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Mechanics
Mechanics
Prepared by Tendayi CHIHAKA
African Virtual university
Université Virtuelle Africaine
Universidade Virtual Africana
African Virtual University Notice
This document is published under the conditions of the Creative Commons
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http://creativecommons.org/licenses/by/2.5/
License (abbreviated “cc-by”), Version 2.5.
African Virtual University Table of Contents
I.
Mechanics_________________________________________________ 3
II.
Prerequisite Course or Knowledge_ _____________________________ 3
III. Time_____________________________________________________ 3
IV. Materials__________________________________________________ 3
V.
Module Rationale_ __________________________________________ 3
VI. Content___________________________________________________ 4
6.1 Overview ____________________________________________ 4
6.2 Outline_ _____________________________________________ 4
6.3 Graphic Organizer______________________________________ 6
VII. General Objectives_ _________________________________________ 7
VIII. Specific Learning Objective(s)__________________________________ 7
IX. Teaching and Learning Activities________________________________ 8
X.
Learning Activities__________________________________________ 14
XI. Glossary of Key Concepts___________________________________ 141
XII. List of Compulsory Readings_ _______________________________ 147
XIII. Compiled List of (Optional) Multimedia Resources________________ 148
XIV. Compiled list of Useful links_________________________________ 149
XV. Synthesis of the Module____________________________________ 151
XVI. Summative Evaluation______________________________________ 152
XVII.References_ _____________________________________________ 163
XVIII. Main Author of the Module _________________________________ 164
African Virtual University I. Mechanics
By Mr. Tendayi Chihaka
II. Prerequisite Courses or Knowledge
Linear Algebra 1 and Calculus 3 is prerequisite.
III.Time
The total time for this module is 120 study hours.
IV.Material
Students should have access to the core readings specified later. Also, they will need
a computer to gain full access to the core readings and web links in the materials.
V.Module Rationale
This topic of mechanics has been treated as that branch of mathematics that mathematically attempts to explain our physical environment. It is that branch of mathematics
that spans the divide between our natural sciences and mathematics.
The module has infused as its basis, both modern mathematics and traditional
mathematics as it explores the basic concepts of Kinematics – which is the study
of motion without referrence to forces which cause motion of particles; Kinetics
which relates the action of forces on particles and bodies to their resulting motions;
Dynamics – the study of the general causes of motion; and Statics – the equillibrium
mechanics of stationary bodies
Practical examples and their implications for classroom practice have been suggested
whenever and wherever appropriate in the module in order to help the teacher to
impart the knowledge of mechanics to pupils.
African Virtual University VI.Content
6.1
Overview
This module is a first degree course in Mechanics. The module starts with a treatment
of vectors and vector calculus and attempts to explain all the topics in mechanics
from this basis. It is hoped that the students enrolling for this course will acquaint
themselves with the basic notions of force and the resultant motion of its application.
Four areas of mechanics: Statics, Dynamics, Kinetics and Kinematics of particles
and rigid bodies have been treated in this module.
The student is strongly advised to consult Physics sources on Mechanics in conjunction with this module for practical examples which are mathematically modelled in
the module.
6.2
Outline: Syllabus
Unit 1: Force, Energy and Motion
Level 1. Priority A. Linear Algebra 1 and Calculus 3 is prerequisite. (Note overlap with Vector Analysis)
Vectors, Velocity and Acceleration: Dot product, cross product, triple products. Derivatives of vectors, Integrals of vectors. Relative velocity and acceleration. Tangential
and normal acceleration, Circular motion. Gradient, divergence and curl. Line integrals
and independent of path.
Newton’s Law of Motion - Work, Energy and Momentum: Work, power, kinetic
energy. Conservative force fields, potential, conservation of energy. Impulse, Torque
and angular momentum, conservation of momentum.
Motion in a Uniform Force Field. Falling bodies and projectiles. Potential and potential energy in a uniform force field. Projectiles, motion in a resisting medium.
Constrained motion. Friction. Because of the volume of content in this unit, we have
seen it prudent to divide this unit into two parts – 1a and 1b.
African Virtual University Unit 2: Oscillations
Level 1. Priority A. Mechanics 1 is prerequisite.
The Simple Harmonic Oscillator and the Simple Pendulum: Energy of a simple harmonic oscillator. Over-damped, critically-damped and under-dumped motion. Simple
pendulum. The two and three dimensional harmonic oscillator.
Unit 3: Dynamics
Level 2. Priority B. Mechanics 2 is prerequisite.
Central Forces and Planetary Motion: Equations of motion for a particle in a central
field; Potential energy of a particle in a central field; Conservation of energy. Kepler’s
laws of planetary motion.
Moving Coordinate Systems: Rotating coordinate systems, Derivative Operations,
Velocity, and Acceleration in a moving system. Coriolis and centripetal acceleration
(and force). Motion of a particle relative to the earth.
Systems of Particles: Conservation of momentum, angular momentum, external torque. Kinetic energy, work, potential energy. Principle of virtual work; D’ALembert’s
principle.
Rockets and Collisions: Problems involving changing mass. Rockets, Collisions
(direct and Oblique).
Unit 4: Rigid Bodies and Energy
Level 3. Priority C. Mechanics 3 is prerequisite.
Plane Motion of Rigid Bodies: Euler’s Theorem. Chasle’s Theorem. Moment of
inertia. Radius of gyration. Parallel axis theorem. Perpendicular axis theorem. Couples. Kinetic energy and angular momentum about a fixed axis. Principle of angular momentum. Principle of conservation of energy. Principle of virtual work and
D’Alembert’s principle. Principle of minimum potential energy.
African Virtual University 6.3
Graphic Organiser
RIGID
BODIES
DYNAMICS
OSCILLATIONS
NEWTON’S
LAWS OF
MOTION
VECTOR
ALGEBRA
PLANE AND
CURVILINEAR
MOTION
African Virtual University VII. General Objective(s)
By the end of the module, the trainee should be able to:
o Relate mathematical notions to physical quantities such as force and motion.
o Model some physical phenomena mathematically as required for effective
teaching of mechanics in the secondary school.
o Relate traditional mechanics operations to vector calculus and vice-versa.
VIII.
Specific Learning Objectives You should be able to:
1. Be equipped as a student and as a teacher with vector operations.
2. Be infused with the basic tools of analysis into vector quantities.
3. Infuse the basic tools of analysis into various types of motion e.g. Simple
Harmonic Motion.
African Virtual University IX.Teaching and Learning Activities
Basic Algebraic Ideas Test
Rationale: To check learner’s familiarity with some concepts assumed in the module.
Questions
1. Velocity is the :
a. rate of change of displacement;
b. rate of change of speed;
c. rate of change of distance;
d. rate of change of time.
2. Which of the following is group of vectors?
a. speed, acceleration and time;
b. displacement, velocity and acceleration;
c. direction, displacement and velocity;
d. force, velocity and time.
3. The resultant of velocities 8ms-1 and 6ms-1 at right angles is :
a. 12ms-1;
b. 10ms-1;
c. 7ms-1;
d. 9ms-1.
4. The momentum of a body is the :
a. mass of a body times its speed;
b. weight of a body times its velocity;
c. mass of a body times its velocity;
d. weight of a body times its speed.
African Virtual University 5. Newton’s first law of motion states that :
a. If a body is at rest it might remain at rest or if it is in motion it moves with a
speed until it stops;
b. If a body is at rest it remains at rest or remains in motion until it is acted on
by a resultant force;
c. If a body is at rest it remains at rest or if it is in motion it moves with uniform
velocity until it is acted on by a resultant force;
d. If a body a body is at rest it remains at rest.
6. A car of mass 1.0 × 103 kg travelling at 72 km h-1 on a horizontal road is brought
to rest in a distance of 40 m by the action of the brakes and frictional forces. Find
the average stopping force.
a. 5.0 × 102 N ;
b. 5.0 × 103 N ;
c. 5.0 × 104 N ;
d. 5.0 × 101 N .
7. A scalar quantity has :
a. direction only;
b. magnitude only;
c. direction and magnitude;
d. none of the above.
8. A train which is moving with uniform acceleration is observed to take 20 s and
30 s to travel successive 400 metres. How much farther will it travel before
coming to rest if the acceleration remains uniform?
a. b.
c.
d.
163.3 m;
963.3 m;
800 m;
663.3 m.
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9. The following is NOT an equation of motion in a straight line.
a. v = u + at ;
b. x = ut + 12 at 2
;
c. v = u + 2ax ;
2
2
d v = u + at 2 .
10. Power :
a. is the ability to have energy;
b. is the ability to run;
c. is the ability to have speed;
d. is the ability to do work.
11. A mass of 5 kg moves on smooth horizontal plane with a speed of 8 m/s, being
attached to a fixed point on the plane by a string of length 4 m. find. The tension
in the string is :
a. 16 N;
b. 40 N;
c. 80 N;
d. 20 N.
12. Impulse is defined as :
a.
b. c. d. the product of force and distance;
the product of force and mass;
the product of force and time ;
the product of force and velocity.
13. The angular velocity of the particle is :
a. b. c. d. radius of a circle over the speed of a particle;
radius of a circle over the velocity of a particle;
speed of a particle over the radius of a circle;
velocity of a particle over the radius of a circle.
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14. The force that must be exerted by the rails towards the centre of the circle is :
a. b. c. d. the moment about that axis of the internal forces acting on the body.
the moment about that axis of the external forces acting on the body.
the moment about that axis of the speed of the body.
the moment about that axis of the acceleration of the body.
15. A particle of mass 3 kg, resting on a smooth table and attached to a fixed point on
the table by a rope 1.2 m long, is making 300 rev/min. Find The angular velocity
of the particle is :
a. 10 rev/s;
b. 10π rad/s;
c. 5 rev/s;
d. 5π rad/s.
16. An engine, of mass 80 Mg, is moving in arc of a circle of radius 240 m, with a
speed of 48 km/h. The force that must be exerted by the rails towards the centre
of the circle is
a. b.
c.
d.
0.59×105 N;
0.59×104 N;
0.59×103 N;
0.59×102 N.
17. A particle is said to move with simple harmonic motion if:
a. the particle moves so that its acceleration along its path is directed towards a
fixed point in that path, and varies inversely as its distance from this fixed point;
b. the particle moves so that its acceleration along its path is directed towards a
fixed point in that path, and varies directly as its distance from this fixed point;
c. the particle moves so that its speed along its path is directed towards a fixed
point in that path, and varies directly as its distance from this fixed point;
d. the particle moves so that its acceleration along its path is directed towards a
fixed point in that path, and varies directly as its velocity from this fixed point.
18. A simple pendulum:
a. consists of a heavy paticle or bob attached to a fixed point by a heavy string
and swinging in a vertical plane;
b. consists of a heavy paticle or bob attached to a fixed point by a weightless
string and swinging in all directions;
c. consists of a heavy paticle or bob attached to a fixed point by a heavy string
and swinging in all directions;
d. consists of a heavy paticle or bob attached to a fixed point by a weightless
string and swinging in a vertical plane.
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19. Which of the following does not typically denote a vector:
(a) -5
(b) (1, 2, 3) (c) A (d) ⎡4⎤
⎢ ⎥
⎢8⎥
⎢ −3⎥
⎣ ⎦
20. A Subspace of a vector space:
(a) Is also a vector space;
(b) Is not a vector space;
(c) Is not a linear space;
(d) Is half of a vector space.
Answer Key
1. a ((b),(c),(d) have scalar quantities speed, distance and time respectively. So
since velocity is a vector quantity a is correct)
2. b (for (a) speed and time are not vectors, for (c) direction is not a vector, for (d)
time is not a vector )
3. b (using the pythagorus theorem take 8ms-1 to be on the opposite side and 6ms-1
to be on the adjacent side then the resultant side is 10ms-1 )
4. c (momentum is the product of mass and velocity since the particle is moving
in a particular direction. So (a), (b) and (d) are not correct)
5. c ((c) is correct because the body is moving in a particular direction and only
stops when a force e.g. friction is applied on it)
6. b (The initial velocity is 72 km h-1 or 20 m/s and final velocity 0 m/s and so
acceleration is 5 ms-2. thus force is mass times acceleration giving the answer in
(b))
7. b (a vector quantity is the one with magnitude and direction so (b) is correct)
8. a
9. d ((d) is not the equation of a sraight line because of the t2 in the equation. The
other equations are correct)
10. d (power is force times velocity or the rate of doing work so (d) is correct)
11. c (here we are looking at motion in a circle so acceleartion towards the fixed point
is (velocity)2/radius=64/4=16 ms-2. therefore tension is mass times acceleration=
5×16= 80 N)
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12. c
13. d
14. b
15 b ( motion is in a circle so we multiply the revolutions made per second by 2π
since each revolution is made up of 2π )
16. a (48 km/h= 40/3 ms-1 and force exerted is mv2/r= 80 000×(40/3)2/240)
17. b
18. d
19. a ((c) normally denotes a vector or matrix, (b) and (d) denote vectors )
20. a
Pedagogical Comment For Learners
The pre-assessment has been designed in such a way that it introduces the student to
basic notions of kinetics and kinematics. It covers such concepts as identifying the
equations of motion in a straihgt line,basic notions familiarity with basic algrebraic
processes .A score of 50 % or less should be a cause for concern and will require the
student to revisit his “O” level work on Algebra and its processes. It is essential that
the student reads widely on that content they are not familiar with as it is important
to have this prior knowledge before embarking on the following units.
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X. Learning Activities
Unit 1a : Vectors, Vector Calculus, Velocity and Acceleration
Specific learning objectives
At the end of these activities you should be able to:
• Define vectors and perform operations on vectors.
• Differentiate and Integrate vector valued functions.
• Define velocity and acceleration in terms of vectors and describe the relationships between velocity and acceleration.
• Given appropriate situations, define and calculate the relative velocities and
accelerations of bodies in motion.
• Describe circular motion and calculate tangential and normal acceleration of
particles moving in circular motion.
• Define and apply the concepts of gradient, divergence and curl.
• Define and evaluate line integrals and independence of paths.
Summary of the learning activity
In this activity you will familiarise yourself with elementary vector calculus and its
application to motion in two and three dimensions.
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Relevant Links and Resources
Vectors
http://en.wikipedia.org/wiki/Vector_(spatial)
Vector-valued function
http://en.wikipedia.org/wiki/Vector-valued_function
Acceleration
http://en.wikipedia.org/wiki/Acceleration
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Velocity
http://en.wikipedia.org/wiki/Velocity
Divergence
http://en.wikipedia.org/wiki/DIVERGENCE
Curl
http://en.wikipedia.org/wiki/CURL
Gradient
http://en.wikipedia.org/wiki/Gradient
Gradient
http://hyperphysics.phy-astr.gsu.edu/hbase/gradi.html
Key Words (for description/definitions see glossary)
• Scalar
• Vector
• Velocity
• Acceleration
• Function
Bridging The Gap
An army sergeant gives the following order to one group of new army recruits on
parade.
“Walk for five hours.”
To another group he orders:
“Run for five kilometres”
To a third he barks:
“On- the- double at ten kilometres per hour!”
As you may well be aware, army recruits are not allowed to question orders
given by a senior.
Describe the situation at the parade ground immediately after these orders and the
possible questions each recruit might be asking himself/herself.
What would happen if the sergeant had given similar instructions to a group of pilots
with the appropriate terminology and figures?
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Detailed description of the activities
In this activity, we revisit the ideas of vectors and vector valued functions with the
intention of exploring vector calculus. We then use the results of vector calculus
to define velocity acceleration force and in Activity 2 finally discuss the motion of
particles and bodies in various settings in Activity 3. You will have opportunity to
examine various problem and solution situations as well as opportunities to solve
problems on your own..
1a.1
Vectors and Scalars
http://en.wikipedia.org/wiki/Vector_(spatial)
1a.1.1 Examples of Scalar Quantities
Vectors are quantities that require not only a magnitude, but a direction to specify
them completely. Let us illustrate by first citing some examples of quantities that are
not vectors. The number of litres of petrol in the fuel tank of your car is an example
of a quantity that can be specified by a single number---it makes no sense to talk
about a «direction» associated with the amount of petrol in a tank. Such quantities,
which can be specified by giving a single number (in appropriate units), are called
scalars. Other examples of scalar quantities include the temperature, your weight,
or the population of a country; these are scalars because they are completely defined
by a single number (with appropriate units).
1a.1.2
Examples of Vector Quantities
However, consider a velocity. If we say that a
car is going 70 km/hour, we have not completely
specified its motion, because we have not specified
the direction that it is going. Thus, velocity is an
example of a vector quantity. A vector generally
requires more than one number to specify it; in
this example we could give the magnitude of
the velocity (70km/hour), a compass heading to
specify the direction (say 30 degrees from North),
and an number giving the vertical angle with
respect to the Earth’s surface (zero degrees except in chase scenes from
action movies!). The adjacent figure shows a typical coordinate system for
specifying a vector in terms of a length r and two angles, θ and ϕ
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1a.1.3
Vectors in in 2-d and 3-d
Definition: The component forms of a vector v in 2-d and 3-d whose initial point
(
)
(
)
is the origin and whose terminal points are x1 , x2 and x1 , x2 , x3 respectively are
given by
(
v = x1 , x2
(
)
v = x1 , x2 , x3
)
Definition: The length of a vector v will be defined as
v =
x12 + x22 for 2-d and
v =
x12 + x22 + x32 for 3-d
Definition: If v is a non-zero vector in the 2-d or 3-d space, then the vector
u=
v
1
=
v
v
v
has length 1 in the direction of v.
Definition: The standard unit vectors (1, 0) and (0, 1) in 2-d and
(1, 0, 0 ) , ( 0,1, 0 ) and ( 0, 0,1) in 3-d are
i = (1, 0) and j = (0, 1) and
i = (1, 0, 0 ) , j = ( 0,1, 0 ) and k = ( 0, 0,1)
It is the assumption here that you have done work on vector addition and scalar
multiplication and the only operations we will discuss here are the dot product and
cross product Your core text, however, has sections that treat these operations as
indicated below. .
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Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin,
Texas.UTP pp 34-38
N.B. You will have recognized that the vector addition described is the component
wise addition that you are familiar with.
Vector magnitude P. 35
Pythagoras Theorem (3.6) P. 35
N.B. Note that the vector magnitude is also called the modules of the vector.
Equation (3.6) and (3.7)
Scalar Multiplication P. 35
Component wise scalar multiplication 3.8 P. 36
Diagonals of a parallelogram P. 36 – 38
N.B. Equations 3.9 – 3.13
Teaching Tip
This gives a geometrical interpretation of vector addition and scalar multiplication
which is a very useful tool when teaching secondary school students.
1a.1.4 The Dot Product u = ( v1 , v2 ) and v = ( v1 , v2 ) is
The Dot product of
u • v = ( u1 v1 + u2 v2 )
The Dot product of u = ( u1 ,u2 ,u) and v = ( v1 , v2 , v3 ) is
u • v = ( u1 v1 + u2 v2 + u3 v3 )
v• v = v
N.B.
2
Theorem: If θ is the angle between two non-zero vectors u and v, then
cos θ =
u• v
u v
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Definition: The work done, W, by a force acting along the line of motion of an
object is given by
W= Force × distance = F
PQ
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp 40
N.B. The scalar product is also referred to as the “DOT” product. The geometrical
figure given in Figure 15 explains this vector operation very well.
1a.1.5
The cross product
Definition: The cross product of vectors u= v1i + v2 j + v3 k and v= v1i + v2 j + v3 k
is u × v = (u2 v3 − u3 v2 )i − ( u1 v3 − u3 v1 ) j + (u1 v2 − u2 v1 )k )
A more convenient way is to write it as
i
u × v = u1
v1
j
u2
v2
k
u3
v3
which is the determinant of a 3x3 matrix. Remember to refer to your module on linear
algebra to refresh yourself on properties of determinants.
African Virtual University 20
DO THIS
Verify that the two definitions are indeed the same
N.B. The cross product is not defined for vectors in 2-d
Theorem: Algebraic properties of the cross product
Let u and v be vectors and c be a scalar:
1.
2.
3.
4.
5.
u×v=v×u
u × (v + w) = (u × v) + (u × w)
c(u × v)= cu × v= u × cv
u×0=0×u=0
u × u =0
6. u •(v + w) = (u × v) • w
Theorem: Geometric properties of the cross product
Let u and v be non-zero vectors and θ be the angle between them.
1. u x v is orthogonal to both u and v.
2. u × v = u v sinθ
3. u x v = 0 if and only if one is a scalar multiple of the other
4. u × v = area of parallelogram of sides u and v
N.B. The cross product can be used torque- the moment M of a force F about a
point P.
Example: If the point of application of the force is Q, the moment of F about P is
M= PQ × F
The magnitude of the moment F measures the tendency of the vector PQ to rotate
counter clockwise about an axis directed along the vector M
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DO THIS
Exercise: Prove that u x v = u v sinθ , where u and v are vectors, θ the angle
between them and x is the cross product.
(Answer): Hint. Remember that sinθ = 1 − cosθ and thatcosθ =
1a.1.6
(u • v)2
u
2
v
2
The Triple Scalar Product
Definition: The triple scalar product is the dot product of u and v + w u •(v + w)
Theorem: For u = u1i + u2 j + u3 k , v= v1i + v2 j + v3 k and w= w1i + w2 j + w3 k
u1
u •(v + w) = v1
w1
u2
v2
w2
u3
v3
w3
DO THIS
Prove the theorem
1a.1.7
Vector Valued Functions
Definition: A vector valued function is a function where the domain is a subset of
the real numbers and the range is a vector In other words, vector valued functions
assign a vector to a number. More specifically,
In 2-d
r(t) = x(t)i + y(t)j or r(t) = (x(t) , y(t)), or
In 3-d
r(t) = x(t)i + y(t)j + z(t)k or r(t) = (x(t) ,y(t) ,z(t))
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You will notice the strong resemblance to parametric equations. In fact there is an
equivalence between vector valued functions and parametric equations.
Example
Sketch the graph of r(t) = (t - 1)i + t2 j Solution (Hide)
We draw vectors for several values of t and connect the dots. Notice that the graph
is the same as y = (x + 1)2 1a.1.8
Calculus of Vector Valued Functions
The formal definition of the derivative of a vector valued function is very similar to
the definition of the derivative of a real valued function.
1a.1.9
The Derivative of a Vector Valued Function
Let r(t) be a vector valued function, then
= x’(t)i + y’(t)j
Because the derivative of a sum is the sum of the derivative, we can find the derivative
of each of the components of the vector valued function to find its derivative.
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Go to this link: http://en.wikipedia.org/wiki/Vector-valued_function
Examples
d/dt (3i + sintj) = costj
d/dt (3t2 i + cos(4t) j + tet k) = 6t i -4sin(t)j + (et + tet) k
1a.1.10
Properties of differentiation of Vector Valued Functions
All of the properties of differentiation still hold for vector values functions. Moreover
because there are a variety of ways of defining multiplication, there is an abundance
of product rules.
Suppose that v(t) and w(t) are vector valued functions, f(t) is a scalar function, and
c is a real number then
1. d/dt(v(t) + w(t)) = d/dt(v(t)) + d/dt(w(t))
2. d/dt(cv(t)) = c d/dt(v(t))
3. d/dt(f(t) v(t)) = f ‘(t) v(t) + f(t) v’(t)
4. (v(t) . w(t))’ = v’(t) . w(t)+ v(t) . w’(t)
5. (v(t) × w(t))’ = v’(t) × w(t)+ v(t) x w’(t)
6. d/dt(v(f(t))) = v’(f(t)) f ‘(t)
1a.1.11
Integration of vector valued functions
Definition: We define the integral of a vector valued function as the integral of each
component. This definition holds for both definite and indefinite integrals.
1.If r(t) = x(t)i +y(t)j, where x and y are continuous on [a , b] then
∫ r (t)dt = ⎡⎣ ∫ x(t)dt ⎤⎦ i + ⎡⎣ ∫ y(t)dt ⎤⎦ j and
b
⎡b
⎤ ⎡b
⎤
r
(t)dt
=
x(t)dt
⎢∫
⎥i + ⎢ ∫ y(t)dt ⎥ j
∫a
⎣a
⎦ ⎣a
⎦
2. If r(t) = x(t)i + y(t)j + z(t)k, and x, y, and z are continuous on [a , b] then
∫ r (t) = ⎡⎣ ∫ x(t)dt ⎤⎦ i + ⎡⎣ ∫ y(t)dt ⎤⎦ j + ⎡⎣ ∫ z(t)dt ⎤⎦ k
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Example
Evaluate
∫(sin t)i + 2t j - 8t3 k dt
Solution
Just take the integral of each component
(∫ (sin t)dt i) + ( ∫2t dt j) - ( ∫8t3 dt k)
= (-cost + c1)i + (t2 + c2)j + (2t4 + c3)k Notice that we have introduce three different constants, one for each component and
that the three scalar constants produce one vector constant Activity 1a.2
Velocity and Acceleration
•
Define velocity and acceleration in terms of vectors and describe the relationships
between velocity and acceleration.
•
Given appropriate situations, define and calculate the relative velocities and
accelerations of bodies in motion
1a.2.1
Velocity
Go to this link: http://en.wikipedia.org/wiki/Velocity
Definition: Velocity and Speed
In single variable calculus the velocity is defined as the derivative of the position function. For vector calculus, we make the same definition for both 2-2 and 3-d space.
Let r(t) be a differentiable vector valued function representing the position vector of
a particle at time t. Then the velocity vector is the derivative of the position vector.
In 2-space:
r(t) = ×(t)i + y(t)j and Velocity = v(t) = r'(t) = ×'(t)i + y'(t)j
Speed = v(t) = r '(t) =
2
2
[ x'(t)] + [ y'(t)]
In 3-space: r(t) = ×(t)i + y(t)j + z(t)k and velocity = v(t) = r’(t) = ×’(t)i + y’(t)j
+ z’(t)k
Speed = v(t) = r '(t) =
Example
2
2
2
[ x'(t)] + [ y'(t)] + [ z'(t)]
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Find the velocity vector v(t) if the position vector is r(t) = 3ti + 2t2j - sin t k We just take the derivative
v(t) = 3i + 4tj + cos t k N.B. When we think of speed, we think of how fast we are going. Speed should not
be negative. In one variable calculus, speed was the absolute value of the velocity. For vector calculus, it is the magnitude of the velocity.
1a.2.2 Motion in one dimension
Compulsory reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp18-31
This section introduces the concepts of displacement, velocity and acceleration and
the motion of a particle varying velocities such as constant, uniform and so on
You should be able to relate the ideas of the position vector to the important concept
of displacement
Discussion: Are the two concepts of displacement and position vector the same?
1a. 2.3 Motion in three dimensions
Compulsory reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP p 33-52
N.B. You should note hear the introduction of the three dimensional Cartesian plane
to provide a suitable frame of reference to describe motion in three dimensions.
For practical examples you can use the idea of an aircraft taking off from an airport.
At any given time, its position with reference to the airport can be described by
asking the questions:
How far north of the airport is the plane?
How far east?
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1a.2.4 Acceleration
Go to this link: http://en.wikipedia.org/wiki/Acceleration
In one variable calculus, we defined the acceleration of a particle as the second derivative of the position function. Nothing changes for vector calculus.
1a.2.5 Definition of Acceleration
Let r(t) be a twice differentiable vector valued function representing the position
vector of a particle at time t. Then the acceleration vector is the second derivative
of the position vector.
In 2-space, r(t) = ×(t)i + y(t)j and acceleration = a(t) = r’’(t) = ×’’(t)
In 3- Space r(t) = ×(t)i + y(t)j + z(t)k
and acceleration =a(t) = r’’(t) = x’’(t)i + y’’(t)j + z’’(t)k
Example
Find the velocity and acceleration of the position function
r(t) = 4t i + t2 j when t = -1. Then sketch the vectors.
Solution
The velocity vector is
v(t) = r’(t) = 4 i + 2t j
Plugging in -1 for t gives
v(-1) = 4 i - 2j
Take another derivative to find the acceleration.
a(t) = v’(t) = 2j
Below is a picture of the vectors.
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DO THIS
Sketch the path of motion of an object whose position vector is
r(t) = (t 2 − 4)i + j
DO THIS
An object starting from rest at P(1, 2, 0) has acceleration
a(t) = j + 2k
where a(t) is measured in m s −2 Find the location of the object after 2seconds
Do not look over the page until you are finished!
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Answer
We have initial conditions as v(0) = 0 and r(0) = x(0)i + y(0)j + z(0) k
i.e. r(0) = 1i + 2j + 0k = i + 2j
v(t) = ∫ a(t)dt = ∫ ( j + 2k)dt = tj + 2tk + CwhereC = C 1i + C 2 j + C 3 k
When t = 0, v(0) = C 1i + C 2 j j + C 3 k = 0 ⇒ C 1 = C 2 = C 3 = 0
Thus the velocity at any time t is
v(t) = tj + 2tk
1 2 2
t + t k + CwhereC = C 4 i + C j + C 6 k
2
Also r (0) = C 4 i + C 5 j + C 6 k = i + 2 j ⇒ C 4 = 1,C 5 = 2,C 6 = 0
Now r (t) = v(t)dt = ( j + 2k)dt =
∫
∫
⎛1
2
⎞
Thus r (t) = i + ⎜ t 2 + 2 ⎟ j + t 2 k
⎝
⎠
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
• Motion with constant velocity.p23
• Motion with constant acceleration p24
• Equations of motion in a straight line p26
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1a.2.6 Motion of Falling Bodies
Aristotelian Physics tell us that the speed of falling of a body depends
completely on its weight, so a one Kilogramme stone will fall faster than
a half-kilogramme stone. Galileo denied this, arguing that everybody falls
down with the same speed and acceleration, for example, if you have got a
stone in your hand, and you suddenly stop holding it, it will fall to the ground
with speed V. And now, you have a paper, and you throw it to the ground,
its falling speed will be now v (a smaller speed), but if you make a little
«ball» with the paper, its speed will be V. (the same than the stone). From
here he postulated that the speed does not depend on the weight, every time
the acceleration is the same, but in the case of the plain paper the air has
more resistance and that’s the cause of the smaller speed. This experiment
was done by Galileo in Pisa’s Tower.
1a.2.7
Experiment
Purpose
In this experiment you’ll be able to see the acceleration of different objects and
compare them such as Galileo (the precursor of Einstein) did.
Materials
A tennis ball
• A football
• A notebook
• A sheet
Procedure
•
1.
2.
3.
4.
5.
6.
7.
8.
9.
Take both balls.
Hold them at the same level, as high as you can (shoulder, face, etc.)
Drop them to the ground at the same time
Both reach the soil at the same time.
Do you think this only happens because they have the same shape. Next try
this with the notebook and tennis ball.
Both reach the soil at the same time!
Now, try with the notebook and the sheet of paper. What happens? Why do
you think this happens?
Now make a little «ball» with the sheet of paper and repeat the experiment.
They should reach the ground at the same time. Why
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Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Free fall under gravity p 26-28
Examples p28-31
DO THIS
Exercise
A child leans out of a window of a building from a height 10m above the ground.
She throws a ball vertically upwards with an initial velocity of 12m/s. What is the
maximum height above the ground reached by the ball and the total amount of time
it takes to strike the ground.
Answer : You should come up with these answers.
Maximum height = 17.4m and total time elapsed = 3.11s
1a.2.8
Projectile Motion
N.B. This should be read together with the section on Newton’s laws in Activity 2.
As mentioned earlier it will be assumed that the only force acting on the projectile
after its launching is the force of gravity. Thus the motion occurs in a vertical plane.
For a projectile of mass m, the foce due to gravity is
F = - mgj
Comparing this with
F = ma (from Newton’s second law of motion) we have
a = -gj which becomes our acceleration vector.
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Now, as shown in the diagram above, if the projectile is launched with initial velocity
v0 from a position r0 then
v(t) = a(t)dt = −gjdt = − gtj + C 1
∫
∫
r(t) = v(t)dt = (−gtj + C 1 )dt = −
∫
∫
1 2
gt j + C 1t + C 2
2
Now, v(0) = v0 ands(0) = s0 and this gives rise to
C 1 = v0 andC 2 = r0
And therefore
r(t)=
1 2
gt j + tv0 + r0 which gives our position vector.
2
Recall that
v0 = xi + yj = ( v0 cosθ)i + ( v0 sinθ) j = v0 cosθi + v0 sinθ j
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Substituting into the above, we have
1
r(t) = − gt 2 j + tv0 cosθi + tv0 sinθ j + hj h being the initial height above the
ground 2
Rearranging , we have The position Function of a Projectile as
⎡
⎣
r(t) = (v0 cosθ)ti + ⎢ h + (v0 sinθ)t −
1 2⎤
gt ⎥ j
2
⎦
DO THIS
A catapult throws a stone from 3m above the ground at an angle of 45 o from the
horizontal at 100m per second. Find the stone’s maximum height. Will the stone go
over a 10m wall located 300m from the point of projection
Do not look over the page until you are finished!
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Answer
We are given h=3, v0 = 100, andθ = 45 0 =
π
4
⎡
⎣
r(t) = (100 cos )ri + ⎢ 3 + (100 sin
π
. Using g = 9.8 m s −2
4
π
⎤
)t − 4.9t 2 ⎥ j
4
⎦
= (50 2t)i + (3 + 50 2t − 4.9t 2 ) j .
The maximum height occurs when the vertical component of v is 0.
That is:
y'(t) = 50 2 − 9.8t = 0 which means that t=
25 2
seconds.
4.9
⎛ 25 2 ⎞
25 2
Maximum height is y = 3 + 50 2(
) − 4.9 ⎜
⎟
4.9
⎝ 4.9 ⎠
2
DO THIS
Simplify the above equation and find the actual value of y
For the wall, x(t) = 300 = 50 2t
Which means t = 3 2 and
y = 3 + 50 2(3 2) − 4.9(3 2 )2 = 3 + 300 -88.2 = 214.8
This means that the stone clears the wall.
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Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Projectile motion P41-44
1a.2.9
Circular Motion
In general, circular motion is rotation along a circle, a circular path or a circular orbit.
The rotation around a fixed axis of a three-dimensional body involves circular motion
of its parts. We can talk about circular motion of an object if we ignore its size, so
that we have the motion of a point mass in a plane.
Examples of circular motion are: an artificial satellite orbiting the Earth in geosynchronous orbit, a stone which is tied to a rope and is being swung in circles (cf. hammer
throw), a racing car turning through a curve in a racetrack, an electron moving perpendicular to a uniform magnetic field, a gear turning inside the gearbox of a car.
A special kind of circular motion is when an object rotates around its own center of
mass. This can be called spinning motion, or rotational motion and will be discussed
in later module. I am sure you can come up with your own examples of motion in
a circle.
Circular motion involves acceleration of the moving object by a centripetal force
which pulls the moving object towards the center of the circular orbit. Without this
acceleration, the object would move inertially in a straight line the tangent the circle,
according to Newton’s first law of motion. Circular motion is accelerated even though
the speed is constant, because the object’s velocity vector is constantly changing
direction. You should deduce from this that the acceleration vector and the velocity
vector are orthogonal
Compulsory readings
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
•
•
•
•
•
•
•
•
•
•
•
Introduction P. 136
Uniform Circular Motion P. 136 – 138
N.B. Equations 7.1 – 7.11
Centripetal Acceleration P. 138 – 141
N.B. Equations 7.12 – 7.17
Definition : Centripetal Acceleration (7.15) p139
The Conical Pendulum p 141 – 142
N.B. Equations 7.18 – 7.25
Non uniform circular motion P. 143 – 147
Explanations: Radial unit vector P. 143
Tangential unit vector P. 143
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•
•
•
•
•
•
•
Radial velocity and tangential velocity P. 144
Radial acceleration and tangential acceleration P. 144
N.B. Equations to note(7.26) – (7.45)
The Vertical Pendulum P. 148 – 150
N.B. Equations to note 7.46 – 7.53
Animated picture)
N.B. Equations 7.54 – 7.64
You need to note the different approach that is not based on vector algebra and
the traditional equations that are employed in your compulsory reading text
Example
Find the velocity vector, speed, acceleration vector of the circle
t
t
r (t) = 2 sin i + 2 cos j
2
2
And sketch the circle.
Solution.
The velocity vector is
t
t
v(t) = r / (t) = 2 cos i − 2 sin j
2
2
The speed at any time is:
r / (t) = cos 2
t
t
+ sin 2 = 1
2
2
The acceleration vector is
1
t
1
t
a(t) = r / / (t) = − sin i − cos j
2
2
2
2
N.B. The parametric equations for the curve are
t
x = 2 sin and
2
t
y = 2 cos
2
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DO THIS
Exercise
Verify that the rectangular equation of the circle is
x 2 + y2 = 4
DO THIS
Exercise
A particle starts from rest at point P(1,2,0) with acceleration
a(t) = j + 2k
in the usual units. Find the position of the particle after 2 seconds.
Do not look over the page until you are finished!
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Answer
You should be able to deduce that
v(0) = 0
and
r (0) = x(0)i + y(0) j + z(0)k
= 1i + 2 j + 0k
= i +2j
To find the position function, you have to integrate twice, each time using one of the
initial conditions to find the constants of integration. Thus
v(t ) = ∫ a (t )dt = ∫ (j +2k)dt = tj + 2tk + C
Where
C = C 1i + C 2 j + C 3k
W hen t = 0 and v(0) = 0 you get
v(0) = C 1i + C 2 j + C 3k = o ⇒ C 1 = C 2 = C 3 = 0
Thus the velocity at any time t is
v(t ) = t(j) + 2t(k)
When we integrate once more we produce
r (t ) = ∫ v(t )dt = ∫ (tj + 2tk)dt =
t2
j + t 2k + C
2
Where C = C 4i + C 5 j + C 6k
When t = 0 and r (0) = i + 2j we have
r (0) = C 4i + C 5 j + C 6k = i + 2j ⇒ C 4 = 1,C 5 = 2,C 6 = 0
Thus the position vector is
⎛ t2
⎞
r (t ) = i + ⎜ + 2⎟ j + t 2k
⎝2
⎠
The position of the particle after 2 seconds is
r (2) = i + 4j + 4k given by the coordinates (1,4,4)
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Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP p 136-160
Note that in our compulsory reading we introduce notions of angular velocity, centripetal acceleration and so on. This section should be read in conjunction with the
sections on curvilinear motion below and you should be able to draw parallels between
circular motion and curvilinear motion which is an extension of secular motion.
1a.2.10 Relative Velocity
We have been able to describe the position of a body and its velocity with reference
to the origin of a given co-ordinate system. Ordinarily, this origin is taken to be fixed
in some other body may be in motion relative to a third and so on. For example, when
we speak of the velocity of a car, we usually mean the velocity of the car relative
to the earth… But the earth is in motion relative to the sun… the sun is in motion
relative to some other star…. And so on…..
Suppose a long train is moving towards the right along a straight level track and an
athlete is running on the train towards the right.
Diagram of a flat car train with man running on the train. In the diagram,
uTE represents the velocity of the train T relative to the earth E
uAT represents the velocity of the athlete A relative to the train T
The velocity of the athlete relative to the earth, uAE ,
is evidently equal to the sum of uAT and uTE :
uAE = uAT + uTE
N.B. The velocity uAE is the algebraic sum of uAT and uT E and when combining
relative velocities
•
•
•
Write each velocity with a double subscript in the proper order meaning “velocity of (first subscript) relative to (second subscript)
When velocities are added, the first letter of any subscript is to be the same
as the last letter of the second subscript
The first letter of the subscript of the first velocity in the sum, and the second
letter of the subscript of the last velocity, are the subscripts, in that order, of
the relative velocity represented by the sum.
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These three lengthy and cumbersome statements are very important to consider
when teaching vector addition in general where the subscripts actually represent the
vectors themselves.
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp 44-48 (Relative velocity)
Worked examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp 48
1a.2.11
Curvilinear Motion
1a.2.12
Tangent Vectors and Normal Vectors
We have in the last section shown that the velocity vector always points\in the direction
of motion. In this section we use this observation to extend this to a concept which
applies to any smooth curve which is not necessarily described in terms of time
1a.2.13
The Unit Tangent Vector
The derivative of a vector valued function gives a new vector valued function that
is a tangent to the defined curve. The analogue to the slope of the tangent line is the
direction of the tangent line. Since a vector contains a magnitude and a direction,
the velocity vector contains more information than we need. We can strip a vector
of its magnitude by dividing by its magnitude. 1a.2.14
Definition of the Unit Tangent Vector
Let r(t) be a differentiable vector valued function and v(t) = r’(t) be the velocity
vector. Then we define the unit tangent vector as the unit vector in the direction of
the velocity vector.
v(t)
T(t) = , v(t) ≠ 0
||v(t)||
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Example
Let r(t) = t i + et j - 3t2 k
Find the T(t) and T(0).
Solution
We have v(t) = r’(t) = i + et j - 6t k
and To find the unit tangent vector, we just divide
To find T(0) plug in 0 to get
DO THIS
Show that the unit tangent vector to the curve given by
r (t) = ti + t2
j when t = 1 is given by
1
(i + 2 j)
5
N.B. In this exercise the direction of the unit tangent vector is determined by the
orientation of the curve. Sketch the curve and verify that the unit tangent vector to
the curve
r (t ) = −(t − 2)i + (t − 2)2 j at point (1,1)
would still be the same but will point in the opposite direction.
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DO THIS
Find T (t ) to the curve given by
π
and
4
show that the parametric equations for the tangent line are:
r (t ) = 2costi + 2sintj + tk at the pointwhen t =
x = x1 + as = 2 − 2s
y = y1 + bs = 2 + 2s
z = z1 + cs =
π
+s
4
π
using the point (x1,y1,z1) = ( 2, 2, )
4
1a.2.15
The Principal Unit Normal Vector
A normal vector is a perpendicular vector. Given a vector v in space, there are infinitely many perpendicular vectors. Our goal is to select a special vector that is normal
to the unit tangent vector. Geometrically, for a non straight curve, this vector is the
unique vector that points into the curve or, in other words, the one that points towards
the concave side of the curve. Algebraically we can compute the vector using the
following definition.
Definition: Let r(t) be a differentiable vector valued function and let T(t) be the unit
tangent vector. Then the principal unit normal vector N(t) is defined by
T’(t)
N(t) = ||T’(t)||
Comparing this with the formula for the unit tangent vector, if we think of the unit
tangent vector as a vector valued function, then the principal unit normal vector is
the unit tangent vector of the unit tangent vector function. You will find that finding
the principal unit normal vector is almost always cumbersome. The quotient rule of
differentiation usually appears to complicate this process!. Example
Find the unit normal vector for the vector valued function
r(t) = ti + t2 j and sketch the curve, the unit tangent and the principal unit normal vectors when t = 1.
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Solution
First we find the unit tangent vector
Now use the quotient rule to find T’(t)
Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor:
The first factor gets rid of the denominator and the second factor gets rid of the
fractional power. We have
Now we divide by the magnitude (after first dividing by 2) to get Now plug in 1 for both the unit tangent vector to get
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The picture below shows the graph and the two vectors.
DO THIS (with a colleague)
Given the curve
r (t ) = 3ti + 2tj
Find N (t ).
What is the value of N (t ) when t = 1
Do not turn the page until you have finished the problem!
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Solution
r / (t ) = 3i + 4j and r / (t ) = 9 + 16t 2
⇒ T (t ) =
r / (t )
1
=
(3i + 4tj)
r / (t )
9 + 16t 2
1
⇒ T / (t ) =
9 + 16t
2
12
=
3
(4j) −
16t
3
(3i + 4tj)
(9 + 16t 2 )2
(−4ti + 3j)
(9 + 16t 2 )2
T / (t ) = 12
⇒ N (t ) =
9 + 16t 2
12
=
2 3
(9 + 16t )
9 + 16t 2
T / (t )
1
=
(−4ti + 3j)
/
T (t )
9 + 16t 2
Substituting for t = 1
1
N (1) = (−4i + 3j)
5
1a.2.16
Tangent and Normal Components of Acceleration
Imagine yourself driving down the hill along a curving road towards and having your
brakes fail. As you are riding you will experience two forces (other than the force
of terror) that will change the velocity. The force of gravity will cause the car to increase in speed. A second change in velocity will be caused by the car going around
the curve. The first component of acceleration is called the tangential component
of acceleration and the second is called the normal component of acceleration. As
you may guess the tangential component of acceleration is in the direction of the
unit tangent vector and the normal component of acceleration is in the direction of
the principal unit normal vector. Once we have T and N, it is straightforward to find
the two components. Definition: The tangential component of acceleration is aT = a • T =
v• a
v
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and the normal component of acceleration is a N = a × N =
v× a
v
and
a = aNN + aTT
N.B. The normal component of the acceleration is also called the centripetal component of the acceleration
Example
Prove that the acceleration vector a(t) is in the plane that contains T(t) and N(t)
Proof
First notice that v = ||v|| T and T’ = ||T’|| N
(N.B.We have simplified notation here and used, for example, T for T (t) and so
on)
Taking the derivative of both sides gives
a = v’ = ||v||’ T + ||v|| T’ = ||v||’ T + ||v|| ||T’ || N This tells us that the acceleration vector is in the plane that contains the unit tangent
vector and the unit normal vector. Example Find the tangential and normal components of acceleration for the prior example
r(t) = ti + t2
j Solution
Taking two derivatives, we have
a(t) = r’’(t) = 2j
We dot the acceleration vector with the unit tangent and normal vectors to get
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DO THIS
Exercise: Show that
If r (t ) = 3ti − tj + t 2k then aT =
4t
10 + 4t
2
and aN =
2 10
10 + 4t 2
Exercise
The position function for a projectile is given by:
r (t) = (50 2t)i + (50 2t − 16t2 ) j
Find the tangent component of acceleration when:
t = 0,1, and
25 2
16
Do not turn the page until you have finished the problem!
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Solution
v(t) = 50 2i + (50 2 − 32t) j
v(t) = 2 502 − 16(50) 2t + 162
Activity 1a.3
Gradient, Divergence And Curl
This section is fully covered in the analysis one module of this course. Students can
refresh their understanding of this work by referring to the compulsory readings given
in the analysis one module.
1a.3.1
Divergence
In vector calculus, the divergence is an operator that measures a vector field’s tendency to originate from or converge upon a given point. For instance, for a vector
field that denotes the velocity of air expanding as it is heated, the divergence of the
velocity field would have a positive value because the air is expanding. Conversely,
if the air is cooling and contracting, the divergence would be negative.
A vector field which has zero divergence everywhere is called solenoidal.
Let x, y, z be a system of Cartesian coordinates on a 3-dimensional Euclidean space,
and let i, j, k be the corresponding basis of unit vectors.
The divergence of a continuously differentiable vector field F = F1 i + F2 j + F3 k
is defined to be the scalar-valued function:
Although expressed in terms of coordinates, the result is invariant under orthogonal
transformations, as the physical interpretation suggests.
The common notation for the divergence ∇·F is a convenient mnemonic, and an abuse
of notation, where the dot denotes something just reminiscent of the dot product:
take the components of ∇, apply them to the components of F, and sum the results.
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Physical interpretation
In physical terms, the divergence of a three dimensional vector field is the extent to
which the vector field flow behaves like a source or a sink at a given point. Indeed,
an alternative, but logically equivalent definition, gives the divergence as the derivative of the net flow of the vector field across the surface of a small sphere relative
to the volume of the sphere. (Note that we are imagining the vector field to be like
the velocity vector field of a fluid (in motion) when we use the terms flow, sink and
so on.) Formally,
where S(r) denotes the sphere of radius r about a point p in R3, and the integral is a
surface integral taken with respect to n, the normal to that sphere.
In light of the physical interpretation, a vector field with constant zero divergence
is called incompressible – in this case, no net flow can occur across any closed surface.
The intuition that the sum of all sources minus the sum of all sinks should give the
net flow outwards of a region is made precise by the divergence theorem.
Go to this link
http://en.wikipedia.org/wiki/DIVERGENCE
1a.3.2 Curl
http://en.wikipedia.org/wiki/CURL
1a.3.3 Gradient
In vector calculus, the gradient of a scalar field is a vector field which points in the
direction of the greatest rate of increase of the scalar field, and whose magnitude is
the greatest rate of change.
Consider a room in which the temperature is given by a scalar field φ, so at each
point (x,y,z) the temperature is φ(x,y,z) (we will assume that the temperature does
not change in time). Then, at each point in the room, the gradient at that point will
show the direction in which the temperature rises most quickly. The magnitude of the
gradient will determine how fast the temperature rises in that direction.
Consider a hill whose height above sea level at a point (x,y) is H(x,y). The gradient
of H at a point is a vector pointing in the direction of the steepest slope or grade at
that point. The steepness of the slope at that point is given by the magnitude of the
gradient vector.
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The gradient can also be used to measure how a scalar field changes in other directions, rather than just the direction of greatest change, by taking a dot product.
Consider again the example with the hill and suppose that the steepest slope on the
hill is 40%. If a road goes directly up the hill, then the steepest slope on the road
will also be 40%. If instead, the road goes around the hill at an angle with the uphill
direction (the gradient vector), then it will have a shallower slope. For example, if
the angle between the road and the uphill direction, projected onto the horizontal
plane, is 60°, then the steepest slope along the road will be 20% which is 40% times
the cosine of 60°.
This observation can be mathematically stated as follows. If the hill height function
H is differentiable, then the gradient of H dotted with a unit vector gives the slope
of the hill in the direction of the vector. More precisely, when H is differentiable the
dot product of the gradient of H with a given unit vector is equal to the directional
derivative of H in the direction of that unit vector.
Go to the links below
http://en.wikipedia.org/wiki/Gradient
Worked Examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp154 – 159
Exercise
DO THIS
Find the magnitude of the acceleration a s from the diagram below. What will be the
magnitude and direction of a s if r = 100 m and v = 15.0 m/s (about 34 mph)?
as
Do not turn the page until you have finished the problem!
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Solution
Since the car moving on a properly banked road is equivalent to the bob of a conical
pendulum, the conditions governing the vector a s , are the same as those governing
the equivalent vector a s for the conical pendulum. That is, the horizontal and vertical
v2
components of a s must be given by a sx =
and a sy = g .
r
To find the magnitude of the vector a s , you use the Pythagorean theorem:
a s = (a sx 2 + a sy 2 )1/ 2
1/ 2
⎡⎛ v2 ⎞ 2
⎤
= ⎢⎜ ⎟ + g2 ⎥
⎢⎝ r ⎠
⎥
⎣
⎦
⎛ v4
⎞
= ⎜ 2 + g2 ⎟
⎝r
⎠
1/ 2
For the numbers given, you have the magnitude
1/ 2
⎡ (15.0 m/ s)4
⎤
2 2
as = ⎢
+
(9.80
m/
s
)
⎥
2
⎣ (100 m)
⎦
= 10.1 m/ s2
v2
or about 3 percent more than the acceleration of gravity. Using the equation tanθ =
rg
you can solve for θ to calculate the ideal banking angle. You have
θ = tan −1
v2
rg
θ = tan −1
(15.0 m/ s)2
= 12.90
2
100 m× 9.80 m/ s
Thus the direction of a s is about 130 from the vertical. This is a relatively steep
angle of bank under ordinary roadway conditions.
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Unit 1b: Newton’s Laws of Motion
Specific learning objectives
By the end of this activity you should be able to:
• State Newton’s Laws of motion and apply them.
• Define the concepts Work, Energy, Power and Momentum, and use them to
solve related problems.
• State the Laws of conservation of energy and momentum
• Define the terms impulse and torque.
• Describe the motion of falling bodies and projectiles in a uniform force
field.
• Describe the motion of bodies in a resting medium.
• Define Friction and describe the motion of a body subject to given
constraints.
• Describe potential energy as energy due to position and derive potential energy
as mgh
• Describe kinetic energy as energy due to motion and derive kinetic energy as
mv2/2
• State conservation of energy laws and solve problems where energy is conserved
• Define power as rate of energy transfer
• Define couple, torque and calculate work done by a variable force or torque
• Solve problems where energy is lost due to friction
Summary of the learning activity
In this activity you will acquaint yourself with Newton’s three laws of motion and
their application in the related fields of energy and momentum
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
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Relevant Links and Resources
Friction
http://en.wikipedia.org/wiki/Friction
Newton’s Laws of Motion
http://en.wikipedia.org/wiki/Newton’s_laws_of_motion
Newton’s Laws of Motion
http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html
http://www.waukesha.k12.wi.us/South/physics1/1.4/notes.html
Work
http://en.wikipedia.org/wiki/Work
Work
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html
Kinetic Energy
http://en.wikipedia.org/wiki/Kinetic_energy
Kinetic Energy
http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html
Power
http://en.wikipedia.org/wiki/Power
Potential Energy
http://en.wikipedia.org/wiki/Potential_energy
Potential Energy
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html
List of relevant MULTIMEDIA resources
Hyper physics Link
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
This link has introductory content very good diagrams on Newton’s laws and
their applications which will provide you with the required insights for your
progression through the activity.
Simulations and Experiments on Newton’s Laws
www.compadre.org/precollege/static/unit.cfm?sb=3
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/newtltoc.html
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Key Words
Impulse
Torque
Momentum
Energy
Power
Work
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1b. Learning Activity: Newton’s Laws of Motion
Introduction
In this activity, you will familiarise yourself with three laws that generally govern
the motion of particles and bodies in space- Newton’s three laws of motion and be
able to apply them to real life situations. Through related questions, exercises and
experiments, you will then be taken through the central theme that runs through all the
concepts in this activity- the fundamental notion of force. The relationships between
force, motion, energy, momentum and other related notions will be dealt with through
your involvement with everyday examples and their mathematical models.
Bridging The Gap
•
Bedside Wallet example;
You go to sleep at night and place your wallet on the bed side table. All things
being equal, you should be able to find your wallet on the same bedside table.
If on the other hand, you find the wallet in the bathroom what would have happened? Obviously the wallet cannot move on its own!
•
What would happen if, when approaching a bus stop, on a straight road, the
brakes of the bus failed?
o
•
A driver wakes up in the morning to find his car battery flat – he requires a
push:
o
o
o
What effect do the brakes of the bus have on its motion?
One person gives him a push and the car won’t move.
Two persons give him a push, the car moves, coughs but won’t start
Three persons give him a push and the vehicle starts.
Describe the effect of the three situations on the movement of the car with the help
of a colleague.
♦ Here is an interesting scenario: you and your friend want to collect fruit from
a tree but you are too short and need something to stand on. You get a stool and
succeed in obtaining the fruit. Your friend tries the same thing but is unsuccessful
as the stool breaks.
What could be the possible reasons for your success and your friend’s failure? The
stool does not break in your case but does so in your friend’s.
♦ You need to feed in order to do work. Why?
♦ You place a book on a table. What makes the book to stay on the table and not
fall through the table top?
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♦ Two vehicles collide and there is minimal damage. What happens immediately
after the collisions? What would an investigating police officer ask about the
movement of the cars?
If you have not been able to answer or explain these scenarios then go through the
activity and attempt to answer them.
1b.1
Newton’s Laws of Motion
Visit the links below
http://en.wikipedia.org/wiki/Newton’s_laws_of_motion
http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html
1b.1.1
Newton’s First Law
Consider the section on bridging the gap above and this time the motion, of a cart
driven by a span of two oxen. If the cart is detached from the oxen that were pulling it would quickly come to rest because there is no tractive force pulling it along.
However, other forces also act on the cart such as the force of friction, which helps
slow down the cart.
Newton’s first law summarizes these ideas and is stated thus.
Every object continues in a state of rest, or of uniform motion in a straight line, unless
it is compelled to change that state by forces acting on it.
The implications of this law are that.
• an object at rest remains at rest, and
• an object in motion continues to move with constant velocity.
Inertia - the tendency of an object to maintain its current state of motion, whether
at rest or in motion.
Because of inertia, only external forces can alter an object’s state of motion; internal
forces cannot.
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Internal force - forces generated by one part of a system of objects on another part
of the same system.
Example. Your left hand pushes on your right hand. The engine of a car
pushes on the transmission which pushes on the wheels.
External force - forces generated by something outside the system of objects.
Example. A tennis racket hits a tennis ball. Your foot kicks a chair.
The key point to remember is that no change of motion can occur unless a force from
outside the object causes it to change.
N.B. Care must be taken when applying this law as it only holds for measurements
made in an absolute frame of reference but are subject to correction depending
on the relative position of the observer.
1b.1.2
Newton’s Second Law of Motion
The acceleration of a particle is proportional to the resultant force acting on it and is
in the direction of the force.
Other interpretations are
• When there is a net force acting on an object the object undergoes acceleration
in the same direction as the force.
• The acceleration and the force are proportional in magnitude.
The amount of inertia an object possesses is indicated by its mass.
Mass - A measure of an object’s resistance to change in motion.
The larger the mass of an object, the more it resists a change in its motion. Smaller
masses have less resistance to change in motion. This is an operational definition of
mass, it requires direct measurement of an object’s change in motion to determine
its value.
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If an object experiences an external force, not equal to zero, it will accelerate proportional to the magnitude of the force, but inversely proportional to its own mass.
In equation form: FT = Ma where Force (F) is in Newtons,
mass (M) is in Kg, and acceleration (a) in m/s2
This means that the acceleration of an object depends on the ratio of the total force
to the mass of the object. The direction of the acceleration is the same as that of the
total force which caused it.
Newton’s second law is the work horse of our mechanical world; it connects the
changes in motion of all objects with the forces needed to make that change. Its a
simple equation to work, but its application can become complicated as you will see
in the next section.
Some examples:
a) If a total force of 10 N is placed on 2 Kg ball, what is its acceleration?
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b) If you push with a force of 25 N on a 2 Kg box to slide it across the floor,
what is the acceleration of the box if the frictional force is 10 N?
This law led to the equation;
F = ma , F = Force , a = acceleration , m = mass of object
Another way of stating this law is:
The rate of change of momentum of a body is proportional to the resultant force and
occurs in the direction of the force.
Mathematically modelling this law we have, if the body changes velocity from u to
v in t seconds
changeof momentum = mv − mu
mv − mu
t
m(v − u)
=
t
F α ma ⇒ F = kma
rate of change of momentum =
Thus and it follows, by defining the unit of Force as the Newton,
F = ma
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1b.1.3
Newton’s Third Law of motion
The forces of action and reaction between interacting bodies are equal in magnitude,
opposite in direction and are co-linear.
Simply put, forces are social beings, they always operate in pairs. It is not possible
to have an isolated, or single force. Isaac Newton expressed it a bit more formally:
If an object expresses a force upon another object, then the latter of the two
exerts a force of equal magnitude back upon the first.
We call these action and reaction forces. It doesn’t make any difference which
you call the action and which the reaction since they happen concurrently and are
reversible.
Example. Your body pushes on the chair, the chair pushes on your body.
A tennis racket hits the ball, the ball hits the racket.
Point to remember: Although these two forces are equal, but oppositely directed,
they do not cancel each other out. They operate on different objects so they cannot
be added together.
If you kick a chair, your foot exerts a force on the chair. The chair exerts a force on
your foot.
See, one force is on the chair, the other force is on your foot; two different objects.
The only force which can cancel any applied force is another force on the same object. The only force which can cancel your foot’s force on the chair is another force on
the chair, like friction between the chair and the floor.
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Remember inertia? How an internal force cannot cause a change in motion? Here’s
how it works in the real world.
Stand up and jump straight up into the air. If you think about what you did you bent
your knees, pushed on the floor, and went up into the air. We say “I jumped”.
Actually, you cannot do anything to yourself to make you go into the air from rest
on the floor. There has to be a force on your from outside of you to make you go
into the air.
Why do you push on the floor? To get the floor to push on you! That’s the action/reaction part of the operation. The floor’s force on you is what’s needed to make you go
up. Pretty tricky, huh? That’s why it took you a while to learn to walk and jump.
More simply put, if body A exerts a force on body B, then body B an equal and opposite force on body A.
Thus forces never occur singly but in pairs as a result of the interaction between the
bodies.
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N.B. You will note that in all the systems in which we will study Newton’s Second
Law, we will also be implicitly and explicitly studying and using Newton’s Third
Law.
Worked Examples
Two shot put balls, one weighing 0.70 kg and the other 7.0 kg both fall towards the
grant with the same downward acceleration of magnitude 9.8m/ s2 . Find the force,
w, exerted by gravity on each ball.
M = 0.7
M = 7.0
W
W
Fig: Two masses with the force of gravity acting on them
By Newton’s Second Law,
For the smaller shot
w = mg = 0.20 × 9.8
= 6.9 Newtons
and for the larger shot-put
W = Mg = 7.0 × 9.8
= 69 Newtons
Example
A Block slides down a long frictionless plane which is at an angle of 690 with respect
to the horizontal. Find the acceleration of the block and the distance the block has
moved 3 seconds from rest.
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Solution
The components of W parallel and perpendicular to the plane are
Wx = mgsinθ
and
Wy = mgcosθ
Since the place is frictionless then N x = 0
And N y = N (when N x and N y are the reactions of the plane to the block parallel
and perpendicular to the plane).
And therefore, the only force that is working on the block is Wx = ma x
⇒ mgsinθ = ma
⇒ a = gsinθ
= gsin 69°
Using S = ut + 12 at 2
and putting u = 0 and S = x
Then
x = 12 ax(30)2
= 12 gsin 69 ⋅ 9
= 92 gsin 69 metres
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Do this
Exercice
A block of mass m slides down a plane supported from the earth at an incline the
angle θ between the plane and the horizontal is adjusted until the block slides with
constant speed. Find the co-efficient of friction between the block and the slope. If
θ is 350, what is the value of μ.
Do not turn the page until you have completed the solution!
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Solution
Since the block is sliding down the plane, then
FR
= μ , where F R is the Frictional force and N the normal reaction
N
We know that N = mgcosθ
and since
F R = μN
then
F R = μmgcosθ
Since motion is down the slope, when the block begins to move
mgsinθ = μmgcosθ
⇒ μ = tanθ
⇒ μ = tan 35°
= 0.70
1b.1.4
Application Of Newtons Laws
The most widely applied Newton’s Law is the Second Law of motion,
F = ma
which is applied in the study of motion of bodies in a variety of systems.
N.B. The quantity F in the equation is the Net force of all the forces acting on the
parts of the system and thus F is the vector sum of all the forces acting on the
part. The question then to answer is:
What are the magnitude and directions of all the forces?
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1b.2
Motion of Connected bodies
The contraption consists of two bodies of mass m1 and m2 attached to the
ends of weightless and inextensible string which runs over a smooth, weightless
pulley. If m1 and m2 are not equal the system will begin to move as soon as it is
released.
If T1 and T2 are the forces exerted by the pulley on the weights and, m1 g and m2 g
are the gravitational forces exerted on the two then, if m1 > m2
m1 g − T1 = m1 a1 and
T2 − m2 g = m2 a2
But surely
T1 = T2 and
a1 = a2
then
T1 − m2 g = m2 a1
and
m1 g − T1 = m1 a1
Adding
(m1 − m2 )g = (m1 + m2 )a1
Thus
a1 =
m1 − m2
g
m1 + m2
and this gives us the equation of motion for the system.
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DO THIS
Exercice
In the diagram, the string and the pulley have negligible mass and the friction in the
pulley system is negligible. If the two bodies are initially at rest at the same level,
how long will it be before the vertical separation between them is 1,5m? How fast
will they be moving?
Do not turn the page until you have completed the solution!
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Answer
t = 2.5s and v = 0.60s
1b.3
Work
Visit the links below:
•
•
http://en.wikipedia.org/wiki/Work
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html
Definition
The work done, dU , by a force F during a small displacement ds of its point of
application is given by
dU = F ⋅ ds
which is the dot product between F and ds.
The magnitude of the work is given by
dU = F ⋅ dscosα
From the equation
dU = F ⋅ ds
Then
U = ∫ F ⋅ ds =
( ∫ F dx + ∫ F dy + ∫ F dz) x
y
z
= ∫ Ft ds
Example
The most common example of work done on a body is the action of a body on a
spring to which it is attached.
If the spring constant is K then
F = −kx
The force exerted on the spring whether it is tension or compression is opposite to
the displacement and so does negative work on the body
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Thus
x2
x2
U = − ∫ F dx = − ∫ kxdx = − 12 k(x22 − x12 )
x
x
1
1
Next we consider
the work
done on a particle of mass m moving along a curved path
under the action of the Force, F. The resultant force is
∑F .
• the position of m is described by the position vector r.
• the displacement during time dt is given by the change dr in its position vector.
now
dU = F ⋅ dv
and
x2
U = ∫ F ⋅ dv =
∫ F dx
t
x1
Given that F = ma
then
but
U = ∫ F ⋅ dv = ∫ ma ⋅ dv a ⋅ dv = at ds
and
at ds = v ⋅ dv
thus
U = ∫ F ⋅ dv = ∫ mv ⋅ dv
= 12 m(v22 − v12 )
which is the integration carried out between the change v1 to v2 in the velocity.
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1b.4
Kinetic Energy
Visit the links below:
• http://en.wikipedia.org/wiki/Kinetic_energy
• http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html
The Kinetic Energy of a particle is defined as
K .E . = T = 12 mv2
and is the work done on a particle to bring it to a velocity V from rest and is always
a positive quantity.
The unit is the Joule.
thus the equation
U = 12 m(v22 − v12 )
may be written as
U = ΔJ
N.B. This is called the Work- Energy equation of a particle and states that
the total work done by all forces acting on a particle during an interval of its motion
i.e. equal to the change in the Kinetic energy of the particle.
Another way of putting it is that the Final K.E., T2, is the sum of the Initial K.E., T1
and the work done U.
T2 = U + T1
The application of the work-energy equations requires an isolation of the particle or
system under consideration.
1b.5
Power
Visit the link below:
•
http://en.wikipedia.org/wiki/Power
Definition: Power is the capacity of a machine to do work and is measured by the
time rate in which it can do the work or derive energy.
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P=
dU
dx
=F⋅
= F ⋅v
dt
dt
and is measured in watts.
Worked Example
In the diagram m = 50kg , μ = 0.30 , θ = 15 , u1 = 4m/ s . Calculate the velocity
of the crate at B.
N.B. Can you explain each step of the solution?
U = F s = ⎡⎣50(9.81)sin15 − 142 ⎤⎦ 10
= −152J
Change in K .E . = T2 − T1 = ΔT .
T = 12 mv2 , ΔT = 12 50(v2 − 42 ) , U = ΔT
∴ −152J = 12 50(v2 − 42 )
∴ v = 3.15m/ s2
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DO THIS
Exercise
1. A truck, carrying a concrete block of 80kg, starts from rest and attains a speed
of 72km/h after travelling for 75m on a road with constant acceleration. Find the
work done by the Friction working on the block during this motion if the static
and kinetic coefficients of friction between the block and truck bed are:
(a) 0.30 and 0.28 respectively, and
(b) 0.25 and 0.20 respectively
2. The position vector r, of a particle is given by r = 8ti + 1.2t 2 j − 0.5(t3 − 1)k , t
is time in seconds from the start of motion. Determine the power P developed by
the force F = 10i − 5 j − 9k Newtons which acts on the particle when t = 6s .
Do not turn the page until you have completed the solution!
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Answers
(a) 16.0 KJ
(b) 8.66 KJ
1b.6
Potential Energy
Visit the links below:
•
•
http://en.wikipedia.org/wiki/Potential_energy
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html
Definition: The Potential Energy of a system is the energy the system has because
of the relative positions of its points and, that is, due to its configuration.
It arises when a body experiences a force in a field such as the earth’s gravitational
field and the potential energy is regarded as the joint property of the body-earth system
and not of either body separately.
1b.7
Conservation of Energy
The full definition of conservation of energy is given in the link below:
http://en.wikipedia.org/wiki/Conservation_of_energy
1b.8
Impulse and Torque
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP pp 271-272
For information on Torque also see the link below:
http://en.wikipedia.org/wiki/Angular_momentum
1b.9
Friction
Definition: Friction is the force that opposes the relative motion or tendency toward
such motion of two surfaces in contact. It is not a fundamental force, as it is made
up of electromagnetic forces between atoms. When contacting surfaces move relative to each other, the friction between the two objects converts kinetic energy into
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thermal energy, or heat. Friction between solid objects and fluids (gases or liquids)
is called drag.
For further reading please refer to the following link:
http://en.wikipedia.org/wiki/Friction
Note that in our reading we use the traditional rather than the vector approach in the
treatment of the topic and that we rely very heavily on adaptations of equations of
uniform motion in a straight line to come up with projectile motion equations. Of
particular importance are the derivations of equations such as the range, the maximum
range, the greatest height attained and so-on…
It is advisable that you DO NOT memorize these equations and that you should be
able to work from first principles through given situations
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Unit 2 : Oscillations
Specific learning objectives
By the end of the unit you should be able to:
o
o
o
o
Describe Simple Harmonic Motion
Model motion that is Over-damped, Under-damped and Critically-damped
Describe the Simple pendulum
Derive the two and three dimensional oscillator
Summary
In this unit we look at the motion of a body when the resultant force acting on it is
not constant, but varies during the motion. Naturally, there is an infinite number of
ways in which a force may vary; hence no general expressions can be given for the
motion of a body when acted on by a variable force, except that the acceleration at
each instant must equal the force at that instant divided by the mass of the body. We
will look at motion which is oscillatory or periodical using examples like simple
pendulums and masses suspended on a string to illustrate this motion.
Required Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Relevant Links and Resources
Simple harmonic motion
http://en.wikipedia.org/wiki/Simple_harmonic_motion
Simple harmonic motion
www.phy.ntnu.edu.tw/java/shm/shm.html
Simple harmonic motion
http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html
Simple harmonic motion
www.kettering.edu/~drussell/Demos/SHO/mass.html
Simple harmonic motion
http://theory.unwinnipeg.ca/physics/shm/nodes2.html
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Classical Mechanics Lecture notes
http://farside.ph.utexas.edu/teaching/301/lectures/lectures.html
Classical Mechanics: An Introductory Course
http://www.lulu.com/content/159798
Key Words
•
•
•
•
•
•
Harmonic Oscillator
Simple Pendulum
Natural Circular frequency
Angular Harmonic Motion
Damped Motion
Hooke’s Law
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Learning Activity 2: Oscillations
2.1
Bridging the Gap
Two children are playing on a swing- one in the sit and the other ‘pushing’ And
waiting for her turn on the seat. Have you ever wondered why
• After the hardest push, the swing always continues with its usual motion of
coming back to its resting place and through it until it reaches its highest
backward position and returns. Why does the swing not coil itself on the
horizontal support or, worse still , go off its supports ?
• If the ‘pusher’ stops and the child on the seat stops ‘kicking’ what happens to
the motion of the swing ?
Fig 2.1
Parents with smiling child on swing
Take a spiral spring, fix one end of the spring, pull the free end slightly and then
release it.
•
Describe the action of the spring.
Introduction
Consider the motion of the child on the swing above. Before she is pushed or «kicks»
herself in the swing, the swing hangs vertically downwards. In this position, we say
that the swing is in its equilibrium position. When the swing is pushed, then it is
displaced from its equilibrium position but the tendency has always been to return
back to its equilibrium position. What causes this tendency?
There surely must be some force that causes it to go back-and forth motion past the
equilibrium position and this force has been aptly named the restoring force.
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DO THIS
Try it yourself!
•
•
•
•
•
Tie a stone on one end of the string
Suspend the stone from the open door frame by fixing one end of the stone
to the top of the door frame
In the equilibrium position the stone hangs straight down
Give it a little displacement from this position by pushing it, say, to the left
What happens to the stone?
The stone does not simply return to the equilibrium position but instead swings back
and forth, past the equilibrium position in a regular repetitive manner.
Can you give other examples of this motion which is defined as periodic or oscillatory motion?
2.1
The Simple Harmonic Oscillator
Fig 2.2
Simple harmonic Oscillator
Consider the movement of a mass m attached to the end of a spring, the other end
which is held stationary at A. We will describe the position of the body with the coordinate x, where x denotes the displacement from the equilibrium position O, the
position of zero spring deflection. We will restrict our discussion to a linear spring.
Such a spring exerts a restoring force -kx, on the mass which means that when the
mass is displaced to the right, the spring force is to the left and vice versa.
k is known as the spring constant or modulus or stiffness of the spring
Thus:
F = −kx
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Note here that the magnitude of the restoring force is directly proportional to the
displacement of the system from equilibrium.
That is F α x .
From Newton’s 2nd Law
F =ma
and this implies that
– KX
or +KX = 0
This differential equation is known as the Simple Harmonic Equation
Read this link
http://en.wikipedia.org/wiki/Simple_harmonic_motion
DO THIS
Exercice
Show that the solution of the equation is
x = a cos(ωt − ϕ) where ϕ, ω, a are constants and ω =
k
m
Hint: We are solving a Second Order Differential Equation which is Homoge2
neous.
Alternatively, and with the same conditions,
+ kx=0 is normally written as + ω2x = 0
ω is called the natural circular frequency and has units in radians per second
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Note that the displacement x oscillates between
x = − a and x = + a, where a is the amplitude of the oscillation
the number of complete cycles per minute, the natural frequency f =
and the time required for one complete motion’s cycle is given by
τ=
ω
n
2π
1 2π
=
f
ω
is the period of the motion
Compulsory Reading Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
pp 237-238
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DO THIS
Exercice
Fig 2.3
Model for Periodic motion
A spring is mounted as shown in the diagram above. Through the attachment of a
spring balance to the free end, it is determined that the force is proportional to the
displacement, a force of 4 N causing a displacement of 0.02 m. A body of 2 kg mass
is then attached to the free end, pulled for a distance of 0.04 m and released.
a)
b)
c)
d)
e)
Find the force constant of the spring
Find the period and frequency of vibration
Compute the maximum velocity attained by the oscillating body
Compute the maximum acceleration
Compute the velocity and acceleration when the body has moved halfway in
towards the centre from its initial position
f) How much time is required for the body to move halfway in to the centre from
its initial position
Do not turn the page until you have completed the solution!
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Solution
Before we answer this question, we need to establish the basic equations of simple
harmonic motion.
We know from above that
x&&== a = −
k
x
m
That is, the acceleration at each instant is proportional to the negative of the instant.
When x has its maximum positive value A, the acceleration has its maximum negative
value −k
A
and at the instant the particle passes the equilibrium position x = 0, the
m
acceleration is zero. However, its velocity at this position is not zero at this point.
Looking back, we can use the Conservation of Energy Principle to analyze some
aspects of simple Harmonic Motion. In a spring, the restoring force is a conservative
force (Remember this from previous activities) and the work done can be stated in
terms of potential energy.
P.E. of Conservative Force =
1 2
kx
2
1
The Kinetic Energy at any point x is again = mv2 and by the Principle of Conser2
vation of Energy, the total Energy.
E =
1 2 1 2
mv + kx is a constant
2
2
Also, when the particle reaches its maximum displacement ± A it stops and turns
back to its equilibrium position. At the instant of timing, V = 0 and there is no Kinetic
Energy and total Energy
E =
1 2
kA
2
thus we have
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1 2 1 2 1 2
mv + kx = kA
2
2
2
v=±
k
A2 − x 2
m
DO THIS
Exercice
When does the speed have its maximum value vm ax (Discuss with a friend)
Do not turn the page until you have completed the solution!
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Answer
The speed has its maximum value at the mid-point because at the mid-point the energy
is all Kinetic and at the end point the energy is all Potential.
Thus:
1
mvm ax 2 = E
2
⇒ vm ax =
2E
m
1 2 1
kA = mvm ax 2 = E
2
2
⇒ vm ax =
k
A
m
Looking at the equation
v=±
k
A2 − x 2 = ±ω A2 − x 2
m
We cannot tell where the particle is at any given time and to be able to have a complete description of motion we need to know the position, velocity, and acceleration
at any given time.
We can enliken the simple harmonic motion of the particle at the end of the spring
to a circle of reference given in the Diagram below.
Fig 2.4
Uniform circular motion
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Suppose point Q moves anticlockwise around a circle of radius A with constant an-
uuur
gular velocity (remember this?) W. The vector OQ represents the position of point
Q relative to O and θ is the angle this vector makes with the positive x- axis.
The horizontal component represents the actual movement depicted by the spring
and particle. You will realize that as Q moves, P represents the actual movement of
the particle and spring.
The displacement OP at any time t is x and
x = Acosθ
and if Q is at the extreme right of the diameter at time t = 0, then
θ = ωt
and hence
x = Acosωt
Thus, the velocity at any time t
And acceleration
You will notice that what we have done here is to assume that when t = 0, x = A, is
positive maximum displacement.
At a different initial position, say, if t = 0 OQ makes and angle θ0 while the positive
x-axis then θ at time t is given by
θ = θ 0 + ωt
And the equations become
x = Acos(θ 0 + ωt)
v = −ω Asin(θ 0 + ωt)
a = −ω 2 Acos(θ 0 + ωt) = −ω 2 x
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By using these results you should be able to verify that the answers to the question
above are
(a) k = 200 Nm-1
(b) τ = 0.628 s, f = 1.59 Hz and ω = 10 s-1
(c) vmax = ± 0.4 ms-1
(d) amax = 4.0 ms-2
(e) v = -0.346 ms-1, a = -2.0 ms-2
(f) t = π/30 s
2.2
The Torsion Pendulum (Angular Harmonic Motion)
Fig 2.5
A torsion pendulum
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
p241-242
The system shown in the compulsory reading is quite self-explanatory. In other
instances Angular Harmonic motion results when a body that is pivoted about an
axis experiences a restoring torque proportional to the angular displacement from
its equilibrium position. The balance wheel from a kinematic watch (obviously not
a battery operated watch) is a very good example of angular harmonic motion.
Note that in this case the restoring force becomes the restoring torque
The Restoring torque is proportional to angular displacement and is given by
Γ = − k ʹθ
where k’ is the torque constant
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The moment of inertia I of the pivoted body corresponds to the mass of a body in
linear motion. Thus the period of angular Harmonic motion is
I
kʹ
τ = 2π
and angular frequency is
ω=
kʹ
I
with the period of oscillation being
τ=
2.3
2π
I
= 2π
ω
kʹ
The Simple Pendulum
Compulsory Reading Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
p242-244.
Fig 2.6
Simple Pendulum
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The simple pendulum is a simplified model of a more complex system. It consists
of a mass suspended by a weightless rod or a weightless and inextensible string in
a uniform gravitation field. When pulled to one side of its equilibrium position and
released, the rod (mass) vibrates about this position.
Does the motion constitute simple Harmonic motion?
We need to establish whether the restoring force is directly proportional to the coordinate x and oppositely directed. The path of the rod is not a straight line but an
arc of a circle of radius L, the length of the string or rod. The coordinate x refers to
distances measured along this arc.
Then if F = -kx the motion will be simple harmonic.
The tangential components of the net force acting on the pendulum bob is given by
F = −mgsinθ
This means that the restoring force is proportional to sin θ and not to θ and therefore
the motion is not simple harmonic.
How do we get around this problem?
DO THIS
Exercice
Demonstrate that if θ is small, then sin θ is very nearly equal to θ.
If you have demonstrated this then
Lθ = x and θ =
x
L
and therefore
F = −mgθ = −mg
x
mg
=−
x
L
L
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Thus the restoring force is proportional to the coordinate for small displacements and
the force constant k is represented by
Period = τ = 2π
= 2π
mg
leading to the following results
L
m
mL
= 2π
k
mg
L
g
The frequency relations are f =
1
2T
g
and ω =
L
g
L
Thus the period of oscillation is entirely determined by the length 2 of the pendulum
and only for small oscillations does the motion of the pendulum approximate Simple
Harmonic Motion.
A more analytical way to describe pendulum motion is to obtain an expression for
aT, the tangential component of the acceleration of the pendulum
Fig 2.7
Motion of Simple Pendulum
If we use a small displacement vector ds in a corresponding small interval of time
dt
then v=
ds
dt
But, from the diagram above
ds = ldφ
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Therefore
v=
ds ldφ
dφ
=
=l
dt
dt
dt
The rate of change of speed is
⎛ dφ ⎞
d⎜ l ⎟
⎝ dt ⎠
dv
=
dt
dt
d ⎛ dφ ⎞
=l ⎜ ⎟
dt ⎝ dt ⎠
d 2φ
=l 2
dt
Which gives the tangential acceleration aT of the bob
Thus
d 2φ
aT = l 2
dt
Now, from Newton’s second law, the tangential component of the force is
FT = maT
but, from above
Ft = F = −mgsinφ
Thus
d 2φ
−mgsinφ = ml 2
dt
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Thus
d 2φ
g
= − sinφ
2
l
dt
Compare this with
(equation 11.23) Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.UTP
on page 244]
Taking α = −
g
l
then the pendulum equation is
d 2φ
= α sinφ
dt2
The argument goes on to say that, for small angles, sinφ is the approximately the
same as φ if φ is small compared to 1 radian.
Thus
d 2φ
= −αφ
dt2
2.4
The Damped Oscillator
In classical mechanics, a harmonic oscillator is a system which, when displaced
from its equilibrium position, experiences a restoring force F proportional to the
displacement x according to Hooke’s law:
where k is a positive constant.
If F is the only force acting on the system, the system is called a simple harmonic
oscillator, and it undergoes simple harmonic motion: sinusoidal oscillations about
the equilibrium point, with a constant amplitude and a constant frequency (which
does not depend on the amplitude).
If a frictional force (damping) proportional to the velocity is also present, the harmonic oscillator is described as a damped oscillator. In such situation, the frequency
of the oscillations is smaller than in the non-damped case, and the amplitude of the
oscillations decreases with time.
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If an external time-dependent force is present, the harmonic oscillator is described
as a driven oscillator.
Mechanical examples include pendula (with small angles of displacement), masses
connected to springs, and acoustical systems.
Go to this link
http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html
DO THIS
Set up a simple pendulum as in an earlier activity and set it in motion: Does the pendulum continue oscillating? If it does eventually stop, why does it stop?
A vehicle hits a hump on the road. It is quiet safe to say that the spring will perform
oscillations which would continue if there was no opposing force to stop then, where,
what provides this force?
In the two cases cited above, the amplitude of the oscillations of the pendulum and
the springs of the car gradually decreases to zero due to the sensitive force that arises from the act, for the pendulum, and that is provided by the shock-absorber. The
motion is therefore not a perfect Simple Harmonic Motion and is said to be damped
by a resistance, friction or the shock absorber.
The behaviour of any mechanical system depends on the extent of damping. Underdamped oscillations are said to be free and perform near Simple Harmonic Motion.
The graph of this motion is
Fig 2.8
A lightly –Damped or Under-Damped motion
However, every mechanical system possesses some inherent degree of friction, which
acts as a consumer of mechanical energy.
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The viscous damper (or dashpot), used as a basis for shock absorbers in vehicles
and related machines is a device in intentionally added to system for the purposes
of limiting or reducing vibratory oscillations. The basic compounds are a cylinder
filled with a viscous fluid and a piston which allows the fluid to flow from one side
of the piston to the other.
Schematically, the diagram below illustrates this system.
Fig 2.9
Viscous Damper
The dashpot exerts a force F whose magnitude is proportional to the velocity of the
mass. The constant of proportionally cv is known as viscous damping co-efficient.
The direction of the damping force as applied to the mass is opposite to that of the
velocity . Hence the force on the mass is
.
From Newton’s Second Law, the equation of motion of the mass is
Alternatively
−kx − cv
dx
d2 x
=m 2
dt
dt
If we substitute wn =
cv
k
and introduce a constant ζ =
. The quantity ζ
m
2mwn
(Zeta) is called the Viscous damping factor damping ratio and is a measure of the
severity of the damping.
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Example
Fig 2.12
Viscous Damping
An 8 kg body is moved 0.2 m to the right of the equilibrium body, from rest at time t
= 0 s. Determine its displacement at time t = 2 s. The viscous damping co-efficient
c is 20 Nsm-1 and the spring co-efficient K is 32 Nm-1
Solution
We must first determine the degree of damping by computing the damping ratio ζ
ωn =
k
32
=
= 2 radians / s
m
8
c
20
=
= 0.625
2mω n 16 ⋅ 2
ζ=
Since ζ is less than 1 then the system is underdamped. The damped natural frequency is
ω n 1 − ζ 2 = 2 1 − (0.625)2 = 1.561 rad / s
The motion is given by
When t =0,
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x0 = E sinψ = 0.2
x0 = −1.25E sinψ + 1.561cosψ = 0
Solving for
E and ψ we have
E = 0.256 mand
ψ = 0.896 rad
Therefore Displacement
x = 0.256 me−1.25t sin(1.561t + 0.896)
When t=2
x = -0.0162
DO THIS
Exercice
Verify that
ζ is non-dimensional.
Thus the equation becomes
Alternatively
d2 x
dx
+ 2ζ wn
+ wn 2 x = 0
2
dt
dt
Hint: This is a Second Order Homogeneous Differential Equation
Do not turn the page until you have completed the solution!
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Solution
This is a differential equation with a possible solution of
x = Aeλ t
Substituting we have
λ 2 + 2ζ wn λ + wn 2 = 0
giving
λ1 = wn [−ζ + ζ 2 − 1] or
λ2 = wn [−ζ − ζ 2 − 1]
By inspection, the general solution is
λ t
λ t
x = Ae 1 + Be 2
= Ae⎝
⎛
⎞
2
⎜ − ζ + ζ − 1 ⎠⎟ ω n t
⎛
⎞
2
⎜ − ζ − ζ − 1 ⎠⎟ ω n t
+ Be⎝
The discriminant ζ 2 − 1 determines the extent of the damping by considering the
values it can take.
(i) If ζ > 1 : Remember from your school days that the roots λ1 and λ2 are real
and distinct negative numbers and the motion depicted below delays so that x
approaches zero at t = ∞ for large value of t. There is no oscillation and no
period associated with the motion. The motion is OVERDAMPED (HEAVILY
DAMPED).
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Fig 2.10
Overdamped and Critically Damped Motion
(ii) If ζ = 1 : The roots are equal real negative numbers ie λ1 = λ2 = −ω n and the
solution to the equation in this case is given by
− ω nt
x = ( A + Bt)e
As depicted in the diagram above, the motion decays, x approaches zero for t = ∞
and the motion is non-periodic. This motion is called CRITICALLY DAMPED
and will approach equilibrium factor than will be an overdamped system.
(iii)If ζ < 1 : The roots are complex for this UNDERDAMPED (LIGHTLY
DAMPED) and transforms the equation to
{
2
i 1− ζ ω n t
x = Ae
2
− i 1− ζ ω n t
+ Be
}
− ζω n t
e
where
Setting a new variable ω k to represent ω n 1 − ζ 2
We have
{
iω k t
x = Ae
− iω k t
+ Be
}e
− ζω n t
i − −1
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Employing Euler’s formula
e± ix = cos x ± i sin x
We have
{
}
− ζω n t
x = A(cosω k t + i sinω k t) + B(cosω k t − i sinω k t) e
= ⎡⎣( A + B)cosω k t + i( A − B)sinω k t ⎤⎦ e
− ζω n t
− ζω k t
= ⎡⎣C cosω k t + D sinω k t ⎤⎦ e
{
}
− ζω k t
= E sin(ω k t + ψ ) e
− ζω k t
= Ee
sin(ω k t + ψ )
DO THIS
Exercice
Show that
C cosω k t + D sinω k t can be written as E sin(ω k t + ψ )
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2.5
Underdamped Oscillator
When a damped oscillator is underdamped, it approaches zero faster than in the case
of critical damping, but oscillates about that zero.
Fig 2.11
Underdamped Oscillator
The frequency ω k = ω n 1 − ζ 2
τd =
and the damped period is given by
2π
2π
=
ωk ω 1− ζ 2
n
DO THIS
Exercise
Which of the motions discussed above best describes the action of shock absorbers
on a vibrating vehicle?
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Unit 3
Dynamics
Specific learning objectives
By the end of this unit you should be able to:
o
o
o
o
Derive the equation of motion for a particle in a central field.
Find the potential energy of a particle in a central field.
State the laws of conservation of energy.
State, derive and apply Kepler’s laws of planetary motion in solving related
problems.
o Find velocity and acceleration of moving systems.
o Define conservation of momentum and angular momentum.
o Solve problems involving changing mass.
Summary
In this module you will familiarize yourself with historical developments concerning
planetary motion beginning with early attempts to explain this phenomenon. You will
also acquaint yourself with early and modern mathematical models that exemplified
this phenomenon resulting in the famous Kepler’s Laws.
Required Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Relevant Links and Resources
Classical Mechanics Lecture notes
http://farside.ph.utexas.edu/teaching/301/lectures/lectures.html
Newtonian Physics
http://www.lightandmatter.com/arealbook1.html
Planetary motion
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/kepler6.html
Planetary motion
http://www-istp.gsfc.nasa.gov/stargaze/kep3laws.htm
Planetary motion
http://en.wikipedia.org/wiki/Kepler’s_laws_of_planetary_motion
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Planetary motion
www.windows.ucar.edu/tour/link=/the_universe/uts/planets.html
Coriolis Effect
http://en.wikipedia.org/wiki/Coriolis_effect
Coriolis Force
http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html
Centripetal Force
http://en.wikipedia.org/wiki/Centripetal_force
Angular Momentum
http://en.wikipedia.org/wiki/Angular_momentum
D’Alembert’s Principle
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle
Key Words
•
•
•
•
•
Central forces
Planetary motion
Central field
Kepler’s laws
Coriolis
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Learning Activity 3 Dynamics
3.1
Bridging the Gap
What stories were you told when you were young about
•
•
•
the rising and setting of the sun
the rising and setting of the moon
the general movement of the sun, moon and the earth ?
Consult some of the elders in your local community and hear their views in these
issues.
Have you ever wondered why, when you look at the diagrams of thr planetary system
why
•
•
the planets don’t just ‘fall off, into space
The planets and the sun maintain their distance apart and do not collide
Fig 3.1
Solar system
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3.2
Central Forces and Planetary Motion
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 262-275
3.2.1
The concept of central force
It is with Kepler that we see the beginning of the concept of central force, that is a
force which acts continuously on the planets to keep them moving in closed, stable
paths from one orbit to the next. It was apparent to Kepler that the force was directed
towards the focus of the ellipse, but he could not describe the nature of the force.
3.2.2
Motion in a general central force-field
Go to this link
http://farside.ph.utexas.edu/teaching/336k/lectures/node53.html
3.3
Kepler’s Laws Of Planetary Motion
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 260
3.3.1
Law 1
The orbit of a planet/comet about the Sun is an ellipse with the Sun’s center of mass at one focus
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This is the equation for an ellipse:
x2 y2
+
=1
a 2 b2
3.3.2
LAW 2: A line joining a planet/comet and the Sun sweeps out equal
areas in equal intervals of time
3.3.3
LAW 3: The squares of the periods of the planets are proportional to the
cubes of their semimajor axes:
Ta2 / Tb2 = Ra3 / Rb3
•
•
•
3.3.4
Square of any planet's orbital period (sidereal) is proportional to cube of its
mean distance (semi-major axis) from Sun
Mathematical statement: T = kR3/2 , where T = sideral period, and R = semimajor axis
Example - If a is measured in astronomical units (AU = semi-major axis of
Earth's orbit) and sidereal period in years (Earth's sidereal period), then the
constant k in mathematical expression for Kepler's third law is equal to 1, and
the mathematical relation becomes T2 = R3
Examples of Kepler’s Third Law
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
P (yr) a (AU)
0.24
0.39
0.62
0.72
1.00
1.00
1.88
1.52
11.9
5.20
29.5
9.54
T2
0.06
0.39
1.00
3.53
142
870
R3
0.06
0.37
1.00
3.51
141
868
In simple terms Kepler’s laws are stated as follows:
1. The motion of planets and other bodies subject to the same force is in orbits
that are «conic sections»: ellipses or hyperbolae or in very special circumstances parabolas (all with the sun as a focus), or straight lines.
2. The area swept out per unit time in any orbit is constant.
3. There is a certain specific relation between the period of an elliptical orbit
and a measure of its radius, which relation we will not discuss further.
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Gravitational Potential Energy
Definition
The Gravitational potential energy is the work done by raising a mass of m Kg, h
above a reference point where the work done is taken to be zero.
Potential Energy is then written as
V g = mgh
This work is taken as energy because the work done may be converted to energy if
the body was allowed to work on a supporting body to return to its original position.
Thus change in potential energy for a body moving from initial height h1 to height
h2 is
ΔV g = mg(h2 − h1 ) = mgΔh
The attendant work done by the gravitational force is
−mgΔh
which is the negative of the change in potential energy.
Newton’s Law of Gravitational governing the attraction between two bodies of mass
m1, m2 is given by
F =G
where
m1 m2
r2
F = force of attraction between two particles
G =Universal Constant of gravitation
= 6.673 × 10−11 m3 / kg ⋅ s2
r =distance between the centres of the particles.
3.3.5
Examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 275-278
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3.3.6
Example
Consider a particle of mass m, moving under the action of the central gravitational
attraction.
F =G
mm0
r2
Where m0 is the mass of the attracting assumed to be fixed, G is the universal gravitational constant, and r is the distance between the centers of the masses. The particle
of mass m could represent the earth moving about the sun, the moon moving about
the earth, or a satellite in its orbital motion about the earth above the atmosphere.
The most convenient coordinate system to use is polar coordinates in the plane of
motion since F will always be in the negative r-direction and there is no force in the
θ − direction . This implies that
The second of the two equations when multiplied by r/m is seen to be the same as
, which is integrated to give
, a constant
This is reduced to
And
Or
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After substitution we get
−Gm0 u2 = −h2 u2
d2u 1 2 4
− hu
dθ 2 u
Or
Gm
d2u
+ u = 20
2
dθ
h
Which is a non-homogeneous linear differential equation.
3.3.7
DO THIS
3.3.7.1
Exercise
Verify that the solution of this familiar second-order equation is
Gm
1
u = C cos(θ + δ ) + 2 0
r
h
3.3.7.2
Exercise
An artificial satellite is launched from point B on the equator by its carrier rocket
and inserted into an elliptical orbit with a perigee altitude of 2000 km. if the apogee
altitude is to be 4000 km, compute (a) the necessary perigee velocity vp and the corresponding apogee velocity vA, (b) the velocity at point C where the altitude of the
satellite is 2500 km, and (c) the period τ for a complete orbit.
Do not turn the page until you have completed the solution!
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Solution
(a) The perigee and apogee velocities for specified altitudes are given by
vP = R
g 1+ e
g rm ax
=R
a 1− e
a rm in
vA = R
g 1− e
g rm in
=R
a 1+ e
a rm ax
Where
rm ax = 6371 + 4000 = 10371km
rm in = 6371 + 2000 = 8371km
a = (rm in + rm ax ) / 2 = 9371km
Thus
vP = R
g rm ax
9.825
10371
= 6371(103 )
a rm in
9371(103 ) 8371
= 7261 m/ s or
vA = R
26140km/ h
g rm in
9.825
8371
= 6371(103 )
3
a rm ax
9371(10 ) 10371
= 5861 m/ s or
21099 km/ h
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(b) For an altitude of 2500 km the radial distance from the earth’s center is r = 6371
+ 2500 = 8871 km. The velocity at point C becomes
⎛1 1 ⎞
⎛ 1
1 ⎞ 1
vC2 = 2gR 2 ⎜ − ⎟ = 2(9.825)[(6371)(103 )]2 ⎜
−
⎝ r 2a ⎠
⎝ 8871 18742 ⎟⎠ 103
= 47.353(106 )(m/ s)2
vC = 6881 m/ s or
24773 km/ h
(c) The period of the orbit is given by
τ = 2π
= 2π
a3/ 2
R g
[(9371)(103 )]3 / 2
(6371)(103 ) 9.825
or
= 9026 s
τ = 2.507 h
3.4
Moving Co-Ordinate Systems
3.4.1
The Coriolis Effect
The Coriolis effect is the apparent deflection of objects from a straight path if the
objects are viewed from a rotating frame of reference.
See link below
http://en.wikipedia.org/wiki/Coriolis_effect
To get more information on the Coriolis Force visit this site
http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html
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3.4.2
Centripetal Acceleration (and Force)
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 138
The rate of change of the object’s velocity vector is the centripetal acceleration.
The centripetal force is the external force required to make a body follow a circular
path at constant speed.
See link below
http://en.wikipedia.org/wiki/Centripetal_force
3.4.3
Examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 155-157
3.4.4
Relative velocity in a moving coordinate system
A man walking at a rate of 4 km per hour toward the forward car of a train when the
latter is traveling at a speed of 50 km per hour is actually traveling at the speed of 54
km per hour over the round. His relative speed with respect to a fixed point on the
moving train is 4 km per hour. This simple concept of superposition of speeds can
be extended readily to the case of differing directions making use of the reference
system velocity and the relative velocity by use of vector addition.
3.4.4.1
Example
The compass of an aircraft indicates that it is headed due north, and its air speed
indicator shows that it is moving through the air at 200 km/h. If there is a wind of 80
km/h from west to east, what is the velocity of the aircraft relative to the earth?
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Solution
Let A refer to the aircraft, and F to the moving air. E refers to the earth. We have
given
VAF = 240 km/h,
due north
VFE = 100 km/h,
due east
and we wish to find the magnitude and direction of VAE:
vAE = vAF 2 + vF E 2
vAE = 2402 + 1002
= 260 km/ h ,
22.60 North East
More exemples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 48-52
DO THIS
3.4.4.2
Exercice
Consider the first example above, in what direction should the pilot head in order to
travel due north? What will then be his velocity relative to the earth?
N.B. The magnitude of his airspeed, and the wind velocity, are the same as in the
first example.
Do not turn the page until you have completed the solution!
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Solution
We now have given:
VAF = 240 km/h, direction unknown
VFE = 100 km/h, due east,
and we wish to find VAE, whose magnitude is unknown but whose direction is due
north.
We find
vAE = 2402 − 1002
= 218.2 km/ h
℘ = sin −1
100
= 24.60
240
Thus , the pilot should head 24.60 North West, and his ground speed will be 218.2
km/h.
3.5
Systems of Particles
3.5.1
Angular momentum
In physics, the angular momentum of an object rotating about some reference point
is the measure of the extent to which the object will continue to rotate about that point
unless acted upon by an external torque.
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Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 204-214
For more information visit the site below
http://en.wikipedia.org/wiki/Angular_momentum
3.5.2
Examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 214-216
3.5.3
Conservation of angular momentum and Torque
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 271-272
For information on Conservation of angular momentum and Torque also see the link
below:
http://en.wikipedia.org/wiki/Angular_momentum
3.5.4
D’Alembert’s Principle
The principle states that the sum of the differences between the generalized forces
acting on a system and the time derivative of the generalized momenta of the system
itself along an infinitesimal displacement compatible with the constraints of the system
(a virtual displacement), is zero.
That is, at equilibrium,
Since
, where F i are generalized forces and
are generalized momenta
See link
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle»
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3.5.5
Example
Each of three balls has a mass m and is welded to the rigid equi-angular frame of
negligible mass. The assembly rests on a smooth horizontal surface. If a force F is
suddenly applied to the one bar determine (a) the acceleration of point O and (b) the
angular acceleration of the frame.
Solution
(a) Point O is the mass center of the system of the three balls, so that its acceleration
is given by
∑F
= ma
F i = 3ma
a = a0 =
F
i
3m . To find H G we note
(b) We determine from the moment principle
that the velocity of each ball relative to the mass center O as measured in the
non-rotating axes x-y is
where is a common angular velocity of the spokes.
The angular momentum of the system about O is the sum of the moments of the
relative linear momenta as shown by the expression below.
And from
we have
So
DO THIS
3.5.6
Exercise
Consider the same conditions as for the example above except that the spokes are
freely hinged at O and so do not constitute a rigid system. Explain the difference
between the two problems.
Do not turn the page until you have completed the solution!
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Solution
The generalized Newton’s second law holds for any mass system, so that the acceleration a of the mass center G is the same as with example above, namely,
a=
F
i
3m
N.B. Although G coincides with O at the instant represented, the motion of the hinge
O is not the same as the motion of G since O will not remain the center of mass as
the angles between the spokes change. Both
have the same values
from the two problems at the instant represented. However, the angular motions of
the spokes in this problem are all different and are not easily determined.
3.6
Rockets And Collisions
Compulsory Reading
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page 115-129
Now that people have begun exploration of space, rockets have become a subject of
general interest. The central feature of the rocket, which underlies its unique capabilities, is its reaction engine. As the rocket engine operates, the mass of the rocket
changes rapidly. Consequently, Newton’s second law of motion in its general form
does not lead to F = ma. We therefore begin the analysis of the motion of a rocket
from the basic principle of momentum conservation.
3.6.1
Examples
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Page129-135
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Example
The ram of a pile driver has a mass of 800 kg and is released from rest 2 m above the
top of the 2400 kg pile. If the ram rebounds to a height of 0.1 m after impact with
the pile, calculate (a) the velocity v'p of the pile immediately after impact. (b) the
coefficient of restitution e, and (c) the percentage loss of energy due to the impact.
Solution
Conservation of energy during free fall gives the initial and final velocities of the ram
from v = 2gh . Thus
vr = 2(9.81)(2) = 6.26 m/ s vr' = 2(9.81)(0.1) = 1.40 m/ s
(a) Conservation of momentum (ΔG y = 0) for the system of the ram and pile
gives
800(6.26) + 0 = 800(−1.40) + 2400v'p
Thus
v'p = 2.55 m/ s
(b) The coefficient of restitution gives
e=
rel. vel. seperation
rel. vel. approach
e=
2.55 + 1.40
= 0.63
6.26 + 0
(c) The kinetic energy of the system just before impact is the same as the potential
energy of the ram above the pile and is
T = V g = mgh = 800(9.81)(2) = 15700 J
The kinetic energy T ' just after impact is
T ' = 12 (800)(1.40)2 + 12 (2400)(2.55)2 = 8620 J
The percentage loss of energy is therefore,
15700 − 8620
(100) = 45.1%
15700
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N.B. The impulses of the weights of the ram and pile are very small compared with
the impulses of the impact forces and thus are neglected during the impact.
DO THIS
3.6.2
Exercise
A cannon is mounted inside a railroad car, which is initially at rest but can move
frictionlessly. It fires a cannonball of mass m = 5 kg with a horizontal velocity of
v = 15 ms-1 relative to the ground at the opposite wall. The total mass of the cannon
and the railroad car is 15 000 kg. (Assume that the mass of the exhaust gases is
negligible.)
(a) What is the velocity V of the car while the cannonball is in flight?
(b) If the cannonball becomes embedded in the wall, what is the velocity of the car
and ball after impact?
Do not turn the page until you have completed the solution!
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Solution
When fired, the cannon exerts a force to the right on the ball. The ball exerts an equal
and opposite force on the cannon, so the car and the cannon recoil to the left. The net
momentum is conserved because there is no external frictional force. The momentum
before firing is zero, so after firing the momentum of the ball to the right must be equal
in magnitude to that of the car and the cannon to the left. Thus mv = MV and
V =
mv (5 kg)(15 ms−1 )
=
M
15000 kg
5 × 10−3 ms−1
This gives
N.B. The recoil speed of the car and the cannon is very small because of their large
mass.
(b) As the ball becomes embedded in the wall, it exerts a force on the wall to the
right. The wall, in turn, exerts a force to the left on the ball. The ball and car both
stop moving when this happens, since the net momentum is still zero. Meanwhile
the car will have rolled to the left as the ball traveled to the right.
DO THIS
3.6.3 Exercise
A baseball initially at rest is struck with a bat. The velocity of the 0.15-kg ball just after
it is hit is 40 ms-1. If the impact time is 10-3s, what is the average force on the ball?
Do not turn the page until you have completed the solution!
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Answer
The initial momentum p of the ball is zero, since it starts from rest; the final momentum
is p’ = mv’. Thus, from F Δt = pʹ − p , the average force F on the ball is
F =
mvʹ (0.15kg)(40ms−1 )
=
= 6000N
Δt
(10−3 s)
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Unit 4: Rigid bodies and Energy
Assumed Knowledge
Unit 3 of this module is the prerequisite, particularly particle kinematics and kinetics
of systems of particles.
Objectives of the unit
The student should be able to:
a) Write and explain the equations of rigid body motion and apply them to general
planar motion of a rigid body.
b) Calculate relative velocity and relative acceleration for points on a rigid body,
and also calculate the instantaneous centre of zero velocity for a rigid body.
c) Apply the equations of planar kinetics to analyze the motion of rigid bodies
undergoing translation, rotation about a fixed and general plane motion.
d) Apply the work-energy principle as well as the conservation of energy principle to solve problems in planar kinetics.
e) Calculate both the linear and angular momentum of a rigid body and apply
them to solve problems in planar kinetics.
Summary of the learning activity
In this unit students will learn the equations of rigid body motion and apply them
to general planar motion of a rigid body. The students will learn how to calculate
relative velocity and relative acceleration for points on a rigid body, and also learn
how to calculate the instantaneous centre of zero velocity for a rigid body.
The students will learn how to apply the equations of planar kinetics to analyze
the motion of rigid bodies undergoing translation, rotation about a fixed point, and
general plane motion.
The student will learn how to apply the work-energy principle as well as the conservation of energy principle to solve problems in planar kinetics. The student will
learn how to calculate both the linear and angular momentum of a rigid body and
apply them to solve problems in planar kinetics. Students will also learn other better
methods of solving mechanical problems other than the Newton laws. They will look
at other principles of mechanics like the D’Alembert’s principles.
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Suggested ICT methods
Use of the Internet to get more information on the relevant topic, Software and Animations. Virtual Science laboratory,
http://www.smartscience.net/SmartScience/SmartScience.html
Content: Plane motion of rigid bodies: Moment of inertia. Radius gyration, Parallel axis theorem, Perpendicular axis theorem, Couples. Kinetic energy and angular
momentum about a fixed axis, Principle of angular momentum, Principle of conservation of energy, Principle of virtual work and D’Alembert’s principle, Principle of
minimum potential energy.
Compulsory reading text
Fitzpatrick, Richard, Classical Mechanics: An Introductory Course, Austin, Texas.
Relevant Links and Resources
Our links are from Wikipedia.
Dynamics of Rigid Bodies
http://www.engin.brown.edu/courses/en4/notes/RigidKinematics/rigkin.htm
Key Words
•
•
•
•
•
•
Mechanical system
Inertia
Degrees of freedom
Kinematic constraints
Rectilinear translation
Curvilinear translation
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Learning Activity 4 : Rigid bodies and Energy
4.1Plane Motion Of Rigid Bodies
Compulsory Reading
Mariam J.L. and Kraige L.G. Engineering Mechanics Volume 2: Dynamics
(page 291-459)
Ruina, A. and Pratap, R. Introduction to Statics and Dynamics
Dynamics of Rigid Bodies
http://www.engin.brown.edu/courses/en4/notes/RigidKinematics/rigkin.htm.
http://en.wikipedia.org/wiki/Rigid_body_dynamics
Remark
On the topic plain motion of rigid bodies, I feel guilty not to recommend
students to read Mariam J.L. and Kraige L.G. Engineering Mechanics Volume 2: Dynamics. All other sources are heavily dependent on this book.
http://www.engin.brown.edu/courses/en4/notes/RigidKinematics/rigkin.htm
Dynamics of Rigid Bodies
http://www.engin.brown.edu/courses/en4/notes/RigidKinematics/rigkin.htm
4.1.1 Introduction
Dynamics is the branch of mechanics which deals with the laws of motion of material
bodies subjected to the action of forces. The motion of bodies from a purely geometrical point of view is discussed in kinematics. Unlike in kinematics, in dynamics the
motion of material bodies is investigated in connection with the acting forces and the
inertia of the material bodies themselves. Inertia is the property of material bodies
to resist change in their velocity under the action of an applied force. For example,
the velocity of one body changes slower than that of another body subjected to the
same force, the former is said to have a greater inertia.
A rigid body is an idealization of a body that does not deform or change shape.
Formally it is defined as a collection of particles with the property that the distance
between particles remains unchanged during the course of motions of the body. Like
the approximation of a rigid body as a particle, this is never strictly true. All bodies
deform as they move. However, the approximation remains acceptable as long as the
deformations are negligible relative to the overall motion of the body.
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Schematic showing a planar rigid body.
In the following, we will restrict attention to the planar motion of rigid bodies. In
particular, we will take all rigid bodies to be thin slabs with motion constrained to lie
within the plane of the slab. Unless otherwise indicated, we will assume basis vectors
of the form {i, j, k}, such that i and j lie in the plane, with k is the plane normal.
DO THIS
Exercise: Kinematics of Rigid Body Motion
1. What do you understand by kinematics of rigid bodies?
2. Explain the following types of motion
i) Rectilinear translation
ii) curvilinear translation
2.2Dynamics Of A System And Rigid Body
Compulsory Reading
Moment of inertia
http://en.wikipedia.org/wiki/Moment_of_inertia
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4 .2.1
Mechanical Systems. External and Internal Forces
A mechanical system is defined as such a collection of material points (particles) or
bodies in which the position or motion of each particle or body of the system depends
on the position and motion of all the other particles or bodies. A classical example
of a mechanical system is the solar system, all the component bodies of which are
connected by the forces of their mutual attraction. Other examples of mechanical
systems are machines or any mechanism whose members are connected with pins,
rods, cables, belts, belts, etc. In this case the bodies of the system are subjected to the
reciprocal compressive or tensile forces transmitted through the constraints.
A collection of bodies not connected by interacting forces does not comprise a mechanical system (e.g, a group of lying aircraft). In this book we shall consider only
mechanical systems, calling them just «systems» for short. It follows from the above
that the forces acting on the particles or bodies of a system can be subdivided into
external and internal forces.
External forces are defined as the forces exerted on the members of a system by
particles or bodies not belonging to the given system.
Internal forces are defined as the forces of interaction between the members of the
same system. We shall denote external forces by the symbol Fe , and internal
forces by the symbol Fi .
Internal forces possess the following properties:
The geometrical sum (the principal vector) of all the internal forces of a system is
zero. This follows from the third law of dynamics, which states that any two particles
of a system (Fig . 15) act on each other with equal and oppositely directed forces Fi12
and Fi21, the sum of which is zero. Since the same is true for any pair of particles of
a system,
∑F
i
k
=0
Fig 15
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The sum of the moments (the principal moment) of all the internal forces of a system
with respect to any centre or axis is zero. For if we take an arbitrary centre O , it
is apparent from Fig. 15 that m0 (F12i ) + m0 (F 21i ) = 0 . The same result holds for the
moments about any axis. Hence, for the system as a whole we have:
∑ m (F
0
i
k
)=0
or
∑m (F
x
i
k
)=0
It does not follow from the above, however, that the internal forces are mutually
balanced and do not affect the motion of the system, for they are applied to different
particles or bodies and may cause their mutual displacement. The internal forces
will be balanced only when a given system is a rigid body.
4.2.2
Mass of a System. Centre of Mass
The motion of a system depends, besides the acting forces, on its total mass and the
distribution of this mass. The mass of a system is equal to the arithmetical sum of
the masses of all the particles or bodies comprising it.
M = ∑ mk
We shall conventionally denote mass by the same letter M as the moment of a force.
Any possibility of confusion is precluded by the fact that when the symbol M denotes
moment of a force, it is always provided with a subscript (e.g. M c ,
Mx,
M t ).
In a homogeneous field of gravity, where g = const., the weight of every particle of a
body is proportional to its mass, hence the distribution of mass can be judged according to the position of the centre of gravity. The equations defining the coordinates
of the centre of gravity can be written as
xc =
∑m x
k
M
k
, yc =
∑m y
k
M
k
, zc =
∑m z
The equations include only the masses
k
M
k
(1)
mk of the material points (particles) of
the body and their coordinate (xk , yk , zk ). Hence, the position of point C ( insert
equation) gives the distribution of mass in the body or in any mechanical system,
where mk and (xk , yk , zk ). are the masses and coordinates of the system’s respective points.
The geometric point C whose coordinates are given b Eqs. (1) is called the centre
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of mass, or centre of inertia of a mechanical system. If the position of the centre of
mass is defined by is radius vector rc we can obtain from Eqs. (1) the following
expression:
rc =
∑m r
k k
M
where rk
is the radius vector of particle k of the system.
Although in a homogenous gravitational field the Centres of mass and gravity coincide, the two concepts are not identical. The concept of centre of gravity, as the point
through which the resultant of the forces of gravity passes, has meaning only for a
rigid body in a uniform in field of gravity. The concept of centre of mass, as characteristic of the distribution of mass in a system, on the other hand, has meaning for
any system of particles or bodies, regardless of whether a given system is subjected
to the action of forces or not.
4.2.3
Moment of Inertia of a Body about an Axis: Radius of Gyration
The position of centre of mass does not characterize completely the distribution of
mass in a system. For if in the system in Fig.16a the distance h of each of the two
identical spheres A and B from the axis Oz is increased by the same quantity, the
location of the centre of mass will not change, though the distribution of mass will
change and influence the motion of the system (all other conditions remaining the
same, the rotation about axis Oz will be slower).
Fig 16a
Accordingly, another characteristic of the distribution of mass, called the moment of
inertia, is introduced in mechanics. The moment of inertia of a body (system) with
respect to a given axis Oz (or the axial moment of inertia) is defined as the quantity
equal to the sum of the masses of the particles of the body (system) each multiplied
by the square of its perpendicular distance from the axis:
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J z = ∑ mk hk2
(2)
It follows from the definition that the moment of inertia of a body (or system) with
respect to any axis is always positive.
It will be shown further on that axial moment of inertia plays the same part in the
rotational motion of a body a mass does in translational motion, i.e. moment of inertia
is a measure of a body’s inertia in rotational motion.
By Eq. (2), the moment f inertia of a body is equal to the sum of the moments of
inertia of all its parts with respect to the same axis.
For a material point located at distance h from an axis, J z = mh2 . The unit for the
moment of inertia in the SI system is 1 kg-m2, and in the mkg(f) system 1 kgf-m-s.
In computing the axial moments of inertia the distances of the points from the axis
(
can be expressed in terms of their coordinates xk , yk , zk
distance from axis Ox is
) (e.g., the square of the
yk2 + zk2 , etc.
Then the moments of inertia about the axes Oxyz will be given by the following
equations:
J x = ∑ mk ( yk2 + zk2 ), J y = ∑ mk (xk2 + zk2 ), J z = ∑ mk ( yk2 + xk2 ).
(3)
The concept of the radius of gyration is often employed in calculations. The radius
of gyration of a body with respect to an axis Oz is a linear quantity ρ g defined by
the equation
J z = M ρ g2
where M is the mass of the body.
(4)
It follows from the definition that geometrically the radius of gyration is equal to the
distance from the axis Oz to a point, such that if the mass of the whole body were
concentrated in it, the moment of inertia of the point would be equal to the moment
of inertia of the whole body.
Knowing the radius of gyration, we can obtain the moment of inertia of a body from
Eq. (4) and vice versa.
Eqs. (2) and (3) are valid for both rigid bodies and systems of material points. In
the case of a solid body, dividing it into elementary parts, we find that in the limit
the sum in Eq. (2) becomes an integral. Hence, taking into account that dm = ρdV
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, where ρ is the density and V the volume, we obtain:
J z = ∫ h2 dm, or
J z = ∫ ρh2 dV
(5)
The integral extends over the whole volume V of the body, and the density and
distance H depend on the coordinates of the points of the body. Similarly, for solid
bodies Equations. (3) take the form:
J x = ∫ ρ( y 2 + z 2 )dV , etc. (5’)
Equations. (5) and (5’) are useful in calculating the moments on inertia of homogeneous bodies of geometric shape. As in that case the density is constant, it can be
taken out of the integral sign.
Let us determine the moments of inertia of some homogeneous bodies.
4.2.4
Thin Homogeneous Rod of Length L and Mass M.
Fig. 16b
Let us find its moment of inertia with respect to an axis Az perpendicular to the rod
(Fig. 16b). If we lay off a coordinate axis Ax along AB, for any line element of length
dx we have h = x and its mass dm = ρ1 dx,
unit length of the rod, and Eq. (5) gives.
l
l
where ρ1 = M / l , is the mass of a
l3
J A = ∫ x dm = ρ1 ∫ x dx = ρ1 .
3
0
0
Substituting the expression for ρ1 , we obtain finally
2
JA =
2
1 2
Ml
3
(6)
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(2) Thin Circular Homogeneous Ring of Radius R and Mass M.
Let us find its moment of inertia with respect to an axis Cz perpendicular to the plane
of the ring through its centre.
Fig 17.
As all the points of the ring are at a distance from axis Cz, Eq. (2) gives
J c = ∑ mk R 2 = (∑ mk )R 2 = MR 2
Here and further on J denotes the moment of inertia with respect to an axis through
A perpendicular to the plane of the cross section in the diagram.
Hence, for a ring J c = MR 2
(7)
It is evident that the same result is obtained for the moment of inertia of a cylindrical
shell of mass M and radius R with respect to its axis.
4.2.5
Circular Homogeneous Disc or Cylinder of Radius R and Mass M
Let us compute the moment of inertia of a circular disc with respect to an axis Cz
perpendicular to it trough its centre (Fig. 17b).
Consider an elemental ring of radius r and width dr. Its area
2πrdr , and its mass dm = ρ2 2πrdr , where ρ2 =
M
π R2
area of the disc. From Eq. (7) we have for elemental ring
is the mass of a unit
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dJ c = r 2 dm = 2πρ2 r 3 dr ,
and for the whole disc
R
1
J c = 2πρ2 ∫ r 3 dr = πρ2 R 4
2
0
Substituting the expression for ρ2 we obtain finally
JC =
1
MR 2
2
(8)
It is evident that the same formula is obtained for the moment of inertia J of a homogeneous circular cylinder of mass M and radius R with respect to its axis Cz (Fig.
17c)
4.2.6
Rectangular Lamina, Cone, and Sphere.
Omitting the computations, here are the equations of the moments of inertia of several
bodies (the students is invited to deduce these formulas independently).
(a) uniform rectangular lamina of mass M with sides of length a and b (axis x is
coincident with side a, axis y with side b):
Jx =
1
1
Mb2 , J y = Ma 2 ;
3
3
(b) uniform right circular cone of mass M and base radius R (axis z is coincident
with the axis of the cone):
J z = 0.3MR 2
(c) uniform sphere of mass M and radius R (axis z is coincident with a diameter):
J z = 0.4 MR 2
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4.2.7
Moments of Inertia of a Body about Parallel Axes
The Parallel Axis (Huygens’) Theorem
In the most general case, the moments of inertia of the same body with respect to
different axes are different. Let us see how to determine the moment of inertia of a
body with respect to any axis if its moment of inertia with respect to a parallel axis
through the body is known.
Fig 18
Draw through the centre of mass of a body C arbitrary axes Cx' y' z' , and through
an arbitrary point 0 on axis Cx axes 0xyz, so that Oy / /Cy' and Oz / /Cz' (Fig.
18). Denoting the distance between axes Cz’ and 0z by d, from Eqs. (3)
J O C = ∑ mk (xk2 + yk2
J O Z ' = ∑ mk (x'2k + y'2k )
But it is apparent from the drawing that for any point of the body
xk = x'k − d, or
xk2 = x'2k + d 2 − 2x'k d and yk = y'k .Substituting these ex-
pressions for xk and yk into the expression for JOZ and taking the common
multipliers d and 2d outside the parentheses, we obtain:
J O Z = ∑ mk (x'2k + y'2k ) + (∑ mk )d 2 − 2d(∑ mk x'2k ).
The first summation in the right member of the equation is equal to JCz’ , and the second
to the mass M of the body. Let us find the value of the third summation. From Eq. (1)
we know that, for the coordinates of the centre of mass,
in our case point C is the origin, x'C = 0,
∑ m x'
k
k
, and consequently
= Mx'C . But since
∑ m x'
k
k
=0
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We finally obtain
J oz = J C z ' + Md 2 (9)
Equation (9) expresses the parallel axis theorem enunciated by Huygens*: The moment of inertia of a body with respect to any axis is equal to the moment of inertia
of the body with respect to a parallel axis through the centre of mass of the body
plus the product of the mass of the body and the square of the distance between the
two axes.
It follows from Eq. (9) that Joz = JCz’ . Consequently, of all the axes of the same
direction, the moment of inertia is least with respect to the one through the centre
of mass.
The parallel axis theorem can also be used to determine the moment of inertia of a body
with respect to a given axis Oz1 if its moment of inertia with respect to any parallel
axis Az2 and the distances d1 and d2 of each axis from the body’s centre of mass
are known. Hence, knowing J Az and d2
2
we obtained
J C z ' from Eq. (9) and,
applying the same formula, determine the required moment of inertia J O z
1
DO THIS
Exercise
1
Determine the moment of inertia of a thin rod with respect to an axis C perpendicular to it through its centre of mass.
2
Determine the moment of inertia of a cylinder with respect to an axis through
its generator (see Fig. 16b).
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4.3
Product of Inertia
4.3.1
Principal Axes of Inertia of a Body
The moment of inertia of a body with respect to an axis also does not completely
characterize the distribution of mass of the system. For example, if the rod DE in
Fig. below
is turned in plane 0yz so as to make other than a right angle with axis 0z and the distance h of spheres A and B from the axis is kept the same by moving them outward,
neither the location of the centre of mass nor moment of inertia of the spheres with
respect to axis 0z will change. Yet the distribution of mass will have changed (the
symmetry with respect to axis 0z being disturbed), and this will affect the system’s
rotation about the axis (additional lateral stresses will appear in the bearing).
Accordingly, the concept of the product of inertia is introduced as characterizing such
asymmetry in the distribution of mass. Drawing coordinate axes 0xyz through point
0, the products of inertia with respect to those axes are the quantities J xy , J yz , J zx
given by the following equations:
J xy = ∑ mk xk yk , J yz = ∑ mk yk zk , J zx = ∑ mk zk xk ,
where mk
is the mass of the points and
xk , yk , zk
(10)
are their coordinates.
Obviously J xy = J yx etc , .
For solid bodies, Eqs. (10) by analogy with (5’) take the form:
J xy = ∫ ρxydV (10’)
V
unlike axial moments of inertia, products of inertia can be either positive, negative
or, in selected coordinates, zero.
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Consider a homogenous body having an axis of symmetry. Draw the coordinate
axes 0xyz so that axis z is directed along the axis of symmetry. By virtue of sym-
(
)
metry, to each point of mass mk with coordinates xk , yk , zk , there corresponds
(
another point of equal mass with coordinates −xk ,− yk ,−zk
∑m x z
k
k
k
∑m y z
= 0 and
k
k
k
)
. Consequently,
= 0 , and taking into account Eqs. (10), we
obtain:
J xz = 0, J yz = 0
(11)
Thus, symmetry in the distribution of mass with respect to axis 0z is characterised by
two centrifugal moments of inertia, J xz and J yz , becoming zero. Axis 0z with
respect to which the products of inertia Jxz whose subscripts contain the notation of
that axis, are zero is called the principal axis of inertia of the body with respect to
point 0
It follows from what has been said that if a body has an axis of symmetry, that axis
is the principal axis of inertia with respect to any of its points.
The principal axis of inertia is not necessarily the axis of symmetry.
Consider a homogenous body having a plane of symmetry. Drawing axes 0xy in
that plane and axis 0z perpendicular to them, by virtue of symmetry, to every point
(
)
with coordinates ( x , y ,−z ) ,. Consequently, as in the previous case, we find that
of mass mk with coordinates xk , yk , zk , there corresponds a point of same mass
k
∑m x z
k
k
k
= 0 and
k
k
∑m y z
k
k
k
= 0 , or J xz = 0, J yz = 0 .
Thus, if a body has a plane of symmetry, any axis perpendicular to that plane is the
principal axis of inertia of the body with respect to the point 0 at which the axis
intersects the plane.
Eqs. (11) express the conditions that axis 0z is the principal axis of inertia of a
body with respect to point 0 (the origin of the coordinate system). Similarly, if
J xy = 0, J xz = 0
, axis 0x will be the principal axis of inertia with respect
to
point 0, etc. Consequently, if all the products of inertia are zero, i.e.,
J xy = 0, J yz = 0, J zx = 0 ,
(12)
each of the coordinate axes 0xyz is a principal axis of inertia with respect to point 0
(the origin of the coordinate system).
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Let us show that principal axes of inertia exist at any point of a body. For this let us
first prove the following theorem:
If the moment of inertia with respect to an axis 0z is greater or smaller than the moment of inertia with respect to any neighboring axis through 0, that axis (0z) is the
principal axis of inertia of the body with respect to 0.
4.4
D’alembert’s Principle
Compulsory Reading
Classical Mechanics Page 260
D’Alembert’s Principle
http://www. en.wikipedia.org/wiki/D%27Alembert%27s_principle
4.4.1
Forces Acting on the Axis of a Rotating Body
D’Alembert’s Principle
All the methods of solving the problems of dynamics examined up till now were
based on equations derived either directly from Newton’s laws or from the general
theorems, which are corollaries of those laws. However, the equations of motion
or equilibrium of other general propositions called the principles of mechanics. We
shall see that in many cases application of those principles offers better methods of
problem solutions. In this case we shall examine one of the general principles of
mechanics known as D’ Alembert’s principle.
Let there be a system of n material particles. Selecting any particle of mass mk , assume
it to be acted upon by external and internal forces F kext and F kint (which include
both active forces and the reactions of constraints), which impart it an acceleration
ωk
with respect to an inertial reference frame.
Let us introduce the quantity
F ki = −mk ω k (96)
with the dimension of force. The vector quantity equal in magnitude to the product
of the particle’s mass and acceleration and directed in the opposite sense of the acceleration and directed in the opposite sense of the acceleration is called the force of
inertia of that particle (sometimes the D’Alembert inertia force).
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The motion of a particle, we then find, satisfies the following D’Alembert’s principle
for a material particle: If, at any moment of time, to the effective forces F kext and
F kint acting on the particle is added the inertia force F ki , the resultant force system
will be in equilibrium, i.e.,
F kext + F kint + F ki = 0 (97)
It will be readily observed that D’Alembert’s principle is equivalent to Newton’s
second law, and vice versa. For Newton’s second law gives for this particle
mk ω k = F kext + F kint . Transferring mk ω k to the right-hand side of the equation, and
taking into account the notation (96), we arrive at Eq. (97), we obtain the formula
expressing Newton’s second law.
Reasoning similarly for all the particles of the system, we arrive at the following result,
which expresses D’Alembert’s principle for a system: If, at any moment of time,
to the effective external and internal forces acting on every particle of a system are
added the respective inertia forces, the resultant force system will be in equilibrium,
and the equations of statistics will apply to it.
Mathematically D’Alembert’s principle is expressed by a set of n simultaneous vector equations of the form (97) which, apparently, are equivalent to the differential
equations of motion of a system. D’Alembert’s principle can be used to derive all
the general theorems of dynamics. The value of D’Alembert’s principle is that, when
directly applied to the problems of dynamics, the equations of motion of a system
can be written in the form of the well-known equations of equilibrium; this makes
for uniformity in the approach to problem solutions and usually greatly simplifies the
computations. Furthermore, when used in conjunction with the principle of virtual
displacement, which will be examined in the next section, D’ Alembert’s principle
yields a new general method of solution of problems of dynamics
In applying D’Alembert’s principle it should be remembered that, like the fundamental
law of dynamics, it refers to motion considered with respect to an inertial frame of
reference. That means that acting on the particles of the mechanical system whose
motion is being investigated are only the external and internal forces that appear as
a consequence of the interactions of the particles of the system among themselves
and with bodies not belonging to a system; it is under the action of those forces that
the particles of the system are moving with their respective accelerations ω k . The
inertia forces mentioned in D’Alembert’s principle do not act on the moving particles
[otherwise, by Eqs. (97), the points would be at rest or in uniform motion in which
case, as is apparent from Eq. (96), there would be no inertia forces]. The introduction of inertia forces is but a device making it possible to examine the equations of
dynamics by the simpler methods of statics.
We know from Statics that the geometrical sum of balanced forces and the sum of their
moments with respect to any centre 0 are zero; we know, further, from the principle
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of solidification that this holds not only for forces acting on a rigid body but for any
deformable system. Thus, according to D’Alembert ‘s principle, we must
∑ (F
⎫⎪
⎬
∑ ⎡⎣ mo (F kext ) + mo (F kint ) + mo (F ki ) ⎤⎦ = 0 ⎭⎪
ext
k
+ F kint + F ki = 0
(98)
let us introduce the following notation:
R i = ∑ F ki ,
M oi = ∑ mo (F ki ).
(99)
The quantities R i and M oi are respectively the principal vector of the inertia
forces and their principal moment with respect to a centre 0. Taking into account
that the sum of the internal forces and the sum of their moments are each zero we
obtain:
∑F
ext
+ R i = 0,
∑ m (F
ext
) + M oi
(100)
The use of Eqs. (100), which follow from D’Alembert’s principle, simplifies the
process of problem solution because the equations do not contain the internal forces.
Actually, Eqs. (100) are equivalent to the equations expressing the theorems of the
change in the momentum of a system, differing from them only in form.
k
o
k
Eqs. (100) are especially convenient in investigating the motion of a rigid body or
a system of rigid bodies. For the complete investigation of any deformable system
these equations, however, are insufficient*).
For the projections on a set of coordinate axes, Eqs. (100) give equations analogous
to the corresponding equations of statics. To use these equations for solving problems
we must know the principal moment of the inertia forces.
4.4.2
The Principal Vector and the Principal Moment of the Inertia Forces of a Rigid Body
It follows from Eqs. (99) that a system of inertia forces applied to a rigid body can
be replaced by a single force equal to Ri and applied at the centre 0, and a couple of
moment M oi . The principal vector of a system, it will be recalled, does not depend on
the centre of reduction and can be computed at once. As F ki = −mk ω k
account the equation of motion of centre of mass
∑mω
k
k
taking into
= Mω c we will have
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R i = − ∑ mk ω k = − Mω c
(101)
Thus, the principal vector of inertia forces of a moving body is equal to the product
of the mass of the body and the acceleration of its centre of mass, and is opposite in
direction to the acceleration.
If we resolve the acceleration ω c into its tangential and normal components, then
vector R i will resolve into components.
Rτi = − Mω cτ
and
R ni = − Mω cn
Let us determine the principal moment of the inertia forces for particular types of
motion.
Translational Motion. In this case a body has no rotation about its centre of mass
C, from which we conclude that
∑ m (F
C
ext
k
) = 0 and Eq. (100) gives M C = 0 .
Thus, in translational motion, the inertia forces of a rigid body can be reduced to a
single resultant R i
through the centre of mass of the body.
Plane Motion. Let a body have a plane of symmetry, and let it be moving parallel
to the plane. By virtue of symmetry, the principal vector and the resultant couple of
inertia forces lie, together with the centre of mass C, in that plane.
Therefore, placing the centre of reduction in point C, we obtain form Eq. (100)
M Ci = − ∑ mc (F kext ) (insert Eq.) On the other hand (see 156), from the last of Eqs.
(68)
∑ m (F
C
ext
k
) = J C ε . We conclude from this that
M Ci = − J C ε (102)
Thus, in such motion of a system of inertia forces can be reduced to a resultant force
R i [Eq. (101)] applied at the centre of mass C (Fig. 22) and a couple in the plane of
symmetry of the body whose moment is given by Eq. (102). The minus sign shows
that the moment M Ci is in the opposite direction of the angular acceleration of the
body.
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4.4.3
Fig 22
Rotation About an Axis Through the Centre of Mass.
Let a body have a plane of symmetry, and let the axis of rotation Cz be normal to
the plane through the centre of mass. This case will thus be a particular case of the
previous motion. But here ω C = 0 and consequently R i = 0 , Thus, in this case a
system of inertia forces can be reduced to a resultant force couple in the plane of
symmetry of the body of moment
M zi = − J z ε
(102’)
in applying Eqs. (101) and (102) to problem solutions, the magnitudes of the respective
quantities are computed and the directions are shown in fig. 22
DO THIS
Exercise
Use D’Alembert’s principle to solve the following:
a) A load of weight P attached to a string of length I is displaced through an angle
α from the vertical to a position M 0 and released from rest ( see Fig below).
Determine the tension in the thread when the load is in the lowest position
M1
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Fig. 19
(a)
b) Two weights P1 and P2 are connected by a thread and move along a horizontal
plane under the action of forces Q applied to the first weight. The coefficient
of friction of the weights on the plane is f. Determine the accelerations of the
weights and the tensions in the thread
Fig 20
c) Wound on a drum of weight P and radius r is a string carrying a load A of weight
Q. Neglecting the mass of the string and friction, determine the angular acceleration of the drum when the load falls, the radius of gyration of the drum with
respect to its axis isρ. Also determine the tension in the thread.
d) A homogeneous rod bent at a right angle as shown in Fig 21 rotates in a horizontal
plane about its end A with an angular velocity ω and angular accelerationε. The
respective distances AB = a and BC = b; the mass of unit length of the rod is
ρ1. Determine the stresses at a cross section of the rod at point D at a distance h
from end B.
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Fig. 21
References
Michon,G. P, Rigid Bodies
http://home.att.net/~numericana/answer/rigid.htm
Ruina, A. and Pratap, R. Introduction to Statics and Dynamics
Targ, S. Theoretical Mechanics : A Short Course Mir Publishers Moscow (1988)
Mariam J.L. and Kraige L.G. Engineering Mechanics Volume 2: Dynamics.
D’Alembert’s Principle
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle
Moment of inertia
http://en.wikipedia.org/wiki/Moment_of_inertia
Dynamics of Rigid Bodies
http://www.engin.brown.edu/courses/en4/notes/RigidKinematics/rigkin.htm
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XI. Compiled List of All Key Concepts (Glossary)
Note: this glossary contains only those basic concepts that span the module. Most of
the important concepts have been singularly defined and explained in the relaevant
units. Should the student have any problem with the definitions, they are advised to
visit the site Wikipedia [www.en.wikipedia.org]. If this is being viewed as a computer
file, key words can be clicked to link with the relavent Wikipedia entry.
Central field
In atomic physics, the central field approximation for many-electron atoms takes
the combined electric fields of the nucleus and all the electrons acting on any of the
electrons to be radial and to be the same for all the electrons in the atom. That is,
every electron sees an identical potential U(r) that is only a function of its distance
from the nucleus. This facilitates an approximate analytical solution to the eigenvalue
problem for the Hamiltonian operator.
Circular Motion
In physics, circular motion is rotation along a circle: a circular path or a circular
orbit. The rotation around a fixed axis of a three-dimensional body involves circular
motion of its parts. We can talk about circular motion of an object if we ignore its
size, so that we have the motion of a point mass in a plane.
Examples of circular motion are: an artificial satellite orbiting the Earth in geosynchronous orbit, a stone which is tied to a rope and is being swung in circles (cf.
hammer throw), a racecar turning through a curve in a racetrack, an electron moving
perpendicular to a uniform magnetic field, a gear turning inside a mechanism.
Circular motion involves acceleration of the moving object by a centripetal force
which pulls the moving object towards the center of the circular orbit. Without this
acceleration, the object would move inertially in a straight line, according to Newton’s first law of motion. Circular motion is accelerated even if the speed is constant,
because the object’s velocity vector is constantly changing direction.
Coriolis Force and Effect
The Coriolis effect is an apparent deflection of moving objects from a straight path
when they are viewed from a rotating frame of reference. The effect is named after
Gaspard-Gustave Coriolis, a French scientist who described it in 1835, though the
mathematics appeared in the tidal equations of Pierre-Simon Laplace in 1778. The
Coriolis effect is caused by the Coriolis force, which appears in the equation of motion
of an object in a rotating frame of reference. Sometimes this force is called a fictitious
force (or pseudo force), because it does not appear when the motion is expressed in
an inertial frame of reference, in which the motion of an object is explained by the
real impressed forces, together with inertia. In a rotating frame, the Coriolis force,
which depends on the velocity of the moving object, and centrifugal force, which
does not, are needed in the equation to correctly describe the motion.
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Force: this is the vector action of one body upon another.
Harmonic Oscillator and Simple Harmonic Motion
In order for mechanical oscillation to occur, a system must posses two quantities:
elasticity and inertia. When the system is displaced from its equilibrium position, the
elasticity provides a restoring force such that the system tries to return to equilibrium.
The inertia property causes the system to overshoot equilibrium. This constant play
between the elastic and inertia properties is what allows oscillatory motion to occur.
The natural frequency of the oscillation is related to the elastic and inertia properties
by:
The simplest example of an oscillating system is a mass connected to a rigid foundation by way of a spring. The spring constant k provides the elastic restoring force,
and the inertia of the mass m provides the overshoot. By applying Newton’s second
law F=ma to the mass, one can obtain the equation of motion for the system:
where
is the natural oscillating frequency. The solutions to this equation
of motion takes the form
where xm is the amplitude of the oscillation, and φ is the phase constant of the oscillation. Both xm and φ are constants determined by the initial condition (intial displacement and velocity) at time t=0 when one begins observing the oscillatory motion.
Kepler’s Three Laws
1. The orbit of every planet is an ellipse with the sun at one of the foci. An ellipse
is characterized by its two focal points; see illustration. Thus, Kepler rejected
the ancient Aristotelean, Ptolemaic,and Copernican belief in circular motion.
2. A line joining a planet and the sun sweeps out equal areas during equal intervals
of time as the planet travels along its orbit. This means that the planet travels
faster while close to the sun and slows down when it is farther from the sun.
With his law, Kepler destroyed the Aristotelean astronomical theory that planets
have uniform velocity.
3. The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axes (the «half-length» of the ellipse) of their orbits. This
means not only that larger orbits have longer periods, but also that the speed of
a planet in a larger orbit is lower than in a smaller orbit.
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Mass: this is defined as the quantitative measure of inertia or resistance to change in
the motion of a body and also gives rise to Gravitational attraction.
Mechanical system A mechanical system is defined as such a collection of material
points (particles) or bodies in which the position or motion of each particle or body
of the system depends on the position and motion of all the other particles or bodies.
A classical example of a mechanical system is the solar system, all the component
bodies of which are connected by the forces of their mutual attraction. Other examples
of mechanical systems are machines or any mechanism whose members are connected
with pins, rods, cables, belts, belts, etc.
Natural Circular Frequency
The fundamental tone, often referred to simply as the fundamental and abbreviated
fo, is the lowest frequency in a harmonic series.
The fundamental frequency (also called a natural frequency) of a periodic signal is
the inverse of the pitch period length. The pitch period is, in turn, the smallest repeating unit of a signal. One pitch period thus describes the periodic signal completely.
The significance of defining the pitch period as the smallest repeating unit can be
appreciated by noting that two or more concatenated pitch periods form a repeating
pattern in the signal. However, the concatenated signal unit obviously contains
redundant information.
In terms of a superposition of sinusoids (for example, fourier series), the fundamental
frequency is the lowest frequency sinusoidal in the sum.
Newton’s Laws of Motion
Newton’s laws of motion are three physical laws which provide relationships between
the forces acting on a body and the motion of the body. They were first compiled by
Sir Isaac Newton in his work Philosophiae Naturalis Principia Mathematica (1687).
The laws form the basis for classical mechanics and Newton himself used them to
explain many results concerning the motion of physical objects. In the third volume
of the text, he showed that these laws of motion, combined with his law of universal
gravitation, explained Kepler’s laws of planetary motion.
Briefly stated, the three laws are:
1. A physical body will remain at rest, or continue to move at a constant velocity,
unless an unbalanced net force acts upon it.
2. The net force on a body is equal to its mass multiplied by its acceleration.
3. For every action there is an equal and opposite reaction.
Particle: is a body of negligible dimensions. Also when the dimensions of a body
are irrelevant to the description of its motion or the action of forces upon it, the body
may be treated as a particle.
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Position: Position in space is determined relative to some geometric referrence system
by means of linear and/or angular measurements. The absolute frame of referrence
is an imaginary set of rectangular axes assumed to have no translation or rotation
effect in space for other motions, such as rocket motion, relative motion of the earth
has to be taken into account and relative corrections made to equations describing
the laws of Newtonian mechanics
Rectilinear translation Motion in which every line in the body remains parallel to
its original position. The motion of the body is completely specified by the motion
of any point in the body. All points of the body have the same velocity and same
acceleration.
Rigid Body
In physics, a rigid body is an idealization of a solid body of finite size in which deformation is neglected. In other words, the distance between any two given points of
a rigid body remains constant in time regardless of external forces exerted on it. In
classical mechanics a rigid body is usually considered as a continuous mass distribution, while in quantum mechanics a rigid body is usually thought of as a collection
of point masses. For instance, in quantum mechanics molecules (consisting of the
point masses: electrons and nuclei) are often seen as rigid bodies (see classification of
molecules as rigid rotors). The position of a rigid body is determined by the position
of its center of mass and by its orientation (at least six parameters in total).
Simple Pendulum
A pendulum is an object that is attached to a pivot which the pendulum can swing
freely. This object is subject to a restoring motion that will accelerate it toward an
equilibrium position. When the pendulum is displaced from its place of rest, the restoring force will cause the pendulum to oscillate about the equilibrium position.
A basic example is the simple gravity pendulum or bob pendulum. This is a weight
(or bob) on the end of a massless string, which, when initially displaced, will swing
back and forth under the influence of gravity over its central (lowest) point.
The regular motion of pendulum can be used for time keeping, and pendulum are
used to regulatee pendulum clocks.
Space: This is the geometric region occupied by bodies. Position in space is determined relative to some geometric reference system by means of linear and angular
measurements.
Time: time is the measure of the succesion of events which is considered to be an
absolute quantity in Newtonian mechanics.
Vectors and Scalars
Scalars have magnitude only. Temperature, speed, mass, and volume are examples
of scalars.
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Vectors have magnitude and direction. The magnitude of is written | | v . Position, displacement, velocity, acceleration and force are examples of vector quantities.
Vectors have the following properties:
1. Vectors are equal if they have the same magnitude and direction.
2. Vectors must have the same units in order for them to be added or subtracted.
3. The negative of a vector has the same magnitude but opposite direction.
4. Subtraction of a vector is defined by adding a negative vector:
- = + (- )
5. Multiplication or division of a vector by a scalar results in a vector for
which
(a) only the magnitude changes if the scalar is positive
(b) the magnitude changes and the direction is reversed if the scalar is negative.
6. The projections of a vector along the axes of a rectangular co-ordinate system
are called the components of the vector. The components of a vector completely define the vector.
Figure 3.1
cos
Projections of a vector in 2-D.
=
Ax = Acos
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sin =
Ay = Asin
We can invert these equations to find A and as functions of Ax and Ay . By Pythagoras we have,
A=
and from the diagram,
tan =
=
.
7. To add vectors by components:
=
+
+
+...
(a) Find the components of all vectors to be added.
(b) Add all x components to get Rx = Ax + Bx + Cx + ...
(c) Add all y components to get Ry = Ay + By + Cy +...
Then:
| |=
=
.
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XII. Compiled List of Compulsory Readings
Fitzpatrick, R. (2001). Classical Mechanics: An Introductory Course. Austin, Texas.
UTP
Rationale
Fitzpartrick has the traditional approach to mechanics and dwells only briefly on
vector algebra and calculus which forms the modern basis for teaching mechanics.
He has adequately covered some aspects of the content in our curriculum and will
make very good basic reading for the student. However, there are no exercises for
the student and only a few examples are given at the end of each chapter.
The student should also note the interchangeable use of Imperical Units and S.I.
Units.
We have, however, supplemented this compulsory reading material by including those
topics which are not found in the compulsory reading book.
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XIII. Compiled List of (Optional) Multimedia Resources
Reading # 1: Wolfram MathWorld (visited 03.11.06)
Complete reference : http://mathworld.wolfram.com
Abstract : Wolfram MathWorld is a specialised on-line mathematical encyclopedia.
Rationale: It provides the most detailed references to any mathematical topic. Students should start by using the search facility for the module title. This will find a
major article. At any point students should search for key words that they need to
understand. The entry should be studied carefully and thoroughly.
Reading # 2 : Wikipedia (visited 03.11.06)
Complete reference : http://en.wikipedia.org/wiki
Abstract : Wikipedia is an on-line encyclopedia. It is written by its own readers.
It is extremely up-to-date as entries are contunally revised. Also, it has proved to be
extremely accurate. The mathematics entries are very detailed.
Rationale: Students should use wikipedia in the same way as MathWorld. However,
the entries may be shorter and a little easier to use in the first instance. Thy will,
however, not be so detailed.
Reading # 3 : MacTutor History of Mathematics (visited 03.11.06)
Complete reference :
http://www-history.mcs.standrews.ac.uk/Indexes
Abstract : The MacTutor Archive is the most comprehensive history of mathematics
on the internet. The resources are organsied by historical characters and by historical
themes.
Rationale: Students should search the MacTutor archive for key words in the topics
they are studying (or by the module title itself). It is important to get an overview
of where the mathematics being studied fits in to the hostory of mathematics. When
the student completes the course and is teaching high school mathematics, the characters in the history of mathematics will bring the subject to life for their students.
Particularly, the role of women in the history of mathematics should be studied to
help students understand the difficulties women have faced while still making an
important contribution.. Equally, the role of the African continent should be studied
to share with students in schools: notably the earliest number counting devices (e.g.
the Ishango bone) and the role of Egyptian mathematics should be studied.
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XIV. Compiled List Of Useful Links
A. Circular motion
http://en.wikipedia.org/wiki/Circular_motion
Rationale : This link is useful because it gives the definition of circular motion
and discusses on other related issues.
B. Circular motion
http://hyperphysics.phy-astr.gsu.edu/HBASE/circ.html
C. Circular motion
www.phy.ntnu.edu.tw/java/shm/shm.html
D. Planetary motion
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/kepler6.html
Rationale : This link is useful because it give animations on planetary motion which help the learner to picture how the motion of planets occurs.
E. Planetary motion
http://www-istp.gsfc.nasa.gov/stargaze/kep3laws.htm
F. Planetary motion
http://en.wikipedia.org/wiki/Kepler’s_laws_of_planetary_motion
Rationale : This link is useful because it gives the definition of planetary
motion and helps the learner to understand some issues related to planetary motion.
G. Planetary motion
www.windows.ucar.edu/tour/link=/the_universe/uts/planets.html
H. Simple harmonic motion
http://en.wikipedia.org/wiki/Simple_harmonic_motion
Rationale : This link is useful because it gives the definition of simple
harmonic motion and goes on to explain some issues related to simple
harmonic motion.
I. Simple harmonic motion
www.phy.ntnu.edu.tw/java/shm/shm.html
J. Simple harmonic motion
http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html
K. Simple harmonic motion
www.kettering.edu/~drussell/Demos/SHO/mass.html
L. Simple harmonic motion
http://theory.unwinnipeg.ca/physics/shm/nodes2.html
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M. Classical Mechanics Lecture notes
http://farside.ph.utexas.edu/teaching/301/lectures/lectures.html
N. Newtonian Physics
http://www.lightandmatter.com/arealbook1.html
O. Classical Mechanics: An Introductory Course
http://www.lulu.com/content/159798
P. Centripetal Force
http://en.wikipedia.org/wiki/Centripetal_force
Q. Angular Momentum
http://en.wikipedia.org/wiki/Angular_momentum
R. D’Alembert’s Principle
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle
T. Vectors and Scalars
http://en.wikipedia.org/wiki/Vector_(spatial)
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XV. Synthesis of The Module
In this module you familiarized yourself, initially, with an introduction to vector
calculus which led to the vector definitions of the basic concepts of plane and curvilinear motion, position, vector and acceleration. In most of this section you considered
kinematics, the motion particles /bodies without referring to the force producing the
motion.
You should, by now be convercant with Newton’s laws of motion which has the
concept of inertia as is concept and their applications to all the other sections of
mechanics such as oscillations, rigid bodies and dynamics.
The crutial concept that runs through this module is that of force and the resultant
sections of different forces.
By this time you should also be able to model real-life situations in mathematical
terms and solve the equations thus produced. Whenever and wherever possible, we
also tried to relate the content to high school mathematics and started with the section
headed ‘’Bridging the gap’’ of common place situations that we should be able to
explain mathematically. You should now be comfortable with the rubic of mechanics
and be able to teach the relevant content to High School students.
Good Luck!
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XVI. Summative evaluation
Dynamics
Question 1
A car travelling at 10ms-1 collides with a tree. (a) A passenger without a seat belt strikes the wind shield headfirst and comes to rest in 0.002 s. The contact area between
the head and the windshield is 6 × 10−4 m2 , and the mass of the head is 5 kg. Find
the average force and the force per unit area exerted on the head. (b) A passenger of
mass 70 kg wearing a shoulder harness comes to rest in 0.2 s. The area of the harness
in contact with the passenger is 0.1 m2. Find the average force and the average force
per unit area.
Question 2
A car of mass m = 1000 kg moving at 30 ms-1 collides with a car of mass M = 2000
kg travelling at 20 ms-1 in the opposite direction. Immediately after the collision, the
1000-kg car moves at right angles to its original direction at 15 ms-1. Find the velocity
of the 2000-kg car right after the collision.
Oscillations
Question 3
An object is connected to one end of a horizontal spring whose other end is fixed.
The object is pulled to the right (in the positive x direction) by an externally applied
force of magnitude 20 N causing the spring to stretch 1.000 cm.
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(a) Determine the value of the force constant.
(b) If the mass of the object is 4 kg determine with which it oscillates if the applied
force is suddenly removed.
(c) Determine the frequency of the oscillation.
(d) Determine the angular frequency of the oscillation.
(e) Determine the position of the object 0.75 s after it begins oscillating.
(f) Determine the velocity of the object at t = 0.75 s.
(g) Determine the acceleration at t = 0.75 s.
(h) Determine the force exerted on the body by the spring at t = 0.75 s
Force Energy And Motion
Question 4
Give the approximate value of the speed, in m/s, for each object listed below. Also
express each speed as a fraction of the speed of light, which is 3.0 × 108 ms−1 . In
cases where full information is not provided, make reasoned estimates in order to
obtain your results.
(a) An ant crawling
(b) A person walking to work at a comfortable pace
(c) An athlete running the mile
(d) A car on a super highway
(e) A cruising jet (650 miles/ hr, or about 90 % the speed of sound).
(f) A satellite orbital radius of 7000 km, (orbital period of about 90 minutes).
(g) The moon in its orbit around the earth (orbital radius of about 380 000 miles,
orbital period of 27.3 days).
(h) The earth in its orbit around the sun (orbital radius of approximately 93 000
000 miles, orbital period of 365.3 days).
N.B. [1 inch = 2.54 cm, 1 mile = 1.61 km and 1 foot = 0.3048 m]
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Question 5
Two dormitory roommates, Hugh and Lou, decide to play an unusual game of catch.
Hugh stands on a balcony, and Lou stands on the ground directly below. The balcony
is 10 m above the ground. Hugh and Lou each through the ball directly towards each
other with an initial speed of 15 m/s.
(a) How long does the ball take to travel from Lou up to Hugh? How fast is the
ball travelling when Hugh catches it ?
(b) How long does it take for the ball to descend from Hugh to Lou? How fast is
the ball travelling when Lou catches it ?
(c) Compare the round-trip travel time with the time that would be required if
the ball traveled with a constant speed of 15 m/s in both directions.
(d) Alice, another resident at the dormitory, lives 15 m above the two roommates.
She decides to douse Hugh and Lou with two water filled balloons. She watches
them toss the ball back and forth several times and learns to anticipate when
Hugh will release the ball. Alice wants to drop the balloons so that Hugh and
Lou will be hit simultaneously, just as the ball is reaching Lou. When should
Alice drop each balloon? Must she drop both the balloons before Hugh throws
the ball downward? How long after Hugh throws the ball will the first balloon
fall past him? How long after that will the second balloon strike him?
(e) Alice’s diabolical spot works perfectly. However, Hugh and Lou are accustomed to Alice’s pranks, and each of them has rotten tomato handy. They hold
the tomatoes , count to three, and simultaneously hurl the tomatoes at Alice.
Each tomato has an initial speed of 30 m/s. How soon after the tomatoes are
thrown does Alice need to be out of the way?
(f) Both tomatoes miss on the way up, but then Alice makes a mistake. She leans
out to gloat at her wet victims. Both tomatoes strike her on the back of the
head. Whose tomato hits Alice first? When does it hit her, and how fast is it
travelling? When does the other tomato hit, and how fast is it travelling?
Question 6
A 3000-pound vehicle is negotiating a circular interchange of radius 300 feet at 30
miles per hour. Assuming the roadway to be level, find the force between the tires
and the road so that the car stays on the circular path and does not skid. (Use F = ma,
where m = 3000/32). Find the angle at which the roadway should be banked so that
no lateral frictional force is exerted on the tires of the vehicle.
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Question 7
The path of a shot thrown at an angle θ is
⎡
⎤
1
r (t) = (v0 cosθ)ti + ⎢ h + (v0 sinθ)t − gt2 ⎥ j
2
⎣
⎦
where v0 is the initial speed, h is the initial height, t is the time in seconds, and g is the
acceleration due to gravity. Verify that the shot will remain in the air for a total of
t=
v0 sinθ + v0 2 sin 2 θ + 2gh
g
seconds
and will travel a horizontal distance of
v0 2 cosθ ⎛
2gh ⎞
2
sinθ
+
sin
θ
+
⎜
⎟ feet .
g ⎝
v0 2 ⎠
N.B. [1 inch = 2.54 cm, 1 mile = 1.61 km and 1 foot = 0.3048 m]
Question 8
Determine the moments of inertia of the homogeneous rectangular parallelepiped of
mass m about the centroidal x0 and z − axes and about the x − axis through one
end.
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Solutions
Solution Question 1
(a) Again we use F Δt = pʹ − p . The final momentum is zero, since the windshield
is stationary, and the initial momentum is the head mass times the velocity. Thus
the magnitude of the average force is
p (5kg)(10ms−1 )
F =
=
= 25000N
Δt
(0.002s)
The average force per unit area is
F
25000N
=
= 4.16 × 107 Nm−2
−4
2
A 6 × 10 m
This is a very large force per unit area and is certain to cause serious injury.
(b) The average force is found from the change in momentum of the entire body as
the speed of the car changes from 10 ms-1 to zero. Thus the magnitude of F is
F =
p (70kg)(10ms−1 )
=
= 3500N
Δt
(0.2s)
This is much smaller than the force exerted on the head of the unrestrained passenger
in part (a). The average force per unit area is
F 3500N
=
= 3.5 × 104 Nm−2
A 0.1 m2
Since this is smaller than the force per unit area on the unrestrained passenger by a
factor 1200, the chance of serious injury is much smaller for this passenger.
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Solution Question 2
Let us take the x and y axes where the x component of the total momentum of the
two vehicles is conserved, so
mvx + MV x = mvx' + MV x'
Since vx' = 0 , we can solve for V x' and substitute:
V x' =
=
mvx + MV x mvx
=
+ Vx
M
M
1000 kg
(30 ms−1 ) + (−20 ms−1 ) = −5 ms−1
2000 kg
The initial momentum components along the y axes were zero, so mv'y + MV y' = 0 ,
and
Vy = −
'
mv'y
M
=−
1000 kg
(−15 ms−1 ) = 7.5 ms−1
2000 kg
Solution Question 3
Solution
(a) the force constant is the k in Hooke’s law
F = −kx
The force F produced by the spring is F = −20.00 N , where the minus sign means
that the force acts to the left (in the negative x direction). Since x = 1.000 × 10−2
you have
k=
m,
F
−20.00 N
=−
= 2.000 × 103 N / m
−2
x
1.000 × 10
m
(b) The ratio α of the force constant to the mass determines the period T, which
k 2.000 × 103 N / m
α= =
= 5.000 × 102 N / (m⋅ kg)
m
4.000 kg
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Thus the period is
T =
2π
α
=
2π
= 0.2810 s
5.000 × 102 N / (m⋅ kg)
(c) Since
ν=
1
, you have
T
ν=
1
= 3.559 cycles / s = 3.559 Hz
0.2810 s
(Since a cycle is simply a count, it is a dimensionless number and can be inserted at
will in the units of the answer.)
(d) you have
ω = 2πν = 2π rad / cycle × 3.559 cycles / s = 22.36 rad / s
(e) the oscillation is described by
x = Acos(ωt + δ )
If A and δ are adjusted to fit the initial conditions and if the proper value of ω is
used. The initial conditions (dx / dt)0 = 0 and x0 = 1.000 × 10−2 m require that
δ = 0 and
A = 1.000 × 10−2 m
So x = 1.000 × 10−2 m× cos(22.36 rad / s × t)
where the value of the ω is that found in part d. the position at t = 0.7500 s is
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x = 1.000 × 10−2 m× cos(22.36 rad / s × 0.7500 s)
= 1.000 × 10−2 m× cos(16.77 rad)
= 1.000 × 10−2 m× (−0.4871) = −4.871 × 10−3 m = −0.4871 cm
The argument of the cosine is larger than 2π rad because the oscillation has passed
beyond its first cycle. The minus sign means that the body is to left of its equilibrium
position, and the spring is compressed. Note that for the purpose of evaluation x it
is most convenient to express the cosine in terms of the angular frequency ω rather
than the frequency ν .
(f) evaluating the terms you have
dx
= − Asin(ωt)
dt
= 1.000 × 10−2 m× 22.36 rad / s × sin(16.77 rad)
= −1.000 × 10−2 m× 22.36 rad / s × (−0.8733)
= 0.1953 m/ s = 19.53 cm/ s
(A radian is a dimensionless number, being defined as the ratio of two lengths, so it can
be deleted at will from the units.) The positive velocity means the body is moving to
the right. Thus although the spring is compressed, the compression is decreasing.
(g) the acceleration is
d2 x
= − Aω 2 cos(ωt)
2
dt
= 1.000 × 10−2 m× (22.36 rad / s)2 × cos(16.77 rad)
= −1.000 × 10−2 m× (22.36 rad / s)2 × (−0.4871)
= 2.435 m/ s2 = 243 cm/ s2
The body is accelerating to the right since the acceleration is positive.
(h) one way to do this is to apply Newton’s Second law:
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d2 x
F =m 2
dt
= 4.000 kg × 2.435 m/ s2
= 9.740 N
d2 x
where the value of
is that found in part g. the force has a positive value, and
dt2
so
acts to the right since the acceleration is to the right.
Another way of finding F is to apply Hooke’s Law:
F = −kx
= −2.000 × 103 N / m× (4.871 × 10−3 m)
= 9.740 N
where the values of k and x are those found in part a and e .
Solution Question 4
Estimates
(a) 0.1 m/s = 3.3 × 10-10 c
(b) to (h) Many possible solutions.
Solution Question 5
(a) 0.981 s, 5.39 m/s
(b) 0.563 s, 20.5 m/s
(c) Actual round-trip time is 1.54 s, while constant-speed time would be 1.33 s
(d) Balloon intended for Lou must be dropped 1.70 s before Hugh throws the
ball; balloon intended for Hugh must be dropped 1.19 s before Hugh throws
the ball, Yes. 0.05 s, 0.51 s
(e) 0.55 s
(f) Lou’s tomato strikes first, 5.13 s after it is thrown (and 4.13 s after missing
on the way up); it hits her at 20.3 m/s. Hugh’s tomato hits 0.44 s later (5.57 s
after it is thrown); it hits at 24.6 m/s
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Solution Question 8
A transverse slice of thickness dz is selected as the element of volume. The moment
of inertia of this slice of infinitesimal thickness equals the moment of inertia of the
area of the section times the mass per unit area ρdz . Thus the moment of inertia of
the transverse slice about the y'− axis is
⎛ 1
⎞
dI y ' y ' = ρdz ⎜ ab3 ⎟
⎝ 12
⎠
( )
and that about the x'− axis is
⎛ 1
⎞
dI x ' x ' = ρdz ⎜ a 3 b⎟
⎝ 12
⎠
( )
As long as the element is a plate of differential thickness, we have
ab
( ) 12
(a
dI zz = dI x ' x ' + dI y ' y ' = ρdz
2
+ b2
)
where m is the mass of the block.
By interchanging symbols the moment of inertia about the x0 − axis is
I 0 x0 =
1
m(a 2 + l 2 )
12
The moment of inertia about the x − axis may be found by the Parallel-axis theorem.
Thus
2
I xx = I x
0 x0
⎛l⎞
1
+ m⎜ ⎟ = m a 2 + 4l 2
12
⎝ 2⎠
(
)
This last result may be obtained by expressing the moment of inertia of the elemental
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slice about the x − axis and integrating the expression over the length of the bar.
Again by the parallel-axis theorem
dI xx
⎛ a2
⎞
⎛ 1 3 ⎞
2
= dI x ' x ' + z dm = ρdz ⎜ a b⎟ + z ρabdz = ρab⎜ + z 2 ⎟ dz
⎝ 12
⎠
⎝ 12
⎠
2
( )
Integrating gives the result obtained previously
1
⎛ a2
⎞
ρabl ⎛ 2 a 2 ⎞ 1
2
2
I xx = ρab∫ ⎜ + z 2 ⎟ dz =
⎜ l + 4 ⎟ = 12 m a + 4l
12
3
⎠
⎝
⎠
0 ⎝
The expression for I xx may be simplified for a long prismatical bar or slender rod
(
)
whose transverse dimensions are small compared with the length. In this case a 2
may be neglected compared with 4l 2 , and the moment of inertia of such a slender
1
bar about an axis through one end normal to the bar becomes I = ml 2 . By the
3 to the bar
same approximation the moment of inertia about a centroidal axis normal
is I =
1 2
ml .
12
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XVII. References
Duncan, T (1981) Advanced Physics, London. John Murray.
Edwards and Penney (1982) Calculus and Analytic Geometry, New Jersey. Prentice-Hall.
Eisberg, R.M and Lerner, L.S. Physics: Foundations and Applications, New York.
McGraw Hill.
Fadell, E.R. Fadell, A.G. (1970) Calculus. New York .Van Nostrand Reinhold.
Kane, J.W. and Sternheim (1978) Physics (2nd Ed). New York ,John Wiley and
Sons.
Learson, Hostetter and Edwards (1998) Calculus (6th Ed) ,Boston. Houghton
Miflin.
Meriam, J.L. and Kraige L.G. (1987) Engineering Mechanics Vol 2 Dynamics,
New York. John Wiley and sons.
Sears, F.W., Zemansky M.N. and Young, H.D. (1982) College Physics (5th Ed),
Reading. Addison-Wesley.
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XVIII. Main Author of the Module
Team
Mr. Tendayi Chihaka (main author)
Mr. Blessing Mufoya (co-author)
Mr. Admire Kurira (co-author)
This module is a product of collaborative work by a team of three members of the
Department of Science and Mathematics Education, at the University of Zimbabwe.
The main Author, Tendayi Chihaka, holds a Masters in Mathematics Education and
has vast experience in pre-service teacher education, having trained pre-service
secondary mathematics teachers from 1978 to 1997 – a period of twenty years. He
joined the Department of Science and Mathematics Education in 2003 where he
was elected chairman of Department in 2004, the position he currently holds. He is
married and has three children.
Mr Kurira holds a Masters in Mathematics and Physics and a Post Graduate Certificate
in Education. He has taught Engineering at Harare Polytechnic 1996-1999. He joined
the University of Zimbabwe, Department of Science and Mathematics in August 1999
as an ICT and Mathematics lecturer. He is married with 3 children.
Mr Mufoya holds a Masters in Mathematics and a Post Graduate Diploma in Education. He joined the University of Zimbabwe, Department of Science and Mathematics
in November 2005. He in currently single.