of Mass (kilogram): It is the mass of a cylinder of platinum-iridium that kept at the Bureau of Metric Standards. Changing units:

Recall: the prefixes of the SI Units:

Factor 109 106 103 10-2 10-3 10-6 10-9 10-12

Prefix Giga- Mega- Kilo- Centi- Milli- Micro- Nano- Pico-

Symbol G M K c m μ n p

Check point:

6- How many microns make up 1Km?

10 μm10 m3

1Km 1m1Km = 1Km (1) = 1Km ( ____ ) ( _____ ) = 109 μm

Dimensions: The dimension of any physical quantity can be written as [M]α [L]β [T]γ where α, β, and γ are powers of their respective dimension. For example, the dimension of acceleration a is 2]][[/ −=⎥⎦

⎤⎢⎣⎡= TL

TTLa

Note:

The dimension of a quantity ≠ its units

Dimensional Analysis: 1- Dimensional analysis is a powerful tool that can be used to determine whether the result of a calculation has even the possibility of being correct or not. I.e, for any equation: the dimensions on the LHS = the dimension of all physical quantities on the RHS.

Check point:

Is the following relation correct or not? 2/12

⎟⎟⎠

⎞⎜⎜⎝

⎛=

c

ec R

gRv 2- Dimensional analysis can also be used as to obtain a rough relationship between physical quantities. Consider the simple pendulum; It consists of a small mass m attached to the end of a massless string of length l. When displaced from its equilibrium position it swings because gravity g tries to bring back the mass to its minimum height above the ground. The time it takes to perform one complete cycle of its motion is called its period τ. The question now is: In the absence of friction, air resistance and all other dissipative forces, how does the pendulum period τ depend on any physical parameters that characterize the pendulum and its environment? Let assume that

γβατ glm∝ whose dimensional relationship must be

[T] 1 = [M]α [L]β ([L]γ [T] −2γ) Matching the dimensions of RHS with LHS we find that;

α = 0, γ = -1/2 and, β + γ = 0, or β = 1/2. Thus, we conclude that

gl

∝τ

Note: When trigonometric functions are involved, their lack

of dimensionality prevents dimensional analysis.

1.3. Vectors: Physical

quantitiesVectors Scalars

A scalar can be completely defined by magnitude.

Examples: mass, density, volume, and energy.

A vector needs both magnitude and direction to be defined.

Examples: displacement, velocity, acceleration, and force.

Its value is coordinate system dependent.

Mathematically, vectors need a special treatment known as Vectors Algebra.

Its value is independent of any chosen coordinates.

Mathematically, scalars obey the normal algebraic rules of addition, multiplication and so on.

Note:

It is customary to use a distinguishing mark, such as an arrow ( A ) or bold litter (A), to

indicate a vector.

Vector Algebra A given vector A can be specified either by: - Its magnitude (A) and its direction (angle) relative to some arbitrarily chosen coordinate system. - or by the set of its components, or projections onto the coordinate axes , i.e. in 2D:( Ax , Ay). Rules of Vector Algebra:

B1- Equality of Vectors; The equation A = B or (Ax, Ay, Az) = (Bx, By, Bz) is equivalent to the three equations Ax=Bx Ay=By Az=Bz Note:

Two vectors are equal if, and only if, their respective components are equal. 2- Vector Addition: The addition of two vectors is defined by the equation

A + B = (Ax + Bx, Ay + By, Az + Bz) (1)

3- Vectors addition has two important properties:

Commutative Law A+B = B+A

Associative Law A+(B+C) = (A+B)+C

4- Vector Subtraction: Subtraction of a given vector B from the vector A is equivalent to adding -B to A. 5- The Null Vector: The vector O = (0,0,0) is called the null vector. 6- Multiplication by a Scalar: If c is a scalar and A is a vector,

cA = c(Ax, Ay, Az) = (cAx, cAy, cAz) = Ac

Geometrically, the vector cA is parallel to A and is c times the length of A. When c = -1, the vector -A is one whose direction is the reverse of that of A. Joining (2) & (6), we get what called the distributive law: c(A + B) = c(Ax + Bx, Ay + By, Az + Bz) = (c(Ax + Bx), c(Ay + By), c(Az + Bz» = (cAx + cBx, cAy + cBy, cAz + cBz) = cA + cB

7- Magnitude of a Vector; The magnitude of a vector A, denoted by |A| or by A, is defined as,

222zyx AAAAA ++== (2)

8- The Unit Vectors: A unit vector is a vector whose magnitude is unity. Unit vectors are often assigned by the symbol e, The three unit vectors

ex = (1,0,0) ey = (0,1,0) ez = (0,0,1)

x y i

z

j

k

The common used unit vectors for Cartesian coordinates are i, j, and k, where, ex = i ey = j ez = k

1.4. The Scalar Product: Given two vectors A and B, the scalar or ” dot " product , A.B is the scalar defined by the equation

A.B = AxBx + AyBy + AzBz (3)

Main properties of the scalar product: It is commutative,

A.B = B.A (4)

It is distributive, A . (B + C) = A . B + A . C (5)

It is a scalar, so its value is independent of choice of coordinates. If vector A is expressed as (A, 0, 0) and the vector B as (B cosθ, B sinθ, 0), then,

Α . Β = ΑxΒx = Α(Β cos θ) = |A| |B| cosθ (6)

Hence, The geometrical interpretation of A . B is that it is the projection of A onto B times the length of B.

The cosine of the angle between the two vectors is given by:

ABBA.B A.B A

==θcos (7)

If A. B = 0 , and neither A nor B is null, then (cosθ = 0)

and A is ⊥ to B. Similarly; i.i =j.j = k.k= 1

i.j =j.k = k.i= 0

1.5. The Vector Product: Given two vectors A and B, the vector or ” cross " product , A x B is a vector whose components are given by the equation

A x B = (AyBz – AzBy , AzBx – AxBz , AxBy – AyBx ) (8) Main properties of the vector product:

It is anti-commutative, A x B = – B x A (9)

It is distributive, A x (B + C) = A x B + A x C (10)

Multiplication by a Scalar: n (A x B) = (n A) x B = A x (n B) (11)

It is a vector, so it can be expressed in ijk form as; A x B = i (AyBz – AzBy ) + j (AzBx – AxBz ) + k (AxBy – AyBx ) Each term between brackets is equal to a determinant,

yx

yx

xz

xz

zy

zy

BBAA

BBAA

BBAA

k j iBA ++=×

Or;

kjiBA

zyx

zyx

BBBAAA=×

(12)

The magnitude of the cross product can be calculated

by squared both sides of (8). This can be reduced to; | Α x Β | = ΑΒ(1− cos2 θ)1/2 = |A| |B| sinθ (13) Where θ is the angle between A and B. Note:

The cross product of two parallel vectors is null.

Hence, i x i =j x j = k x k= 0

The direction of the resultant vector is ⊥ to the plane

containing A and B. Therefore, from (13) we can write Α x Β = (A B sinθ) n (14) Where n is a unit vector normal to the plane containing A and B. The sense of n is given by the right –hand rule. Hence,

i x j =k , j x k =i , k x i= j

Note: The cross product A x B has 1- A magnitude of A B sinθ which is equal to the area of the parallelogram with sides A and B shown by the shaded area in the Figure. 2- A direction ⊥ to the plane containing A and B.

A x B

B θ

θA

1.7. Triple Products: The expression

zyx

zyx

zyx

CCCBBBAAA

=×C)A.(B (15)

is called scalar triple product of A, B and C. While, the expression A x (B x C) = B(A.C) – C(A.B) (16) is called the vector triple product of A, B and C. Note:

Eq. (16) can be remembered easily as The “ back minus cab” rule.

1.9. Derivative of a vector: Previously we were concerned mainly with vector algebra. Now, we will begin to study the calculus of vectors and its use in the description of the motion of particles.

Consider a vector A, whose components are functions of a

single variable u which is usually the time t, i.e,

A(u) = i Ax(u) + j Ay(u) + k Az(u) The derivative of A with respect to u is defined by

dudA

dudA

dudA

dud zyx kjiA

++= (17)

Note:

dud

dud

dud B. A B .AA.B +=)(

dud

dud

dud B ABABA ×+×=× )(

The derivative of a vector is a vector whose components

are ordinary derivatives.

Some vector calculus rules: (18)

a- dud

dud

dud BABA +=+ )( c-

b- dudn

dudnn

dud AAA +=)( d-

Notice that it is necessary to preserve the order of the terms in the derivative of the cross product.

1.10. Velocity & Acceleration In rectangular coordinates: In a given reference system, the position of a particle can be specified by a single vector. This vector is called the position vector of the particle. In rectangular coordinates, the position vector is simply

r = i x + j y + k z (19) For a moving particle, these components are functions of the time. The time derivative of r is called the velocity, (v), which is given by;

zyxdtd

&&& k j irv ++== (20)

The vector dr/dt expresses both the direction and the rate of motion. As Δt approaches zero, the point P' approaches P, and the direction of the vector Δr/Δt approaches the direction of the tangent to the path at P, which is dr/dt.

Note:

The velocity vector is always tangent to the path of motion.

The magnitude of the velocity is called the speed (v). In rectangular components the speed is just

( ) 2/1222 zyxdtdsv &&& v ++=== (21)

where s is the distance.

The time derivative of the velocity is called the acceleration (a). Hence;

zyxdtd

dtd

&&&&&& k j irva ++=== 2

2

(22)

Some different types of motion

Projectile Motion

Circular Motion

Rolling Wheel

The position vector

)2

(2gtct-bt j ir +=

tbtb j ir ωω cossin +=

21 rrr += where; j ir1 btb += ω

tbtb j ir2 ωω cossin +=

The velocity

)(c-gtb j iv +=

tsbtcb in j os iv ωωωω −= tbtcbb j os ( ivvv 21

ωωωωω sin) −+=+=

The

acceleration

g ja −=

tbtb j ia ωωωω cossin 22 −−=

r a 2ω−=

tbtsb j in ia ωωωω cos22 −−=

The path

Parabola

Circle

Cycloid

1.11. Velocity and Acceleration in Plane Polar Coordinates: To express the position of a particle moving in a plane, it is often convenient to use polar coordinates : r, θ. Using Vectors, the position of the particle can be written as

the product of the radial distance r by a unit radial vector er:

r = r er

which are both functions of the time.

If we differentiate with respect to t, we get the velocity ;

dtdrr

dtd r

reerv +== & (23)

A study of the figure shows that;

Δer ≅ | Δer| eθ ≅ |er| Δθ eθ ≅ Δθ eθ

dtd

dtd r θ

θee=Dividing by Δt and taking the limit gives;

Hence, we can rewrite eq.( 23 ) as

(24)

where, is known as the radial component of the velocity

vector, and as the transverse component.

θθ e ev && rr r +=

r&

θ&r

To find the acceleration vector, we take the derivative of the

velocity with respect to time. This gives

( ) ( ) θθθθ e e va &&&&&&& rrrrdtd

r 22 ++−== (25)

where, the radial component of the acceleration vector is

(26) 2θ&&& rrar −=

and the transverse component is

( )θθθθ&&&&& 212 r

dtd

rrra =+= (27)

Note:

Path r θ dr/dt dθ/dt ar aθ Circle

const.

(b) vari. 0 θ& 2θ&b

toward the center θ&&b

⊥ r Radial line vari. const. r& 0 r&& 0 If r and θ both vary, then the general expression (25) gives the acceleration.

1.12 Velocity and Acceleration in Cylindrical & Spherical Coordinates: Cylindrical Coordinates The position vector of a particle can be described in cylindrical coordinates (R, φ , z) as r = ReR + zez (28) In view of the facts that; deR /dt = eφ dφ /dt , deφ /dt = -eR dφ /dt and dez /dt = 0 the velocity and acceleration vectors can be given by the following equations:

zR zRR ee ev &&& ++= φφ (29)

( ) ( ) zR zRRRR ee e a &&&&&&&&& +++−= φφφφ 22 (30)

Spherical Coordinates

The position vector of a particle can be described in Spherical coordinates (r, θ , φ ) as r = rer (31)

where the direction of er is specified by the two angles φ and θ. Thus we introduce two more unit vectors, eφ and eθ, as shown in the Figure . Because any vector can be expressed in terms of its projections onto the x,y,z axes as;

)()()( .kek.jej.ieie rrrr ++= (32) The relationships between the ijk and er eφ eθ can be derived as the followings;

)(cos)sin(sin)cos(sin θφθφθ kjie ++=r )(sin)sin(cos)sin(cos θφθφθθ kjie −+=

)(cos)(sin φφφ jie +−= Using the differentiating of the above equations, the velocity and acceleration vectors can be given by

θφ θθφ e e ev &&& rrr r ++= sin (33)

( ) ( )( ) φ

θ

θφθθφθφ

θθφθθθφθ

e

e e a

cos2sin2sin

cossin2sin 2222

&&&&&&

&&&&&&&&

rrr

rrrrrr r

+++

−++−−= (34)

2.1 Newton's laws of Motion: In 1687, Isaac Newton laid down three fundamental laws of motion, which are: 1st . Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. 2nd . The change of motion is proportional to the net force impressed and is made in the direction of that force. 3rd . To every action there is always an equal reaction;

Newton's First law (Inertial Reference Systems): The first law describes a common property of matter, known as inertia. What is Inertia? It is the resistance of a matter to change its state of motion. This means that; in the absence of applied forces, matter simply persists in its current velocity state-forever. Inertial Frames of Reference: - A mathematical description of the motion of a particle requires the selection of a frame of reference, or a set of coordinates according to which the position, velocity, and acceleration of the particle can be specified. - Uniformly moving reference frames (e.g. those considered at 'rest' or moving with constant velocity in a straight line) are called inertial reference frames. - If we can neglect the effect of the earth’s rotations, a frame of reference fixed in the earth is an inertial reference frame. - Newton’s laws are only applicable to inertial reference frames.

Newton's Second and Third Laws (Mass and Force): - The physical quantity that measures inertia is called mass. - The more massive an object is, the more resistive it is to acceleration.

Suppose we have two masses m1, m2 on a frictionless surface. Now imagine someone pushing the two masses together, and then suddenly releasing them so that they fly apart, achieving speeds v1 and v2. The ratio of the two masses can be expressed as;

2

1

vv

=1

2

mm

Or; ( ) ( )2211 vv mm Δ−=Δ

The (-) appears because the final velocities v1 and v2 are in opposite directions. If we divide by Δt and take limits as Δt ~ 0, we obtain

( ) ( 2211 vv mdtdm

dtd

−= ) (35)

According to Newton's 2nd law, this “change of motion” is proportional to the force caused it;

( )vF mdtd

∝

Defining the unit in the SI system, Newton's 2nd law can be expressed in the familiar form:

( ) avF mmdtd

== (36)

Note:

The force F on the left side of Equation (36) is the net force Fnet acting upon the mass m.

Equation (35) is equivalent to F1= -F2 (37) Which is Newton's 3rd law, that states; two interacting bodies exert equal and opposite forces upon one another. Linear Momentum: Since, the product of mass and velocity, mv, is called linear momentum, P, the 2nd law can be rewritten as;

dtdpF =

(38) which means that; The time rate of change of an object’s linear momentum is proportional to the applied force, F. Similarly, Equation (35), which describes the behavior of two mutually interacting masses, is equivalent to

( ) 021 =+ ppdtd

or ( ) .conspp =+ 21 (39) In other words, Newton's 3rd law implies that the total momentum of two mutually interacting bodies is a constant.

2.2. Motion with Constant Force (Rectilinear Motion): When a moving particle remains on a single straight line, the motion is said to be rectilinear. In this case, we can choose the x-axis as the line of motion. The general equation of motion is then ( ) xmtxxF &&& =,, (40) The simplest case is when F is constant. In this case a is constant ;

amFx === constant&& (41)

Integrating with respect to time:

0vatvx +==& (42)

002

21 xtvatx ++= (43)

where v0 is the velocity and x0 is the position at t = 0. from (42) & (43 ) we obtain;

(44) 2

02

0 )(2 vvxxa −=−

There are a number of fundamental applications for such case. For example, the acceleration of a body falling freely near the surface of the Earth, neglecting air resistance, is nearly constant.

2.3 Forces that Depend on Position (The Concepts of Kinetic and Potential Energy): If the force is independent of velocity or time, then the differential equation for rectilinear motion is simply

( )

dxdT

dxvdm

dxdvmv

dxxd

dtdxmxmxF

==

=⎟⎠⎞

⎜⎝⎛==

)( 2

21

&&&

(45)

The quantity 2

21 mvT = is called the kinetic energy of the

particle. Taking the integral of (45 ):

(46) ( ) 0

0

TTdxxFWx

x

−== ∫

Where W is the work done on the particle by the impressed force F(x). This work is equal to the change in the kinetic energy of the particle. Let us define another function V(x) such that;

(47) ( ) ( ) 00

00

TT)V(xxVdVdxxFWx

x

x

x

−=+−=−== ∫∫

The function V(x) is called the potential energy. From (47);

( ) ExVT)V(xT ≡+=+ 00 (48)

E is known as the total mechanical energy of the particle.

Note:

CmgxV

mgdxdVF

+=

−=−=

Emgxmv =+221

gxvv 220

2 −=

021 +mv 2

0212 =+ mvmgx

1- E the sum of the kinetic and potential energies is constant throughout the motion of the particle.

2- The force is a function only of the position x. Such a force is said to be conservative. 3- v=0 when V(x)= E . This point known as “the turning point” Exp (2.3.1): Free Fall The motion of a freely falling body is an example of conservative motion. In this case: Hence; We can choose C = 0, which means that V = 0 when x = 0. The energy equation is then For instance, let the body be projected upward with initial speed v0 from the origin x=0. These values give; so; The turning point of the motion, which is in this case the maximum height, is given by setting v = 0. This gives

gvxh2

20

max == (49) Exp (2.3.3): Morse Function The potential energy of a vibrating diatomic molecule as a function of x is given by;

[ ] 02/)(

001)( VeVxV xx −−= −− δ

Show that x0 the separation of the two atoms is min, and its value is V(x0)=-V0 . Solution: V(x) is min when its derivative (w.r.t) x is zero;

( )(

0

0

/)(

/)(/)(0

/)()1ln(01

012

0)()(

0

00

xxxx

e

eeVdx

xdVxF

xx

xxxx

=

)

∴−−==−

=−

=−=

−−

−−−−

δ

δδ

δδ

Substituting in the main equation, the value of the min V(x) can be found as; V(x0)=-V0

2.4. Velocity-Dependent Forces (Fluid Resistance and Terminal Velocity):

The force that acts on a body often is a function of the velocity of the body. For example, the viscous resistance exerted on a body moving through a fluid depends on its velocity. In such case, the differential equation of motion may be written in either of the two forms

or dxdvmvF(v)F

dtdvmF(v)F

=+

=+

0

0

(50)

Here F0 is any constant force that does not depend on v. Since, F(v) is a complex function and must be found through experimental measurements, it can be replaced by the following approximation :

or )( 21

21

vccvF(v)

vvcvcF(v)

+−=

−= (51)

where c1 and c2 are constants whose values depend on the size and shape of the body. For spheres in air,

c1 = 1.55 × 10-4D & c2 = 0.22 D2 where D is the diameter of the sphere in meters. For small v the linear term in F(v) can be used , while the quadratic term dominates at large v.

Note:

1- The absolute-value sign is necessary on the last term of (51) because the force of fluid resistance

is always opposite to the direction of v. To decide whether the case is linear or quadratic, the ratio of the latter to the former usually used;

DvvD.

Dvvvcvvc

-3

4

2

1

2 104.110551

22.0×=

×=

(52)

If the value of v will make the ratio exceeds 1 then it is a quadratic case, otherwise, it is a linear one.

Horizontal Motion through a Fluid 1- With Linear Resistance (Exp.2.4.1) Suppose a block is projected with initial velocity v0 on a smooth horizontal surface and that there is air resistance such that the linear term dominates. Then, in the direction of the motion, F0 = 0, and F(v) = -c1v. The differential equation of motion is then;

dtdvmvc =− 1

By integrating,

)ln(0110

vv

cm

vcmdvt

v

v

−=−= ∫

Solving for v as a function of t gives;

mtcevv /0

1−=A second integration gives

or ( )mtc

tmtc

ec

mvx

dtevx

/

1

0

0

/0

1

1

1 −

−

−=

= ∫

Showing that the block approaches a limiting position given by; xlim = mv0 / c1 2- With Quadratic Resistance (Exp.2.4.2) If the parameters are such that the quadratic term dominates, then

dtdvmvc =− 2

2

Similarly we can get t, v and the position x as a function of time.

Vertical Fall through a Fluid (Terminal Velocity): 1- Linear case: For an object falling vertically in a resisting fluid, the force F0 in (50) is the weight of the object, -mg. For the linear case of fluid resistance, the differential equation of motion is;

dtdvmvcmg =−− 1

Integrating and solving for v, we get

mtcev

cmg

cmgv /

011

1)( −++−=

After a sufficient time (t >> m/c1), the velocity approaches a limiting value ( -mg/c1). This limiting velocity of a falling body is called the terminal velocity (vt). Hence the terminal speed is;

1cmgvt =

Note:

1cm

gvt ==τ

2cmgvt =

gcm

gvt

2

==τ

At the velocity vt the force of resistance is just equal and opposite to the weight of the body so that the net force is zero, and so the acceleration is zero. The value of vt/g is known as the characteristic time of the motion (τ). I.e , 2- Quadratic case: In this case F(v) ∝ v2 and the differential equation of motion is;

dtdvmvcmg =−− 2

2

Similarly, the terminal speed is ; And the characteristic time is;

3.1 Introduction to Oscillations:

In order to describe the complicated forms of periodic motion around us, usually we start by an analysis of the simplest form of oscillations, the simple harmonic motion.

The main two characteristics of the simple harmonic motion are; (1) It obeys the Superposition principle . Which means it is described by a second-order, linear differential equation with constant coefficients. Thus, if two particular solutions are found, their sum is also a solution. (2) It has Amplitude-independent periods .That is, the periodic time of the motion, is independent of the maximum displacement from equilibrium (the amplitude) .

3.2. Linear Restoring Force: Harmonic Motion - Consider a mass m on a frictionless surface attached to a wall by means of a spring. - Let Xe is the unstretched length of the spring. This position represents the equilibrium position where the potential energy is a minimum.

- If the mass is pushed or pulled away from this position, the spring will be either compressed or stretched, and then exert a force on the mass. - This force will always attempt to restore it to its equilibrium position. - To calculate the motion of the mass, we need an expression for this restoring force . According to Hooke's law The spring's restoring force is given by ; F(x) = -kx (53) where k is the spring constant. In fact, this law is valid only for small displacements from equilibrium, where the restoring force is linear. Newton's second law of motion can now be written as

0=+ xmkx&& (54)

)sin( 00 φω += tAx

Equation (54) can be solved in many ways. But we are looking for a solution which can show that the motion is both periodic and bounded. Sine and cosine functions both can exhibit that sort of behavior. Thus, a possible solution is; (55)

m0k

=ωwhere (56) is the angular frequency of the system.

The motion represented by Eq. (55) is shown in the following Fig.; The motion exhibits the following features: (1) The motion repeats itself after a time T0 known as the period of the motion. Which is the time required for a phase advance of 2π , and is given by

0

00000 2)(

02ωπ

πφωφω

=

++

T

= tTt + +Or;

AxA +≤≤−(2) The motion is bounded; that is, it is confined within the limits . Where, A is called the amplitude of the motion and it is independent of ω0 .

)sin()0( 0φAtx ==

(3) The phase angle φ0 is the initial value of the sine function. It determines the value of the displacement x at time t = 0. I.e,

(4) The term frequency , f0 , refer to the reciprocal of the period of the oscillation or

mkf

Tf

π

πω

21

21

0

0

00

=

==

(57) The unit of frequency (cycles per second, or s-l) is called the hertz (Hz).

0

000

000

00

tan

)cos()0()sin()0(

vx

vAxxAx

ωφ

φωφ

=∴

====

&

(5) Constants of the Motion A and φ0 , can be determined from the initial conditions as follows: (58) Assuming that the mass initially displaced from equilibrium to a position x0 where it is then released with initial velocity v0 . 3.2.1 Simple Harmonic Motion as the Projection of a Rotating Vector Imagine a vector A rotating at a constant angular velocity ω0 . Let this vector denote the position of a point P

moves in uniform circular motion. The projection of A traces out simple harmonic motion. Suppose the vector A makes an angle θ with the x-axis at some time t, as shown in the Figure. Since

and 000 θωθωθ +== t& Thus, the projection of P onto the x-axis is given by ) 00cos(cosθ ω== tAAx +θ Or with the equivalence expression:

) 00sin( φω += tAx Where φ0 −θ0 = π/2 Or we could use a sum of sine and cosine functions to represent the general solution for harmonic motion:

tAtAx 0000 sincoscossin ωφωφ +=

C tDt 00 sincos ωω + (59)

=

Note that:

DCADC 222

0tan +== , φ

3.2.2 Effect of a Constant External Force When a light spring supports a block of mass m vertically, the total force acting on the mass is given by adding the weight mg to the restoring force. I.e, F(x) = -k(X - Xe)+ mg (60) where the positive direction is down. Example (3.2.1): Suppose that X-Xe=D1 . If the block is furthermore pulled downward a distance D2 from the equilibrium position and then released at time t = 0, find: (a) the resulting motion. (b) the velocity of the block when it passes back upward through the equilibrium position. (c) the acceleration of the block at the top of its oscillatory motion. Solution: First, for the equilibrium position we have Fx = 0 = -kD1 + mg

This gives us the value of the spring constant k:

1D

mgk =

From this we can find the angular frequency of oscillation:

1

0 Dg

mk

==ω

We will express the motion in the form tDtCx 00 sincos ωω +=Then, tDtCx 0000 cossin ωωωω +−=& From the initial conditions (58) we find x0 = D2 = C & v0 = 0 = Dω0 D = 0 The motion is, therefore, given by

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛= t

DgDx

12 cos

Note that the mass m does not appear in the final expression. The velocity is then

⎟⎟⎠

⎞⎜⎜⎝

⎛−= t

Dg

DgDx

112 sin&

and the acceleration

⎟⎟⎠

⎞⎜⎜⎝

⎛−= t

Dg

DgDx

112 cos&&

As the block passes upward through the equilibrium position, (one-quarter period), its velocity is;

(b) 1

2 DgDx −=& (center)

At the top of the swing, (one-half period), the acceleration is;

(c) 1

2 DgDx =&& (top)

In the case Dl = D2 the downward acceleration at the top of the swing is just g. This means that the block, at that particular instant, is in free fall; that is, the spring is exerting zero force on the block. Example (3.2.2): The Simple Pendulum The simple pendulum consists of a small mass m swinging at the end of a light string of length l. The motion is along a circular arc defined by the angle θ, as shown in the Fig.

The restoring force is ; Fs = -mg sin θ. Therefore, the differential equation of motion is; θsinmgsm −=&& Since s = lθ and, for small θ, sinθ =θ, we can write the differential equation of motion as follows:

or (61) 0=+ slgs&& 0=+ θθ

lg&&

Although the motion is along a curved path, the differential equation is mathematically identical to that of the linear harmonic oscillator; 0=+ x

mkx&&

Thus, for the angles that the approximation sin θ = θ is valid, we can conclude that the motion is simple harmonic with angular frequency

(62)

lg

=0ω

and period

(63)

glT π

ωπ 22

00 ==

3.3. Energy Considerations in HM: Consider a particle under the action of a linear restoring force Fx = -kx. Let us calculate the work done by an external force Fext in moving the particle from the equilibrium position (x = 0) to some position x. We have, Fext = -Fx = kx, so

2

00 2xkdxkxdxFW

xx

ext === ∫∫

This work is stored in the spring as potential energy: V(x), where 2

21)( kxxV =

The total energy, is given by the sum of the kinetic and potential energies, namely, (64) 2

212

21 kxxmE += &

The total energy is constant if there are no other forces except the restoring force acting on the particle.

maxvx ±=&

0=x Ax ±=&

This constant energy E is totally in the form of kinetic energy at the center, where x = 0 and ,and it is

all potential energy at the extrema, where &

Thus, we can write; (65) 2

212

max21 kAmvE ==

Example (3.3.1): The Energy Function of the Simple Pendulum The potential energy of the simple pendulum is given by the expression V = mgh where h is the vertical distance from the reference level. From the figure we see that h = l - l cosθ, so V(θ) = mgl (1 - cosθ) For small θ we have approximately cosθ=1− θ 2/2. This gives 2

21)( θθ mglV =

or, equivalently, because s = lθ,

2

21)( s

lmgsV =

Thus, the total energy is given by

(66) 2

212

21 s

lmgsmE += &

3.4. Damped Harmonic Motion : Consider an object of mass m is supported by a light spring of constant k. We assume that there is a viscous retarding force (-cv) that is a linear function of the velocity.

The differential equation of motion is, therefore, 0=++ kxxcxm &&& Or,

(67) 0=++ xmkx

mcx &&&

If we introduce the damping factor (γ) defined as

mc

2=γ (68)

Eq.( 67 ) may rewritten as

(69)

02 20 =++ xxx ωγ&&&

Simple sine or cosine solutions do not work, because of the presence of the velocity-dependent term. The suitable solution for this case is;

(70) tqtq eAeAtx )(2

)(1)( +−−− += γγ

where

20

2 ωγ −=q (71) There are three possible situations: I. q real > 0 (Overdamping) II. q real = 0 (Critical damping ) III. q imaginary (Underdamping) I. Overdamped case: Both exponents in Eq. (70) are real. The constants A1 and A2 are determined by the initial

conditions. The motion is an exponential decay with two different decay constants, (γ - q) and (γ + q). The mass will be prevented from oscillating by the strong damping force. II. Critical damping case: Here q = 0. The two exponents in Eq. (70) are each equal to γ. The general solution for such case is given by; tt BeAtetx γγ −− +=)(

As in case I, the motion is a returning to equilibrium with no oscillation. III. Underdamping case: If the constant γ is small enough that q in Eq. (71) is imaginary. The motion, in this case, is oscillatory but with an ultimate death. Let introduce the constant ωd such that; q = iωd Then;

2

222

0 4mc

mk

d −=−= γωω (72)

Which is known as the angular frequency of the under-damped oscillator. The solution for the underdamped oscillator could be; (73) )(cos)( 0θωγ += − tAetx d

t

Or; )(sin)( 0φωγ += − tAetx dt

Note: The solution for the underdamped oscillator is nearly the same as that of the undamped oscillator but with two differences:

1- The presence of the real exponential factor e-γ t leads to a gradual death of the oscillations. 2- The underdamped oscillator vibrates a little more slowly than the undamped oscillator does. I.e, ωd < ω0 because of the presence of the damping force.

The period of the underdamped oscillator is given by

220

22γω

πωπ

−==

ddT (74)

Thus, in one complete period the amplitude diminishes by a factor dTe γ−

Energy Considerations: The total energy of the damped harmonic oscillator is given by the sum of the kinetic and potential energies:

2212

21 kxxmE += &

For the undamped oscillator E is constant. To find E in the damped harmonic oscillator let us differentiate the above expression with respect to t:

xkxxmxkxxxmdtdE

&&&&&&& )( +=+=

Using the differential equation of motion Eq. (67), we can write;

2xc

dtdE

&−= (75)

Because this is always either zero or negative, the total energy continually decreases. Quality Factor: The ratio of the energy stored in the oscillator to that lost in a single period of oscillation is characterized by a parameter Q, called the quality factor. This factor is related to ωd by the relation;

γω2

dQ = (76)

3.6. Forced Harmonic Motion:(Resonance) Let us exert an external periodic force upon a damped harmonic oscillator. When the driving frequency is close to the natural frequency ω0 of the oscillator, a remarkable phenomenon, called resonance , occurs.

tieF ω0Suppose that the applied force has the form of .

Thus, the equation of motion is (77) tieFxckxxm ω

0+−−= &&&

Mathematically, the variable x is now a complex number. Therefore, for the steady-state condition, we shall try a solution of the complex exponential type; (78) )()( φω −= tiAetx where the amplitude A and phase difference φ are constants to be determined. Substituting from (78 ) into Eq.( 77 ), and equating the real and imaginary parts, yields the two following equations φω cos)( 0

2 FmkA =−

φω sin0FAc = From these two equations, we can obtain the following relations;

2tanω

ωφmk

c−

=

We can then solve for A, the amplitude of the steady-state oscillation, as a function of the driving frequency;

( ) 2222

0)(ωω

ωcmk

FA+−

=

In terms of ω0 =k/m and γ = c/2m, we can write the above expressions in another form as follows:

220

2tanωω

γωφ−

= (79)

( ) 222220

0

4

/)(ωγωω

ω+−

=mFA (80)

A plot of A(ω) versus ω shows that the amplitude presumes a maximum value at a certain applied frequency ωr called the amplitude resonant frequency, or resonant frequency for short. To find ωr we calculate dA/dω from Eq. (80 ) and set the result equal to zero. Thus;

or; (81) 222

220

2 2

γωω

γωω

−=

−=

dr

r

When the damping is weak: If γ << ω0 , the resonant frequency ωr , the freely running damped oscillator frequency ωd , and the natural frequency ω0 of the oscillator are essentially identical. I.e; ωr ≈ ωd ≈ ω0 When the damping is strong: If γ2 > ω0

2/2 , no amplitude resonance occurs, because the amplitude then becomes a totally decreasing function of ω0. 3.6.1. Amplitude at the Resonance Peak The steady-state amplitude at the resonant frequency, which we call Amax , is obtained from Eqs. (80) and (81). The result is

220

0max

2/

γωγ −=

mFA (82)

In the case of weak damping, we can neglect γ2 and write

0

0max 2 ωγm

FA ≈

In mechanical systems large resonant amplitudes may or may not be desirable.

3.6.2. Sharpness of the Resonance: (Quality Factor) Let us consider the case of weak damping (ωr ≈ ωd≈ ω0). Then, the expression for steady-state amplitude, Eq. (80), can be written as:

( ) 220

max)(γωω

γω+−

≈AA

The above equation shows that when |ω0–ω|=γ , or equivalently, if ω = ω0 ± γ ,then 2

max212 AA =

This means that 2γ is a measure of the width of the resonance curve Δω at half-energy points. Recall, that the quality factor Q is given as;

γω2

dQ =

Or, for weak damping case;

γω2

0≈Q

Thus, the width of the resonance curve Δω is;

Q02 ωγω ≈=Δ (83)

Note: A high Q results in sharp resonance.

3.6.3. Phase Angle: The phase difference between the applied driving force and the steady-state response is given by Eq. (79);

220

2tanωω

γωφ−

= (84)

The plot of this relation shows φ as a function of the driving frequency ω. 1- For small ω: the phase Difference is zero (φ =0) and remains small. So; the response is in phase with the driving force. 2-Near the resonance frequency: Actually at ω = ωr, the phase angle φ increases to π /2 and so; the response is 90o out of phase at this frequency. 3-For large values of ω: the value of φ approaches π, hence; the motion of the system is just 180o out of phase with the driving force .

3.6.4. Electrical-Mechanical Analogs There is an exact analogy between a moving mechanical

system of masses and springs with frictional forces and

an electric circuit containing; inductance, capacitance,

and resistance. In fact, inductance L and charge q are

analogous to mass m and displacement x, respectively,

and potential difference is analogous to force. Similarly,

we see that the reciprocal of C is analogous to the

stiffness constant k of a spring, and the resistance R is

in analogy with the damping constant c for a mechanical

system. The following table summarizes the situation:

Mechanical Electrical Displacement x Charge q Velocity dx/dt Current dq/dt Mass m Inductance L Stiffness constant k Reciprocal of capacitance C-1 Damping constant c Resistance R Force F Potential difference V

General Motion of a Particle in 3D 4.1.General Principles: In 3D, the vector form of the equation of motion is ;

dtdPF = (85)

where p = mv is the linear momentum of the particle. This vector equation is equivalent to three scalar equations in Cartesian coordinates; zmFymFxmF zyx &&&&&& === There is no general method for obtaining an analytic solution to the above equations of motion. In order to develop a powerful analytical technique that can be applied in such case, we need to introduce some main principles; 1- The Work Principle: If we take the dot product of both sides of Eq. (85) with the velocity v:

vv(vPvF .).dtmd

dtd

==⋅

But , then; vv.(v.v)/ &2=dtd

dtdTm

dtd

==⋅ v.v)vF 21(

in which T is the kinetic energy. Substituting by v = dr/dt, then integrating, we obtain;

TdTd Δ==⋅ ∫∫ rF (86) The line integral, on the left-hand side, represents the work done on the particle by the net force F as the particle moves along its trajectory from point A to point B. The right-hand side of the equation is the net change in the kinetic energy of the particle. Hence, Eq. (86) states that; The work done on a particle by the net force acting on it, in moving from one position in space to another, is equal to the difference in the kinetic energy of the particle at those two positions. 2- Conservative Forces If the force acting on a particle were conservative, it could be defined as the derivative of a scalar potential energy function V, i.e; F = -dV(r)/dr. Hence the work done by such a force in moving a particle from point A to point B is )()()( BVAVVdVd −=Δ−=−= ∫∫ rr F. (87)

Thus, we no longer need a detailed knowledge of the motion of the particle from A to B to calculate the work done on it by a conservative force.

All what we need to know is the potential energy at the endpoints of the motion. But the work done was also equal to the change in kinetic energy of the particle, W=ΔT = T(B) - T(A) Therefore, we find that; Etot = V(A) + T(A) = V(B) + T(B) = constant (88) This is known as the conservation of total energy principle, which is applicable only in case of conservative force. What is the conservative force? A conservative force is a force whose work is path-independent. In other words, in moving an object from point A to point B, the total work done is independent of the path that the object took. Therefore, when F is conservative the work, W, is zero for any simple closed path:

0. =∫ drF The most familiar conservative forces are gravity, the electric force, and spring force.

4.2.The Del Operator; We can now express a conservative force F vectorially as

zV

yV

xV

∂∂

−∂∂

−∂∂

−= kjiF (89)

This equation can be written as; V−∇=F (90) Where we have introduced the vector operator del;

zyx ∂∂

+∂∂

+∂∂

=∇ kji

The expression V∇ is also called the gradient of V and is sometimes written grad V. Taking the curl for both sides of Equation (90), gives; ( ) 0=∇×∇−=×∇= Vcurl FF since the curl of any grad is identically zero. Hence, the condition that a force be conservative is;

0=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=×∇yF

xF

xF

zF

zF

yF xyzxyz kjiF

Similarly, if , then F can be derived from a scalar function V by the operation

0=×∇ FV−∇=F .

4.3. Independent (Separable) Forces: Suppose that we have a force where the coordinates x, y, and z are independent variables: F = iFx(x) + jFy (y) + kFz(z) Forces of this type are separable. The curl of such a force is identically zero:

)()()(

///zFyFxFzyx

kji

zyx

∂∂∂∂∂∂=×∇ F

Therefore, the force is conservative because each partial derivative is of the mixed type and vanishes identically. In this case the equations of motion for each component can be solved by the methods described under rectilinear motion in Chapter 2. 4.3.1. (Projectile Motion)

No Air Resistance For simplicity, we first consider the case of a projectile moving with no air resistance. Only one force, gravity, acts on the projectile, and, as we shall see, it affects only its vertical motion.

Choosing the z-axis to be vertical, we have the following equation of motion:

mgdtdm kr

−=2

2

(91)

Let us assume g is constant. Then, the force is conservative and of the separable type. Let v0 is the initial speed of the projectile, and the origin of the coordinate system is its initial position. We can calculate the velocity of the projectile at any instant of time by integrating Eq. (91);

0v krv +−== gtdtd

(92)

In terms of unit vectors: The initial velocity is

α+α= sinvcosv 000 k iv (93)

Hence, the velocity at any instant is

( )gtvv −+= αα sincos 00 k iv (94)

And the position vector is;

( ) ( )( )221

00 sincos gttvtv −+= αα k ir (95)

In terms of components: the components of the velocity at any instant v are; ( )cosv

( ) 0

0

0

gtsinvv

vv

z

y

x

−α=

=α= (96)

And the components of the position of the projectile at any instant are;

( )

( ) 221

0

0

sin0

cos

gttvzy

tvx

−=

==

α

α

(97)

Since the only applied force is conservative, the speed of

the projectile can be calculated as a function of its

height, z, using the conserved energy equation Eq. (42);

(98) gzvv 220

2 −=

Using the first of Equations (97) to solve for t and then

substitute the resulting expression in the third of

Equations (97), we get;

( ) 222

0 cos2tan x

vgxz ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

αα (99)

We can now see that the path of the projectile is a

parabola as it is shown in the previous Figure.

Main properties of the projectile motion: (1) The maximum height:

α

R

zmax

Using Eq. (98) and noting that at maximum height the vertical component of the velocity of the projectile is zero so that its velocity is in the horizontal direction and equal to the constant horizontal component, v0 cos α. Thus;

max2

022

0 2cos gzvv −=αThen;

gvz

2sin22

0max

α= (100)

(2) The time it takes to reach maximum height: The time it takes to reach maximum height can be obtained by setting vz=0 in the last of Equations (96); 0sin max0 =− gtv α Hence;

gsinv α

= 0tmax

(101)

(3) The flight time: We can obtain the total time of flight T of the projectile by setting z = 0 in the last of Equations (97), which yields

gvT αsin2 0= (102)

This is twice the time it takes the projectile to reach maximum height. (4) The range: Substituting the total time of flight, T, into the first of Equations (97), gives the range of the projectile ;

gvR α2sin2

0= (103)

At α= 45° , R has its maximum value

gvR =

20

max (104)

With Air Resistance In this case, the motion does not conserve total energy, which continually diminishes during the flight of the projectile. Contrary to the case of zero air resistance the path of the projectile is not a parabola, but

rather a curve that lies below the corresponding parabolic trajectory.

4.4. The Harmonic Oscillator in 2D and 3D: Consider the motion of a particle attached to a set of elastic springs as shown in the Figure. This is the three-dimensional generalization of the linear oscillator studied earlier. The differential equation of the motion simply can be expressed as;

r r kdtdm −=2

2

(105)

4.4.1 The Two-Dimensional Isotropic Oscillator First, we will consider the motion of the isotropic oscillator, in

which the restoring force is independent of the direction of the displacement. In the case of motion in a single plane, 2D, Eq.

(105) is equivalent to the two component equations

kyymkxxm −=−= &&&& These are separated, so their solutions can be in the form

)cos()cos( βωαω +=+= tBytAx

With mk /=ω . The constants A, B, α, and β are determined

from the initial conditions .

To find the equation of the path, we eliminate the time t between the x & y equations. We then have a quadratic equation in x and y;

Δ=+Δ

− 22

2

2

2

sincos2By

ABxy

Ax

Where Δ = β−α is the phase difference. Recall: The general quadratic equation; ax2 + bxy + cy2 + dx + ey = f represents; An ellipse, if b2 - 4ac < 0 A parabola, if b2 - 4ac = 0 or A hyperbola, if b2 - 4ac > 0 In our case the discriminant is equal to -(2sinΔ/AB)2, which is negative, so the path is an ellipse. If the phase difference Δ is equal to π/2, then the equation of the path reduces to the equation

12

2

2

2

=+By

Ax

which is the equation of an ellipse whose axes coincide with the coordinate axes.

If the phase difference Δ is equal 0 or π, then the equation of the path reduces to that of straight line;

xABy ±=

Similarly for 3D isotropic oscillator the motion will take place totally in a single plane, and the path of the particle in that plane is an ellipse. 4.4.2. 3D Non-isotropic Oscillator: If the magnitudes of the components of the restoring force depend on the direction of the displacement, we have the case of the non-isotropic oscillator. The differential equations for this case can be written as; zkzmykymxkxm 321 −=−=−= &&&&&& Here we have three different frequencies of oscillation, ω1 , ω2 and ω3 , and the motion is given by the solutions

)cos()cos()cos(

3

2

1

γωβω

αω

+=+=+=

tCztBytAx

The resulting oscillation of the particle lies completely within a rectangular box (whose sides are 2A, 2B, and 2C) centered on the origin.

If ω1 , ω2 and ω3 are commensurate-that is,

3

3

2

2

1

1

nnnωωω

==

Where n1, n2 and n3 are integers, the path is closed, i.e. the

particle returns to its initial position and the motion is

repeated. Such a path called Lissajous figure.

On the other hand, If ω1 , ω2 and ω3are not commensurate, the

path is not closed and the path will completely fill the

rectangular box.

4.4.3. Energy Considerations

For the general 3D case, it is easy to verify that

232

1222

1212

1),,( zkykxkzyxV ++=

because Fx = -dV/dx = -k1 x, and similarly for Fy and Fz.

If k1 =k2 = k3 = k, we have the isotropic case, and

( ) 221222

21),,( krzyxkzyxV =++=

The total energy in the isotropic case then given by; 2

1 2212 krmvE += (106)

Lissajous Figures

ω1 : ω2 = 1:2 ω1 : ω2 = 3:2 ω1 : ω2 = 5:4

4.5. Motion of Charged Particles in Electric & Magnetic Fields: When a charged particle is surrounded by other electric charges, it will experience a force. This force F is caused by the electric field E, which arises from these other charges. We write F = qE (107) where q is the electric charge. The equation of motion of the particle is then

Er qdtdm =2

2

or, in component form, zyx qEzmqEymqExm === &&&&&& Let us consider a case of a uniform constant electric field which is directed along the z-axis. Then Ex = Ey = 0, and E = Ez. The differential equations of motion of a particle of charge q moving in this field are then

.00 consm

qEzyx z ==== &&&&&&

These are of the same form as those for a projectile in a uniform gravitational field. Therefore, the path is a parabola, if vx and vy if are not both zero initially. Otherwise, the path is a straight line, as with a body falling vertically.

According to the electromagnetic theory, if E is due to static charges then; 0E =×∇ (108) This means that motion in such a field is conservative, and that there exists a potential function Φ such that; Φ−∇=E (109)

From (107) and (109), the force F caused by the electric field E may be rewritten as; Φ−∇== qqEF Then, the potential energy of a particle of charge q in such a field is then qΦ. The total energy, hence, is constant and is equal to Φ+ qmv2

21 (110)

In the presence of a static magnetic field B, called the magnetic induction, the force acting on a moving particle is conveniently expressed by means of the cross product, namely, ( )BvF ×= q (111) where v is the velocity and q is the charge. The differential equation of motion of a particle moving in a purely magnetic field is then

( Bvr×= q

dtdm 2

2

) (112)

Note: Equation (112) states that the acceleration of the particle is always ⊥ to the direction of motion. This means that the tangential component of the acceleration is zero, and so the particle moves with constant speed. The path is a helix , and if there is no component of the velocity in the z direction, the path is a circle. 4.6. Constrained Motion of a Particle When a moving particle is restricted geometrically, i.e. it must stay on a certain definite surface or curve, the motion is said to be constrained. Examples of constrained motion: - A piece of ice sliding around a bowl. (one-sided constraint) - A bead sliding on a wire. (complete constraint)

4.6.1. The Energy Equation for Smooth Constraints The total force acting on a particle moving under constraint can be expressed as the vector sum of the net external force F and the reaction force of the constraint R. The equation of motion may, therefore, be written

RFv+=

dtdm (113)

If we take the dot product with the velocity v, we have

.vR F.v.vv+=

dtdm (114)

In the case of a smooth constraint, the reaction R is normal to the surface while the velocity v is tangent to the surface. Hence, R is ⊥ to v, and R . v = 0. Eq.(114) then reduces to

( ) F.vv.v =mdtd

21

Therefore, if F is conservative, we can integrate, as in Eqs. (86) & (87), and show that the total energy of the constrained particle is constant. EXAMPLE 4.6.1 A particle is placed on top of a smooth sphere of radius a. If the particle is slightly disturbed, at what point will it leave the sphere?

Solution: The forces acting on the particle are the downward force of gravity mg and the reaction R of the spherical surface. The equation of motion is

Rgv+= m

dtdm

According to the chosen coordinate the potential energy is then mgz, and the energy equation is; Emgzmv =+2

21

But from the initial conditions (v = 0 for z = a) we have E = mga, so; mgamgzmv =+2

21

As the particle slides down, its speed is given by; ( )zagv −= 22

Taking radial components of the equation of motion, we can write the force equation as

RazmgRmg

amv

+−=+=− θcos2

Hence,

( )aza

mgR 23 −=

Thus, R vanishes when az 3

2= . At this point the particle leaves the sphere.

EXAMPLE 4.6.2 Constrained Motion on a Cycloid Consider a particle sliding under gravity in a smooth cycloidal channel. Such motion represented by the parametric equations

( )( )φ

φφ2cos1

2sin2−=

+=AyAx

where φ is the parameter. Now the energy equation for the motion is

( ) mgzyxmzVmvE ++=+= 22221

2)( &&

Doing some algebra, we can express the energy in terms of φ as; φφφ 2222 sin2cos8 mgAmAE +=Let s = 4A sin φ , then, the energy equation can be written as

2212

21

4s

Amgsm ⎟

⎞⎜⎛+= &E ⎠⎝

This is just the energy equation for harmonic motion in the single variable s. Thus, the particle undergoes periodic motion whose frequency is independent of the amplitude of oscillation.

5.1. Accelerated Coordinate Systems: Recall: Uniformly moving reference frames (e.g. those considered at 'rest' or moving with constant velocity in a straight line) are called inertial reference frames. Sometimes it is necessary, to employ a coordinate system that is not inertial. Let us first consider the case of a coordinate system that undergoes pure translation. Assume Oxyz are the primary fixed coordinate axes, and O'x'y'z' are the moving axes. In the case of pure translation, the respective axes Ox and O'x', and so on, remain parallel. The position vector of a particle P is denoted by r in the fixed system and by r' in the moving system. The displacement OO' of the moving origin is denoted by R0. Thus, from the triangle OO'P, we have r = R0 + r' (115) Taking the first and second time derivatives gives v = V0 + v' (116)

a = A0 + a' (117)

in which V0 and A0 are, respectively, the velocity and acceleration of the moving system, and v' and a' are the velocity and acceleration of the particle in the moving system. If the moving system is not accelerating, i.e. it is also inertial, so that A0 = 0, then a = a' In this case we cannot specify a unique coordinate system, because Newton's laws will be the same in both systems. For example, Newton's second law in fixed system F = ma becomes F' = ma' in the moving system. On the other hand if the moving system is accelerating, then Newton's second law becomes F = mA0 + ma' or F - mA0 = F' (118) where (-mA0) is known as the inertial term or inertial force. Such "force" is not due to interactions with other bodies; rather, it happens as a result of the acceleration of the reference system. 5.2. Rotating Coordinate Systems In this section, we show how velocities, accelerations, and forces transform between an inertial frame of reference and a noninertial one that is rotating.

Assume that the axes of the both coordinate systems have a common origin. Let the rotation of the rotated system takes place about some specific axis of rotation, whose direction is designated by a unit vector, n. The angular velocity of the rotating system then is; ω = ωn The direction of the velocity vector is given by the right-hand rule. The position of any point P in space can be designated by the vector r in the fixed system and by the vector r' in the rotating system. Because the coordinate axes of the two systems have the same origin, these vectors are equal, that is, r = r' (119) or; ix + jy+ kz = i'x' + j'y'+ k' z' When we differentiate with respect to time to find the velocity, we must keep in mind the fact that the unit vectors i', j', and k' are not constant. Thus, we can write the velocity vector v in the fixed system as;

dtkdz

dtjdy

dtidx

′′+

′′+

′′+′= vv (120)

Where v' is the velocity in the rotating system. From the definition of the cross product, we can write;

idtd ′×=

′ωi

, jdt

d ′×=′

ωj and k

dtd ′×=

′ωk

.

Hence;

( ) ( ) (

rω

kωjωiωkji

′

)

×=

′×′+′×′+′×′=′

′+′

′+′

′ zyxdtdz

dtdy

dtdx

This is the velocity of P due to rotation of the coordinate system. Accordingly, Eq. (120 ) can be rewritten as rωvv ′×+′= (121) Taking the first time derivatives gives the acceleration in the fixed system in terms of the position, velocity, and acceleration in the rotating system; ( )rωωv2ωrωaa ′××+′×+′×+′= & (122) If the moved system is undergoing both translation and rotation, the general equations for transforming from a fixed system to a moving and rotating system will be: 0Vrωvv +′×+′= (123) And;

( ) 0Arωωv2ωrωaa +′××+′×+′×+′= & (124)

The term 2ω × v' is known as the Coriolis acceleration, which appears whenever a particle moves in a rotating coordinate system except when the velocity v' is parallel to the axis of rotation.

The term ω×(ω× r') is called the centripetal acceleration. which is the result of the particle being carried around a circular path in the rotating system. It is always directed toward the axis of rotation and is perpendicular to the axis as shown in the figure.

The term is called the transverse acceleration, because it is perpendicular to the position vector r'. It appears whenever the rotating system has an angular acceleration, i.e. if the angular velocity vector is changing in either magnitude or direction, or both.

rω ′×&

5.3. Dynamics of a Particle in a Rotating System: It is well known that the equation of motion of a particle in an inertial frame of reference is ; F = ma where F is the sum of all real, physical forces acting on the particle. According to Eq.(124) , we can write the equation of motion of a particle in a noninertial frame of reference as ; ( ) aArωωvω2rωF ′=−′××−′×−′×− mmmmm 0& (125) All inertial forces have names corresponding to their relevant accelerations. Thus;

The force -m rω ′×& is called the transverse force, because it is perpendicular to the position vector r'. It is present only if there is an angular acceleration (or deceleration) of the rotating coordinate system.

The force -2mω×v' is the Coriolis force, which appears whenever a particle moves in a rotating coordinate system. Its direction is always perpendicular to v', thus it seems to deflect the moving particle at right angles to its direction of motion.

The force - mω×(ω× r') is the centrifugal force, which is the result of the particle being carried around a circular path in the rotating system. It is directed outward away from the axis of rotation and is perpendicular to that axis. If r' is perpendicular to ω, the magnitude of the centrifugal force is mr'ω2. A no inertial observer in an accelerated frame of reference must include all, or some, of these inertial forces along with the real forces F to calculate the correct motion of the particle. In other words, such an observer writes the fundamental equation of motion as; F' = ma' in which the sum of the vector forces F' acting on the particle is given by F' = Fphysical + F'trans + F'Cor + F'cel1trif – mA0 F (or Fphysical) forces are the only forces that a no inertial observer claims are actually acting upon the particle.

5.4. Effects of Earth's Rotation

Consider a coordinate system that is moving with the Earth. Because the angular speed of Earth's rotation is 2π radians per day, the effects of such rotation is relatively small.

Nevertheless, it is the spin of the Earth that makes the equatorial radius is some 13 miles greater than the polar radius, i.e. equatorial bulge. 5.4.1. Static Effects: The Plumb line Let us describe the motion of the plumb bob in a local frame of reference whose origin is at the position of the bob. Our frame of reference is attached to the surface of the Earth, so it is undergoing translation as well as rotation.

The translation of the frame takes place along a circle whose radius is ρ = re cosλ, where re is the radius of the Earth and λ is the geocentric latitude of the plumb bob. Hence; A0 =ω2ρ =ω2 re cosλ (126)

Its rate of rotation is ω, the same as that of the Earth about its axis. Let us now examine the terms of Eq. (124):

The force - ma' is zero, because the bob is at rest in the local frame of reference, i.e. a' = 0.

The Coriolis force -2mω × v' is zero, because v'= 0.

The transverse force -m rω ′×& is zero, because ω is constant.

The centrifugal force - m ω×(ω× r') is zero, because the origin of the local coordinate system is centered on the bob. I.e, r'= 0. The only surviving terms in Eq. (124) are the real forces F and the inertial term –mA0, which arises because the local frame of reference is accelerating. Thus, F – mA0 = 0 (127) In other words, the rotation of the Earth causes the acceleration A0 of the local frame. The bob does not hang on a line pointing toward the center of the Earth because the inertial force –mA0 throws it outward, away from Earth's axis of rotation. The magnitude of this force is; mω2 re cosλ It is a maximum when λ = 0 at the Earth's equator and a minimum at either pole.

The tension T in the string balances out the real gravitational force mg0 and the inertial force –mA0, i.e; (T + mg0) – mA0 = 0 (128) Now, when we hang a plumb bob, we normally think that the tension T balances out the local force of gravity, which we call mg. We can see from the Figure that: mg = mg0 – mA0

or g = g0 –A0 (129) As can be seen the inertial reaction –mA0, directed away from Earth's axis, causes the direction of the plumb line to deviate by a small angle ε away from the direction toward Earth's center. We can easily calculate the value of the angle ε. From the Figure we have;

mgrm e

λλω

ε sincos

sin2 =

or, because ε is small

λωεε 2sin2

sin2

gre=≈ (130)

Thus, ε vanishes at the equator (λ = 0) and the poles (λ =±90). The maximum deviation of the direction of the plumb line from the center of the Earth occurs at λ = 45°

where; oe

gr 1.0

2

2

max ≈=ωε

5.4.2. Dynamic Effects: 1- Falling Body A body that is dropped from a height h above the ground, as it falls, it will drift to the east. The eastward drift is given by:

λω cosghx ⎟⎟

⎠

⎞⎜⎜⎝

⎛=′

3

31 8

(131)

For a height of 100 m at latitude of 45°, the drift is 1.55 cm

2- Deflection of a Rifle Bullet If we fire a projectile to east with high initial speed v0 , the projectile will bend to the south. If H is the horizontal range of the projectile, the transverse deflection is then;

λω sinvH

0

2

≈Δ (132)

This is the same for any direction in which the projectile is initially aimed, provided the trajectory is flat.

6.1. Newton’s Law of Gravitation:

During his study of the motions of the planets and of the moon, Newton discovered the fundamental charter of the gravitational attraction between any two bodies. In 1687, Newton published the law of gravitation which may be stated as follows: Every particle of matter in the universe attracts every other particle with force (Fg) that is directly proportional to the product of the masses (m1, m2) of the particles and inversely proportional to the square of the distance between them (r). Or;

221

rmGmF =g (131)

G is a fundamental physical constant called the gravitational constant. The numerical value of G (in SI units ) is G = 6.67×10-11 N.m2/kg2

Equation (131) tells us that if the distance r is doubled,

the force is only one-fourth as great, and so on. Although many of the stars in the night sky are more massive than the sun, they are so far away that their gravitational force on earth is negligible.

Gravitational forces always act along the line joining the two particles, which form an action-reaction pair. Even when the masses of the particles are different, the two interaction forces have equal magnitude.

Gravitational forces combine vectorially. If each of two masses exerts a force on a third, the total force on the third mass is the vector sum of the individual forces of the first two. This property is often called superposition of forces.

The earth's gravitational force on a body of mass m at

any point outside the earth is given by ; Fg = GmEm/r2, where mE is the mass of the earth and r is the distance of the body from the earth's center. Therefore, we can express the gravitational potential energy (U) in more general form as;

rU E

g −= (132) mGm

6.2. Kepler’s laws and the motion of planets:

One of the great intellectual events of the 16th and 17th centuries was the threefold realization; 1- that the earth is also a planet, 2- that all planets orbit the sun, 3- and that the apparent motions of the planets as seen from the earth can be used to determine the orbits of the planets precisely. The first and second of these ideas were published by Nicolaus Copernicus in 1543. The determination of planetary orbits was carried out between 1601 and 1619 by the German astronomer and mathematician Johannes Kepler.

By trial and error, Kepler discovered three observed laws that accurately described the motions of the planets. Two hundred years later, Newton discovered that each of Kepler's laws can be derived using Newton's laws of motion and the law of gravitation. First law: Law of Ellipses. Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse. The following Figure shows the geometry of the ellipse with its main properties;

* The longest dimension 2a is the major axis, with half-length a known as the semi-major axis. * S and S' are the foci (plural of focus). The sun is at S, and the planet is at P. * The sum of the distances from S to P and from S' to P is the same for all points on the curve. * The distance of each focus from the center of the ellipse is ea, where e is a dimensionless number between 0 and 1 called the eccentricity. If e = 0, the ellipse is a circle. The actual orbits of the planets are somewhat circular; their eccentricities range from 0.007 for Venus to 0.248 for Pluto. The earth's orbit has e = 0.017. * The point in the planet's orbit closest to the sun is the perihelion, and the point most distant from the sun is the aphelion. Scecond law: Law of Equal Areas A line from the sun to a given planet sweeps out equal areas in equal times.

In a small time interval dt, the line from the sun S to the planet P turns through an angle dθ. The area swept out is the dA = ½ r2 dθ .The rate at which area is swept out, dA/dt, is called the sector velocity :

dtdr

dtdA θ2

21=

The real meaning of Kepler's second law is that the sector velocity has the same value at all points in the orbit. When the planet is close to the sun, r is small and dθ/dt is large; when the planet is far from the sun, r is large and dθ/dt is small. Third law: Harmonic Law The square of the period of a planet is directly proportional to the cube of the semi-major axis of the plant’s orbit. This law can be expressed as; 32 kaT = If the distance measured in astronomical units (1AU=1.5 ×108 km) and periods are measured in Earth years then: k=1.

EXAMPLE (6.6.1): Find the period of a comet whose semi-major axis is 4 AU. Solution: With T measured in years and a in astronomical units, we have

T2 = a3= (4)3 =64 yrs2

T = 8 yrs

However, using Newton’s laws of motion and the inverse-square law of gravity, one can found (see problem 6.5) that the constant k in SI unit is equals to;

sGmk

24π=

Where ms is the sun's mass. Kepler’s 3rd law can be then rewritten as

sGmT 2π=

a 2/3 (133)

Note:

( )( ) years 5.751038.21099.11067.6

)10672 2 9

3011

2/312

=×=××

×=

−s.(T π

The period does not depend on the eccentricity e. I.e. an asteroid in an elliptical orbit with semi-major axis a will have the same orbital period as a planet in a circular orbit of radius a. The key difference is that the asteroid moves at different speeds at different points in its elliptical orbit, while the planet's speed is constant around its circular orbit.

Example : Comet Halley Comet Halley moves in an elongated elliptical orbit around the sun. At perihelion, the comet is 8.75 × 107 km from the sun; at aphelion it is 5.26 × 109 km from the sun. Find the semi- major axis, eccentricity, and period of the orbit. Solution: - The length of the major axis is; 2a = 8.75 × 107 + 5.26 × 109 So; a = 2.67 × 109 km - Since the comet-sun distance at perihelion is given by a - ea = a (1- e) = 8.75 × 107

Then; e = 0.967

- From Eq. (133),