Mechanics II
Prepared by Tilahun Tesfaye, Ph.D.
NOTICE
TABLE OF CONTENTS
FOREWORD
This module has four major sections
The first one is the INTRODUCTORY section that consists of five parts vis:
1.
TITLE:- The title of the module is clearly described
2.
PRE-REQUISIT KNOWLEDGE: In this section you are provided with information
regarding the specific pre-requisite knowledge and skills you require to start the
module. Carefully look into the requirements as this will help you to decide
whether you require some revision work or not.
3.
TIME REQUIRED: It gives you the total time (in hours) you require to complete the
module. All self tests, activities and evaluations are to be finished in this specified
time.
4.
MATERIALS REQUIRED: Here you will find the list of materials you require to
complete the module. Some of the materials are parts of the course package you
will receive in a CD-Rom or access through the internet.
Materials
recommended to conduct some experiments may be obtained from your host
institution (Partner institution of the AVU) or you may acquire borrow by some
other means.
5.
MODULE RATIONALE: In this section you will get the answer to questions like
“Why should I study this module as pre-service teacher trainee? What is its
relevance to my career?”
The second is the CONTENT section that consists of three parts:
6.
OVERVIEW: The content of the module is briefly presented. In this section you will
fined a video file (QuickTime, movie) where the author of this module is
interviewed about this module. The paragraph overview of the module is followed
by an outline of the content including the approximate time required to complete
each section. A graphic organization of the whole content is presented next to
the outline. All these three will assist you to picture how content is organized in
the module.
7.
GENERAL OBJECTIVE(S): Clear informative, concise and understandable
objectives are provided to give you what knowledge skills and attitudes you are
expected to attain after studying the module.
8.
SPECIFIC LEARNING OBJECTIVES (INSTRUCTIONAL OBJECTIVES): Each of
the specific objectives, stated in this section, are at the heart of a teaching
learning activity. Units, elements and themes of the module are meant to achieve
the specific objectives and any kind of assessment is based on the objectives
intended to be achieved. You are urged to pay maximum attention to the specific
objectives as they are vital to organize your effort in the study of the module.
The third section is the bulk of the module. It is the section where you will spend more
time and is referred to as the TEACHING LEARNING ACTIVITIES. The gist of the nine
components is listed below:
9.
PRE-ASSESSMENT: A set of questions, that will quantitatively evaluate your level
of preparedness to the specific objectives of this module, are presented in this
section. The pre-assessment questions help you to identify what you know and
what you need to know, so that your level of concern will be raised and you can
judge your level of mastery. Answer key is provided for the set of questions and
some pedagogical comments are provided at the end.
10.
TEACHING AND LEARNING ACTIVITIES: This is the heart of the module. You
need to follow the learning guidance in this section. Various types of activities are
provided. Go through each activity. At times you my not necessarily follow the
order in which the activities are presented. It is very important to note:

formative and summative evaluations are carried out thoroughly

all compulsory readings and resources are done

as many as possible useful links are visited

feedback is given to the author and communication is done
11.
COMPILED LIST OF ALL KEY CONCEPTS (GLOSSARY): This section contains
short, concise definitions of terms used in the module. It helps you with terms
which you might not be familiar with in the module.
12.
COMPILED LIST OF COMPULSORY READINGS: A minimum of three
compulsory reading materials are provided. It is mandatory to read the
documents.
13.
COMPILED LIST OF (OPTIONAL) MULTIMEDIA RESOURCES: Total list of
copyright free multimedia resources referenced in, and required for completion
of, the learning activities is presented.
14.
COMPILED LIST OF USEFUL LINKS: a list of at least 10 relevant web sites. that
help you understand the topics covered in the module are presented. For each
link, complete reference (Title of the site, URL),a screen capture of each link as
well as a 50 word description are provided.
15.
SYNTHESIS OF THE MODULE: Summary of the module is presented.
16.
SUMMATIVE EVALUATION:Enjoy your work on this module.
I.
MECHANICS II
BY TILAHUN TESFAYE ADDIS ABABA UNIVERSITY ETHIOPIA
Fig. 1 When
the wheel of a bike is spinning the axle turns about the suspension point so that the
plane of rotation of the wheel remains vertical.
II
PREREQUISITE COURSE OR KNOWLEDGE
In order to study this module you need to complete Mechanics I of the AVU Teachers’
Training Module.
III
TIME
This module can be completed in 120hrs.
IV
MATERIALS
The following list identifies and describes the equipment necessary for all of the activities
in this module. The quantities listed are required for each group.
1. Computer: - A personal computer with word processing and spreadsheet software
2. String and Ball: - Rotational motion experiment
3. Meter Stick: -Rotational Motion Experiment.
V
MODULE RATIONALE
Physics is a study of Energy and its transformations. One of the ways of energy
transformation happens when objects are set in motion. Description of motion has been
studied in Mechanics I module. The emphasis on mechanics I was on the kinematic and
dynamic description of particles motion.
This module extends the kinematics and dynamics of particle motion to dynamics of a
system of particles; rotational motion rigid bodies and Gravitation. Hence ability solve
problems using the equation of motion of a rotating rigid body when the motion is about
any fixed axis, as well as when the motion is abut a principal axis. will be developed.
Furthermore the learner will be able to calculate the kinetic energy of rotation of a rotating
rigid body and use this as an additional form of kinetic energy in solving problems using
the conservation of energy.
VI
OVERVIEW
The central concepts of this module (Mechanics II) are dynamics of a system of particles,
rotational motion and Gravitation. The module begins with the study of impulse of a force
and its relation with momentum.
The second activity is the kinematic and dynamic descriptions of rotational motion. New
quantities to describe rotational motion are introduced and used. It will be show that the
equations of motion that describe linear motion possess a rotational counterpart .
The third activity is on Gravitation Up to now we have described various forces from an
entirely empirical point of view. To gain a more unified understanding of such forces and
to achieve greater predictive power, we shall now examine two of the four fundamental
forces which are ultimately responsible for all other forces. Thus in the third activity we’ll
discuss the gravitational force which accounts for the interaction between all astronomical
bodies, the motion of the planets and the moon, the trajectories of space vehicles, the
occurrence of the tides, and the weights of objects.
6.1 OUTLINE
1 Dynamics of a system of Particles
(40 hours)
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


Linear momentum of particle and of a system of particles,
Conservation of linear momentum.
Impulse and linear momentum,
Conservation of linear momentum in collisions and explosions; elastic and inelastic
collisions;
 Collisions in two dimensions.

Centre of mass and motion about centre of mass.
2 Rotational Motion
(45 hours)
 Rotational kinematics, angular variables, relationship between linear and angular
kinematics - fixed axis.
 Rotational dynamics, torque, angular momentum (of a particle and a system of particles),
rotation inertia, rotational kinetic energy,
 Conservation of angular momentum.
3 Gravitation







(35 hours)
The Law of Universal Gravitation,
Planet and satellite motion,
Gravitational field and potential, inertia and gravitational mass.
Variation in gravitational field strength due to latitude, altitude.
Motion of planets and satellites- geostationary orbits.
Relative velocity. Uniform relative translational motion.
The Galilean transformation.
6.2 GRAPHIC ORGANIZER
A. Dynamics of
systems of particles:
C. Gravitation:
Linear momentum of particle and of a system of particles,
conservation of linear momentum.
The Law of Universal Gravitation,
Impulse and linear momentum,
planet and satellite motion,
Conservation of linear momentum in collisions and explosions;
elastic and inelastic collisions; collisions in two dimensions.
gravitational field and potential,
inertia and gravitational mass.
Variation in gravitational field strength
due to latitude, altitude.
Mechanics
II
Centre of mass and motion about centre of mass.
B. Rotational Motion:
Motion of planets and satellitesgeostationary orbits. Relative velocity.
Uniform relative translational motion.
The Galilean transformation.
Rotational kinematics, angular variables,
relationship between linear and angular kinematics - fixed axis.
Rotational dynamics, torque, angular momentum
(of a particle and a system of particles),
rotation inertia, rotational kinetic energy,
conservation of angular momentum.
VII. GENERAL OBJECTIVE(S)
After completing this module you will be able to
 Develop understanding of linear and angular momentum
 Understand dynamics of rotational motion
 Understand gravitational interaction and its application in artificial satellite.
 Develop skills and habits of solving problems in a well reasoned and neat manner
VIII. Specific Learning Objectives (Instructional Objectives)
Content
1.
Dynamics of a system of
Particles (40 hours)
 Linear momentum of particle and of a





2.



3.
system of particles,
Conservation of linear momentum.
Impulse and linear momentum,
Conservation of linear momentum in
collisions and explosions; elastic and
inelastic collisions;
Collisions in two dimensions.
Centre of mass and motion about
centre of mass.
Rotational Motion: (45 hours)
Rotational kinematics, angular
variables, relationship between linear
and angular kinematics - fixed axis.
Rotational dynamics, torque, angular
momentum (of a particle and a
system of particles), rotation inertia,
rotational kinetic energy,
Conservation of angular momentum.
Gravitation: (35 hours)
 The Law of Universal Gravitation,
 Planet and satellite motion,
 Gravitational field and potential,




inertia and gravitational mass.
Variation in gravitational field strength
due to latitude, altitude.
Motion of planets and satellitesgeostationary orbits.
Relative velocity. Uniform relative
translational motion.
The Galilean transformation.
Learning objectives
After Completing this section you
would be able to:














Relate impulse and linear momentum
Solve problems involving elastic and
inelastic collisions in 1 and 2 dimensions
Describe motion of centre of mass and
motion about centre of mass for a system
of particles
Derive and use equations describing
rotational motion
Relate angular and linear quantities for
rotation around a fixed axis
Use T=Ia to solve problems
Define angular momentum and its
conservation
Solve problems in rotational dynamics
Use Newton’s law of universal gravitation
to solve problems
Describe Gravitational field and potential
(never use two verbs in the same obective
Describe the gravitational potential
Distinguish between inertial and
gravitational mass(replace with Distinguish
between inactive and gravitational force)
Calculate escape velocity of satellites
Use Galilean transformation to solve
problems in gravitation
IX.
PRE-ASSESSMENT: Are you ready for Mechanics II?
Dear Learner:
In this section, you will find self-evaluation questions that will help you test your
preparedness to complete this module. You should judge yourself sincerely and do the
recommended action after completion of the self-test. We encourage you to take time and
answer the questions.
Dear Instructor:
The Pre-assessment questions placed here guide learners to decide whether they are
prepared to take the content presented in this module. It is strongly suggested to abide by
the recommendations made on the basis of the mark obtained by the learner. As their
instructor you should encourage learners to evaluate themselves by answering all the
questions provided below. Education research shows that this will help learners be more
prepared and help them articulate previous knowledge.
9.1 SELF EVALUATION ASSOCIATED WITH MECHANICS II
Evaluate your preparedness to take the module on thermal physics. If you score greater
than or equal to 60 out of 75, you are ready to use this module. If you score something
between 40 and 60 you may need to revise your school physics on topics of heat. A score
less than 40 out of 75 indicates you need to physics.
Try the following questions and evaluate where you are in topics related to Mechanics II
1. A person travels a distance d meters given by (5m/s2 )t 2 where t is in seconds.
Which of the following statements are correct? Which of the following statements are
correct?
(a).The distance traveled in 10 seconds is 500m.
(b).The average speed of the person in 10 seconds is 50m/sec.
(c). The average speed of the person in 10 seconds is 100m/sec.
(d).The instantaneous speed of the person at t  10 sec is 100m/sec
2. The position of a particle moving along the x -axis depends on the time as per the
following equation for four seconds
t4
2
3
x  4t  2t 
4
where x is in meters and t in seconds. Choose the correct alternatives:
(a) The particle reaches its maximum x -position. in 2 seconds
(b) The displacement of the particle in 4 seconds is zero.
(c)
The distance covered by the particle in 4 seconds is 8 meters
(d) The particles speed after 2 seconds is zero.
3. Choose the correct statements
(a).A projectile fired from the ground follows a circular path
(b).A projectile fired from the ground follows a parabolic path
(c). The speed of the projectile is minimum at the highest point of its path
(d). The speed of the projectile is maximum at the highest point of its path
4. A particle is acted upon by a force of constant magnitude which is always
perpendicular to the velocity of the particle. The motion of the particle takes place in a
plane. Then the particle.
(a).Has constant velocity
(c). Has constant kinetic energy.
(b).Has constant acceleration.
(d).moves in a circular path.
5. A curve track is banked for a certain velocity when a particle goes round the circular
track with a lesser velocity. The forceof friction perpendicular to direction of motion
(a).is totally absent
(c). acts horizontally outwards
(b).acts along the road outwards
(d).acts along the road inwards
6. If two equal forces have a resultant equal to the magnitude of either of the 2 forces
then the angle between the 2 forces is
(a). 0
(c). 190
(b). 60
(d). 120
7. A solid body floats in a liquid of specific gravity 0.8 with 2/5th of its volume exposed to
air . Its specific gravity is is
(a).0.6
(c). 0.4
(b).0.48
(d).0.32
8. In which of the folowing examples is the motion of a car not accelerated?
(a). A car climbs a steep hill with its speed dropping from 60 km/hr at the top
(b). A car turns a corner at the constant speed of 29km/hr
(c). A car climbs a steep hill at the constant speed of 40km/hr
(d).A car climbs a steep hill and goes over the crest and down on the other side, all at
a speed of 40km/hr.
9. Which law of motion makes swimming possible?
(a).Second
(c). Third
(b).First
(d).None
10. A projectile is fired at an angle of 37 with an initial speed of 100/sec. What is the
approximate vertical component fo its velocity after 2 sec?
(a).80m/sec
(c). 60m/sec
(b).40m/sec
(d).100m/sed
0
11. A projectile is fired at an angle of 37 with an initial speed of 100 m/s. the vertical
component of its velocity after 2s?
(a). 80 m/s
(c). 60 m/s
(b). 40.4 m/s
(d). 29.6 m/s
12. In problem 11 the position above the ground after 3s is approximately
(a). 140 m
(c). 136 m
(b). 200 m
(d). 120 m
13. A projectilie is fired horizontally with an initial speed of 20m/s. Its horizontal speed 3s
later is
(a). 20 m / s
(b). 6.67 m / s
(c). 60 m / s
(d). 29.4 m / s
14. The vertical speed of the above projectile after 3s is approximately
(a). 9.8 m/s
(c). 29.4 m/s
(b). 60 m/s
(d). 20 m/s
15. Which of the following projection angles will result in the greatest range?
0
0
(a). 37
(c). 48
0
0
(b). 20
(d). 60
16. A 10kg block is lifted 20m above the ground in a gravitational field. The work done by
the field is
(a).Negative
(c). Positive
(b).Equal to the final potential energy
(d).A vector quantity
17. A body in equilibrium may not have
(a).Momentum
(c). Acceleration
(b).Velocity
(d). Kinetic energy
18. Watt-sec is a unit of
(a).Momentum
(c). Energy
(b).Force
(d).Power
19. The 2.5kg head fo an ax exerts a force of 80kN as it penetrates 18mm into the trunk
of a tree. The velocity of the axe head when it strikes the tree is
(a).1.2m/s
(c). 3.4m/s
(b). 34m/s
(d). 107m/s
20. A 50 kg mass has a PE of 4.9kJ relative to the ground. The height of the mass above
the ground is
(a).10m
(c). 960m
(b). 98m
(d).245m
9.2 ANSWER KEY:
*One relevant image must be inserted here.
1. A, C, D
6. D
11. B
16. A
2. A, B, C, D
7. B
12. C
17. C
3. B, C
8. C
13. A
18. C
4. C, D
9. C
14. C
19. C
5. B
10. B
15. C
20. A
9.3 PEDAGOGICAL COMMENT FOR THE LEARNER:
Many students have a sad impression that mechanics is difficult to grasp. The single most
resposible factor for this impression is not the lack of information or theoretical concepts
but rather the absence of clear and correct ideas about the relations between the
concepts of physics. Learners often cannot say what forms the basis of a definition, what
is the result of an experiment, and what should be treated as a theoretical generalizsation
of experimental knowledge.
It is important to distinguish whether a presented fact is a self evident corollary of
previousely stated fact or not. It is also important not to regard different formulations of
the same problem as different laws. Here comes the importance of solving as many
problems as possible. Most importantly do exercises and self assessments on schedule;
don't put it off until the last minute (or later).
Extensive research in recent years has shown that the students who do best in physics
(and other subjects) are those who involve themselves actively in the learning process.
This involvement can take many forms: writing lots of questions in the margins of the
module; asking questions by email; discussing physics in the AVU discussion forums etc.
A Final Word…..
Physics is not so much a collection of facts as a way of looking at the world. The author of
this module hopes that this course will not only teach you mechanics, but will also
improve your skills in careful thinking, problem solving, and precise communication. In
this course you will gain lots of experience with qualitative explanations, rough numerical
estimates, and careful quantitative problem solving. When you understand a phenomenon
on all of these levels, and can describe it clearly to others, you are "thinking like a
physicist" (as we like to say). Even if you eventually forget every fact learned in this
course, these skills will serve you well for the rest of your life.
X. TEACHING AND LEARNING ACTIVITIES
ACTIVITY 1: Dynamics of A system of Particles
You will require 45 hours to complete this activity. In this activity you are guided with a
series of readings, Multimedia clips, worked examples and self assessment questions
and problems. You are strongly advised to go through the activities and consult all the
compulsory materials and as many as possible among useful links and references.
Specific Teaching and Learning Objectives



Relate impulse and linear momentum
Solve problems involving elastic and inelastic collisions in 1 and 2 dimensions
Describe motion of center of mass and motion about center of mass for a system of
particles.
Summary of the Learning Activity
Questions related to collisions of bodies are difficult to answer directly by applying
Newton’s second law  F = ma , because the forces acting between the colliding objects
are not fully known. In this activity we’ll that we don’t have to know the forces and the
time of their action to analyze motion of a system of interacting particles. For this purpose
the concept of impulse J  p is defined and used.
To further our comprehension of mechanics we must begin to examine the interactions of
many particles at once. To begin this study, we define and examine a new concept, the
center of mass, which will allow us to make mechanical calculations for a system of
particles.
List of Required Readings
Copyright free readings should also be given in electronic form (to be provided on a CD with the module)
Reading 1: Momentum in One Dimension.
Complete reference : Conservation of Momentum
From html version of Simple Nature, by Benjamin Crowell.
URL : http://www.lightandmatter.com/html_books/0sn/ch03/ch03.html#Section3.1
Accessed on the 20th April 2007
Abstract :
This is part of a book by Benjamin Crowell. It is freely available at
www.lightandmatter.com the part given here the relevant section for this activity.
Rationale:
This section has a well illustrated content on linear momentum. The motion of center of
mass is treated at the end. It provides another way of looking at the theories of collision
and momentum conservation. The examples drawn from nature, like comet, are
interesting and educational reading materials.
Reading 2: Momentum Conservation and Transfer.
Complete reference : Momentum Conservation and Transfer
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m15.pdf
Accessed on the 20th April 2007
Abstract :
In this article, momentum is defined for a single and a system of particles. Using Newton’s
laws and the definition of momentum it is shown that the momentum of an isolated
system of particles remain unchanged with time (i.e. conserved)
Rationale:
This article covers the contents of this activity. It gives another way of looking at the
theories of collision and momentum conservation. Further the sample tests and exercises
given at the end provide good opportunity to use the theories and principles exercised
from different perspectives.
List of Relevant MM Resources
1. Reference http://jersey.uoregon.edu/vlab/Piston/index.html
Date Consulted:-Nov 2006
Description:- This Java applet helps you to do a series of virtual experiments,
you will control the action of a piston in a pressure chamber which is filled with an
ideal gas. The gas is defined by four states: Temperature; Volume or density;
Pressure and Molecular Weight
2. Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted:- August 2006
Description:- This Java applet helps you understand the effect of temperature
and volume on the number of collisions of the gas molecules with the walls. In the
applet, you can change the temperature and volume with the sliders on the left
side. You can also adjust the time for which the simulation runs. The applet counts
all collisions and displays the result after the run. By varying temperature and
volume and keeping track of the number of collisions, you can get a good feeling of
what the main result of kinetic theory will be.
3. Reference: video.google.com
Date Consulted: Nov 2006
Complete Reference:- Computer calculation of Phase Diagrams.
http://video.google.com/videoplay?docid=1397988176780135580&q=Thermodyna
mics&hl=en
Rationale: Thermodynamic models of solutions can be used together with data to
calculate phase diagrams. These diagrams reveal, for a given set of all parameters
(such as temperature, pressure, magnetic field), the phases which are
thermodynamically stable and in equilibrium, their volume fractions and their
chemical compositions...
List of Relevant Useful Links
1. Title:- About Temperature
URL:- http://eo.ucar.edu/skymath/tmp2.html
Abstract:-This document was prepared for the middle school math teachers who
are taking part in Project Skymath. It is provided a detailed description of
temperature.
2. Title: A Brief History of Temperature Measurement
URL: http://thermodynamics-information.net/
Abstract: Brief biography and thermometers created by scientists from Réne
Antoine Ferchault de Réamur (1683-1757) to William John Macquorn Rankine
(1820-1872) are presented. There is also a table that compares the values of the
Fahrenheit, Celsius, and Kelvin temperature for some common reference
temperatures.
3. Title: Why do we have so many temperature scales?
URL: http://www.unidata.ucar.edu/staff/blynds/tmp.html
Abstract:
Detailed Description of the Activity (Main Theoretical Elements)
Introduction
In Mechanics I module you have studied forces acting on bodies and particles and there
effects. These forces act on their points of application for a time sufficiently enough to
measure. There are phenomena in which interaction between bodies is so fast that it is
difficult to measure the forces that are produced between them or the time that the
interaction lasts. For example, how long does the collision between two billiard balls last
for? What force does one ball apply on the other? These questions are, no doubt, difficult
to answer. Should we give up trying to calculate the result of collisions? Should we leave
everything to the billiard player's experience and intuition? No, physics doesn't give up on
trying to explain phenomena that look difficult that easily.
In these cases, the notion of linear momentum and impulse, in addition to the conditions
under which linear momentum is conserved, will allow us to make predictions of the
speed and direction of the movement after the interaction.
The scalar quattities work and energy have no directions associated with them. When two
or more bodies interact with one another, or a single body breaks up into two or more
other bodies, the various directions of motion cannot be related by energy
consideerations alone. The vector quatities called linear momentum and impulse are
important in anlyzing such events.
1.1:
Linear Momentum and Impulse
In the mechanics I, the concepts of work and energy were developed from Newton’s laws
of motion. We shall see next how the concepts of linear momemtum and impulse also
arize from these law.
r
Consider a particle of mass m moving with a velocity v . Suppose a const force F acts
along the line of motion, then
v  v  at
multiply both sides of this equation by m we obtain
mv  mv 0  Ft
 mv  mv 0  F .t
  p  Ft
The linear momentum of a particle is defined as the product of its mass and its linear
velocity
P = mv valid only for v << the speed of light c
p
x
= mv x ; p y = mv y ; p 2 = mv 2
The impulse of a force is defined as the product of a force and the time during which it
acts i.e.
r
r
J  p
Interms of these two newly defined quantities
 p  J  F .t
Note that the above equation holds independently for components. i.e
px  J x  Fx .t
p4  J 4  F4 .t
p2  J 2  F2 .t
Linear momentum of a particle can also be related to the net force acting on the particle
as follows
dv d
dp
F = ma = m
= (mv) =
dt dt
dt
 dp = Fdt
Integrating the last expression, the change in momentum of a particle is
tf
ur
ur
ur
ur
p  p f  p i   F.dt
ti
tf
The quantity
 F .dt is called the impulse
of the force F for t   f  ti
ti
tf
I   F .dt  p
Impulse – momentum Theorem.
ti
Since the force can generally vary in tire, it is convenient to define a time averaged force
tf
 1 
F     F .dt
 t  ti
 I  p  F t
1.2:
Conservation of Linear Momentum
Consider two particles that can interact each other but are isolated from the surrounding
Let
F12  the force on particle 1 due to particle 2
F21  the force on particle 2 due to particle 1
Applying Newton's 2nd law
dp
F12  1 , and
dt
dp
F21  2
dt
These forces could be of any origing (i.e. gravitational, electromagnetic etc.) Since the
two are action reaction pairs
F12  F21  0
dp1 dp2 d

 ( p1  p2 )  0
dt
dt
dt
 total momentum (p)  p1  p2  constant.
 pix  p fx
piy  p fy
piz  p fz
The statement the total momentum of the system remains constant is referred to as the
law of conservation of momentum.
Example 1.1: Linear momentum and impulse
A child bounces a super ball on a side walk. The linear impulse delivered by the side
walk to the super ball is 2 N.s during 1/800 sec of contact. What is the magnitude of
the average force exerted on the super ball?
Solution
Given: I  2Ns and t 
1
sec.
800
By definiton:
I  Fav t
Therefore the average force exerted on the superball is, Fav 
2
1
800
 1600N.
Example 1.2: A 3kg steel ball strikes a massive wall with a speed of 10m/s at an angle of
600 with the angle as shown. If the ball is in contact with the wall for 0.2 sec what is the
average force exerted on the ball by the wall.
Solution:
p
 Fav 
t
p f 2  pi 2  pi p f
t
302  302  2  30  30 cos

0.02
 260 N . in the horizontal direction.
1.3. A System of Particles
The simplification by treating interactions as point particle interactions has its limitations
for two reasons:
 Mostly objects are extended
 Systems in which mass flow exist can not be treated (rocket propulsion, explosion
etc)
In this part of the activity we shall generalize the laws of motion to overcome these
difficulties. We begin by restating Newton’s second law
F  ma 

d
(mv)
dt
 m is constant
dP
dt
This form of 2nd law is preferred to F  ma because it is generalized to complex
systems, and because momentum turns out to be more fundamental than m or v
separately.
Here we define some of the terms used in this module
A system:- is a set of objects or substances which are interdependent and governed by
physical laws, forming a whole. eg. Solar system is governed by Gravitational
law, mass-spring system
Closed System: - is a system that does not interact with its surrounding.
External force: - is a force exerted on or a part of a system by some body or agency
outside the system
Consider a system of N interacting particles with masses m1 , m2 , m3 ... mN . The
position of the j th particle is rj , the force on the j th particle is f
j
and its momentum is
p j  m j rj
The equation of motion for the j th particle is thus
dp
 f jinternal  f jexternal
dt
the force on particle j can be split between two terms (internal and external)
fj
Adding all the equations of motion of all the particles in the system, we obtain
f1int  f1ext 
d p1
dt
f j int  f j ext 
d pj
f N int  f N 
dt
d pN
dt
  f j int   f j ext  
f
ext
j
d pj
dt
,
j  1, 2, 3,
n
 is the sum of all external forces acting on all particles.
= total external force acting on the system
f
=Fext
int
j
= is the sum of all internal forces acting on all the particles
= 0 This follows from newtons third law.
By Newtons 3rd law, the forces between any two paricles are equal and opposite. The
internal forces in the system of particles cancel in pair.
Therefore
f
Denoting
p
F ext 
ext
j
j
dp
 F ext   

 dt 
d
   p j
derivative of a sum is the sum of the derivatives.
 dt 
 p the total momentum of the system the last equation becomes
dp
dt
The total external force acting on a system of particles is equal to the time rate of change
of the total momentum. This is true irrespective of the details of interaction F ext could be a
single force acting on a single particle, or it could be the resultant of many ting
interactions
The Center of Mass of Particles
The concept of the center of mass allows us to describe the movement of a system of
particles by the movement of a single point. We will use the center of mass to calculate
the kinematics and dynamics of the system as a whole, regardless of the motion of the
individual particles.
The center of mass for the simplest possible system of particles, one containing only two
particles, will be difined first and we will generalize for systems containing many particles.
Center of Mass for Two Particles in One Dimension
If a particle with mass m1 has a position of x1 and a particle with mass m 2 has a position
of x 2 , then the position of the center of mass of the two particles is given by:
m x  m2 x 2
x cm  1 1
m1  m 2
Thus the position of the center of mass is a point in space that is not necessarily part of
either particle. This phenomenon makes intuitive sense: connect the two objects with a
light but rigid pole. If you hold the pole at the position of the center of mass of the objects,
they will balance. That balancing point will often not exist within either object.
Center of Mass for Two Particles beyond One Dimension
The concept of the center of mass can be extended to velocity and acceleration: taking a
simple time derivative of our expression for xcm we see that:
m v  m2 v2
vcm  1 1
m1  m 2
Differentiating again, we can generate an expression for acceleration:
m a  m2a 2
a cm  1 1
m1  m 2
With this set of three equations we have generated the necessary elements of the
kinematics of a system of particles.
From our last equation, however, we can also extend to the dynamics of the center of
mass. Consider two mutually interacting particles in a system with no external forces. Let
the force exerted on m 2 by m1 be F21 , and the force exerted on m1 by m 2 by F12 . By
applying Newton's Second Law we can state that F12  m1a1 and F21  m 2 a 2 . We can now
substitute this into our expression for the acceleration of the center of mass:
F  F21
a cm  12
m1  m 2
However, by Newton's Third Law F12 and F21 are reactive forces, and F12  F21 Thus
a cm  0 Thus, if a system of particles experiences no net external force, the center of
mass of the system will move at a constant velocity.
But what if there is a net force? Can we predict how the system will move? Consider
again our example of a two body system, with m1 experiencing an external force of F1
and m 2 experiencing a force of F2 We also must continue to take into account the forces
between the two particles, F21 and F12 . By Newton's Second Law:
F1  F12  m1a1
F2  F21  m 2a 2
Substituting this experession into our center of mass acceleration equation we obtain
F1  F2  F12  F21  m1a1  m 2a 2
 F1  F2  m1a1  m 2a 2   m1  m 2  a cm

F
external
  m1  m 2  a cm
 Ma cm
This equation bears a striking resemblance to Newton's Second Law. The overall
acceleration of a system of particles, no matter how the individual particles move, can be
calculated by this equation. Consider now a single particle of mass M placed at the
center of mass of the system. Exposed to the same forces, the single particle will
accelerate in the same way as the system would. This leads us to an important
statement:
The overall motion of a system of particles can be found by applying Newton's Laws as if
the total mass of the system were concentrated at the center of mass, and the external
force were applied at this point.
Systems of More than Two Particles
A simple extension of our two particle equations to an n particle system will show the
total mass of the system M is
M  m1  m2  m3  L  mn
With this definition we can simply state the equations for the position, velocity, and
acceleration of the center of mass of a many particle system, similar to the two-particle
case. Thus for a system of n particles:
1
x cm 
 mn x n
M
1
vcm 
 mn vn
M
1
a cm 
 mn a n
M
 Fext  Ma cm
These equations require little explanation, as they are identical in form to our two particle
equations. All these equations for center of mass dynamics may seem confusing,
however, so we will discuss a short example to clarify.
Consider a missile composed of four parts, traveling in a parabolic path through the air. At
a certain point, an explosive mechanism on the missile breaks it into its four parts, all of
which shoot off in various directions, as shown below.
Figure: A missile breaking into pieces
What can be said about the motion of the system of the four parts? We know that all
forces applied to the missile parts upon the explosion were internal forces, and were thus
cancelled out by some other reactive force: Newton's Third Law. The only external force
that acts upon the system is gravity, and it acts in the same way it did before the
explosion. Thus, though the missile pieces fly off in unpredictable directions, we can
confidently predict that the center of mass of the four pieces will continue in the same
parabolic path it had traveled in before the collision.
Such an example displays the power of the notion of a center of mass. With this concept
we can predict emergent behavior of a set of particles traveling in unpredictable ways. We
have now shown a way to calculate the motion of the system of particles as a whole. But
to truly explain the motion we must generate a law for how each of the individual particles
react. We do so by introducing the concept of linear momentum in the next section.
Conservation of Momentum (System of Particles)
We have seen that the rate of change of momentum of a body is proportional to the
resultant force acting on the body and is in the direction of the force:
dp
f 
dt
For a system of n particles.
p  p1  p 2 
pn
 m1 v1  m2 v 2 
mn v n  MVcm
 Total momentum of   total mass of   The velocity of the center 
i.e. 

 

 a system of particles   the system   of mass of the system

dv
dP

 M cm  Macm
dt
dt
dP
 Fext 
 0, when no external force acts on the system
dt
When the resusltan external force acting on a system is zero, the total vector momentum
of the system remains constan. This is the principleof conservation of momentum..
System of variable Mass:
y
y
m
M
u
v
x
M-m
v+v
x
Fig: mass m moving with velocity v.
The above figure shows mass m moving with velocity v. After some time a mass m is
ejected out with a velocity u in opposite direction to that of v . For this system of variable
dp
mass we can write Fext 
as an approximate result for
dt
 p p f  pi
Fext

t
t
[( M  M )(v  v)  Mu ]  [ Mv]
v
M
M
 [u  (v  v)]
t
t
as t  0
dM
dM
dM
Fext  M
v
u
dt
dt
dt
Note:- this equation is reduced to the familiar law of conservation of momentum.
Example 1.3: A grenade flying horizontally with a velocity of 12 m/s is explodes in to
two fragments with masses of 10 kg and 5kg. The velocity of the larger fragment is
25m/sec and forms an angle of 330o with the horizontal. Find the magnitude and direction
of the velocity of the smaller fragment.
Solution: The given situation looks like the figure below
 Total momentum   Total Momentum 

=

 before explosion   after explosion 
( Ptot ) before =(Ptot )after
 (15  12)i=(250 cos 30 + 5 v cos  )i  ( 250sin 30  5v sin  ) j
0  5v sin  =125
 vsin  =25
 180 =216.5 + 5v cos   vcos  =-7.3 (2)
(1)  (2)  tan  =25/-7.3 = -3.4247   =-73.720 
tan is - ve in the 4th or 2nd quadrant for the prblem we take the angle in the 2nd
 using the value of  in (1)
25
25
v=

 26m / s
sin 6 sin106.28
Example 1.4: A 4N weight rests on a smooth horizontal plane it is struck with a 2N blow
that lasts 0.02 sec. Three seconds after the start of the first blos a second blos of -2N is
delivered. This lasts for 0.01 sec. What will be the speed of the body after 4 sec?
Solution The forces in this problem are
For any t>3.01 sec is the sum of the two areas
i.e  J  (2  0.02)  (0.01 (2))  0.02 N  sec
 0.02 N .sec 
4
(v  0)  v  0.049m / s
9.8
Example 1.6: A stream of water with cross-sectional area 2000 mm 2 and moving 10m/s
horizontally, strikes a fixed blade curve as show. Assuming the speed of the water relative
to the blade is constant (no friction is considered), determine the horizontal and vertical
components of the force of the blade on the stream of water.
Solution
o
v '  v '' , but directions are different
The mass m of all particles of water in time interval t is
m  Avg t
A  area , v speed, p-density of water
= (2000  106 m 2 )(10m/s)(103 kg/m3 )=20(t)
using J   p in the x and y directions
1
Fx t  m(v11
x  vx )  (20t )( 10 cos 45  10)  (20t )( 17.07)
1
Fy t  m(v11
y  v y )  (20t )( 10 sin 45  0)  (20t )(7.07)
 Fx  341.4 N ; Fy  141.4 N
Example 1.7:
Consider the motion from o to A.
v11  vog  gtob 
tob 
v0 sin 45
g
(1)
x1  vox .tob  (v0 cos 45)(
When
v0 sin 45
)
3
(3)
the
projectile explde at A, the velocity is entirely along x.
m
v0 cos 45m(v0 cos 45)  v f by cons of mom and where v f is the v of the flying frag.
2
 v f  240 cos 45
 x2  (240 cos 45)tob
i.e
since the fine of fall is
v0 sin 45 v 2 0 sin 45cos 45
 (240 cos 45)

g
g
v 2 0 sin 45cos 45 2v 2 0 sin 45cos 45 v 2 0 sin 45cos 45
 xtot  x1  x2 


[1  2]
g
g
g
3v 2 0 sin 45cos 45

 1.055 105 ft
g
Solve the following probles
*Reading assignments, Worked Examples, groupworks, experiments, hands on experiences…..
Task 1.1. Distance moved by a fragment from an explosion
A projectile fined from a gun at an angle of 45o with the horizontal and with muzzle speed
of 1500 ft/sec. At the highest point in its flight the projectile explodes into two fragments of
equal mass- one fragment whose initial speed is zero, falls vertically. How far from the
gun does the other fragment land, assuming a level terrain.?
Answer :  1.055  105 ft
Task 1.2.Mass of a recoiling boat
Four 50kg-girsl simultaneously dive horizontally at 2.5m/sec from the same side of a boat,
whose recoil velocity is 0.1m/sec. what is the mass of the boat?
Answer : 5000kg
Task 1.3.Question for discussion
Discuss the following questions with your colleagues or on the discussion forum of AVU
1. Why does a gun recoil ?
2. Suppose you catch a baseball, and then someone invites you to catch a bullet
with the same momentum or with the same kinetic energy. which would you
choose?
3. It is not the fall that hurts you; it is the sudden stop at the bottom. Discuss?
Task 1.4 Experiment with Billiard balls
Have you ever played a billiard ball? If no try it for experimentation purpose. A ball (b1)
thrown collides with a ball that was at rest (b2). Immediately after collision, b2 moves with
the same velocity as b1 and b1 stops.
Try this with balls of different mass and determine the relation ship between the masses
of the two balls. Use conservation of linear momentum to mathematically support your
answer.
Formative Evaluation 1
1. Blocks A and B have a mass of 10 kg and 20 kg, respectively. If they are
travelling with the speeds shown, determine their common velocity if they
collide and become coupled together
2m/sec
4m/sec
A
10kg
.
a. 2m/s to right
b. 2m/s to left
c. 3.33m.s to left
d. 4m/sec to left
2. Suppose the entire population of the world gathers in one spot and, at the
sound of a prearranged signal, everyone jumps up. While all the people are in
the air, does the Earth gain momentum in the opposite direction?
a. No; the mass of the Earth is so large that the planets change in motion
is imperceptible.
b. Yes; however, because of the much larger mass of the Earth the
change in the planets momentum is much less than that of all the
jumping people.
c. Yes; the Earth recoils with a change in momentum equal to and
opposite that of the people.
d. It depends
3. Suppose rain falls vertically into an open cart rolling along a straight
horizontal track with negligible friction. As a result of the accumulating water,
the speed of the cart:
a. Increases
b. Does not change
c. decreases
4. A person stands under an umbrella during a rain shower. A few minutes later
the raindrops turn to hail though the number of “drops” hitting the umbrella per
time and their speed remains the same. Is the force required to hold the
umbrella in the hail
a. the same as
b. more than
c. less than the force
d. required in the rain?
5.
A 4 N force acts on a 3 kg object moving at 8 m/s for 10 sec. What is the
object's change in momentum? What impulse acts on the object? What is the
object's final speed?
6. A 1000 kg car traveling at 9 m/s, east, strikes a stationary 2000 kg truck. They
interlock as a result of the collision and move off as one. What is their speed?
What is their velocity?
7. A 15,000 kg rocket launcher holds a 5000 kg rocket. The rocket exits the
launcher at +450 m/s. What is the recoil velocity of the launcher?
8. A 100 g ball traveling to the right at 2 m/s strikes a 200 g ball traveling to the
left at 4 m/s. After the collision, the 100 g ball has a velocity of 8 m/s, left. What
is the velocity of the 200 g ball?
9. A 1200 kg car moving at 8m/s, north, strikes a 2000 kg truck moving at 4 m/s,
south. The velocity of the car is 6 m/s, south. What is the velocity of the truck?
10. A 1325 kg car traveling north at 27 m/s collides with a 2165 kg car moving east
at 17 m/s. As a result of the collision, they stick together. What is their velocity
after the collision?
11. A sticky ball with a mass of 200 g is moving to the west at 6 m/s. It collides
with another sticky ball with a mass of 300 g moving north at 5 m/s. The sticky
balls stick together as a result of the collision and move off as one. What is
their velocity?
12. A 6 kg object A moving at 3 m/s, right, collides with a 6 kg object B at rest.
After the collision, A moves at 1.6 m/s, 30°. What is the velocity of B after the
collision?
13. A stationary 0.14 kg ball is struck by a 0.23 kg ball moving east at 2 m/s. After
the collision, the 0.14 kg ball has a velocity of 0.9 m/s, 30°. What is the velocity
of the 0.23 kg ball?
14. A 0.50 kg ball at rest is struck by a 0.30 kg ball moving west at 5 m/s. After the
collision, the 0.30 kg ball has a velocity of 3 m/s, 200°. What is the velocity of
the 0.50 kg ball?
15. A 2 kg object moves at 4 m/s, south. It strikes a 3 kg object at rest. After the
collision, the 2 kg object has a velocity of 2.5 m/s, 300°. What is the velocity of
the 3 kg object?
16. Hockey puck A moves to the right at 50 m/s. It strikes an identical hockey
puck B that is stationary on the ice. After the collision, the velocity of A is 35
m/s, 27.6°. What is the velocity of B? Ans: 24.97 m/s, 40.52° below the
horizontal
17. A 600 g billiard ball moving to the right at 2 m/s collides with an 800 g ball at
rest. After the collision, the 600 g ball is deflected at an angle of 37° above its
original direction at a rate of 0.5 m/s. What is the magnitude and direction of
the 800 g ball’s velocity? Ans: 1.22 m/s, 10.85° below the horizontal
18. A 6000 kg truck traveling north at 5 m/s collides with a 4000 kg car traveling
west at 15 m/s. The two remain locked together after the collision. What is their
velocity after the collision? Ans: 6.71 m/s, 26.6° north of west
19. A 1200 kg car traveling east at 60 km/h collides with a 3000 kg truck traveling
north at 40 km/h. After the collision, they remain joined. What is their velocity?
Answer in km/h. Ans: 33.3 km/h, 59°
20. A 10 kg ball travels west at 4 m/s. It strikes a 12 kg ball at rest. After the
collision, the velocity of the 10 kg ball is 2.5 m/s, 40° below the horizontal.
What is the 12 kg ball’s velocity? Ans: 2.20 m/s, 142.4°
Optional Formative Evaluation 1
Teaching the Content in Secondary School 1
What interests students about collision and momentum conservation ? This question is a
good starting point to prepare a lesson on conservation of momentum. . Students may be
allowed to list their ideas about momentum, how it describes the ‘’quantity of motion’’ and
how it provides another view of Newton’s laws This activity will give the teacher some
information about learners’ current knowledge and understanding of motion in general. It
will also tell something about what students do not know and how well they are able to
pose interesting questions that can be answered mathematically.
With this inventory, it is possible to prepare a series of activities that will guide learners
toward momentum and its conservation principle
ACTIVITY 2: Rotational Motion
You will require 30 hours to complete this activity. In this activity you are guided with a
series of readings, Multimedia clips, worked examples and self assessment questions
and problems. You are strongly advised to go through the activities and consult all the
compulsory materials and use as many as possible useful links and references.
Specific Teaching and Learning Objectives





Derive and use equations describing rotational motion
Relate angular and linear quantities for rotation around a fixed axis
Use   I to solve problems
Define angular momentum and its conservation.
Solve problems in rotational Dynamics.
Summary of the Learning Activity
Rotational motion is very common in nature as compared to linear motion. Bodies ranging
from celestial objects to subatomic particles, like electrons, are in a state of rotational
motion. The kinematic and dynamic descriptions of rotary motion are the contents of this

activity. You will need to define new quantities like angular displacement   , angular
velocity   , angular acceleration    , moment of inertia  I  , torque    , and angular
momentum  L  , for the description of rotary motion. The interesting thing here is there is
strong parallelism between linear and angular quantities. Hence the equations of motion
for rotational quantities are similar in form to that of equations describing angular
quantities.
List of Required Readings
Reading 3: Angular Acceleration.
Complete reference : Angular Acceleration in circular motion
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m33.pdf
Accessed on the 24th April 2007
Abstract :
In this article, the two agents of change, angular and linear acceleration, producing
angular acceleration and constant torque case rotational kinematics are treated.
Rationale:
This article covers the contents of this activity. It gives another way of looking at the
theories of collision and momentum conservation. Further the sample tests and exercises
given at the end provide good opportunity to use the theories and principles exercised
from different perspectives.
Reading 4: Momentum in One Dimension.
Complete reference : Conservation of Momentum
From html version of Simple Nature, by Benjamin Crowell.
URL : http://www.lightandmatter.com/html_books/0sn/ch04/ch04.html
Accessed on the 20th April 2007
Abstract :
This is part of a book by Benjamin Crowell. It is freely available at
www.lightandmatter.com the part given here the relevant section for this activity.
Rationale:
This section has a well illustrated content on angular momentum. Theories of angular
momentum in two dimensions and in three dimensions is well developed. Further there is
a good deal of theory on rigid body rotation.
Reading 5: Torque and Angular Momentum.
Complete reference : Torque and Angular Momentum in circular motion
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m34.pdf
Accessed on the 24th April 2007
Abstract :
In this article, the two agents of change, angular and linear acceleration, producing
angular acceleration and constant torque case rotational kinematics are treated.
Rationale:
This article covers topics on torque and angular momentum, system of particles,
conservation of angular momentum, nonplanar rigid bodies. The problem supplement and
the model exam at the end makes this site popular.
Reading 2: Torque and Angular Momentum.
Complete reference : Torque and Angular Momentum in circular motion
From Project PHYSNET PDF Modules
URL http://35.9.69.219/home/modules/pdf_modules/m34.pdf
Accessed on the 20th April 2007
Abstract :
This article discusses the core subject of this activity..
Rationale:
This module treats angular acceleration and constant torque equation of a rotating object
Reading 3: Moments of Inertia.
Complete reference : Moments of Inertia Principal moments
From Project PHYSNET PDF Modules
URL http://35.9.69.219/home/modules/pdf_modules/m34.pdf
Accessed on the 20th April 2007
Abstract :
This article discusses the last part of the activity
Rationale:
This module treats angular acceleration and constant torque equation of a rotating object.
It has a good collection of exercises and exam samples at the end.
List of Relevant MM Resources
1. Reference http://jersey.uoregon.edu/vlab/Piston/index.html
Date Consulted:-Nov 2006
Description:- This Java applet helps you to do a series of virtual
experiments, you will control the action of a piston in a pressure chamber
which is filled with an ideal gas. The gas is defined by four states:
Temperature; Volume or density; Pressure and Molecular Weight
2. Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted:- August 2006
Description:- This Java applet helps you understand the effect of
temperature and volume on the number of collisions of the gas molecules with
the walls. In the applet, you can change the temperature and volume with the
sliders on the left side. You can also adjust the time for which the simulation
runs. The applet counts all collisions and displays the result after the run. By
varying temperature and volume and keeping track of the number of collisions,
you can get a good feeling of what the main result of kinetic theory will be.
3. Reference: video.google.com
Date Consulted: Nov 2006
Complete Reference:- Computer calculation of Phase Diagrams.
http://video.google.com/videoplay?docid=1397988176780135580&q=Thermod
ynamics&hl=en
Rationale: Thermodynamic models of solutions can be used together with
data to calculate phase diagrams. These diagrams reveal, for a given set of all
parameters (such as temperature, pressure, magnetic field), the phases which
are thermodynamically stable and in equilibrium, their volume fractions and
their chemical compositions...
List of Relevant Useful Links
1. Title:- About Temperature
URL:- http://eo.ucar.edu/skymath/tmp2.html
Abstract:-This document was prepared for the middle school math teachers
who are taking part in Project Skymath. It is provided a detailed description of
temperature.
2. Title: A Brief History of Temperature Measurement
URL: http://thermodynamics-information.net/
Abstract: Brief biography and thermometers created by scientists from Réne
Antoine Ferchault de Réamur (1683-1757) to William John Macquorn Rankine
(1820-1872) are presented. There is also a table that compares the values of
the Fahrenheit, Celsius, and Kelvin temperature for some common reference
temperatures.
3. Title: Why do we have so many temperature scales?
URL: http://www.unidata.ucar.edu/staff/blynds/tmp.html
Abstract:
Detailed Description of the Activity (Main Theoretical Elements)
Introduction
In your school physics, you have done a little on rotation, discussing objects moving in a
circle at constant speed, and learning about centripetal force. Not surprisingly, life can be
more complicated. Objects moving in a circle don't have to move at constant speed. They
can have what is called tangential acceleration. This means that the angular speed
(measured for example in revolutions per second) is changing. For example, a CD player
tangentially decelerates as the laser moves to the outside, in order to preserve the same
speed in m/s.
To study real problems of the sort mentioned above, it is very useful to introduce a new
way of describing angular distance and speed. Degrees are fairly inconvenient: the fact
that the circle is divided into 360 degrees is completely arbitrary and not very intuitive.
What is natural is the fact that the circumference of a circle is 2 times the radius. This
suggests that instead of dividing a circle into 360 parts, we divide it into 2 parts. These
parts are called radians. Thus 90o corresponds to  radians, 45o corresponds to
2
 radians etc. Therefore we define and use the radian measure to handle real rotation
4
problems.
2.1:
Rotational Kinematics:
Angular Variables:
Angular displacement   :
S
r

r  S
S  C  2 r
   2
Conversion: 360o  2 radians
o
Angular velocity  
The angular speed is defined just like ordinary speed: an object which moves from an
angle  i to an angle  f has average angular speed  :
  f  i

t
t
Instantaneous angular speed is obtained by taking t small. An object moving 45
revolutions per minute (rpm) has   90 radians/sec . As it is seen from the definition of
the radian measure of an angle an object moving in a circle at constant speed travels
a distance 2 radians per revolution. Moreover, one revolution corresponds to 2
radians. Thus the speed v is related to the angular speed  by the simple formula:
o v  r
This formula is only valid if  is expressed in radians. Notice that radians really aren't
a unit, in the sense that v is still measured in m/s or mph or whatever: we don't need
to put a radian. Now we can write centripetal acceleration as:
v2  2r 2

  2r
o ac 
r
r
Notice that centripetal acceleration is directed perpendicular to the velocity at evry
point of the path and is responsible for the change of direction of the velocity. Not the
magnitude!
o

Angular acceleration   :
The direction of centripetal acceleration and force is inward (radial). If there's an
acceleration tangential to the circle, then the angular speed must change. Angular
acceleration  is defined asthe time rate of change of angular velocity:


t
For constant angular speed,   0 and hence   0 . For constant angular
acceleration the tangential acceleration is given by:
aT   r
(again  must be measured in radians). The direction of tangential acceleration is
tangential to the circle and so is always perpendicular to the centripetal acceleration.
Kinematic Relations
Once we define the angular quantities, we can easily do the kinematics of angular motion,
just like we did the kinematics of linear motion in Mechanics I. From the definition of
angular speed, we see that for constant angular acceleration,
  o   t
All the linear formulas have angular versions. You can easily notice the similarity of the
above equation with v  vo  at For example, to find the angular position of a particle
undergoing a constant angular acceleration, you can use:
1
2
  ot   t 2
The other equations also apply if you replace x with  , v with  and a with 
Example 2.1: The angular speed of a helicopter blade increases from 1 rad/s to 64 rad/s
in 3 seconds with constant angular acceleration. What angle has the blade turned through
in this time? What is the angular acceleration?
Solution -:
The average angular acceleration is given by:
 64 rad/s -1rad/s

 21rad/s
t
3s
To get the angular displacement, use
1
  ot   t 2  98rad ; 15revolutions.
2

Example 2.2: On a bicycle, the gears next to the petal are a radius r1 , and on the back
wheel, they are a radius r2 . The wheel is a radius rw Relate the linear speed v of the
bicycle to the angular speed at which you pedal.
Solution -:
A bike is built so that the front gears go at the same angular speed as the petals, while
the back gears go at the same angular speed as the wheel when you are pedaling. (If you
stop pedaling, the gears disengage from the wheel.) The chain connects the front gears
to the back gears, so the chain must have the same linear velocity at both the front and
the back gears. Call this linear velocity vchain Then we have
vchain   petal r1
for the front gear, and
vchain  wheel r2
Combining the two means that  petal r1  wheel r2 Now we need to get the speed of the bike.
The linear speed of the bike is related to the angular speed of the wheels by:
v  wheel rw
Thus
rw r1
r2
Changing the gears on the bike changes the ratio r1 r2 Changing to a smaller gear in the
v   petal
front, or a larger one in back, (i.e. decreasing r1 r2 ) makes it easier to petal. The reason is
for a fixed angular speed  petal it makes v smaller. The lower v , the less work you’re
doing.
Example 2.3: Linear momentum and impulse
A ceiling fan is rotating at 0.90 rev/s. When turned off, it slows uniformly to a stop in 2.2
min. (a) How many revolutions does the fan make in this time? (b) Using the result from
part (a), find the number of revolutions the fan must make for its speed to decrease from
0.90 rev/s to 0.45 rev/s.
Solution
The ceiling fan rotates about its axis, slowing down with constant angular acceleration
before coming to rest.
Use the kinematic equations for rotation to find the number of revolutions through which
the fan rotates during the specified intervals. Because the fan slows down at a constant
rate of acceleration, it takes exactly half the time for it to slow from 0.90 rev/s to 0.45 rev/s
as it does to come to a complete stop.
(a)
1
1
    o  t   0  0.90rev/s   2.2 min 60sec/ min  59rev
2
2
(b)
1
1
    o  t   0.45  0.90rev/s  1.1min 60sec/ min  45rev
2
2
Example 1.2: A 3kg steel ball strikes a massive wall with a speed of 10m/s at an angle of
600 with the angle as shown. If the ball is in contact with the wall for 0.2 sec what is the
2.2:
Rotational Dynamics
The description of rotational motion is anlogous to the description of linear motion as we
have seen it in the previous section. The analogy can be extended to dynamics of
rotation.
Displacement (x)  Angular accelaration ( )
Speeed (v)  Angular velocity ( )
Acceleration (a)  Angular acceleration ( )
Force (F)  Torque ( )
Mass (m)  Moment of Inertia (I )
Momentum (p)  Angular Momentum (L)
Moment of Inertia:
Moment of inertia (I) is the rotational analogue of mass. The greater the moment of inertia
of a body the greater its resistance to a change in its angular velocity. The value of the
moment of inertia I of a body about a particular axis of rotation depends not only upon the
body’s mass but also upon how the mass is distributed about the axis.
Torque
You might have noticed that it’s hard to open a door if you push close to the hinges. The
farther you are from the hinges, the easier it is. i.e. it requires more force to give the door
the same angular speed if you’re pushing near the hinges than if you’re at the other end.
Similarly, if you’re using a wrench, it takes less force to loosen the bolt if you’re pushing
the wrench farther away from the bolt.
Therefore we need to introduce something beyond force, to understand rotation fully and
to take into account the effect of different radii.
The idea behind a torque is that applying forces can cause rotation. In other words, just
like applying a force causes linear acceleration, applying a torque causes an angular
acceleration. To define torque, consider applying a force a distance r from the center of
rotation. The magnitude of torque is then
  FT r
FT is the component of the force perpendicular to the radius (also referred to as tangential
force). The reason we need to only include the tangential component of the force is fairly
obvious. Torque is a vector and has direction. You need to be aware that the direction of
a torque is taken care by assigning a positive or negative sign to it. Just like for linear
motion the sign of the velocity meant the direction of the motion, for rotation, the sign of
the torque indicates the direction of rotation. By convention, we choose a counterclockwise rotation to be positive torque, and clockwise rotation to be negative torque.
Notice that the torque depends on the tangential force. Thus the torque is completely
unrelated to the centripetal force. The centripetal force is what keeps the object moving in
a circle. The torque is related to whether the angular speed is increasing or decreasing.
Another thing to notice about torque is that the larger the radius, the larger the torque.
Angular Momentum
A particle of mass m and velocity v has linear momentum p  mv . The particle may also
have angular momentum L with respect to a given point in space. If r is the vector from
the point to the particle, then
r
r r
L  rp
The angular momentum is always a vector perpendicular to the plane defined by the
r
r
r
vectors r and p (or v ). For example, if the particle (or a planet) is in a circular orbit, its
angular momentum with respect to the centre of the circle is perpendicular to the plane of
the orbit and in the direction given by the vector cross product right-hand rule, as shown
below.
Figure 10: The angular momentum L of a particle traveling in a circular orbit.
Since in the case of a circular orbit, r is perpendicular to p (or v ), the magnitude of L is
simply
L  rp  mvr
The significance of angular momentum arises from its derivative with respect to time,
r
dL
d r r
d r r

 r  p  m  r  v ,
dt
dt
dt
r
r
where p has been replaced by mv and the constant m has been factored out. Using the
product rule of differential calculus,
r
r
d r r
dr r r dv
r

v


v

r



dt
dt
dt
r
r
r r
In the first term on the right-hand side, dr dt is simply the velocity v , leaving v  v . Since
the cross product of any vector with itself is always zero, that term drops out, leaving
r
r dv
d r r
 r  v  r 
dt
dt
r
r
Here, dv dt is the acceleration a of the particle. Thus, if both sides of the above equation
r
are multiplied by m , the left-hand side becomes dL dt and the right-hand side may be
r
r
r
r
written r  ma . Since, according to Newton's second law ma ,is equal to F , the net force
acting on the particle, the result is
r
dL r r
 rF
dt
The above equation means that any change in the angular momentum of a particle must
r
be produced by a force that is not acting along the same direction as r .
One particularly important application is the solar system. Each planet is held in its orbit
by its gravitational attraction to the Sun, a force that acts along the vector from the Sun to
the planet. Thus the force of gravity cannot change the angular momentum of any planet
with respect to the Sun. Therefore, each planet has constant angular momentum with
respect to the Sun. This conclusion is correct even though the real orbits of the planets
are not circles but ellipses.
r r
r
The quantity r  F is called the torque  . Torque may be thought of as a kind of twisting
force, the kind needed to tighten a bolt or to set a body into rotation. Using this definition,
the above equation may be rewritten
r
r r r
dL
  rF
dt
This equation means that if there is no torque acting on a particle, its angular momentum
is constant, or conserved.
Suppose, however, that some agent applies a force Fa to the particle resulting in a torque
r
r r
equal to r  Fa . According to Newton's third law, the particle must apply a force Fa to
r r
the agent. Thus there is a torque equal to  r  Fa acting on the agent. The torque on the
r r
particle causes its angular momentum to change at a rate given by dL dt  r  Fa
r
r r
However, the angular momentum La of the agent is changing at the rate dLa dt   r  Fa
Therefore, dLa dt  dL dt  0 , meaning that the total angular momentum of particle plus
agent is constant, or conserved.
This principle may be generalized to include all interactions between bodies of any kind,
acting by way of forces of any kind. Total angular momentum is always conserved. The
law of conservation of angular momentum is one of the most important principles in all of
physics.
Example 2.3: Moment of Inertia of a Rod of Uniform Mass Density, Part I
Consider a thin uniform rod of length and mass . In this problem, we will calculate the
moment of inertia about an axis perpendicular to the rod that passes through the center of
mass of the rod. A sketch of the rod, volume element, and axis is shown in below.
Solution:
Choose Cartesian coordinates, with the origin at the center of mass of the rod, which is
midway between the endpoints since the rod is uniform. Choose the x  axis to lie along
the length of the rod, with the positive x -direction to the right, as in the figure.
Identify an infinitesimal mass element dm  dx , located at a displacement x from the
center of the rod, where the mass per unit length   m L is a constant, as we have
assumed the rod to be uniform. When the rod rotates about an axis perpendicular to the
rod that passes through the center of mass of the rod, the element traces out a circle of
radius r  x We add together the contributions from each infinitesimal element as we
go from x  L 2 to x  L 2 . The integral is then
L2
Icm 
 r 

body
m  L 2

L
3
3
2
x3
dm     x 2  dx  
3
L 2
m   L 2

L
3
L2
L 2
3

1
mL2
12
By using a constant mass per unit length along the rod, we need not consider variations
in the mass density in any direction other than the x - axis. We also assume that the
width is the rod is negligible. (Technically we should treat the rod as a rectangle in the
x  y plane if the axis is along the z axis. The calculation of the moment of inertia under
this assumption would be more complicated.)
Solve the following problems
*Reading assignments, Worked Examples, group works, experiments, hands on experiences…..
Task 2.1. Uniformly accelerated Engine
An engine requires 5s to go from its idling speed of 600 rev/min.to 1200 rev/min (a) What
is its angular acceleration? (b) How many revolutions does it make in this period?
Answer :
(a) 12.6 rad/s
(b) 75.2 rev.
Task 2.2. Acceleration of a body rolling down an inclined plane:
The radius of gyration of a body about a particular axis is the distance from that axis to a
point at which the body’s entire mass may be considered to be concentrated. Thus the
moment of inertia of a body of mass M and radius of gyration K is I  Mk 2 (a) The radius
of gyration of a hollow sphere of radius R and mass M is k  2R 3 . What is its moment
of inertia? (b) Find the radius of gyration of a solid sphere.
2
(a) I  MR 2
3
Answer :
2
(b) k 
R
5
Task 2.3.Question for discussion
Discuss the following questions with your colleagues or on the discussion forum of AVU
1. When milk is churned, what force separated the cream from the milk?
2. Two identical balls move down an inclined plane. Ball A slides down without
friction and ball B rolls down. Do the balls reach the bottom together? If not
which is first? Why?
3. A solid cylinder and a hollow cylinder of the same mass and diameter, both
initially at rest, roll down the same inclined plane without slipping.
a. Which reaches the bottom first?
b. How do their kinetic energies compare at the bottom?
4. An aluminium cylinder of radius R, and a lead cylinder of radius R and lead
cylinder of radius 2R all roll down the same inclined plane. In what order will
they reach the bottom.
5. A square and a rectangle of the same mass are ut from a sheet of metal.
Which has the greater moment of inertia about a perpendicular axis through its
centre
Task 2.4 Thought Experiment
In a science fiction story the Earth rotational velocity was changed by launching a
projectile along a tangent to the equator. What should the difference be between the
velocities of light and the projectile in order to stop the Earth rotating around its axis? The
Earth’s radius is M E  6370 km , and its mass is ME  6  1024 kg . The earth’s moment of
inertia relative to its rotational axis, with its density no uniformity properly incorporated, is
quite accurately given by the formula
M E R 2E
I
3
Compare kinetic energies of the projectile and the Earth’s rotation. The projectile mass
can be taken be m  106 kg .
Formative Evaluation 2
1. A 45-kg, 5.0-m-long uniform ladder rests against a frictionless wall and makes an
angle of 60°with a frictionless floor. Can an 80-kg person stand safely on the ladder,
2.0 m from the top,without causing the ladder to slip if a second person exerts a
horizontal force of 500 N toward the wall at a point 3.5 m from the top of the ladder?
(Note: All distances are measured along the ladder.)
2. The combination of an applied force and a frictional force produces a constant torque
of 36.0 N ×m on a wheel rotating about a fixed axis. The applied force acts for 6.00 s,
during which time the angular speed of the wheel increases from 0 to 10.0 rad/s. The
applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the
moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the
total number of revolutions of the wheel. (38.)
3. A 48.0-kg diver stands at the end of a 3.00-m-long diving board. What torque does the
weight of the diver produce about an axis perpendicular to and in the plane of the
diving board through its midpoint?
4. Two children sit on a seesaw such that a 400-N child is 2.00 m from the support (the
fulcrum). Where should a second child of weight 475 N sit in order to balance the
system if the support is at the center of the plank?
5. The 400-N child of Problem 4 decides that she would like to seesaw alone. To do so,
she moves the board such that its weight is no longer directly over the fulcrum. She
finds that she will be balanced when she is 1.5 m to the left of the fulcrum and the
center of the plank is 0.50 m to the right of the fulcrum. What is the weight of the
plank?
6. A dormitory door 2.50 m high and 1.00 m wide weighs 250 N, and its center of gravity
is at its geometric center. The door is supported by hinges 0.250 m from top and
bottom, each hinge carrying half the weight. Determine the horizontal component of
the forces exerted by each hinge on the door.
7. A racing car has a mass of 1600 kg. The distance between the front and rear axles is
3 m. If the center of gravity of the car is 2 m from the rear axle, what is the normal
force on each tire?
8. An iron trapdoor 1.25 m wide and 2.00 m long weighs 360 N and is hinged along the
short dimension. Its center of gravity is at its geometric center. What force applied at
right angles to the door is required to lift it (a) when it is horizontal and (b) when it has
been opened so that it makes an angle of 30° with the horizontal? (Assume that the
force is applied at the edge of the door opposite the hinges.)
9. A 4.50-kg ball on the end of a chain is whirled in a horizontal circle by an athlete. If the
distance of the ball from the axis of rotation is 2.50 m, find the moment of inertia of the
ball, assuming it can be considered as a point object.
10. What torque must the track star exert on the ball of Problem 9 to give it angular
acceleration of 2.00 rad/s2?
11. (a) Find the moment of inertia of a solid cylinder of mass 1.50 kg and radius 30.0 cm
about an axis through its center. (b) Repeat for a solid sphere of the same mass and
radius about an axis through its center.
12. The cylinder of Problem 11(a) is rotating at an angular velocity of 2.00 rev/s. What
torque is required to stop it in 15.0 s?
13. An automobile tire, considered as a solid disk, has a radius of 35.0 cm and a mass of
6.00 kg. Find its rotational kinetic energy when rotating about an axis through its
center at an angular velocity of 2.00 rev/s.
14. An automobile engine part is in the shape of a thin rod of mass 100 g and length 5.00
cm. When the rod is rotating at an angular velocity of 3.00 rad/s, find its kinetic energy
when (a) rotating about an axis through a point 2.50 cm from each end. (b) Repeat
when it is rotating about an axis through one end.
15. If the system of masses shown in the fig below is set into rotation about the x axis
with an angular velocity of 2.5 rad/s, (a) find the kinetic energy of the system. (b)
Repeat the calculation for the system in rotation at the same speed about the y axis.
Optional Formative Evaluation 2
Teaching the Content in Secondary School 2
ACTIVITY 3: Gravitation
You will require 30 hours to complete this activity. In this activity you are guided with a
series of readings, Multimedia clips, worked examples and self assessment questions
and problems. You are strongly advised to go through the activities and consult all the
compulsory materials and use as many as possible useful links and references.
Specific Teaching and Learning Objectives





Use Newton’s law of universal gravitation to solve problems
Describe gravitational field and potential
Distinguish between inertial and gravitational mass
Calculate escape velocity of satellites.
Use Galilean transformation to solve problems in gravitation.
Summary of the Learning Activity
According to Newton's postulated law of universal gravitation, two bodies of mass m1 and
m2 separated by a distance r . exert equal attractive forces on each other of magnitude
given by F  Gm1m 2 r 2 . , G is a universal constant, applying to all bodies, whatever their
constitution. Gravitation is by far the weakest known force in nature and due to its long
reach and universality, however, gravity shapes the structure and evolution of stars,
galaxies, and the entire universe. The trajectories of bodies in the solar system are
determined by the laws of gravity, while on Earth all bodies have a weight, or downward
force of gravity, proportional to their mass, which the Earth's mass exerts on them.
The observation of Galileo that all bodies in free-fall accelerate equally implies that the
gravitational force causing acceleration bears a constant relation to the inertial mass.
In astronomy and space exploration, artificial satellites are sent in to space. The velocity
that is sufficient for a satellite to escape from a gravitational centre of attraction without
undergoing any further acceleration is referred to as escape velocity. . Escape velocity
decreases with altitude and is equal to the square root about 1.414 times the velocity
necessary to maintain a circular orbit at the same altitude. At the surface of the
Galilean transformation also called Newtonian Transformation, is a set of equations in
classical physics that relate the space and time coordinates of two systems moving at a
constant velocity relative to each other. Adequate to describe only low-speed
phenomena, Galilean transformations formally express the ideas that space and time are
absolute; that length, time, and mass are independent of the relative motion of the
observer.
List of Required Readings
Reading 6: Universal Gravitation:.
Complete reference : Newton’s law of Universal Gravitation
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m101.pdf
Accessed on the 23rd April 2007
Abstract :
In this article, historical account on the discovery of the law, the center of mass and
effects of extended objects are discussed. Determination of G is described in three ways..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important..
Reading 7: Orbital Motion:.
Complete reference : Orbital motion in an inverse-square law force field
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m102.pdf
Accessed on the 23rd April 2007
Abstract :
This document has a good summary of the theories developed to account for the motion
of planets, Copernicus’ proposal of heliocentric solar systerm, Kepler’s law of planetary
motion, Newton’s interpretation of planetary motion and motion of satellites are
discussed..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important.
Reading 8: Gravitational Phenomena:.
Complete reference : Orbital motion in an inverse-square law force field
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m107.pdf
Accessed on the 23rd April 2007
Abstract :
This document has a good summary of the theories developed to account for the motion
of planets, Copernicus’ proposal of heliocentric solar system, Kepler’s law of planetary
motion, Newton’s interpretation of planetary motion and motion of satellites are
discussed..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important.
Reading 3: Gravitational Potential Energy.
Complete reference : Gravitational Potential Energy
From Project PHYSNET PDF Modules
URL: http://35.9.69.219/home/modules/pdf_modules/m107.pdf
Accessed on the 20th April 2007
Abstract :
This article discusses the last part of the activity
Rationale:
This module treats angular acceleration and constant torque equation of a rotating object.
It has a good collection of exercises and exam samples at the end.
List of Relevant MM Resources
1. Reference http://jersey.uoregon.edu/vlab/Piston/index.html
Date Consulted:-Nov 2006
Description:- This Java applet helps you to do a series of virtual experiments,
you will control the action of a piston in a pressure chamber which is filled with an
ideal gas. The gas is defined by four states: Temperature; Volume or density;
Pressure and Molecular Weight
2. Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted:- August 2006
Description:- This Java applet helps you understand the effect of temperature
and volume on the number of collisions of the gas molecules with the walls. In the
applet, you can change the temperature and volume with the sliders on the left
side. You can also adjust the time for which the simulation runs. The applet counts
all collisions and displays the result after the run. By varying temperature and
volume and keeping track of the number of collisions, you can get a good feeling of
what the main result of kinetic theory will be.
3. Reference: video.google.com
Date Consulted: Nov 2006
Complete Reference:- Computer calculation of Phase Diagrams.
http://video.google.com/videoplay?docid=1397988176780135580&q=Thermodyna
mics&hl=en
Rationale: Thermodynamic models of solutions can be used together with data to
calculate phase diagrams. These diagrams reveal, for a given set of all parameters
(such as temperature, pressure, magnetic field), the phases which are
thermodynamically stable and in equilibrium, their volume fractions and their
chemical compositions...
List of Relevant Useful Links
1. Title:- About Temperature
URL:- http://eo.ucar.edu/skymath/tmp2.html
Abstract:-This document was prepared for the middle school math teachers who
are taking part in Project Skymath. It is provided a detailed description of
temperature.
2. Title: A Brief History of Temperature Measurement
URL: http://thermodynamics-information.net/
Abstract: Brief biography and thermometers created by scientists from Réne
Antoine Ferchault de Réamur (1683-1757) to William John Macquorn Rankine
(1820-1872) are presented. There is also a table that compares the values of the
Fahrenheit, Celsius, and Kelvin temperature for some common reference
temperatures.
3. Title: Why do we have so many temperature scales?
URL: http://www.unidata.ucar.edu/staff/blynds/tmp.html
Abstract:
Detailed Description of the Activity (Main Theoretical Elements)
Introduction
3.1:
The Law of Universal Gravitation
Sir Isaac Newton thought that celestral and terrestrial motions might obey the same law
i.e the centripetal acceleration of the moon in its orbit and the downward acceleration  g 
of a falling apple must have the same origin. His arguments and calculations are as
follows:

Assuming the noom’s orbit to be circular and knowing the distance of the moon
from the earth ( R  3.84 108 m ) and its period of rev ( T  27.3d ) Newton calculated
a as follows
v 2 (2 R / T ) 2 4 2 R


 2.72  103 m / s 2
R
R
T2
g  9.8m/s 2
a 
a
g

1
3606

From the knowledge:
radius of earth (R E )
1
 , he observed
radius of moon's orbit (R) 60
2
a1  RE 
2
2

  gRE  a R
g  R 
 The acceleration of a body and hence the force is inversely proportional
to the square of the distance from the center of the earth
This led Newton to postulate gravitational force obays the inverse square law. He went
a step further and argued Fg m1m2 by 2nd and 3nd law
Any two particles of matter anywhere in the universe attract each other with a force which
is directly proportional to the product of their masses and inversely proportional to the
square of the distance between them; the direction of the force being along the line
joining them
Gm1m2
rˆ12
r2
Gm1m2
Gm1m2
F21 
rˆ21 
rˆ12   F12
2
r
r2
G = Gravitational constant
F12  
r 21  r 12
 6.67 1011 Nm 2 /kg 2
Note:
i.
ii.
iii.
Gravitational forces between two particles are an acriton- and- reaction pair
G is universal.
Force of gravitational interaction between two bodies is independent of the
presence of other bodies or the properties of the intervening space. i.e. no gravity
screen
iv.
v.
The gravitational force obeys the law of superposition i.e. the force due to a
collection of particles is the vector sum of the foresees exerted by the particles
individually.
The gravitation force of a sphere of radius R is
F  G
Mm
rˆ
r2
if r  R
0
if r  R
Gravitational Potential Energy
The gravitational potential energy of a mass m at a distance r from another mass M is
defined as the amount of work done in bringing the mass m from infinity to a distance r
dW  Fdr
r
r
GmM
dr  GmM
r2

 S   Fdr  

r

1
dr
r2
1
GmM
 GmM 

r
r
r
Thus PE 
GmM
r
Note (i) We can use the above equation to compute the amount of work done against the
gravitational force of mass M to move the mass m from position r1 to r2
1 1
W  U  GmM   
 r1 r2 
 If r1  r2 W will be positive
r1  r2 W will be negative
Gravitational Potential
The gravitational potential V , at point the gravitational field a mass is defined as the work
done in moving a unit mass from infinity to that point. Thus is m  1
V
GM
r
Example 3.1: Gravitational field due to a non-homogenous spherical mass
A non-homogenous spher of radius R has the following density variation:
p  0 for :r<
0
R
3
R
3R
<r
2
3
4

R
 0 for : < r  R
8
3

for :
What is the gravitational field due to the sphere at r 
Solution:
R
,
4
R 5R
,
and 2R
2
6
g
Gm GpV
 2
r2
r
4
3
G 0   R 4 

3
 g  R 4 
 G 0 R
2
3
 R 4
1
3
3
3 

R 3    R 2    R 2   
4 


R
2
g    G 0  

2
3 
( R / 2)
2



1
2
3
1
 G 0  4 3  R  1 3  1 2    4
2
2

4
 G 0 R  2 1 27  1 8  0.432 G  0 R
3
1
1
3
3
3
3
3 

R 3    3R 4    R 3     5 R 6    3 R 4   


4
 8

 5R 
2
g

  G 0  
2
3 
( R / 2)
 6 



 0.48 G  0 R
1
1
3
3
3
3

Similarly g(2R)=G 0  4 3  R 1 3   3 4   1 3   1   3 4    4


2

 8
 0.1 G  0 R
Example 3.2: Gravitational Potential
Prove that there is no gravitational field due to mass of a spherical shell inside .
Solution:
Let
p= be any point inside a spherical sheet
  mass per unit area of the shell
Consider two cones, with theis apices at P, intercepting small areas S and S’ on the shell
as shown draw a plane x-y through , 1 the diareter through p
S  r 2 .
S cos   r 2
S '  r '2 .
S cos   r '2 w
mass of S = r 2 / cos 
mass of S = r '2  / cos 
 intensity at p due to S=
r 2
G
G
2
cos r
cos 
and
intensity at P due to S' =
r'2
G
G
2
cos r '
cos 
These two intensities at P being equal and opposite, their resultant is zero. Similar is the
care for all other, pairs of comes on opposite sides of xy in to which the shell may be
divided so that the resultant intensity or field at p due to the whole shell zero.
Example 3.3: Gravitational torque
For a body of mass M , and pivoted at o as shown, show that the gravitational torque
acts on it as if the entire mass is concentrated at the center of mass.
Solution
Let us consider the body as being made up of a large number of point masses and one
such mass be mi
   i   ri  mi g  ( mi ri )  g
g is constant
By defintion of center of mass
m r
i
i
 R cm  mi  R cm M
R cm  the position vector of the center of mass
Using the above two expressions it follows.
 =R cm  (mg )
Thus we find that the torque gets as if the entire mass M of the body were conc at its cm.
3.2:
Motion of Planets and Satellites
All planets move in elliptical orbits, the sun being at one focus. We may for simplicity
consider the orbits to be very nearly circular, with the sun at the centre. The centripetal
force required for the circular attractions. Even though both the bodies revolve around
their common centre of mass, if one is of much greater mass than the other, the heavier
body may be considered to be at rest. This may be applied also to the case of the earth
and the artificial satellites.
If M is the mass of the heavy body at the centre of the circle of radius r , and m the mass
of the lighter body,
GM
GM
 m 2 r or
  2r
2
r
r2
GM
4 2
2



r2
T2
From this, we may deduce the period of the planet or the satellite.
4 2 2
T2 
r
GM
Kepler’s Law
1. The path of a planet is an elliptical orbit, with the sun at one of its foci.
2. The radius vector drawn from the sun to a planet sweeps out equal areas in equal
intervals of time.
3. The square of the planet’s period is proportional to the cube of the semi major axis
of its orbit.
Orbital Velocity
If a satellite is to keep moving in a circular orbit round the earth, at a distance h from the
surface of the earth,
mv0 2
GM

( RE  h) 2 ( RE  h)
where RE is the radius of the earth
v0 
GM
( RE  h)
If h  RE , v0  gRE
The amount of work required to move a body from the earth to infinity is given by the
formula
GE Mm
RE
This is about 0.6 107 joules/kg. If we could provide a projectile more than this energy at
the surface of the earth, it would escape from the earth. The critical initial speed v0 called
escape velocity is given by the formula
1 2 GM Em
mv0 
2
RE
v0 
2GM E
 11.2 km/second
RE
Synchronous Satellite
This is an artificial satellite whose period is 24 hours, so that it is always vertically over
the same place on the earth. For this to be possible,
( RE  h)3
T  2 r
 86400seconds.
GM
`h’ may be calculated. It is about 22,700 miles above the earth’s surface.
Example 3.4: Acceleration due to gravity at an altitude
Find the acceleration due to gravity at an altitude of 1000km.
Solution::
The gravitational force of the earth on an object of mass m at a distance r from the
earths center is equal to the object’s weight mg at that distance,
GM E m
GM E
 mg
g 
2
r
r2
At the earth's surface, wher g=g o  9. 8m/s 2 and r  rE  6400km
From these fromulas for g and g o we find that, at the distance r from
the earth's center, the acceleration of gravity is
2
r
g=  E  g o
 r 
using r=rE +h=6400km + 1000km =7400km
Thus g=7.3m/s 2
3.3. Relative Velocity
Principle of Relativity
For observers moving in two different reference frames, no mechanical force can
distinguish which observer is at rest and which observer is moving Einstein extended this
to all physical phenomena.
The laws of physics are the same in all relatively inertialreference frames, In particular,
Einstein extended the Galilean principle ofrelativity to electromagnetism and optics which
describethe theory of light
Galilean Transformation
Consider two frames of reference the O-frame (label events according to t,x,y,z) and the
O'-frame (label events according to t',x',y',z') moving at a constant velocity V, with respect
to each other at let the origins coincide at t= t' = 0. In the Galilean transformations the O
and O' frames are related as follows:
Example 3.5: Objects 1 and 2 move with constant speeds in the same direction taken to
be positive where object 2 is in the lead. With Vrel  V1  V2 , the two objects will collide if
1. Vrel  0
2. Vrel  0
3. Vrel  0
Solution:
V
V
For object 1 to overtake object 2 and thus collide, 1 must be greater than 2 This gives
Vrel  0 . Since both V1 and V2 are positive.
Example 3.6: A passenger on a train traveling 40 mph North. If he walks 5 mph toward
the front of the train, what is your speed relative to the ground? A) <40 mph B) 40 mph
C) >40 mph
Solution: The given situation that both velocities are in the same direction. Therefore the
velocity of the passenger and the velocity of the train add with respect to the ground.
Hence the relative velocity of the passenger relative to the ground is 45 mph
Example 3.4: A 4N weight rests on a smooth horizontal plane it is struck with a 2N blow
that lasts 0.02 sec. Three seconds after the start of the first blos a second blos of -2N is
delivered. This lasts for 0.01 sec. What will be the speed of the body after 4 sec?
Solution The forces in this problem are
Solve the following probles
*Reading assignments, Worked Examples, group works, experiments, hands on experiences…..
Task 3.1. Distance moved by a fragment from an explosion
A projectile fined from a gun at an angle of 45o with the horizontal and with muzzle speed
Answer :  1.055  105 ft
Task 3.2.Mass of a recoiling boat
Four 50kg-girsl simultaneously dive horizontally at 2.5m/se……………….
Answer : 5000kg
Task 3.3.Question for discussion
Discuss the following questions with your colleagues or on the discussion forum of AVU
6. Why does a gun recoil ?
7. Suppose you catch a baseball, and then someone invites you to catch a
bulle…
8. t It is not the fall that hurts you; it is the sudden stop at the bottom. Discuss?
Task 3.4 Experiment with Billiard balls
Have you ever played a billiard ball? If no tryntation purpose. A ball (b1) …..
Try this with balls of different massd determine the relation ship between the masses of..
Formative Evaluation 3
16. Blocks A and B have a mass oey collide and become coupled together.
a. 2m/s to right
b. 2m/s to left
c. 3.33m.s to left
d. 4m/sec to left
17. A 4 N force acts on a 3 kg object moving at 8 m/s for 10 sec. What is the object's
change in momentum? What impulse acts on the object? What is the object's final
speed?
Optional Formative Evaluation 3
Teaching the Content in Secondary School 3
This activity introduces the concepts of thermal physics using microscopic point of view.
It expands the concept of conservation of energy, and prepares students for the study of
kinetic theory and gas laws presented in activity 3 and 4 of this module. The first law can
be treated as an extension of the conservation of energy with heat defined as the transfer
of energy by means of a diffeence in temperature.
The difference in specific heat capacity can be observed by heating a block of iron and an
equal mass of water simultaneously over hot plates and observe the difference in
XI
COMPILED LIST OF ALL KEY CONCEPTS (GLOSSARY)
1. Momentum: The momentum of a particle, a vector quantity, is the product of the
particle’s mass and velocity.
P  mv
In terms of momentum, Newton’s second law for a particle may be expressed as
dp
F  dt
2. Impulse: When a constant force acts for a certain time interval, the impulse of the
force is the product of force and the time interval. The change of momentum of a
body or system equals the impulse of the net force acting on it.
j  F (t 2  t 1)  F t
if the net force varies with time, impulse is
t2
J    Fdt
t1
The change in momentum of a body in any time interval equals the impulse of the
net force that acts on the body during that interval:
o
J  P2  P1
The momentum of a body equals the impulse that accelerated it from rest to its
present speed.
3. Internal Force: An internal force is a force exerted by one part of a system on
another. An external force is a force exerted on a part of a system by something
outside the system. An isolated system is one with no external forces.
4. Total Momentum:- The total momentum of a system of particles A , B , C ,
vector sum of the momenta of the individual particles:
o P  PA  PB  PC 
 mAv A  mBv B  mCvC 
is the
5. Conservation of Momentum: In any system of two or more particles in which the net
force on each particle is due only to interactions with the other particles of the
system, the total momentum (vector sum of the momenta of the particles) is constant
or conserved.
6. Elastic Collision: A collision in which total kinetic energy is conserved is called an
elastic collision. When kinetic energy is not conserved, the collision is inelastic.
7. Center of Mass:- The center of mass of a system is the average position of the mass
of the system. Its motion under given forces is the same as though all the mass
were concentrated at the center of mass.
8. Rigid Body: A rigid body is a body with a definite and unchanging shape and size.
9. Angular Displacement (  ): -is the measure of change in angular position of a
rotating object.
10. Angular Displacement (  ): is the time rate of change of angular displacement:
d

dt
11. Angular Acceleration   : is the time rate of change of angular velocity

d
dt
12. Moment of Inertia  I  : Is the rotational analog of mass. the greater the moment of
inertia of a body, the greater its resistance to a change in its angular velocity.
Moment of inertia of a body about a particular axis of rotation depends not only upon
the body’s mass but also upon how the mass is distributed about the axis.
13. Torque   : The torque exerted by a force on a body is a measure of its
effectiveness in turning the body about a certain pivot point. The moment arm
of a force F about a pivot O is the perpendicular distance L between the line of
action of the force and O. The torque   excreted by the force about O has
the magnitude:    Force    Moment arm
14. Kinetic Energy of Rotation.: The kinetic energy of a body of moment of inertia I and
angular velocity  (in rad/s) is
KE 
1 2
I
2
15. Rotational Work (W):- The work done by a constant torque   that acts on a bodyh
while it experiences the angular displacement  in rad is
W  
16. Angular Momentum : is the equivalent of linear momentum in rotational motion. The
angular momentum  L  of a rotating body has the magnitude
L   moment of inertia    angular velocity 
 I
17. Law of Universal Gravitation: States every body in the universe attracts every other
body with a force that is directly proportional to each of their masses and inversely
proportional to the square of the distance between them
mm
Graviataional Force  F  G 1 2 2
r
G  6.67 1011 Nm2 /kg 2
18. Gravitational Field: The Newtonian theory of gravity is based on an assumed force
acting between all pairs of bodies--i.e., an action at a distance. When a mass
moves, the force acting on other masses has been considered to adjust
instantaneously to the new location of the displaced mass. The field theory of
electrical and magnetic phenomena, has met empirical success so that most modern
gravitational theories are constructed as field theories. In a field theory the
gravitational force between bodies is formed by a two-step process: (1) One body
produces a gravitational field that permeates all surrounding space but has weaker
strength farther from its source. A second body in that space is then acted upon by
this field and experiences a force. (2) The Newtonian force of reaction is then viewed
as the response of the first body to the gravitational field produced by the second
body, there being at all points in space a superposition of gravitational fields due to
all the bodies in it..
19. Gravitational Potential Energy:- Potential energy arises in systems with parts that
exert forces on each other of a magnitude dependent on the configuration, or
relative position, of the parts. In the case of an Earth-ball system, the force of gravity
between the two depends only on the distance separating them. The work done in
separating them farther, or in raising the ball, transfers additional energy to the
system, where it is stored as gravitational potential energy. Gravitational potential
energy near the Earth's surface may be computed by multiplying the weight of an
object by its distance above the reference point.
20. Gravitational field Strength:- is the gravitational force acting on a unit mass at a
given point in space. It is just acceleration due to gravity and is given by:
GM Earth
g
r2
21. Galilean Transformation: also called Newtonian Transformations, set of equations
in classical physics that relate the space and time coordinates of two systems
moving at a constant velocity relative to each other. Adequate to describe only lowspeed phenomena, Galilean transformations formally express the ideas that space
and time are absolute; that length, time, and mass are independent of the relative
motion of the observer; and that the speed of light depends upon the relative motion
of the observer
XII
COMPILED LIST OF COMPULSORY READINGS
Reading 1: Momentum in One Dimension.
Complete reference : Conservation of Momentum
From html version of Simple Nature, by Benjamin Crowell.
URL : http://www.lightandmatter.com/html_books/0sn/ch03/ch03.html#Section3.1
Accessed on the 20th April 2007
Abstract :
This is part of a book by Benjamin Crowell. It is freely available at
www.lightandmatter.com the part given here the relevant section for this activity.
Rationale:
This section has a well illustrated content on linear momentum. The motion of center of
mass is treated at the end. It provides another way of looking at the theories of collision
and momentum conservation. The examples drawn from nature, like comet, are
interesting and educational reading materials.
Reading 2: Momentum Conservation and Transfer.
Complete reference : Momentum Conservation and Transfer
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m15.pdf
Accessed on the 20th April 2007
Abstract :
In this article, momentum is defined for a single and a system of particles. Using Newton’s
laws and the definition of momentum it is shown that the momentum of an isolated
system of particles remain unchanged with time (i.e. conserved)
Rationale:
This article gives another way of looking at the theories of collision and momentum
conservation. Further the sample tests and exercises given at the end provide good
opportunity to use the theories and principles exercised from different perspectives.
Reading 3: Angular Acceleration.
Complete reference : Angular Acceleration in circular motion
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m33.pdf
Accessed on the 24th April 2007
Abstract :
In this article, the two agents of change, angular and linear acceleration, producing
angular acceleration and constant torque case rotational kinematics are treated.
Rationale:
This article covers the contents of this activity. It gives another way of looking at the
theories of collision and momentum conservation. Further the sample tests and exercises
given at the end provide good opportunity to use the theories and principles exercised
from different perspectives.
Reading 4: Momentum in One Dimension.
Complete reference : Conservation of Momentum
From html version of Simple Nature, by Benjamin Crowell.
URL : http://www.lightandmatter.com/html_books/0sn/ch04/ch04.html
Accessed on the 20th April 2007
Abstract :
This is part of a book by Benjamin Crowell. It is freely available at
www.lightandmatter.com the part given here the relevant section for this activity.
Rationale:
This section has a well illustrated content on angular momentum. Theories of angular
momentum in two dimensions and in three dimensions is well developed. Further there is
a good deal of theory on rigid body rotation.
Reading 5: Torque and Angular Momentum.
Complete
reference
:
Torque
From
Project
URL
:
Accessed on the 24th April 2007
and
Angular
Momentum
in
circular
motion
PHYSNET
PDF
Modules
http://35.9.69.219/home/modules/pdf_modules/m34.pdf
Abstract :
In this article, the two agents of change, angular and linear acceleration, producing
angular acceleration and constant torque case rotational kinematics are treated.
Rationale:
This article covers topics on torque and angular momentum, system of particles,
conservation of angular momentum, nonplanar rigid bodies. The problem supplement and
the model exam at the end makes this site popular.
Reading 6: Universal Gravitation:.
Complete reference : Newton’s law of Universal Gravitation
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m101.pdf
Accessed on the 23rd April 2007
Abstract :
In this article, historical account on the discovery of the law, the center of mass and
effects of extended objects are discussed. Determination of G is described in three ways..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important..
Reading 7: Orbital Motion:.
Complete reference : Orbital motion in an inverse-square law force field
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m102.pdf
Accessed on the 23rd April 2007
Abstract :
This document has a good summary of the theories developed to account for the motion
of planets, Copernicus’ proposal of heliocentric solar systerm, Kepler’s law of planetary
motion, Newton’s interpretation of planetary motion and motion of satellites are
discussed..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important.
Reading 8: Gravitational Phenomena:.
Complete reference : Orbital motion in an inverse-square law force field
From Project PHYSNET PDF Modules
URL : http://35.9.69.219/home/modules/pdf_modules/m107.pdf
Accessed on the 23rd April 2007
Abstract :
This document has a good summary of the theories developed to account for the motion
of planets, Copernicus’ proposal of heliocentric solar system, Kepler’s law of planetary
motion, Newton’s interpretation of planetary motion and motion of satellites are
discussed..
Rationale:
This article covers topics in line with this module and the problem supplement and the
model exam at the end makes this reading very important.
XIII COMPILED LIST OF (OPTIONAL) MM RESOURCES
Resource #1
Title: Motion of Center of Mass
URL: http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html
Screen Capture:
Description: Applet shows the motion of the centre of mass of a dumbbel shaped object.
The red and blue dots represent two masses and they are connected by a
massless rod. The dumbbel's projection velocity can be varied by using the
velocity and angle sliders. The mass ratio slider allows shifting of centre of mass.
Here m1 is the mass of the blue object and m2 is the mass of red object. Check
boxes for path1 and path2 can be used to display or turn off the paths of the two
masses.
Rationale: This applet depicts the motion of centre of mass of two balls (shown in red and blue
colour). The applets speed and angle of projection can be varied..
Resource #2;Hyper Physics
url:-: http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html
Date Consulted:-April 2007
Description:- This Java applet helps you to do a series of virtual experiments, . you can
determine the escape and orbital velocities by varying different parameters of the
projectile.
Resource #3 Rotating Stool
url:- http://hyperphysics.phy-astr.gsu.edu/hbase/rstoo.html#sm
Complete Reference:- Good animation graphics and applet to visualize the dependence
of moment of inertia on distribution of matter on an object..
Rationale: Strengthens what is already discussed in Activity 2.
XIV
COMPILED LIST OF USEFUL LINKS
Useful Link #1 Classical Mechanics
Title: All Thermodynamics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics courses.
The learner can consult chapters 7, 8 and 9 of the book. The pdf version is also available.
Useful Link #2 Universal Gravitation from IWikipedia
Title: Universal Gravitation
URL: http://en.wikipedia.org/wiki/Law_of_universal_gravitation
Screen Capture:
Description: This is a good collectionn of theory and historical account of the newtons low
of universal gravitation.
Rationale: The site provides a detailed description and solved problems on the topic. .
Useful Link #3 From The physics Class room
Title: Universal Gravitation and Planetary Motion
URL: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/circles/u6l3c.html
Screen Capture:
Description: Lecture notes and discussion forum from from the physics class room.
Rationale: Reach in discussion topics and interactive problems.
Useful Link #4 Tutorial on torque from university of Guelph
Title: Torque
URL: http://www.physics.uoguelph.ca/tutorials/torque/index.html
Screen Capture:
Description:The site gives detailed description of torque
Rationale: Here you will find a good collection of tutorial problems on torque..
Useful Link #5 Wikipedia
Title: Gravitational Field
URL: http://en.wikipedia.org/wiki/Gravitational_field
Screen Capture:
Description: Gravitaional field, its meaning in classical mechanics and its meaning in general
relativity are described in this section.
Rationale: Useful for the one who needs to compare many references.
Useful Link #6 Geostationary Orbit
Title: Geostationary orbit
URL: http://en.wikipedia.org/wiki/Geostationary
Screen Capture:
Description: This link Explains geostationary orbit. The animated graphics helps visualization.
Rationale: This supplements the theory given in Activity three..
XV SYNTHESIS OF THE MODULE
Mechanics II:
In this module (Mechanics II) dynamics of a system of particles, rotational motion and
Gravitation are dealt in detail. The module began with the study of impulse of a force and
its relation with momentum. The impulse force relation is generalized for a system of
particle.
In the second activity is the kinematic and dynamic descriptions of rotational motion were
done using new quantities. . It was shown that the equations of motion that describe
linear motion possess a rotational counterpart.
The third activity is on Gravitation Up to now we have described various forces from an
entirely empirical point of view. To gain a more unified understanding of such forces and
to achieve greater predictive power, we shall now examine two of the four fundamental
forces which are ultimately responsible for all other forces. Thus in the third activity we’ll
discuss the gravitational force which accounts for the interaction between all astronomical
bodies, the motion of the planets and the moon, the trajectories of space vehicles, the
occurrence of the tides, and the weights of objects.
EXPECTED SOLUTIONS TO SOME PROBLEMS SET
XVI. Summative Evaluation
Module Writing Tip. A summative evaluation can take different forms: a test (multiple
choice, short answers, etc.), a project, a written production, a problem solving task, etc.
The summative evaluation is usually what is used to provide students with a final mark for
the module. This section is therefore designed to provide information to determine the
amount of learning by a student at the end of the module. Module Developers should
keep in mind that the summative evaluation must be conducted in a distance education
context. They should also carefully consider what should be evaluated, and how it should
be evaluated. Instructions provided to learners in a final evaluation must be clear, concise
and well written. An answer key must be provided by Module Developers. For multiple
choice exams, a significant database of questions (3 to 5 for each topic) must be provided
so that exams can vary significantly from student to student. Also, Module Developers
must indicate how learners will submit their answers to the instructor. Will the summary
evaluation be emailed to the instructor ? Will it be submitted online ? Will there be an
online test ? Will there be an on-site evaluation ? Finally, summative evaluations should
include opportunities for students to respond to the module to provide feedback to the
instructor.
Short answer questions
1. A moving object strikes a stationary one. After the collision, must they move in the
same direction?.
2. A 5-kg rifle and a 7-kg rifle fire identical bullets with the same muzzle velocities.
Compare the recoil memento and recoil velocities of the two rifles.
3. An empty dump truck is coasting with its engine off along a level road when rain
starts to fall
a. Neglecting friction, what (if anything) happens to the velocity of the truck ?
b. The rain stops and the collected water leaks out., What (if anything)
happens to the velocity of the truck now?
4. Can a body move in a curved path without being accelerated?
5. In what ways, if any, do acceleration due to gravity (g) and the universal
gravitational constant (G) change with increasing height above the earth’s surface?
6. Two identical balls move down an inclined plane. Ball A slides down without
friction and ball B rolls down. Do the balls reach the bottom together? If not, which
is first? Why?
7. An aluminium cylinder of radius R, a lead cylinder of radius 2R all roll down the
same inclined plane. In what order will they reach the bottom?
8. A square and a rectangle of the same mass are cut from a sheet of metal. Which
has the greater moment of inertia about a perpendicular axis through its center?
9. A solid cylinder and a hollow cylinder of the same mass and diameter, both initially
at rest, roll down the same inclined plane without slipping.
a. Which reaches the bottom first?
b. How do their kinetic energies at the bottom compare?
10. Strings are wound around a shaft and a sheave of equal mass and a load is
attached to the end of each string (the loads have equalled mass). Which of the
two loads will descend with a greater acceleration and which of the rotating
objects, the shaft or the sheave, has a greater angular acceleration?
Multiple Choice questions
1. The total angular momentum of a system of particles
a. Changes when a net external force acts upon the system
b. Remains constant under all circumstances
c. Changes when a net external torque acts upon the system
d. May or may not change under the influence of a net external torque,
depending upon the direction of the torque
2. Which of the following is not a unit of impulse?
a. N.S
b. lb.s
c. lb.h
d. N.m
3. If the momentum of a body increases by 20%, the percentage increase in its K.E.
is equal to:
a. 44
b. 88
c. 66
d. 20
4. The face of a golf club exerts an average force of 4000 N while it is in contact with
the golf ball. If the impulse is 80 N.S., the time of contact is
a. 2s
b. 0.02 s
c. 0.2 s
d. 0.002 s
5. An irom sphere whose mass is 50 kg has the same diameter as an aluminium
sphere whose mass is 10.5 kg. The spheres are simultaneously dropped from a
cliff. When they are 10 m from the ground, they have identical
a. Acceleration
b. Momenta
c. Potential energies
d. Kinetic energies
6. If a car is to gain momentum it must
a. Lose inertia
b. Accelerate
c. Move rapidly
d. Lose weight
7. If the collision in problem 16 is completely elastic, the speed of the 20 kg cart after
collision will be approximately
a. 3m/s
b. 6m/s
c. 4m/s
d. 2m/s
8. A bomb dropped from an aeroplane explodes in air. Its total
a. Momentum decreases
b. Momentum increases
c. Kinetic energy increases
d. Kinetic energy decreases
9. When an aeroplane loops the loop, the pilot does not fall down because
a. Weight of the pilot provides the necessary centripetal force
b. Weight of the pilot provides the necessary force against gravity
c. Weight of the pilot provides the necessary centrifugal force
d. All the above.
10. A thin circular ring of mass M and radius r is rotating about its axis with a constant
angular velocity  . Two objects each of mass m , are attached gently to the
opposite ends of a diameter of the ring. The wheel now rotates with an angular
velocity.
a.  M ( M  m)
b.  M ( M  2m)( M  2m)
c.  M ( M  2m)
d.  ( M  2m) M
11. The co-efficient of restitution is
a. A number which varies from -1 to 1
b. A negative number which varies from 0 to -1
c. A positive number which varies from 0 to 1
d. A positive number
12. If the polar ice caps melts, the length of the day will
a. Increase
b. Decrease
c. Remain the same
d. Insufficient information to predict
13. In order to cause a moving body to pursue a circular path, it is necessary to apply
a. Gravitational force
b. Inertial farce
c. Centrifugal force
d. Centripetal force
14. The centripetal acceleration of a particle in circular motion is
a. Equal to its tangential acceleration
b. Less than its tangential acceleration
c. More than its tangential acceleration
d. May be more or less than its tangential acceleration
15. If the frequency is 2 rev/s, the angular speed is
a. 4 rad / s
b.  8 rad / s
c. 2 rad/s
d. 4 rad/s
16. In order to cause something to move in a circular path, we must supply
a. Inertial force
b. Centripetal force
c. Centrifugal force
d. Gravitational force
17. The acceleration of a body undergoing uniform circular motion is constant in
a. Magnitude only
b. Both magnitude and direction
c. Direction only
d. Neither magnitude nor direction
18. The centripetal acceleration of a 2 kg swinging in a 0.5 m radius with a linear
speed of 4 m/s is
a. !0m / s 2
b. 40m / s 2
c. 4m / s 2
d. 20m / s 2
19. In an equilibrium problem, the axis about which torques are computed
a. Must pass through the centre of gravity of the body
b. Must pass through one end of the body
c. Must intersect the line of action of at least one force acting on the body
d. May be located anywhere
20. Which of the following objects has the largest moment of inertia assuming they all
have the same mass and the radius?
a. A solid disk
b. A solid sphere
c. A circular hoop
d. A solid cylinder
21. The linear speed of an object swinging in a circular path of radius 2m with a
frequency of 5 rev/s is
a. 4 m/s
b. 20 m/s
c. 3 m/s
d. 10 m/s
22. A point on the edge of a rotating disc of radius 8 m moves through an angle of 2
rad. The length of the are described by the point is
a. 0.25 m
b. 4 m
c. 16 m
d. 4 rad
23. A quantity not directly involved in the rotational motion body is
a. Mass
b. Torque
c. Moment of inertia
d. Angular velocity
24. A full circle contains
a.  /2 radians
b.  /4 radians
c.  radians
d. 2  radians
25. Wheel with angular momentum of 10 kg m 2 / s has a moment of inertia equal to 6.5
kg m 2 . Its angular speed is
a. 5 rad/s
b. 20 rad/s
c. 40 rad/s
d. 0.02 rad/s
26. An object swings the end of a string in uniform circular motion. Which of the
following changes would not case an increased centripetal force?
a. A longer string a
b. A greater linear speed
c. A shorter string
d. A larger mass
27. In a rigid body undergoing uniform circular motion, a particle that is a distance R
from the axis of rotation
a. Has a angular velocity inversely proportional to R
b. Has an angular velocity proportional to R
c. Has a linear speed proportional to R
d. Has a linear speed inversely proportional R
28. The unit for the moment of inertia in the M.K.S. system is
a. kg m3
b. kg m 2
c. kg /m 2
d. kg m
29. An object is travelling in a circle with a constant speed. Its acceleration is constant
in
a. Magnitude only
b. Both magnitude and direction
c. Direction only
d. Neither magnitude nor direction
30. Which of the following formulae for the moment of inertia (M.I.) of some simple
cases is not correct?
a. The M.I. of a uniform rod of mnss M and length
about an axis through its
2
centre and perpendicular to its length is I  M /12
b. The M.I. of a uniform circular disc of Mass M and radius r about
perpendicular to the plane of the disc is I  1/ 2Mr 2
c. The M.I. in case of (b) about the axis through one end and perpendicular to
its length is I  M 2 / 4
d. The M.I. of a sphere of mass M and radius r about any diameter is
I  2 / 5Mr 2
31. For a rigid body rotating about an axis, if I ane  be its moment of inertia and the
angular speed respectively, its angular momentum L about the given axis is given
by
a. L  I
b. L   / I
c. L  I / 
d. L  I
32. If the mass of a rotating body moves towards the axis of rotation, in moment of
intertia
a. Remains the same
b. Decreases
c. Increases
d. May increases of decreases
33. When milk is churned, the cream separates from it due to
a. Gravitational forces
b. Centrifugal forces
c. Cohesive forces
d. Frictional forces
34. When a ball on a string moves in a vertical circle, the tension on the string is
greatest when
a. String is made longer
b. Speed of the all is increased
c. The ball is at the highest point
d. Speed of the ball is decreased
35. One revolution is equivalent to
a. 57.3 radians
b. 6.28 radians
c. 57.3 degrees
d. 6.28 degrees
36. When a car is travelling at constant speed around a circular track, a quantity that is
constant but not zero is its
a. Acceleration
b. Angular velocity
c. Velocity
d. Angular acceleration
37. When the angular position of the swing is 450 , its linear velocity is
a. Negative and increasing
b. Positive and decreasing
c. Positive and increasing
d. Negative and decreasing
38. A solid lead cylinder of radius R, a hollow lead cylinder of radius R/2, and a solid
lead sphere of radius plane at the same time. The one that reaches the bottom first
is the
a. Solid aluminium cylinder
b. Hollow lead cylinder
c. Solid lead cylinder
d. Solid lead sphere
39. Due to redistribution of a mass of rotating body if its moment of inertia increases
then its angular velocity
a. Decreases
b. Remains unchanged
c. Increases
d. None of the above
40. If I and E are the moment of inertia and rotational kinetic energy of body
respectively then its angular momentum L is calculated form
2E 2
a.
2I
b. 2E 2 I
c.
2EI
d. 2EI
41. Physical quantity not directly involved in rotatory motion is
a. Moment of inertia
b. Mass
c. Angular velocity
d. Torque
d 2x
42. Vibration of a body is given by the differential equation
2x  0
2
dt
The amplitude and time period are
a. 6cm and 10 
b. 8cm and 10 
c. 10cm and 0.2 sec
d. 14cm and 2 sec.
43. The acceleration of a body rolling down an inclined plane is given by
g sin 
a.
k2
1 2
R
g sin 
b.
R2
1 2
K
g sin 
c.
k2
R2  2
R
g sin 
d.
k2
K2  2
R
44. If the earth stopped rotating, the weight of objects at the equator would
a. Be greater
b. Be the same as before
c. Be less
d. Vary with latitude
45. The man in exercise 3 could have reached the same destination if he had headed
in. What is single direction from the start?
a. 450 east of north
b. 220 east of north
c. 500 east of north
d. 630 east of north
46. A satellite travels in a circular orbit at a speed of 20, 000 km/hr to stay at a constant
altitude. To escape from the earth, the speed would have to be increased to
a. 28, 000 km/hr
b. 21, 000 km/hr
c. 40, 000 km/hr
d. 64, 000 km/hr
47. The gravitational force with which the earth attracts the moon
a. Is less than the force with which the moon attracts the earth
b. Is the same as the force with which the moon attracts the earth
c. Is more than the force with which the moon attracts the earth
d. Varies with the phase of the moon
48. If the earth stopped rotating, the weight of objects at either pole would
a. Be less
b. Be greater
c. Be the same as before
d. Vary with latitude
49. The gravitational constant is 6.67 1011 N .m2 / kg 2 . What is the gravitational force
between two 4 kg balls separated by 0.2m ?
a. 1.33  108 N
b. 2.67 108 N
c. 5.34 107 N
d. 6.67 108 N
50. The value of the universal gravitation constant G in the M.K.S. system is
a. 6.67 1011 N .m2 kg 2
b. 2.81 2011 N .m2 kg 2
c. 5.68 1011 N .m2 kg 2
d. 3.00 1011 N .m2 kg 2
51. The gravitational force between bodies does not depend upon
a. The product of their n asses
b. Their separation
c. The sum of their masses
d. The constant of gravitation
52. When a space ship is twice of earth radial distant from the center of the earth, its
gravitational acceleration is
a. 9.8m / sec 2
b. 19.6m / sec 2
c. 4.0m / sec 2
d. 2.45m / sec 2
53. A satellite is orbiting close to the earth. In order to make it move to infinity, its
orbital speed is to be increased by
a. 20%
b. 10%
c. 41.4%
d. 100%
54. A 100kg astronaut releases 1g of gas from a special pistol at a speed of 50m/s. As
a result, he moves in the opposite direction at
a. 50 cm / s
b. 5 cm / s
c. 0.5 cm / s
d. 0.05 cm / s
55. Kepler modified the Copernican system by showing that the planetary orbits are
a. Ellipses
b. Circles
c. Combinations of circles forming looped orbits
d. The same distance apart from one another.
56. If there were no atmosphere, the duration of the day on the earth will:
a. Decrease
b. Remain the same
c. Increase
d. Depend upon the weather.
57. If the earth had another satellite at twice the distance of the moon , its period would
be
a. 28days 22/ 3
b. 28days  22/ 3
c. (28days)3/ 2
d. (240, 000)3/ 2
58. An imaginary planet has twice the mass and twice the radius of the earth. The
acceleration of gravity at its surface is
a. 4.9m / sec 2
b. 19.5m / sec2
c. 9.8m / sec 2
d. 39.2m / sec 2
59. If the earth were three times farther from the sun than it is now, the gravitational
force exerted it is now, the gravitational force exerted on it by the sun would be
a. Nine times as large as it is now
b. Three times as large as it is now
c. One third as large as it is now
d. One ninth as large as it is now
60. An artificial satellite is moving in a circular orbit about the earth. If R is the radius of
the earth and h the height of the satellite above the surface of the earth, then which
of the following formulae is used for the orbital velocity of the satellite?
1/ 2
 g 
a. v  R 

 ( R  h) 
 g 
b. v  R 

 ( R  h) 
2
 g 
c. v  R 

 ( R  h) 
d. 29.6 m / s
61. A weight is suspended from the middle of a rope whose ends are at the same
level. In order for the rope to be perfectly horizontal, the forces applied to the ends
of the rope
a. Must be greater than the weight
b. Must be equal to the weight
c. Might be so great as to break the rope
d. Must be infinite
62. A hliow metal sphere is filled with water and is hung by a long thread. It flows
through a small hole in the bottom, how will the period of oscillation be affected?
a. The period will go on decreasing till the sphere is empty
b. The period will go on increasing till the sphere is empty
c. The period will remain unchanged throughout
d. The period will first increases, then it will decrease will it is empty and the
period will be finally same as when the sphere was full of water
63. What must the volume of a bottom be if it is to support total mass of 10000 kg at a
point where the density of air is 1.2kg / m3 ? (The total mass includes that of the
bottom and the helium with which it is filled, as well as the payload)
a. 833 m3
b. 1200 m3
c. 85 m3
d. 29.6 m / s
64. The minimum velocity of projection to go out from the earth’s gravitational field is
known as
a. Projectile velocity
b. Escape velocity
c. Angular velocity
d. Terminal velocity
65. The square of a satellite time of revolution round the earth is
a. Directly proportional to the cube of radius of orbit
b. Inversely proportional to the cube of radius or orbit
c. Directly proportional to radius of orbit
d. Inversely proportional to radius of orbit
66. The earth retains its atmosphere because
a. The earth is sphere
b. The mean speed of molecules
c. The earth has population
d. The escape velocity is more than the mean speed of molecules
67. To find time the astronaut in the earth satellite should use:
a. A spendulum clock
b. A watch having spring to keep it going
c. Either of these two
d. None of the above
68. To hit the target, one has to point his rifle
a. Higher than the target
b. Lower than the target
c. In the same direction
d. Vertically upwards
69. The ratio of orbital velocity and escape velocity is
a. 1: 2
b. 2 :1
c.
2 :1
d. 4 :1
70. Acceleration due to gravity is not affected by which one of the following?
a. Latitude
b. Altitude
c. Longitude
d. Depth
71. A rocket can go vertically upwards earths atmosphere because
a. Of gravitational pull of the sum
b. It is highter than air
c. It has a fan which displaces more air per unit time than the weight of the
rocket
d. Of the force exerted on the rocket by gases ejected by it
72. The escape velocity for a body projected vertically upwards from the surface of the
earth is 11.2 km/s. If the body is projected in a direction making an angle of 450
with the vertical. The escape velocity will be:
a. 11.2  2km / s
1
b. 11.2 
km / s
2
c. 11.2  2km / s
d. 11.2 km / s
XVII. References:
This is a compiled list of the references, like standard reference books for the discipline, used in the development of the
module. (Not for the learner do not have to be copyright free) Atleast 10 in APA style
1. Finn, C. B.P (1993). Thermal Physics , Chapman & Hall, London.
2. Raymond A. Serway (1992). PHYSICS for Scientists & Engineers. Updated Version.
3. Kleppner & Kolenkow An introduction to mechanics.
4. Douglas D. C. Giancoli Physics for scientists and engineers. Vol. 2. Prentice Hall.
5. Sears, Zemansky and Young, College Physics, 5th ed.
6. Sena L.A. (1988) Collection of Questions and Problems in physics, Mir Publishers
Moscow.
7. Nelkon & Parker (1995), Advanced Level Physics, 7th ed, CBS Publishers &
Ditributer, 11, Daryaganji New Delhi (110002) India. ISBN 81-239-0400-2.
8. Godman A, and Payne E.M.F, (1981) Longman Dictionary of Scientific Usage.
Second impression, ISBN 0 582 52587 X, Commonwealth Printing press Ltd, Hong
Kong.
9. Beiser A., (2004) Applied Physics, 4th ed., Tata McGraw_Hill edition, New Delhi,
India
10. Halliday D., Resnick R., and Walker J. (1997), Fundamentals of Physics, 5th ed., John
Wiley and Sons
XVIII. Main Author of the Module
About the author of this module:
Name:Tilahun Tesfaye
Title:
Dr.
Address:
Department of physics, Addis Ababa University,
Ethiopia, East Africa.
P.O.Box 80359 (personal), 1176 (Institutional)
E-mail: dttilahun@yahoo.com; ttesfaye@phys.aau.edu.et.
Tel: +251-11-1418364
Breif Biography: The author is currently the chairperson of the department of physics at
Addis Ababa University. He has authored school text books that are in use
all over Ethiopian schools. His teaching experience spans from junior
secondary school physics to postgraduate courses at the university level.
He also worked as a curriculum development expert and Educational
materials development panel head at Addis Ababa Education Bureau.
You are always welcome to communicate with the author regarding any question, opinion,
suggestions, etc this module.
XX. File Structure
Name of the module (WORD) file :
 Mechanics II.doc
Name of all other files (WORD, PDF, PPT, etc.) for the module.

Compulsory readings MechanicsII.pdf
Abstract : The eight compulsory readings proposed for this module are compiled in one pdf file. .

Read_me.txt
Abstract : In this file you will find information about the other files included in the
Readings Directory (Folder).