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S. Widnall16.07 Dynamics

Fall 2009Version 2.0

Lecture L1 - Introduction

Introduction

In this course we will study Classical Mechanics and its application to aerospace systems. Particle motion in

Classical Mechanics is governed by Newton’s laws and is sometimes referred to as Newtonian Mechanics. The

motion of extended rigid bodies is analyzed by application of Newton’s law to a multi-particle system. These

laws are empirical in that they combine observations from nature and some intuitive concepts. Newton’s laws

of motion are not self evident. For instance, in Aristotelian mechanics before Newton, a force was thoughtto be required in order to maintain motion. Much of the foundation for Newtonian mechanics was laid by

Galileo at the end of the 16th century. Newton, in the middle of the 17th century stated the laws of motion

in the form we know and use them today, and shortly after, he formulated the law of universal attraction.

This led to a complete theory with which he was able to explain many observed phenomena, in particular

the motion of the planets. Nevertheless, these laws still left many unanswered questions at that time, and

it was not until later years that the principles of classical mechanics were deeply studied and rationalized.

In the eighteenth century, there were many contributions in this direction, such as the principle of virtual

work by Bernoulli, D’Alambert’s principle and the theory of rigid body dynamics developed by Euler. In the

nineteenth century, Lagrange and later Poisson, Hamilton and Jacobi developed the so called analytical or

rational mechanics and gave to the theory of Newtonian mechanics a much richer mathematical structure.

Classical Mechanics has its limitations and breaks down where more modern theories such as relativity and

quantum mechanics, developed in the twentieth century, are successful. Newtonian mechanics breaks down

for systems moving at speeds comparable with the speed of light, and also fails for systems of dimensions

comparable to the size of the atom. Nevertheless, for practical engineering applications, Newtonian mechanics

provides a very good model to represent reality, and, in fact, it is hard to nd examples in aerospace where

Newtonian mechanics is not adequate. The most notable perhaps are the relativistic corrections that need

to be made for modeling satellite communications.

16.07’s Place in the Aero-Astro Curriculum

Aero-Astro focuses on the analysis, design and control of aerospace vehicles, both aircraft and space craft

and the environment in which they are used. The place of 16.07 within the overall curriculum is shown in

Figure 1.

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16.07 is a core discipline of aerospace engineering: dealing with the natural dynamics of aero-astro systems.

The complexities of aerodynamic forces (as seen in unied engineering and in 16.100) or structural exibility

(as covered in unied engineering and in 16.20) and their effect on vehicle dynamics are treated very simply,

if at all. Beyond the study of the natural dynamics of aircraft and space craft, with or without aerodynamic

forces and structural exibility, is our need to develop approaches to control the behavior of the system.

Thus 16.06-Automatic Control surrounds this group of courses, moving beyond the natural dynamics to

impose control laws upon the system.

Depending upon our interests, 1) in the dynamic motion of aircraft, where we would want to predict

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position, velocities, and acceleration under the action of forces and moments as well as aircraft stability

or 2) in the motion of spacecraft, orbits, satellite stability, launch dynamics, and orbit transfers,

we will model the system as simply as possible for our purposes. If we are interested in the earth’s motion

about the sun, we will model both the sun and the earth as point masses of no extent. To determine the

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motion of a satellite in orbit, we may model the satellite as a point mass, or if we are concerned whether the

satellite will tip over, we will model it as a body of nite extent.

Particles, Rigid Bodies and “Real Bodies”

In this course real bodies will be idealized either as particles or as rigid bodies.

A particle is a body of negligible dimensions. When the dimensions of the body are unimportant to the

description of its motion, we will idealize the body as a particle.

A rigid body is a body that has a nite size but does not deform. This will be a useful approximation when

the deformation of a body is negligible compared to the overall motion. For instance, we may consider an

aircraft as being a rigid body when considering the behavior of the aircraft along its ight path, even though

under some specic conditions the deection of the wing tips may be considerable. In describing the motion

of a rigid body, we need to be concerned not only with its position but also with its orientation.

On the other hand, real bodies have a nite size and are always deformable under loading. In some situations

it will be required to consider the deformation of the body when considering its dynamic behavior, but this

is outside the scope of this course, except perhaps when we introduce the topics of vibration and waves.

Scalars and Vectors

Scalars and vectors are mathematical abstractions that are very useful to describe many of the concepts in

dynamics. You should already have an intuitive idea of what they are. A scalar is a single number which is

useful to describe the reading of a physical property on a scale. For instance, temperature, length or speed

are scalar quantities. On the other hand, vectors are much richer entities. They exist in a multi-dimensional

space and they have both direction and magnitude . Velocities, forces and electric elds are examples of

vector quantities.

Newton’s Laws

Newton’s three laws of motion are:

1.- A particle in isolation moves with constant velocity.

A particle in isolation means that the particle does not interact with any other particle. Constant

velocity means that the particle moves along a straight line with constant speed. In particular, it

can be at rest. It turns out that the motion (e.g. velocity and acceleration) we observe depends on

the reference frame we use. Therefore, the above law cannot be veried in all reference frames. The

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reference frames for which this law is satised are called inertial reference frames . In some sense, we

can say that Newton’s rst law postulates that inertial reference frames exist.

2.- The acceleration of a particle relative to an inertial reference frame is equal to the force per unit mass

applied to the particle.

In other words, if F represents the (vector) sum of all forces acting on a particle of mass m, any inertial

observer will see that the particle has an acceleration a which is given by,

F = ma . (1)

This equation introduces two new concepts: force and mass. Precise denitions for these concepts are

not easy even though we all have some intuition about both force and mass. Forces result when bodies

interact. Forces are vectors with magnitude and direction. They are measured by comparison, e.g. to

the weight of a standard mass, or to the deformation of a spring. We will assume that the concept

of force is absolute and does not depend on the observer. Once we have dened force, we can denemass as the constant of proportionality between force and acceleration. Mass is a scalar quantity. One

could think of mass as the resistance of bodies to a change in motion. That is, a given force applied to

a body with small mass will produce a large acceleration, whereas the same force on a body of large

mass will produce a small acceleration.

Equation (1) is a vector equation. This means that the force and the acceleration always have the

same direction and the ratio of their magnitudes is m. It turns out that this equation is the basis for

most engineering dynamics.

3.- The forces of action and reaction between interacting bodies are equal in magnitude and opposite in

direction

This law makes explicit the fact that a force is the result of interaction between bodies. This law is

clearly satised when the bodies are in contact and in static equilibrium. The situation for bodies

in motion interacting at a distance, e.g. electromagnetic or gravity interactions, is a little bit more

complicated. We know that electromagnetic signals travel at a nite speed and therefore there is a

time delay whenever two bodies interact at a distance. Unfortunately, Newton did not foresee such a

situation, and in these cases, Newton’s third law breaks down. However, for practical purposes and for

most engineering applications, the error made by assuming that these interactions are instantaneous,and, hence, assuming that Newton’s third law is applicable, is negligible.

Note Units

We shall primarily use two systems of units: the International System , also called SI, and the English System .

The international system is the most widely used system for science and engineering. The English system,

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however, is still in widespread use within the United States engineering community. For this reason we will

use and should feel comfortable with both systems. The SI units are meter (m), kilogram (kg), and second

(s) for length, mass and time, respectively. Acceleration is measured in m/s 2 , and force is measured in kg m·

/s 2 , which is also called a Newton (N) i.e. 1N = 1Kg m/s 2 . In the English system, the units of length,·

mass and time are foot (ft), slug , and second (s), respectively. Acceleration is measured in ft/s2

, and force ismeasured in slug ft/s 2 , also called pound (lb), i.e. 1 pound = 1slug ft/s 2 . We have the following conversion· ·

factors:

1 ft = 0 .348 m

1 slug = 14.5 kg

1 N = 0 .224 lb

Example Inertial vs. Non-inertial observers

This example is meant to illustrate the fact that we can easily come up with situations in which Newton’s

second law is not satised for accelerating, non-inertial observers. We will come back to this example later

on in the course.

Consider a rocket sled which can move on a horizontal track as shown. We consider an inertial observer

O which is xed on the ground and an observer O which is on the sled. We also have an accelerometer

mounted on the sled. This consists of a known proof mass m, whose horizontal motion relative to the sled

is constrained by a spring. We assume that the friction between the mass and the sled is negligible. Thismeans that the only mechanism to exert a horizontal force on the mass is through the spring.

We consider two situations:• The engine is off, T = 0, a = 0, the mass is at rest and the spring is uncompressed. Both observers O

and O agree in all their measurements.

• The engine is on, T is a constant non-zero force, the sled has an acceleration a , and the spring is

compressed and exerting a force F on the mass. Both observers are able to measure the force exerted

by the spring by reading on a scale how much the spring deforms. For observer O , the mass is not

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moving (assume that the initial transient oscillation is gone and the spring has settled to a xed position

relative to the sled). For the observer on the ground the mass is accelerating with acceleration a . Since

observer O is an inertial observer, he or she is able to verify that F = ma . On the other hand, the

observer on the sled, O , measures a force F , but observes a zero acceleration. Hence, the observations

of observer O do not satisfy Newton’s second law.

Law of Universal Attraction

The law of universal attraction was proposed by Newton shortly after formulating the laws of motion. The

law postulates that the force of attraction between any two particles, of masses M and m, has a magnitude,

F , given byMm

F = G r 2 (2)

where r is the distance between the two particles, and G = 6 .673(10− 11 ) m 3 /(kg s2 ) is the universal constant·

of gravitation which is determined according to experimental evidence. The direction of the force is parallel

to the line connecting the two particles.

The law of gravitation stated above is strictly valid for point masses. One would expect that when the size

of the masses is comparable to the distance between the masses one would observe deviations to the abovelaw. It turns out that if the mass M is distributed uniformly over a sphere of radius R, the force on a mass

m, outside M , is still given by (2), with r being measured from the sphere’s center.

Weight

The gravitational attraction from the earth to any particle located near the surface of the earth is called the

weight. Thus, the weight, W , of a particle of mass m at sea level is given by

W = − GM e m

e r = − g0 me r = mg 0 .R 2e

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Here, M e ≈ 5.976 × 1024 kg and R e ≈ 6.371 × 106 m, are the mass and radius of the earth, respectively, and

g 0 = − (GM e /R e2 ) e r , is the gravitational acceleration vector at sea level. The average value of its magnitude

is g0 = 9 .825 m/s 2 .

The variation of the gravitational attraction with altitude is easily determined from the gravitational law.

Thus, the weight at an altitude h above sea level is given byM e m R e2 R e2W = − G

(R e + h)2e r = − g0

(R e + h)2me r = m

(R e + h)2g 0 .

It turns out that the earth is not quite spherical and so the weight does not exactly obey the inverse-squared

law. The magnitude of the gravitational acceleration, g0 , at the poles and at the equator, is slightly different.

In addition, the earth is also rotating. As we shall see this introduces an inertial centrifugal force which has

the effect of reducing the vertical component of the weight. We will study these effects later on in the course.

References

[1] R. Dugas, A History of Mechanics , Dover, 1988.

ADDITIONAL READING

J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS , 5th Edition

1/1, 1/2, 1/3, 1/4, 1/5 (Effect or Altitude only), 1/6, 1/7, 3/2

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16.07 DynamicsFall 2009

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S. Widnall 16.07 Dynamics

Fall 2009

Version 1.0

Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion

Degrees of Freedom

Degrees of freedom refers to the number of independent spatial coordinates that must be specied to deter

mine the position of a body. If the body is a point mass, only three coordinates are required to determine

its position. On the other hand, if the body is extended, such as an aircraft, three position coordinates and

three angular coordinates are required to completely specify its position and orientation in space.

Kinematic Constraints

In many situations the number of independent coordinates will be reduced below this number, either because

the number of spacial dimensions is reduced or because there are relationships specied among the spatial

coordinates. When setting up problems for solution it is useful to think of these relationships as constraints.

For example, if a point mass is constrained to move in a plane (two dimensions) the number of spatial

coordinates necessary to describe its motion is two. If instead of being a point mass, this body has extended

dimensions, such as a at plate conned to a plane, it requires three coordinates to specify its position and

orientation: two position coordinates and one angular coordinate.

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If a particle is conned to move on a curve in either two or three dimensions, such as a bead moving on a

wire, the number of independent coordinates necessary to describe its motion is one.

Another source of constraints on the motion of particles is connections between them. For example, the two

particle connected by a cable passing over a pulley are constrained to move in equal and opposite directions.

More complex arrangements are possible and can be analyzed using these ideas. Two gears in contact are

constrain to move together according to their individual geometry.

A cylinder rolling on a plane is constrained in two ways. Contact with the plane reduces the two-dimensional

motion to one spatial coordinate along the plane, and the constraint of rolling provides a relationship be

tween the angular coordinates and the spatial position, resulting in a single degree of freedom system.

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Internal Force-Balance Constraints

Another type of constraint occurs when we consider the of a system of particles and the necessary force

balance that occurs between the parts. These constraints follow directly from Newton’s third law: the force

of action and reaction between two bodies are equal in magnitude and opposite in direction. We will pursue

these ideas in greater depth later in the course. For now, we will give a simple example to illustrate the

principle.

Consider the systems shown in a) and b) .

System a) consists of two masses m in contact resting on a frictionless plane in the presence of gravity. A

force F is applied to mass 1, and it is obvious that the two masses will accelerate at a = F / (2m). If we

look at the two masses separately, we can determine what internal force must exist between them to cause

the motion. It is clear that each mass feels a net force of F/2 , since its acceleration is a = F / (2m). This

net force arises because between the two masses there is an equal and opposite force F/2 acting across the

interface. Another way to look at this is that the interface between the bodies is a “body” of zero mass, and

therefore can have no net force acting upon it otherwise its acceleration would be innite.

System b) is a bit more complex, primarily because the forces between one mass and the mass above it

are shear forces and must be supplied by friction. Assuming that the friction coefficient is large enough to

accelerate the three masses an equal amount given by a = F/ (3m), by the reasoning we have discussed, the

force balance is as sketched in b) : equal and opposite normal forces F 2/3 on the vertical surfaces, and

equal and

opposite

shear

forces

F

1/3

on

the

horizontal

surfaces.

Rectilinear Motion

In many case we can get an exact expression for the position of a particle as a function of time. We start by

considering the simple motion of a particle along a straight line. The position of particle A at any instant

can be specied by the coordinate s with origin at some xed point O.

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The instantaneous velocity is ds

v = = v . ˙ (1)dt

We will be using the “dot” notation, to indicate time derivative, e.g. (˙) ≡ d/dt . Here, a positive v means that the particle is moving in the direction of increasing s, whereas a negative v, indicates that the particle

is moving in the opposite direction. The acceleration is

dv d2 s a = = v̇ = = s̈ . (2)

dt dt 2

The above expression allows us to calculate the speed and the acceleration if s and/or v are given as a

function of t, i.e. s(t) and v(t). In most cases however, we will know the acceleration and then, the velocity

and the position will have to be determined from the above expressions by integration.

Determining the velocity from the acceleration

From a(t)

If the acceleration is given as a function of t , a(t), then the velocity can be determined by simple integration

of equation (2),

t

v(t) = v0 + a(t) dt . (3) t 0

Here, v0 is the velocity at time t 0 , which is determined by the initial conditions.

From a(v)

If the acceleration is given as a function of velocity a(v), then, we can still use equation (2), but in this case

we will solve for the time as a function of velocity, v dv

t(v) = t 0 + . (4) a(v)v 0

Once the relationship t(v) has been obtained, we can, in principle, solve for the velocity to obtain v(t).

A typical example in which the acceleration is known as a function of velocity is when aerodynamic drag

forces are present. Drag forces cause an acceleration which opposes the motion and is typically of the form

a(v) ∝

v2 (the sign “∝

” means proportional to , that is, a(v) = κv 2 for some κ, which is not a function of

velocity).

From a(s)

When the acceleration is given as a function of s then, we need to use a combination of equations (1) and

(2), to solve the problem. From dv dv ds dv

a = dt

= ds ∗

dt

= vds

(5)

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we can write

a ds = v dv . (6)

This equation can now be used to determine v as a function of s,

v 2 (s) = v 2 0

+ 2

s

s 0 a(s) ds . (7)

where, v0 , is the velocity of the particle at point s 0 . Here, we have used the fact that,

v v 2 2 2

v dv = d( v2

) = v2 −

v2

0 . v 0 v0

A classical example of an acceleration dependent on the spatial coordinate s, is that induced by a deformed

linear spring. In this case, the acceleration is of the form a(s) ∝

s.

Of course, when the acceleration is constant, any of the above expressions (3, 4, 7), can be employed. In

this case we obtain,

v = v0 + a(t − t 0 ), or v 2 = v0

2 + 2 a(s − s0 ) .

If a = g, this reduces to the familiar

v = v0 + g(t − t 0 ), or v 2 = v0

2 + 2 g(s − s 0 ) .

Determining the position from the velocity

Once we know the velocity, the position can be found by integrating ds = vdt from equation (1). Thus,

when the velocity is known as a function of time we have,

t

s = s0 + v(t) dt . (8) t 0

If the velocity is known as a function of position, then

s ds t = t 0 + . (9)

v(s)s 0

Here, s0 is the position at time t 0 .

It is worth pointing out that equation (6), can also be used to derive an expression for v(s), given a(v),

s v v s − s 0 = ds = a(v) dv . (10) s 0 v0

This equation can be used whenever equation (4) is applicable and gives v(s) instead of t(v). For the case

of constant acceleration, either of equations (8, 9), can be used to obtain,

1 s = s0 + v0 (t − t 0 ) + a(t − t 0 )

2 .2

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In many practical situations, it may not be possible to carry out the above integrations analytically in which

case, numerical integration is required. Usually, numerical integration will also be required when either the

velocity or the acceleration depend on more than one variable, i.e. v(s, t ), or, a(s, v ).

Example

Reentry, Ballistic

Coefficient,

Terminal

Velocity

Terminal Velocity

Terminal velocity occurs when the acceleration becomes zero and the velocity Consider an air-dropped

payload starting from rest. The force on the body is a combination of gravity and air drag and has the form

1 F = mg − ρv

2 (11)∗

C D ∗ A

2

Applying Newton’s law and solving for the acceleration a we obtain

a = g − 21

ρv2 ∗

C D m ∗

A (12)

The quantity C D m ∗ A characterizes the combined effect of body shape and mass on the acceleration; it is an

important parameter in the study of reentry; it is called the Ballistic Coefficient. It is dened as β = C D m ∗ A .

Unlike many coefficients that appear in aerospace problems, the Ballistic Coefficient is not non-dimensional,

but has units of mass/length 2 or kg/meters 2 in mks units. Also, in some applications, the ballistic coefficient

is dened as the inverse, B = C D m ∗ A , so it pays to be careful in its application. Equation 13 then becomes

1 a = g − ρv

2

∗ /β (13)

2

Terminal velocity occurs when the force of gravity equals the drag on the object resulting in zero acceleration.

This balance gives the terminal velocity as

vterminal = 2gm

C D Aρ =

2gβ ρ

(14)

For the Earth, atmospheric density at sea level is ρ = 1 .225kg/m 3 ; we shall deal with the variation of

atmospheric density with altitude when we consider atmospheric reentry of space vehicles. Typical value of

β range from β = 1 (Assuming C D = .5, a tennis ball has β = 35.) to β = 1000 for a reentry vehicle. As an

example, consider a typical case of β = 225, where the various parameters then give the following expression

for the acceleration

a = g − 0.002725v 2 m/s .

Here g = 9 .81m/s 2 , is the acceleration due to gravity and v is the downward velocity. It is clear from this

expression that initially the acceleration will be g. Therefore, the velocity will start to increase and keep on

increasing until a = 0, at which point the velocity will stay constant. The terminal velocity is then given by,

0 = g − 0.002725 v 2 or vf = 60m/s .f

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�

To determine the velocity as a function of time, the acceleration can be re-written introducing the terminal

velocity as, a = g(1 − (v/v f )2 ) . We then use expression (4), and write

1 v dv 1 v 1/ 2 1/ 2 vf vf + v t = = ( + ) d v = ln . g 0 1 − (v/v f )

2 g 0 1 + ( v/v f ) 1 − (v/v f ) 2g vf − v

Solving for v we obtain, 2 gt/v f v = vf

e2 gt/v f

− 1 m/s . e + 1

We can easily verify that for large t, v = vf . We can also nd out how long does it take for the payload to

reach, say, 95% of the terminal velocity,

vf 1.95 t = ln = 11 .21s .2g 0.05

To obtain an expression for the velocity as a function of the traveled distance we can use expression (10)

and write

v 21 v dv vs = g 1 − (v/v f )

2 = − 2f

g ln(1 − (v/v f )2

) . 0

Solving for v we obtain

v = vf 1 − e− 2 gs/v 2 m/s .f

We see that for, say, v = 0 .95vf , s = 427.57m. This is the distance traveled by the payload in 11.21s,

which can be compared with the distance that would be traveled in the same time if we were to neglect air

resistance, sno drag = gt 2 / 2 = 615.75m.

Example Spring-mass system

Here, we consider a mass allowed to move without friction on a horizontal slider and subject to the force

exerted by a linear spring. Initially the system is in equilibrium (no force on the spring) at s = 0. Suddenly,

the mass is given a velocity v0 and then the system is left free to oscillate. We know that the effect of the

spring is to cause an acceleration to the body, opposing the motion, of the form a = −κs , where κ > 0 is a constant.

Using equation (7), we have

v 2 = v02

− κs2 .

The displacement can now be obtained using expression (9),

s ds 1 √ κs t =

= arcsin ,

2 0 v0 − κs

2 √ κ v0

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which gives,

s = √ v0

κ sin √ κt .

Finally, the velocity as a function of time is simply, v = v0 cos √ κt . We recognize this motion as that of an undamped harmonic oscillator.

References

[1] R. Dugas, A History of Mechanics , Dover, 1988.

ADDITIONAL READING


1/1, 1/2, 1/3, 1/4, 1/5 (Effect or Altitude only), 1/6, 1/7, 3/2

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S. Widnall 16.07 Dynamics

Fall 2009

Lecture notes based on J. Peraire Version 2.0

Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and dening appropriate operations between them, physical laws can often be written in

a simple form. Since we will making extensive use of vectors in Dynamics, we will summarize some of their

important properties.

Vectors

For our purposes we will think of a vector as a mathematical representation of a physical entity which has

both magnitude and direction in a 3D space. Examples of physical vectors are forces, moments, and velocities. Geometrically, a vector can be represented as arrows. The length of the arrow represents its magnitude.

Unless indicated otherwise, we shall assume that parallel translation does not change a vector, and we shall

call the vectors satisfying this property, free vectors . Thus, two vectors are equal if and only if they are

parallel, point in the same direction, and have equal length.

Vectors are usually typed in boldface and scalar quantities appear in lightface italic type, e.g. the vector

quantity A has magnitude, or modulus, A = |A |. In handwritten text, vectors are often expressed using the

arrow, or underbar notation, e.g. −A , A.→

Vector Algebra

Here, we introduce a few useful operations which are dened for free vectors.

Multiplication by a scalar

If we multiply a vector A by a scalar α , the result is a vector B = αA , which has magnitude B = |α |A. The

vector B , is parallel to A and points in the same direction if α > 0. For α < 0, the vector B is parallel to

A but points in the opposite direction (antiparallel).

If we multiply an arbitrary vector, A , by the inverse of its magnitude, (1/A ), we obtain a unit vector which

ˆis parallel to A . There exist several common notations to denote a unit vector, e.g. A , eA , etc. Thus, we

have that Â = A /A = A / | | , and A = A ˆ |A | = 1. A A, ˆ

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Vector addition

Vector addition has a very simple geometrical interpretation. To add vector B to vector A , we simply place

the tail of B at the head of A . The sum is a vector C from the tail of A to the head of B . Thus, we write

C = A + B . The same result is obtained if the roles of A are reversed B . That is, C = A + B = B + A .

This commutative property is illustrated below with the parallelogram construction.

Since the result of adding two vectors is also a vector, we can consider the sum of multiple vectors. It can

easily be veried that vector sum has the property of association, that is,

(A + B ) + C = A + ( B + C ).

Vector subtraction

Since A − B = A + ( − B ), in order to subtract B from A , we simply multiply B by − 1 and then add.

Scalar product (“Dot” product)

This product involves two vectors and results in a scalar quantity. The scalar product between two vectors

A and B , is denoted by A B , and is dened as ·

A B = AB cos θ . ·

Here θ, is the angle between the vectors A and B when they are drawn with a common origin.

We note that, since cos θ = cos(− θ), it makes no difference which vector is considered rst when measuring

the angle θ. Hence, A B = B A . If A B = 0, then either A = 0 and/or B = 0, or, A and B are· · ·

orthogonal, that is, cos θ = 0. We also note that A A = A2 . If one of the vectors is a unit vector, say ·

B = 1, then A B̂ = A cos θ, is the projection of vector A along the direction of B̂ .·

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Exercise

Using the denition of scalar product, derive the Law of Cosines which says that, for an arbitrary triangle

with sides of length A, B , and C , we have

C 2 = A2 + B 2 − 2AB cos θ .

Here, θ is the angle opposite side C . Hint : associate to each side of the triangle a vector such that C = A − B ,

and expand C 2 = C C .·

Vector product (“Cross” product)

This product operation involves two vectors A and B , and results in a new vector C = A × B . The magnitude

of C

is

given

by,

C = AB sin θ ,

where θ is the angle between the vectors A and B when drawn with a common origin. To eliminate ambiguity,

between the two possible choices, θ is always taken as the angle smaller than π. We can easily show that C

is equal to the area enclosed by the parallelogram dened by A and B .

The vector C is orthogonal to both A and B , i.e. it is orthogonal to the plane dened by A and B . The

direction of C is determined by the right-hand rule as shown.

From this denition, it follows that

B × A = − A × B ,

which indicates that vector multiplication is not commutative (but anticommutative). We also note that if

A × B = 0, then, either A and/or B are zero, or, A and B are parallel, although not necessarily pointing

in the same direction. Thus, we also have A × A = 0.

Having dened vector multiplication, it would appear natural to dene vector division. In particular, we

could say that “A divided by B ”, is a vector C such that A = B × C . We see immediately that there are a

number of difficulties with this denition. In particular, if A is not perpendicular to B , the vector C does

not exist. Moreover, if A is perpendicular to B then, there are an innite number of vectors that satisfy

A = B × C . To see that, let us assume that C satises, A = B × C . Then, any vector D = C + β B , for

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any scalar β , also satises A = B × D , since B × D = B × (C + β B ) = B × C = A . We conclude therefore,

that vector division is not a well dened operation.

Exercise

Show that |A × B | is the area of the parallelogram dened by the vectors A and B , when drawn with a common origin.

Triple product

Given three vectors A , B , and C , the triple product is a scalar given by A (B × C ). Geometrically, the ·

triple product can be interpreted as the volume of the three dimensional parallelepiped dened by the three

vectors A , B and C .

It can be easily veried that A (B × C ) = B (C × A ) = C (A × B ).· · ·

Exercise

Show that A (B × C ) is the volume of the parallelepiped dened by the vectors A , B , and C , when drawn ·

with a common origin.

Double vector product

The double vector product results from repetition of the cross product operation. A useful identity here is,

A × (B × C ) = ( A C )B − (A B )C .· ·

Using this identity we can easily verify that the double cross product is not associative, that is,

= ( A × B ) × C .A × (B × C )

Vector Calculus

Vector differentiation and integration follow standard rules. Thus if a vector is a function of, say time, then

its derivative with respect to time is also a vector. Similarly the integral of a vector is also a vector.

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Derivative of a vector

Consider a vector A (t) which is a function of, say, time. The derivative of A with respect to time is dened

as,

dA = lim A (t + ∆ t) − A (t) . (1)dt ∆ t 0 ∆ t→

A vector has magnitude and direction, and it changes whenever either of them changes. Therefore the rate

of change of a vector will be equal to the sum of the changes due to magnitude and direction.

Rate of change due to magnitude changes

When a vector only changes in magnitude from A to A + dA , the rate of change vector dA is clearly parallel

to the original vector A .

Rate of change due to direction changes

Let us look at the situation where only the direction of the vector changes, while the magnitude stays

constant. This is illustrated in the gure where a vector A undergoes a small rotation. From the sketch, it

is clear that if the magnitude of the vector does not change, dA is perpendicular to A and as a consequence,

the derivative of A , must be perpendicular to A . (Note that in the picture dA has a nite magnitude and therefore, A and dA are not exactly perpendicular. In reality, dA has innitesimal length and we can see

that when the magnitude of dA tends to zero, A and dA are indeed perpendicular ).

ββ.

An alternative, more mathematical, explanation can be derived by realizing that even if A changes but its modulus stays constant, then the dot product of A with itself is a constant and its derivative is therefore

zero. A A = constant . Differentiating, we have that, ·

dA A + A dA = 2 A dA = 0 ,· · ·

which shows that A , and dA , must be orthogonal.

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��

Suppose that A is instantaneously rotating in the plane of the paper at a rate β ̇ = dβ/dt , with no change in

˙magnitude. In an instant dt, A , will rotate an amount dβ = βdt and the magnitude of dA , will be

dA = |dA | = Adβ = A β̇dt .

Hence, the

magnitude

of

the

vector

derivative

is

dA dt

= A β̇ .

In the general three dimensional case, the situation is a little bit more complicated because the rotation

of the vector may occur around a general axis. If we express the instantaneous rotation of A in terms of

an angular velocity Ω (recall that the angular velocity vector is aligned with the axis of rotation and the

direction of the rotation is determined by the right hand rule), then the derivative of A with respect to time

is simply,dA

= Ω × A . (2)dt

constant magnitude

To see that, consider a vector A rotating about the axis C − C with an angular velocity Ω. The derivative

will be the velocity of the tip of A . Its magnitude is given by lΩ, and its direction is both perpendicular to

A and to the axis of rotation. We note that Ω × A has the right direction, and the right magnitude since

l = A sin ϕ.

x

Expression (2) is also valid in the more general case where A is rotating about an axis which does not pass

through the origin of A . We will see in the course, that a rotation about an arbitrary axis can always be

written as a rotation about a parallel axis plus a translation, and translations do not affect the magnitude

not the direction of a vector.

We can now go back to the general expression for the derivative of a vector (1) and write dA dA dA dA

= + = + Ω × A . dt dt dt dt constant direction constant magnitude constant direction

Note that (dA /dt ) is parallel to A and Ω × A is orthogonal to A . The gure below shows constant direction

the general differential of a vector, which has a component which is parallel to A , dA , and a component

which is orthogonal to A , dA The magnitude change is given by dA , and the direction change is given ⊥.

by dA ⊥.

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Rules for Vector Differentiation

Vector differentiation follows similar rules to scalars regarding vector addition, multiplication by a scalar,

and products. In particular we have that, for any vectors A , B , and any scalar α ,

d(α A ) = dα A + αd A

d(A + B ) = dA + dB

d(A B ) = dA B + A dB· · ·

d(A ×

B ) =

dA

×

B

+

A

×

dB

.

Components of a Vector

We have seen above that it is possible to dene several operations involving vectors without ever introducing

a reference frame. This is a rather important concept which explains why vectors and vector equations are

so useful to express physical laws, since these, must be obviously independent of any particular frame of

reference.

In practice however, reference frames need to be introduced at some point in order to express, or measure,

the direction and magnitude of vectors, i.e. we can easily measure the direction of a vector by measuring the angle that the vector makes with the local vertical and the geographic north.

Consider a right-handed set of axes xyz , dened by three mutually orthogonal unit vectors i , j and k

(i × j = k ) (note that here we are not using the hat (̂ ) notation). Since the vectors i , j and k are mutually

orthogonal they form a basis. The projections of A along the three xyz axes are the components of A in the

xyz reference frame.

In order to determine the components of A , we can use the scalar product and write,

Ax = A i , Ay = A j , Az = A k .· · ·

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��

��

��

��

The vector A , can thus be written as a sum of the three vectors along the coordinate axis which have

magnitudes Ax , Ay , and Az and using matrix notation, as a column vector containing the component

magnitudes.

Ax

Ay

A = A

x + A

y + A

z = A

xi + A

y j + A

z k = .

Az

Vector operations in component form

The vector operations introduced above can be expressed in terms of the vector components in a rather

straightforward manner. For instance, when we say that A = B , this implies that the projections of A and

B along the xyz axes are the same, and therefore, this is equivalent to three scalar equations e.g. Ax = Bx ,

Ay = By , and Az = B z . Regarding vector summation, subtraction and multiplication by a scalar, we have

that, if C = αA + β B , then,

C x = αA x + βB x , C y = αA y + βB y , C z = αA z + βB z .

Scalar product

Since i i = j j = k k = 1 and that i j = j k = k i = 0, the scalar product of two vectors can be · · · · · ·

written as,

A B = ( Ax i + Ay j + Az k ) (Bx i + By j + B z k ) = Ax Bx + Ay By + Az B z .· ·

Note that, A A = A2 = Ax 2 + Ay 2

+ Az 2 , which is consistent with Pythagoras’ theorem. ·

Vector product

Here, i × i = j × j = k × k = 0 and i × j = k , j × k = i , and k × i = j . Thus,

A × B = ( Ax i + Ay j + Az k ) × (Bx i + By j + B z k )

= ( Ay B z − Az By )i + ( Az Bx − Ax B z ) j + ( Ax By − Ay Bx )k = i j k

Ax Ay Az

Bx By B z

.

Triple product

The triple product A (B × C ) can be expressed as the following determinant ·

A (B × C ) = · Ax Ay Az

Bx By B z ,

C x C y C z

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which clearly is equal to zero whenever the vectors are linearly dependent (if the three vectors are linearly

dependent they must be co-planar and therefore the parallelepiped dened by the three vectors has zero

volume).

Vector Transformations

In many problems we will need to use different coordinate systems in order to describe different vector

quantities. The above operations, written in component form, only make sense once all the vectors involved

are described with respect to the same frame. In this section, we will see how the components of a vector

are transformed when we change the reference frame.

Consider two different orthogonal, right-hand sided, reference frames x1 , x 2 , x 3 and X 1 , X 2 , X 3 . A vector A

in coordinate system x can be transformed to coordinate system X’ by considering the 9 angles that dene

the relationships between the two systems. (Only three of these angles are independent, a point we shall

return to later.)

Referring to a) in the gure we see the vector A , the x and X’ coordinate systems, the unit vectors i1 , i2 , i3 of

the x system and the unit vectors i1 , i2 , i3 of the X’ system; a) focuses on the transformation of coordinates

from x to X’ while b) focuses on the ”reverse” transformation from X’ to x .

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In the x coordinate system, the vector A , can be written as

A = A1 i 1 + A2 i 2 + A3 i 3 , (3)

or, when referred to the frame X’ , as

A = A1 i 1 + A2 i 2 + A3 i 3 . (4)

Since the vector A remains the same regardless of our coordinate transformation

A = A1 i 1 + A2 i 2 + A3 i 3 = A1 i 1 + A2 i 2 + A3 i 3 , (5)

We can nd the components of the vector A in the transformed system in term of the components of A in

the original system by simply taking the dot product of this equation with the desired unit vector i j in the

X’ system so that

Aj = A1 i j · i 1 + A2 i j · i 2 + A3 i j · i 3 (6)

where Aj is the jth component of A in the X’ system. Repeating this operation for each component of A

in the X’ system results in the matrix form for A

A1

A2 =

i i 1 i i 2 i1 · 1 · 1 · i 3

A1

A2 .i i 1 i i 2 i2 · 2 · 2 · i 3

A3 i 3 · i 1 i 3 · i 2 i 3 · i 3 A3

The above expression is the relationship that expresses how the components of a vector in one coordinate

system relate to the components of the same vector in a different coordinate system.

Referring to the gure, we see that i j ·i i is equal to the cosine of the angle between i�

j and i i which is θj �i ; in

particular we see that i i 1 = cosθ21 while i i 2 = cosθ12 ; these angles are in general not equal. Therefore, 2 · 1 ·

the components of the vector A are transformed from the x coordinate system to the X’ system through

the transformation

Aj = A1 cos θj 1 + A2 cos θj 2 + A3 cos θj 3 . (7)

where the coefficients relating the components of A in the two coordinate systems are the various direction

cosines of the angles between the coordinate directions.

The above relations for the transformation of A from the x to the X’ system can be written in matrix form as

A1

A2 =

cos(θ11 ) cos (θ12 ) cos (θ13 )

cos(θ21 ) cos (θ22 ) cos (θ23 )

A1

A2 A = . (8)

A3 cos(θ31 ) cos (θ32 ) cos (θ33 ) A3

We use the symbol A’ to denote the components of the vector A in the ’ system. Of course the vector A is

unchanged by the transformation. We introduce the symbol [T ] for the transformation matrix from x to X’ .

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This relationship, which expresses how the components of a vector in one coordinate system relate to the

components of the same vector in a different coordinate system, is then written

A’ = [ T ]A . (9)

where [T ] is the transformation matrix.

We now consider the process that transforms the vector A’ from the X’ system to the x system.

A =

A1

A2 =

cos (Θ11 ) cos (Θ 12 ) cos (Θ 13 )

cos (Θ21 ) cos (Θ 22 ) cos (Θ 23 )

A1

A2 . (10)

A3 cos (Θ31 ) cos (Θ 32 ) cos (Θ 33 ) A3

By comparing the two coordinate transformations shown in a) and b), we see that cos(θ12 )=cos(Θ 21 ), and

that therefore the matrix element of magnitude cos(θ12 ) which appears in the 12 position in the transfor

mation matrix from x to X’ now appears in the 21 position in the matrix which transforms A from X’ to

x . This pattern is repeated for all off-diagonal elements. The diagonal elements remain unchanged since

cos(θii )=cos(Θ ii ). Thus the matrix which transposes the vector A in the X’system back to the x system is

the transpose of the original transformation matrix,

A = [ T ]T A’ . (11)

where [T ]T is the transpose of [T ]. (A transpose matrix has the rows and columns reversed.)

Since transforming A from x to X’ and back to x results in no change, the matrix [T ]T is also [T ]− 1 the

inverse of [T ] since

A = [ T ]T [T ]A = [ T ]− 1 [T ]A = [ I]A = A . (12)

where [I ] is the identity matrix

I =

1 0 0

0 1 0

0 0 1

(13)

This is a remarkable and useful property of the transformation matrix, which is not true in general for any

matrix.

Example Coordinate transformation in two dimensions

Here, we apply for illustration purposes, the above expressions to a two-dimensional example. Consider the

change of coordinates between two reference frames xy, and x y , as shown in the diagram.

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The angle between i and i

is γ . Therefore, i i

= cos γ . Similarly, j i

= cos(π/ 2 − γ ) = sin γ ,· ·

i j = cos(π/ 2 + γ ) = − sin γ , and j j

= cos γ . Finally, the transformation matrix [T ] is · ·

[T ] = cos (θ11 )

cos (θ21 ) cos (θ12 )

cos (θ22 ) =

cos γ

− sin γ sin γ

cos γ ,

and we can write, A1 = [T ]

A1 . A2 A2

and

A1 = [ T ]T A1 .

A2 A2

Therefore,

A1 = A1 cos γ + A2 sin γ (14)

A2 = − A1 sin γ + A2 cos γ. (15)

For instance, we can easily check that when γ = π/ 2, the above expressions give A1 = A2 , and A2 = − A1 ,

as expected.

An additional observation can be made. If in three dimensions, we rotate the x, y, z coordinate system about

the z axis, as shown in a) leaving the z component unchanged,

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the transformation matrix becomes

[T ] =

cos (θ11 ) cos (θ12 ) 0 cos(θ21 ) cos (θ22 ) 0 =

cos γ sin γ 0 − sin γ cos γ 0 .

0 0 1 0 0 1

Analogous results can be obtained for rotation about the x axis or rotation about the y axis as shown in b)

and c) .

Sequential Transformations; Euler Angles

The general orientation of a coordinate system can be described by a sequence of rotations about coordinate

axis. One particular set of such rotations leads to a description particularly convenient for describing the

motion of a three-dimensional rigid body in general spinning motion, call Euler angles. We shall treat

this topic in Lecture 28. For now, we examine how this rotation ts into our general study of coordinate

transformations. A coordinate description in terms of Euler angles is obtained by the sequential rotation of

axis as shown in the gure; the order of transformation makes a difference.

To develop the description of this motion, we use a series of transformations of coordinates. The nal result is

shown below. This is the coordinate system used for the description of motion of a general three-dimensional

rigid body such as a top described in body-xed axis. To identify the new position of the coordinate axes as a result of angular displacement through the three Euler angles, we go through a series of coordinate

rotations.

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First, we rotate from an initial X,Y,Z system into an x , y , z system through a rotation φ about the Z, z

axis.

0

x

y =

cosφ

sinφ

X

X

= [ T 1 ]

0− sinφ cosφ Y Y .

z 0 0 1 Z Z

The resulting x , y coordinates remain in the X, Y plane. Then, we rotate about the x axis into the

x , y , z system through an angle θ. The x axis remains coincident with the x axis. The axis of rotation

for this transformation is called the ”line of nodes”. The plane containing the x , y coordinate is now tipped

through an angle θ relative to the original X, Y plane. coordinates

x

y

=

1 0 0

0

cosθ

sinθ

x

y = [ T 2 ]

x

y

.

z 0 − sinθ cosθ z z

And nally, we rotate about the z , z system through an angle ψ into the x,y,z system. The z axis is

called the spin axis. It is coincident with the z axis.

0

x

y = [ T 3 ]

x

y .

cosψ sinψ x

= 0− sinψ cosψ y

z 0 0 1 z z

The nal coordinate system used to describe the position of the body is shown below. The angle ψ is called

the spin; the angle φ is called the precession; the angle θ is called the nutation. The total transformation is

given by X x

= [ T 3 ][T 2 ][T 1 ]

Y y .

z Z

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Euler angles are not always dened in exactly this manner, either the notation or the order of rotations can

differ. The particular transformation used in any example should be clearly described.

References

[1] J.B. Marion and S.T. Thornton, Classical Dynamics of Particles and Systems , Harcourt Brace, 1995.

[2] D. Kleppner and R.J. Kolenkow, An Introduction to Mechnics , McGraw Hill, 1973.

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S. Widnall, J. Peraire 16.07 Dynamics

Fall 2009 Version 2.0

Lecture L4 - Curvilinear Motion. Cartesian Coordinates

We will start by studying the motion of a particle . We think of a particle as a body which has mass,

but has negligible dimensions. Treating bodies as particles is, of course, an idealization which involves an

approximation. This approximation may be perfectly acceptable in some situations and not adequate in

some other cases. For instance, if we want to study the motion of planets, it is common to consider each

planet as a particle. This simplication is not adequate if we wish to study the precession of a gyroscope or

a spinning top.

Kinematics of curvilinear motion

In dynamics we study the motion and the forces that cause, or are generated as a result of, the motion.

Before we can explore these connections we will look rst at the description of motion irrespective of the

forces that produce them. This is the domain of kinematics . On the other hand, the connection between

forces and motions is the domain of kinetics and will be the subject of the next lecture.

Position vector and Path

We consider the general situation of a particle moving in a three dimensional space. To locate the position of

a particle in space we need to set up an origin point, O, whose location is known. The position of a particle

A, at time t, can then be described in terms of the position vector , r , joining points O and A. In general,

this particle will not be still, but its position will change in time. Thus, the position vector will be a function

of time, i.e. r (t). The curve in space described by the particle is called the path , or trajectory .

We introduce the path or arc length coordinate , s, which measures the distance traveled by the particle along

the curved path. Note that for the particular case of rectilinear motion (considered in the review notes) the

arc length coordinate and the coordinate, s, are the same.

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Using the path coordinate we can obtain an alternative representation of the motion of the particle. Consider

that we know r as a function of s, i.e. r (s), and that, in addition we know the value of the path coordinate

as a function of time t, i.e. s(t). We can then calculate the speed at which the particle moves on the path

simply as v = ṡ ≡ ds/dt . We also compute the rate of change of speed as a t = s̈ = d2 s/dt 2 .

We consider below some motion examples in which the position vector is referred to a xed cartesian coordinate system.

Example Motion along a straight line in 2D

Consider for illustration purposes two particles that move along a line dened by a point P and a unit vector

m . We further assume that at t = 0, both particles are at point P . The position vector of the rst particle is

given by r 1 (t) = r P + m t = ( r P x + mx t)i + ( r P y + m y t) j , whereas the position vector of the second particle

is given by r 2 (t) = r P + m t 2 = ( r P x + m x t2 )i + ( r P y + m y t2 ) j .

Clearly the path for these two particles is the same, but the speed at which each particle moves along the

path is different. This is seen clearly if we parameterize the path with the path coordinate, s. That is,

we write r (s) = r P + m s = ( r P x + m x s)i + ( r P y + my s) j . It is straightforward to verify that s is indeed

the path coordinate i.e. the distance between two points r (s) and r (s + ∆ s) is equal to ∆ s. The two motions introduced earlier simply correspond to two particles moving according to s1 (t) = t and s 2 (t) = t 2 ,

respectively. Thus, r 1 (t) = r (s1 (t)) and r 2 (t) = r (s2 (t)).

It turns out that, in many situations, we will not have an expression for the path as a function of s. It is

in fact possible to obtain the speed directly from r (t) without the need for an arc length parametrization of

the trajectory.

Velocity Vector

We consider the positions of the particle at two different times t and t + ∆ t, where ∆ t is a small increment

of time. Let ∆ r = r (r + ∆ t) − r (t), be the displacement vector as shown in the diagram.

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The average velocity of the particle over this small increment of time is

∆ r v ave = ,∆ t

which is a vector whose direction is that of ∆ r and whose magnitude is the length of ∆ r divided by ∆ t. If

∆ t is small, then ∆ r will become tangent to the path, and the modulus of ∆ r will be equal to the distance

the particle has moved on the curve ∆ s.

The instantaneous velocity vector is given by

v = lim ∆ r dr (t)

≡ ṙ , (1)∆ t 0 ∆ t ≡ dt→and is always tangent to the path. The magnitude, or speed, is given by

∆ s ds v = v = lim s . | | ∆ t → 0 ∆ t ≡ dt ≡ ˙

Acceleration Vector

In an analogous manner, we can dene the acceleration vector. Particle A at time t, occupies position

r (t), and has a velocity v (t), and, at time t + ∆ t, it has position r (t + ∆ t) = r (t) + ∆ r , and velocity

v (t + ∆ t) = v (t) + ∆ v . Considering an innitesimal time increment, we dene the acceleration vector as the

derivative of the velocity vector with respect to time,

∆ v dv d2 r a = lim = . (2)

∆ t 0 ∆ t ≡ dt dt 2 →

We note that the acceleration vector will reect the changes of velocity in both magnitude and direction.

The acceleration vector will, in general, not be tangent to the trajectory (in fact it is only tangent when the

velocity vector does not change direction). A sometimes useful way to visualize the acceleration vector is to

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translate the velocity vectors, at different times, such that they all have a common origin, say, O�. Then,

the heads of the velocity vector will change in time and describe a curve in space called the hodograph . We

then see that the acceleration vector is, in fact, tangent to the hodograph at every point.

Expressions (1) and (2) introduce the concept of derivative of a vector . Because a vector has both magnitude

and direction, the derivative will be non-zero when either of them changes (see the review notes on

vectors) . In general, the derivative of a vector will have a component which is parallel to the vector itself,

and is due to the magnitude change ; and a component which is orthogonal to it, and is due to the direction

change .

Note Unit tangent and arc-length parametrization

The unit tangent vector to the curve can be simply calculated as

e t = v /v.

It is clear that the tangent vector depends solely on the geometry of the trajectory and not on the speed

at which the particle moves along the trajectory. That is, the geometry of the trajectory determines the

tangent vector, and hence the direction of the velocity vector. How fast the particle moves along the

trajectory determines the magnitude of the velocity vector. This is clearly seen if we consider the arc-length

parametrization of the trajectory r (s). Then, applying the chain rule for differentiation, we have that,

dr dr ds v = = = e t v , dt ds dt

where, ṡ = v, and we observe that dr /ds = e t . The fact that the modulus of dr /ds is always unity indicates

that the distance traveled, along the path, by r (s), (recall that this distance is measured by the coordinate

s), per unit of s is, in fact, unity!. This is not surprising since by denition the distance between two

neighboring points is ds, i.e. |dr | = ds.

Cartesian Coordinates

When working with xed cartesian coordinates, vector differentiation takes a particularly simple form. Since

the vectors i , j , and k do not change, the derivative of a vector A (t) = Ax (t)i + Ay (t) j + Az (t)k , is simply

Ȧ (t) = Ȧ x (t)i + Ȧ y (t) j + Ȧ z (t)k . That is, the components of the derivative vector are simply the derivatives

of the components.

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Thus, if we refer the position, velocity, and acceleration vectors to a xed cartesian coordinate system, we

have,

r (t) = x(t)i + y(t) j + z(t)k (3)

v (t) = vx (t)i + vy (t) j + vz (t)k = ẋ (t)i + ẏ(t) j + ż(t)k = ṙ (t) (4)

a (t) = ax (t)i + ay (t) j + az (t)k = v̇x (t)i + v̇y (t) j + v̇z (t)k = v̇ (t) (5)

Here, the speed is given by v = vx 2 + vy 2

+ vz 2 , and the magnitude of the acceleration is a = a2

x + a 2

y + a 2

z .

The advantages of cartesian coordinate systems is that they are simple to use, and that if a is constant, or

a function of time only, we can integrate each component of the acceleration and velocity independently as

shown in the ballistic motion example.

Example Circular Motion

We consider motion of a particle along a circle of radius R at a constant speed v0 . The parametrization of

a circle in terms of the arc length is

s s r (s) = R cos( )i + R sin( ) j .

R R

Since we have a constant speed v0 , we have s = v0 t . Thus,

r (t) = R cos( v0 t

)i + R sin( v0 t

) j . R R

The velocity is

v (t) = dr (t)

= −v0 sin( v0 t

)i + v0 cos( v0 t

) j ,dt R R

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�

�

�

which, clearly, has a constant magnitude |v | = v0 . The acceleration is, a (t) =

dr (t)= −

v02

cos( v0 t

)i − v02

sin( v0 t

) j . dt R R R R

Note that, the acceleration is perpendicular to the path (in this case it is parallel to r ), since the velocity

vector changes direction, but not magnitude.

We can also verify that, from r (s), the unit tangent vector, e t , could be computed directly as

e t = dr (s)

= − sin( s

)i + cos( s

) = − sin( v0 t

)i + cos( v0 t

) j . ds R R R R

Example Motion along a helix

The equation r (t) = R cos ti + R sin t j + ht k , denes the motion of a particle moving on a helix of radius R,

and pitch 2πh , at a constant speed. The velocity vector is given by

dr

v = = −R sin t i + R cos t j + hk ,dt and the acceleration vector is given by,

dv a = = −R cos ti + −R sin t j . dt

In order to determine the speed at which the particle moves we simply compute the modulus of the velocity

vector,

v = |v | = R 2 sin2 t + R 2 cos2 t + h2 = R 2 + h2 .

If we want to obtain the equation of the path in terms of the arc-length coordinate we simply write,

ds = |dr | = vdt = R 2 + h2 dt . Integrating, we obtain s = s0 + √ R 2 + h2 t , where s 0 corresponds to the path coordinate of the particle at time zero. Substituting t in terms of s, we obtain the expression for the position vector in terms of the

arc-length coordinate. In this case, r (s) = R cos(s/ √ R 2 + h2 )i + R sin(s/ √ R 2 + h2 ) j + hs/ √ R 2 + h2 k . The gure below shows the particle trajectory for R = 1 and h = 0 .1.

10.5

00.5

1

10.5

00.5

10

0. 5

1

1. 5

2

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Example Ballistic Motion

Consider the free-ight motion of a projectile which is initially launched with a velocity v 0 = v0 cos φi +

v0 sin φ j . If we neglect air resistance, the only force on the projectile is the weight, which causes the projectile

to have a constant acceleration a = −g j . In component form this equation can be written as dvx /dt = 0 and dvy /dt = −g. Integrating and imposing initial conditions, we get

vx = v0 cos φ, vy = v0 sin φ − gt ,

where we note that the horizontal velocity is constant. A further integration yields the trajectory

x = x0 + ( v0 cos φ) t, y = y0 + ( v0 sin φ) t − 21

gt 2 ,

which we recognize as the equation of a parabola.

The maximum height, ymh , occurs when vy (tmh ) = 0, which gives tmh = ( v0 /g )sin φ, or,

v02

sin2

φ ymh = y0 + .2g

The range, x r , can be obtained by setting y = y0 , which gives t r = (2v0 /g )sin φ, or,

2v02 sin φ cos φ v02

sin(2φ) x r = x0 + = x0 + . g g

We see that if we want to maximize the range x r , for a given velocity v0 , then sin(2φ) = 1, or φ = 45o .

Finally, we note that if we want to model a more realistic situation and include aerodynamic drag forces of

the form, say, −κv2 , then we would not be able to solve for x and y independently, and this would make the

problem considerably more complicated (usually requiring numerical integration).

ADDITIONAL READING


2/1, 2/3, 2/4

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S. Widnall, J. Peraire 16.07 Dynamics

Fall 2008

Version 2.0

Lecture L5 - Other Coordinate Systems In this lecture, we will look at some other common systems of coordinates. We will present polar coordinates

in two dimensions and cylindrical and spherical coordinates in three dimensions. We shall see that these

systems are particularly useful for certain classes of problems.

Polar Coordinates (r − θ) In polar coordinates, the position of a particle A, is determined by the value of the radial distance to the

origin, r , and the angle that the radial line makes with an arbitrary xed line, such as the x axis. Thus, the

trajectory of a particle will be determined if we know r and θ as a function of t , i.e. r (t), θ(t). The directions

of increasing r and θ are dened by the orthogonal unit vectors e r and e θ .

The position vector of a particle has a magnitude equal to the radial distance, and a direction determined

by e r . Thus,

r = r e r . (1)

Since the vectors e r and e θ are clearly different from point to point, their variation will have to be considered

when calculating the velocity and acceleration.

Over an innitesimal interval of time dt , the coordinates of point A will change from (r, θ ), to (r + dr , θ + dθ)

as shown in the diagram.

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We note that the vectors e r and e θ do not change when the coordinate r changes. Thus, de r /dr = 0 and

de θ /dr = 0 . On the other hand, when θ changes to θ + dθ, the vectors e r and e θ are rotated by an angle

dθ. From the diagram, we see that de r = dθe θ , and that de θ = −dθe r . This is because their magnitudes in the limit are equal to the unit vector as radius times dθ in radians. Dividing through by dθ, we have,

de r de θ = e θ , and = −e r . dθ dθ Multiplying these expressions by dθ/dt ≡ θ̇, we obtain,

de r dθ de r de θ dθ dt ≡ dt = θ̇e θ , and dt = −θ̇e r . (2)

Note Alternative calculation of the unit vector derivatives

An alternative, more mathematical, approach to obtaining the derivatives of the unit vectors is to express

e r and e θ in terms of their cartesian components along i and j . We have that

e r = cos θi + sin θ j

e θ = − sin θi + cos θ j .

Therefore, when we differentiate we obtain,

de r de r= 0 , = − sin θi + cos θ j ≡ e θdr dθ de θ de θ = 0 , = − cos θi − sin θ j ≡ −e r . dr dθ

Velocity vector

We can now derive expression (1) with respect to time and write

v = ṙ = ṙ e r + r ė r ,

or, using expression (2), we have

v = ṙ e r + rθ̇ e θ . (3)

Here, vr = ṙ is the radial velocity component, and vθ = rθ̇ is the circumferential velocity component. We

also have that v = vr 2 + vθ

2 . The radial component is the rate at which r changes magnitude, or stretches,

and the circumferential component, is the rate at which r changes direction, or swings.

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Acceleration vector

Differentiating again with respect to time, we obtain the acceleration

a = v̇ = r̈ e r + ṙ ė r + ṙθ̇ e θ + rθ ¨ e θ + rθ̇ ė θ

Using the expressions (2), we obtain,

a = (r̈ − rθ̇2 ) e r + ( rθ ¨ + 2 ṙθ̇ ) e θ , (4)

where a r = (r̈ − rθ̇2 ) is the radial acceleration component, and aθ = ( rθ ¨ + 2ṙθ̇ ) is the circumferential

acceleration component. Also, we have that a = a 2 r + a2 θ .

Change of basis

In many practical situations, it will be necessary to transform the vectors expressed in polar coordinates to

cartesian coordinates and vice versa.

Since we are dealing with free vectors, we can translate the polar reference frame for a given point (r, θ ), to

the origin, and apply a standard change of basis procedure. This will give, for a generic vector A ,

Ar = cos θ sin θ Ax and

Ax = cos θ − sin θ Ar .

Aθ − sin θ cos θ Ay Ay sin θ cos θ Aθ

Example Circular motion

Consider as an illustration, the motion of a particle in a circular trajectory having angular velocity ω = θ̇,

and angular acceleration α = ω̇.

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In polar coordinates, the equation of the trajectory is

1 r = R = constant , θ = ωt + αt 2 .

2

The velocity components are

vr = ṙ = 0 , and vθ = rθ̇ = R(ω + αt ) = v ,

and the acceleration components are, 2

a r = r̈ − rθ̇2 = −R(ω + αt )

2 = − vR

, and aθ = rθ ¨ + 2 ṙθ̇ = Rα = a t ,

where we clearly see that, a r ≡ −an , and that aθ ≡ a t . In cartesian coordinates, we have for the trajectory,

1 1 x = R cos(ωt + αt 2 ), y = R sin(ωt + αt 2 ) .

2 2

For the velocity,

1 1 vx = −R(ω + αt )sin( ωt + αt

2 ), vy = R(ω + αt ) cos(ωt + αt2 ) ,

2 2

and, for the acceleration,

1 1 1 1 ax = −R(ω+ αt )

2 cos(ωt+ αt 2 )−Rα sin(ωt+ αt2 ), ay = −R(ω+ αt )

2 sin(ωt+ αt 2 )+ Rα cos(ωt+ αt 2 ) .2 2 2 2

We observe that, for this problem, the result is much simpler when expressed in polar (or intrinsic) coordi

nates.

Example Motion on a straight line

Here we consider the problem of a particle moving with constant velocity v0 , along a horizontal line y = y0 .

Assuming that at t = 0 the particle is at x = 0, the trajectory and velocity components in cartesian

coordinates are simply,

x = v0 t y = y0

vx = v0 vy = 0

ax = 0 ay = 0 .

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In polar coordinates, we have,

r = v02 t 2 + y02 θ = tan − 1 (

y0 )

v0 t

vr = ṙ = v0 cos θ vθ = rθ̇ = −v0 sin θ a

r = r̈

− rθ̇ 2 = 0 a

θ = rθ ¨ + 2 ṙθ̇ = 0 .

Here, we see that the expressions obtained in cartesian coordinates are simpler than those obtained using

polar coordinates. It is also reassuring that the acceleration in both the r and θ direction, calculated from

the general two-term expression in polar coordinates, works out to be zero as it must for constant velocity-

straight line motion.

Example Spiral motion (Kelppner/Kolenkow)

A particle moves with θ̇ = ω = constant and r = r 0 eβt

, where r 0 and β are constants.

We shall show that for certain values of β , the particle moves with a r = 0.

a = (r̈ − rθ̇2 )e r + ( rθ ¨ + 2 ṙθ̇ )e θ

= ( β 2 r 0 e βt − r 0 e βt ω2 )e r + 2 βr 0 ωeβt

e θ

If β = ±ω, the radial part of a vanishes. It seems quite surprising that when r = r 0 eβt , the particle moves with zero radial acceleration. The error is in thinking that r̈ makes the only contribution to a r ; the term

−rθ̇2 is also part of the radial acceleration, and cannot be neglected.

The paradox is that even though a r = 0, the radial velocity vr = ṙ = r 0 βe βt is increasing rapidly in time.

In polar coordinates

�

vr = ar (t)dt ,

because this integral does not take into account the fact that e r and e θ are functions of time.

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Equations of Motion

In two dimensional polar rθ coordinates, the force and acceleration vectors are F = F r e r + F θ e θ and

a = a r e r + aθ e θ . Thus, in component form, we have,

F r =

m a r = m

(r̈

− rθ

˙2)

F θ = m a θ = m (rθ ¨ + 2 ṙθ̇ ) .

Cylindrical Coordinates (r − θ − z ) Polar coordinates can be extended to three dimensions in a very straightforward manner. We simply add

the z coordinate, which is then treated in a cartesian like manner. Every point in space is determined by

the r and θ coordinates of its projection in the xy plane, and its z coordinate.

The unit vectors e r , e θ and k , expressed in cartesian coordinates, are,

e r = cos θi + sin θ j

e θ = − sin θi + cos θ j

and their derivatives,

ė r = θ̇e θ , ė θ = −θ̇e r , k̇ = 0 . The kinematic vectors can now be expressed relative to the unit vectors e r , e θ and k . Thus, the position

vector is

r = r e r + z k ,

and the velocity,

v = ṙ e r + rθ̇ e θ + ż k ,

v2 2 + v2where vr = ṙ , vθ = rθ̇ , vz = ż, and v =

r + vθ z . Finally, the acceleration becomes

a = (r̈ − rθ̇2 ) e r + ( rθ ¨ + 2 ṙθ̇ ) e θ + z̈k ,

2where a r = r̈ − rθ̇2 , aθ = rθ ¨ + 2 ṙθ̇ , az = z̈, and a =

a 2 + a + a 2 .r θ z

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Note that when using cylindrical coordinates, r is not the modulus of r . This is somewhat confusing, but it

is consistent with the notation used by most books. Whenever we use cylindrical coordinates, we will write

|r | explicitly, to indicate the modulus of r , i.e. |r | = √ r 2 + z2 .

Equations of

Motion

In cylindrical rθz coordinates, the force and acceleration vectors are F = F r e r + F θ e θ + F z e z and a =

a r e r + aθ e θ + az e z . Thus, in component form we have,

F r = m a r = m (r̈ − rθ̇2 )

F θ = m a θ = m (rθ ¨ + 2 ṙθ̇ )

F z = m a z z . = m ¨

Spherical Coordinates (r − θ − φ) In spherical coordinates, we utilize two angles and a distance to specify the position of a particle, as in the

case of radar measurements, for example.

The unit vectors written in cartesian coordinates are,

e r = cos θ cos φ i + sin θ cos φ j + sin φ k

e θ = − sin θ i + cos θ j e φ = − cos θ sin φ i − sin θ sin φ j + cos φ k

The derivation of expressions for the velocity and acceleration follow easily once the derivatives of the unit vectors are known. In three dimensions, the geometry is somewhat more involved, but the ideas are the

same. Here, we give the results for the derivatives of the unit vectors,

ė r = θ̇ cos φ e θ + φ̇ e φ , ė θ = −θ̇ cos φ e r + θ̇ sin φ e φ , ė φ = −φ̇ e r − θ̇

sin φ e θ ,

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and for the kinematic vectors

r = r e r

v = ṙ e r + rθ̇ cos φ e θ + rφ̇ e φ

a

=

(r̈

− rθ

˙2 cos

2

φ

− rφ

˙2) e r

+ (2ṙθ̇ cos φ + rθ ¨ cos φ − 2rθ̇ φ̇ sin φ) e θ + (2ṙφ̇ + rφ̇ 2 sin φ cos φ + rφ̈ ) e φ .

Equations of Motion

Finally, in spherical rθφ coordinates, we write F = F r e r + F θ e θ + F φ e φ and a = a r e r + aθ e θ + aφ e φ . Thus,

F r = m a r = m (r̈ − rθ̇2 cos2 φ − rφ̇

2 )

F θ = m a θ = m (2ṙθ̇ cos φ + rθ ¨ cos φ

− 2rθ̇ φ̇ sin φ)

F φ = m a φ = m (2ṙφ̇ + rφ̇2 sin φ cos φ + rφ̈ ) .

Application Examples

We will look at some applications of Newton’s second law, expressed in the different coordinate systems that

have been introduced. Recall that Newton’s second law

F = ma , (5)

is a vector equation which is valid for inertial observers.

In general, we will be interested in determining the motion of a particle given that we know the external

forces. Equation (5), written in terms of either velocity or position, is a differential equation. In order

to calculate the velocity and position as a function of time we will need to integrate this equation either

analytically or numerically. On the other hand, the reverse problem of computing the forces given motion is

much easier and only requires direct evaluation of (5). Is is also common

mit dynamics

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