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Mechanics II

Prepared by Tilahun Tesfaye

African Virtual universityUniversité Virtuelle AfricaineUniversidade Virtual Africana

Physics Module 2

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Notice

This document is published under the conditions of the Creative Commons http://en.wikipedia.org/wiki/Creative_Commons Attribution http://creativecommons.org/licenses/by/2.5/ License (abbreviated “cc-by”), Version 2.5.


I. MechanicsII______________________________________________ 3

II. PrerequisiteCourseorKnowledge_____________________________ 3

III. Time____________________________________________________ 3

IV. Materials_________________________________________________ 4

V. ModuleRationale __________________________________________ 4

VI. Content__________________________________________________ 4

6.1 Overview____________________________________________ 4 6.2 Outline_____________________________________________ 5 6.3 Graphicorganizer_____________________________________ 6

VII. GeneralObjective(s)________________________________________ 7

VIII. SpecificLearningObjectives__________________________________ 7

IX Pre-assessment ___________________________________________ 9

X. TeachingandLearningActivities_______________________________ 9

XI. KeyConcepts(Glossary)____________________________________ 81

XII. CompulsoryReadings______________________________________ 85

XIII. CompulsoryResources_____________________________________ 89

XIV. UsefulLinks _____________________________________________ 91

XV. SynthesisOfTheModule ___________________________________ 96

XVI. SummativeEvaluation______________________________________ 97

XVII.References _____________________________________________ 113

XVIII.MainAuthoroftheModule________________________________ 114

XIX.FileStructure ___________________________________________ 115

Table of ConTenTs


I. Mechanics IIBy 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 KnowledgeIn order to study this module you need to complete Mechanics I of the AVU Teachers’ Training Module.

III. TimeThis module can be completed in 120hrs.


IV. MaterialsThe 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. StringandBall: - Rotational motion experiment3. MeterStick: -Rotational Motion Experiment.

V. Module RationalePhysics 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. Content

6.1 Overwiew

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

Dynamics of a system of Particle (40 hours)

• 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.

Rotational Motion (35 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 (25 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.

Relativity of Motion (20 hours)

• Relative velocity. • Uniform relative translational motion.• The Galilean transformation.


6.3 Graphic organizer

MechanicsII

A. Dynam ic s o fs ys tem s o f pa rt ic les:

B . R ot at ional Mo t ion:

C. Gra vita tio n:

D. Re lat iv it y of Mo tio n:

L in ear m om entum of part icl e and of a sys tem of part icl es,c onservati on of li n ear m om entum .

Im pul se and li near m om entum ,

Con servation of li near m om entu m in colli sion s and expl osi ons;elast ic and i nelasti c colli s ions; colli s ions in t wo dim ensi ons.

Cent re of m ass and m otion about centr e of m ass.

Rotati onal k inem ati cs, angul ar vari ables,rel ation ship betwee n li near and an gular k in em ati cs - fixed ax i s.

Rotati onal d ynam ic s, t orque, ang ular m omentu m(of a part ic le and a sys tem of p ar ti cl es ),rotat ion iner tia, rot ation al k i netic energ y,

c onservati on of ang ular m omentu m .

The Law of Universal Gravi tati on,

planet an d satelli te m oti on,

grav i tati onal fi e ld and pot ent ia l ,i nert ia and grav it at ional m ass.

V ar iati on i n g rav itati onal fie l d st rengthdue to lati tud e, alt itu de.

Mot ion of pl anets and satelli t es-geostati onary orbi ts. Relati ve vel ocit y.

Relat ive Vel ocit y

Unform Rel ative Tran slati onal Moti on

The Galili an Transform ati on

MechanicsII


VII. General objective(s)

After completing the module you should be able to

• Develop understanding of linear and angular momentum• Understand dynamics of rotational motion• Understand gravitational interaction and its application in artificial satel-

lite.• Develop skills and habits of solving problems in a well reasoned and neat

manner

VIII. specific learning objectives (Instructional objectives)

Content Learning objectives After Completing this section you should be able to:DynamicsofasystemofParticles

(40hours)• Linear momentum of particle and

of a system of particles,• Conservation of linear momen-

tum.• 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.

• Relate impulse and linear momen-tum

• 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


Rotational Motion: (35 hours)tion: (35 hours) (35 hours)

• Rotational kinematics, angular va-riables, 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 momen-tum.

Gravitation: (25 hours)

• The Law of Universal Gravita-tion,

• 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.

Relativity of Motion: (20 hours)

• Relative velocity.,• Uniform relative translational mo-

tion,• The Galilean transformation

• 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 dyna-

mics

• Use Newton’s law of universal gravitation to solve problems

• Describe Gravitational field and potential

• Describe the gravitational poten-tial

• Distinguish between inertial and gravitational mass(replace with Distinguish between inactive and gravitational force)

• Calculate escape velocity of satel-lites

• Describe the relativity of mo-tion

• 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 themsel-ves by answering all the questions provided below. Education research shows that this will help learners be more prepared and help them articulate previous knowledge.

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 Me-chanics II

1. A person travels a distance d meters given by (5m/s2 )t2 where t is in se-conds. 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.

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(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/sec2. The position of a particle moving along the x -axis depends on the time as

per the following equation for four seconds

x = 4t2 − 2t3 +

t4

4where 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 (b). Has constant acceleration.(c). Has constant kinetic energy.(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 (b). acts along the road outwards(c). acts horizontally 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

o

(b). 60o

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(c). 190o

(d). 120o

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(b). 0.48(c). 0.4(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(b). First (c). Third(d). None

10. A projectile is fired at an angle of 37o with an initial speed of 100/sec. What

is the approximate vertical component fo its velocity after 2 sec?

(a). 80m/sec(b). 40m/sec(c). 60m/sec(d). 100m/sed

11. A projectile is fired at an angle of 370 with an initial speed of 100 m/s. the vertical component of its velocity after 2s?

(a). 80 m/s

(b). 40.4 m/s


(c). 60 m/s

(d). 29.6 m/s12. In problem 11 the position above the ground after 3s is approximately

(a). 140 m(b). 200 m(c). 136 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

(b). 60 m/s

(c). 29.4 m/s

(d). 20 m/s

15. Which of the following projection angles will result in the greatest range?

(a). 370

(b). 200

(c). 480

(d). 600

16. A 10kg block is lifted 20m above the ground in a gravitational field. The work done by the field is

(a). Negative (b). Equal to the final potential energy(c). Positive (d). A vector quantity

17. A body in equilibrium may not have

(a). Momentum (b). Velocity


(c). Acceleration(d). Kinetic energy

18. Watt-sec is a unit of

(a). Momentum (b). Force (c). Energy(d). Power

19. The 2.5kg head of 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(b). 34m/s(c). 3.4m/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(b). 98m (c). 960m(d). 245m


Answer Key

1. A, C, D

2. A, B, C, D

3. B, C

4. C, D

5. B

6. D

7. B

8. C

9. C

10. B

11. B

12. C

13. A

14. C

15. C

16. A

17. C

18. C

19. C

20. A


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 for-mulations 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 40 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 di-

mensions• 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

r

F∑ = mr

a , 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 = Δp is defined and used.

To further our comprehension of mechanics we must begin to examine the interac-tions of many particles at once. To begin this study, we define and examine a new concept, the centre 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.

Completereference: ConservationofMomentumFrom html version of SimpleNature, 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.ligh-tandmatter.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.

Completereference: MomentumConservationandTransferfrom 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 (for the Learning Activity)

Software, Interactive online exercises Videos, animations etc

Resource#1

Title: Motion of Centre of Mass

URL: http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html

ScreenCapture:

Description: Applet shows the motion of the centre of mass of a dumbbell sha-ped object. The red and blue dots represent two masses and they are connected by a mass less rod. The dumbbell’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...

List of Relevant Useful Links (for the Learning Activity)

List of links, providing an alternative perspective on the curriculum material, each with “screen capture”


UsefulLink#1ClassicalMechanicsTitle:AllThermodynamicsURL:http://farside.ph.utexas.edu/teaching/301/lectures/

ScreenCapture:

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.

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.


Linear Momentum and Impulse

In the mechanics I, the concepts of work and energy were developed from New-ton’s laws of motion. We shall see next how the concepts of linear momemtum and impulse also arize from these law.

Consider a particle of mass m moving with a velocity vr

. Suppose a const

force r

F acts along the line of motion, then

vr

= vr

0 + ar

t

mvr

= mvr

0 + Fur

t

⇒ mvr

− mvr

0 = Fur

t

⇒Vvr

= Fur

t

The linear momentum of a particle is defined as the product of its mass and its linear velocity

r

P = mr

v valid only for r

v << the speed of light c

⇒ px

= mvx; 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

J = Δr

p

Interms of these two newly defined quantities

Note that the above equation holds independently for components. i.e

Δpx= J

x= F

x.t

Δp4= J

4= F

4.t

Δp2= J

2= F

2.t

Linear momentum of a particle can also be related to the net force acting on the particle as follows

multiply both sides of this equation by m we obtain


r

F = mr

a = mdr

v

dt=

d

dt(m

r

v) =dr

p

dt⇒ d

r

p =r

Fdt

Integrating the last expression, the change in momentum of a particle is

Δpur

= pur

f − pur

i = Fur

ti

t f

∫ .dt

The quantity

Fur

ti

t f

∫ .dt is called the impulse of the force Fur

for Δt = Δ

f− t

i

I = F

ur

ti

t f

∫ .dt = Δp Impulse – momentum Theorem.

Since the force can generally vary in tire, it is convenient to define a time ave-raged force

Fur

=1

Δt⎛

⎝⎜⎞

⎠⎟Fur

ti

t f

∫ .dt

⇒ I = Δr

p = Fur

Δt


Conservation of Linear Momentum

Consider two particles that can interact each other but are isolated from the surrounding

These forces could be of any origing (i.e. gravitational, electromagnetic etc.) Since the two are action reaction pairs

F12+ F

21= 0

dp1

dt+

dp2

dt=

ddt

( p1

uru

+ p2

uru

) = 0

⇒ total momentum (r

p) =r

p1+

r

p2= constant.

⇒ pix= p

fx p

iy= p

fy p

iz= p

fz

The statement the total momentum of the system remains constant is referred to as the law of conservation of momentum.

Example1: 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 Δt =1

800sec.

By definiton:

I = FavΔt

Therefore the average force exerted on the superball is, Fav=

21800

= 1600N.

Example2: 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:

ΔpΔt

= Fav=

pf

2 + pi

2 + pip

f

Δt

=302 + 302 + 2 × 30 × 30cos

0.02= 260N . in the horizontal direction.


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 over-come these difficulties. We begin by restating Newton’s second law

Fur

= mar

=ddt

(mvr

) ∴m is constant

=dP

ur

dt

This form of 2nd law is preferred to

Fur

= mar

because it is generalized to com-plex 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

Asystem:- 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

ClosedSystem: - is a system that does not interact with its surrounding.

Externalforce: - 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, m

2, m

3... m

N.

The position of the j th particle is r

j

ur

, the force on the j th particle is fur

j and

its momentum is r

pj= m

j

r

&rj

The equation of motion for the j th particle is thus

fur

j =d p

ur

dt=

r

fj

internal +r

fj

external

the force on particle j can be split between two terms (internal and external)

Adding all the equations of motion of all the particles in the system, we obtain

r

f1

int +r

f1

ext =d p

1

uru

dtM

r

fj

int +r

fj

ext =d p

j

uru

dtM

r

fN

int +r

fN=

d pN

u ru

dt

⇒r

fj

int∑ +r

fj

ext∑ =d p

j

uru

dt, j = 1, 2, 3, Ln∑

fj

ext∑ = is the sum of all external forces acting on all particles.

= total external force acting on the system

=r

Fext

fj

int∑ = 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 fj

ext∑ = Fur

ext =d p

ur

dt

⎛

⎝⎜

⎞

⎠⎟∑

=ddt

⎛

⎝⎜⎞

⎠⎟pur

∑ jQ derivative of a sum is the sum of the derivatives.

Denoting

pj= p

ur

∑ the total momentum of the system the last equation be-comes

Fur

ext =d p

ur

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 inte-

raction Fur

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 contai-ning many particles.

CenterofMassforTwoParticlesinOneDimension

If a particle with mass m

1 has a position of

x

1 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:

xcm

=m

1x

1+ m

2x

2

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 cen-


ter 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 x

cm we see that:

vcm

=m

1v

1+ m

2v

2

m1+ m

2 Differentiating again, we can generate an expression for acceleration:

acm

=m

1a

1+ m

2a

2

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

m

1 be

F

21, and the force exerted on

m

1 by

m

2 by

F

12. By applying Newton’s Second Law we can state that

F

12= m

1a

1 and

F

21= m

2a

2. We can now substitute this into our expression for the acceleration

of the center of mass:

acm

=F

12+ F

21

m1+ m

2

However, by Newton’s Third Law F

12 and

F

21 are reactive forces, and

F

12= −F

21

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 m

1 experiencing an external force

of F

1 and

m

2 experiencing a force of

F

2 We also must continue to take into ac-

count the forces between the two particles, F

21and

F

12. By Newton’s Second Law:


F1+ F

12= m

1a

1

F2+ F

21= m

2a

2

Substituting this experession into our center of mass acceleration equation we obtain

F1+ F

2+ F

12+ F

21= m

1a

1+ m

2a

2

⇒ F1+ F

2= m

1a

1+ m

2a

2= m

1+ m

2( ) acm

⇒ Fexternal∑ = m

1+ m

2( ) acm

= Macm

This equation bears a striking resemblance to Newton’s Second Law. The overall acceleration of a system of particles, no matter how the in-dividual particles move, can be calculated by this equation. Consider now a single particle of mass M placed at the center of mass of the sys-tem. 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.

SystemsofMorethanTwoParticles

A simple extension of our two particle equations to an n particle system will show the total mass of the system M is

M = m

1+ m

2+ m

3+ L + m

n

With this definition we can simply state the equations for the position, velocity, and acceleration of the center of mass of a many particle sys-tem, similar to the two-particle case. Thus for a system of n particles:

xcm

=1

Mm

nx

n∑

vcm

=1

Mm

nv

n∑

acm

=1

Mm

na

n∑F

ext= 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:

r

f =d

r

pdt

For a system of n particles.


pur

= pur

1 + pur

2 +L pur

n

= m1vr

1 + m2vr

2 +Lmnvr

n = MVcm

i.e. Total momentum of

a system of particles

⎛

⎝⎜⎞

⎠⎟=

total mass of

the system

⎛

⎝⎜⎞

⎠⎟ ×

The velocity of the center

of mass of the system

⎛

⎝⎜⎞

⎠⎟

⇒d

r

Pdt

= Mdr

vcm

dt= M

r

acm

⇒r

Fext

=d

r

Pdt

= 0, when no external force acts on the system

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:

v

M

y

x

y

xv+Δv

M-ΔmΔm

Fig: mass m moving with velocity v.

The above figure shows mass m moving with velocity v. After some time a mass

Δm is ejected out with a velocity u in opposite direction to that of v . For this sys-

tem of variable mass we can write F

ext=

d pur

dtas an approximate result for

Fext

;

Δ pur

Δt=

pur

f − pur

i

Δt; [( M − ΔM )(v + Δv) + ΔMu]− [ Mv]

= MΔvΔt

+ [u− (v + Δv)]ΔMΔt

as Δt → 0

Fext

= MdMdt

+ vdMdt

− udMdt


Note:- this equation is reduced to the familiar law of conservation of momen-tum.

Example3: 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

before explosion

⎛

⎝⎜⎞

⎠⎟=

Total Momentum

after explosion

⎛

⎝⎜⎞

⎠⎟

(Ptot

)before

=(Ptot

)after

⇒ (15× 12)r

i=(250 cos 30 + 5 v cos θ )ir

+ (−250sin30 + 5vsinθ) jr

0 ⇒ 5v sin θ=125

⇒ vsin θ=25

⇒180 =216.5 + 5v cos θ ⇒ vcos θ =-7.3L(2)

(1) ÷ (2) ⇒ tan θ =25/-7.3 = -3.4247 ⇒ θ=-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 θ in (1)

v=25

sin 6=

25

sin106.28= 26m/ s

Example4: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 Jr

∑ = (2 × 0.02) + (0.01× (−2)) = 0.02N − sec

⇒ 0.02N .sec =4

9.8(v − 0) ⇒ v = 0.049m/ s


Example5

A stream of water with cross-sectional area 2000 mm2 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 'ur

= v ''uru

,but directions are different

The mass m of all particles of water in time interval Δt is

m= AvgΔt A = area , v speed, p-density of water

= (2000 ×10−6 m2 )(10m/s)(103kg/m3 )=20(Δt)

using Jr

= Δ pur

in the x and y directions

FxΔt = m(v

x

11 − vx

1 ) = (20Δt)(−10cos45−10) = (20Δt)(−17.07)

FyΔt = m(v

y

11 − vy

1 ) = (20Δt)(−10 sin 45− 0) = (20Δt)(7.07)

∴ Fx= −341.4N ; F

y= 141.4N

Example6

Consider the motion from o to A.

v11= vog − gt

ob⇒

tob=

v0sin45

gL(1)

x1= v

ox.t

ob= (v

0cos45)(

v0sin45

3)L(3)

When the projectile explde at A, the velocity is entirely along x. i.e

v

0cos45m(v

0cos45) =

m2

vf by cons of mom and where

v

f is the v of the flying frag.


⇒ vf= 240cos45

∴ x2= (240cos45)t

ob since the fine of fall is

= (240 cos 45)v

0sin45

g=

v2

0sin45cos45

g

⇒ xtot= x

1+ x

2=

v2

0sin45cos45

g+

2v2

0sin45cos45

g=

v2

0sin45cos45

g[1+ 2]

=3v2

0sin45cos45

g= 1.055×105 ft


Solve the following problems

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 explo-des 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-girls imultaneously 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. Dis-cuss?

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.


FormativeEvaluation1

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 com-mon veloci ty i f they col l ide and become coupled together

.

A10kg

2m/sec 4m/sec

a. 2m/s to rightb. 2m/s to leftc. 3.33m.s to leftd. 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. Increasesb. Does not changec. 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 asb. more thanc. less than the forced. 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 trave-ling 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°

Teaching the Content in Secondary School 1

What interest students about collision and momentum conservation? This ques-tion 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: Rotational Motion

You will require 35 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.


• Derive and use equations describing rotational motion• Relate angular and linear quantities for rotation around a fixed axis

• Use τ = Iα to solve problems• Define angular momentum and its conservation.• Solve problems in rotational Dynamics.


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 θ( ) , angular velocity ω( ) , angular acceleration α( ) , moment of

inertia

I( ) , torque τ( ) , 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.


Reading 3: Angular Acceleration.

Completereference: AngularAccelerationincircularmotionFrom 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.


Completereference: ConservationofMomentumFrom 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:


Rationale:

This section has a well illustrated content on angular momentum. Theories of angular momentum in two dimensions and in three dimensions are well developed. Further there is a good deal of theory on rigid body rotation.

Reading 5: Torque and Angular Momentum.

Completereference:TorqueandAngularMomentumincircularmotionFrom 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, non planar rigid bodies. The problem sup-plement and the model exam at the end make this site popular.


List of Relevant MM Resources (for the Learning Activity) Software, Interactive online exercises Videos, animations etc

Resource #2: 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.

List of Relevant Useful Links (for the Learning Activity).


Useful Link #2 Tutorial on torque from university of Guelph

Title:Torque

URL:http://www.physics.uoguelph.ca/tutorials/torque/index.html

ScreenCapture:

Description: The site gives detailed description of torque

Rationale: Here you will find a good collection of tutorial problems on tor-que...


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 inconve-nient: 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 2π times the radius. This suggests that instead of dividing a circle into

360 parts, we divide it into 2π parts. These parts are called radians. Thus 90o

corresponds to π

2radians, 45o corresponds to

π

4radians etc. Therefore we

define and use the radian measure to handle real rotation problems.

2.1: Rotational Kinematics:

Angular Variables:

Angular displacement θ( ) :

rθ

ΔS

o

Angular velocity ω( )The angular speed is defined just like ordinary speed: an object which moves

from an angle θ i to an angle

θ

f has average angular speedω :


o ω =

Δθ

Δt=θ

f−θ

i

ΔtInstantaneous angular speed is obtained by taking Δt small. An object moving

45 revolutions per minute (rpm) has ω = 90π 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 2π radian per revolution. Moreover, one re-volution corresponds to 2π radians. Thus the speed v is related to the angular speed ω by the simple formula:

o v = ωr

This formula is only valid if ω 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:

o a

c=

v2

r=ω 2r 2

r= ω 2r

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 velo-city. Not the magnitude!

Angular acceleration α( ) :

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 α is defined asthe time rate of change of angular velocity:

α =

Δω

ΔtFor constant angular speed, Δω = 0 and hence α = 0 . For constant angular acceleration the tangential acceleration is given by:

aT= αr

(again α must be measured in radians). The direction of tangential accelera-tion 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,

ω = ωo

+αt

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:

θ = ω

o

t +1

2αt2

The other equations also apply if you replace

x with θ , v with ω and a with α

Example1: 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:

α =

Δω

Δt=

64rad/s -1rad/s

3s= 21rad/s

To get the angular displacement, use

θ = ω

o

t +1

2αt2 = 98rad ;15revolutions.

Example2: 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

v

chain= ω

petalr

1

For the front gear, and

vchain= ω

wheelr

2

Combining the two means that ω

petalr

1= ω

wheelr

2 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 = ωwheel

rw

Thus

v = ω

petal

rwr

1

r2

Changing the gears on the bike changes the ratio r1r

2 changing to a smaller

gear in the front, or a larger one in back, (i.e. decreasing r1r

2) makes it easier

to petal. The reason is for a fixed angular speed ω

petal it makes v smaller. The

lower v , the less work you’re doing.

Example3: 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.

SolutionThe ceiling fan rotates about its axis, slowing down with constant angular acce-leration 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)

Δθ =

1

2ω +ω

o

( )t =1

20 + 0.90rev/s( ) × 2.2min× 60sec/ min = 59rev

(b)

Δθ =

1

2ω +ω

o

( )t =1

20.45+ 0.90rev/s( ) ×1.1min× 60sec/ min = 45rev

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 (θ)

Speeed (v) ↔ Angular velocity (ω )

Acceleration (a) ↔ Angular acceleration (α )

Force (F) ↔ Torque (τ )

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 mo-ment of inertia of a body the greater is 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 hin-ges. 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.


F1 and F2 are equal in magnitude. Which Force produces a better turning effect? F1 or F2?

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

τ = F

Tr

F

Tis 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 counter-clockwise 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 com-pletely 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 MomentumAngular 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

L =r

r ×r

p

The angular momentum is always a vector perpendicular to the plane defined

by the vectors r

r and r

p (or r

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.

Figure10: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,

dr

L

dt=

d

dt

r

r ×r

p( ) = md

dt

r

r ×r

v( ) ,where

r

p has been replaced by mr

v and the constant m has been factored out. Using the product rule of differential calculus,

d

dt

r

r ×r

v( ) = dr

r

dt×

r

v +r

r ×dr

v

dt


In the first term on the right-hand side, dr

r dt is simply the velocity r

v , leaving

r

v ×r

v . Since the cross product of any vector with itself is always zero, that term drops out, leaving

d

dt

r

r ×r

v( ) = r

r ×dr

v

dt

Here, dr

v dt is the acceleration r

a of the particle. Thus, if both sides of the above

equation are multiplied by m , the left-hand side becomes dr

L dt and the right-

hand side may be written r

r × mr

a . Since, according to Newton’s second law mr

a

,is equal to r

F , the net force acting on the particle, the result is

dr

L

dt=

r

r ×r

F

The above equation means that any change in the angular momentum of a particle

must be produced by a force that is not acting along the same direction as r

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.

The quantity r

r ×r

F is called the torque

r

τ . 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

τ =r

r ×r

F =dr

L

dtThis 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 equal to r

r ×r

Fa

. According to Newton’s third law, the particle must apply

a force −

r

Fa to the agent. Thus there is a torque equal to

−

r

r ×r

Fa

acting on the agent. The torque on the particle causes its angular momentum to change at a rate

given by dL dt =

r

r ×r

Fa However, the angular momentum

r

La of the agent


is changing at the rate dL

adt = −

r

r ×r

Fa

Therefore, dL

adt + 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.

Example4: MomentofInertiaofaRodofUniformMassDensity,PartI

Consider a thin uniform rod of length and mass . In this problem, we will cal-culate 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.

SolutionChoose 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 = λdx , located at a displacement x

from the center of the rod, where the mass per unit length λ = m L is a constant, as we have assumed the rod to be uniform. When the rod rotates about an axis per-pendicular 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

Icm

= r⊥( )

body

∫2

dm = λ x2( ) dx = λx3

3− L 2

L 2

∫− L 2

L 2

=m

L

L 2( )3

3

−m

L

− L 2( )3

3=

1

12mL2


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

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) 75.2rev.

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 concen-trated. Thus the moment of inertia of a body of mass M and radius of gyration

K is I = Mk2 (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.

Answer:

(a) I =

2

3MR2 (b) k =

2

5R



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.


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

M

E= 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

I =

MER

E

2

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 sup-port (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.


TeachingtheContentinSecondarySchool2

Rotational motion extends the concepts of linear motion to curvilinear motion. It is not likely that introductory physics students will recognize that the motion of projectiles is the result of an object having linear motion in two directions at the same time. The concept of the independence of thes velocities gives students considerale difficulty. Demonstrations, laboratory exercises, and solutions of problems will help in teaching this concept.

Circular motion is a concept that frequently causes confusion for nonscientists. Centripetal motion must be analyzed from nonaccelerated point of view. Only then can Newton’s second law be applied While teaching circular motion, it is advisable to stress the necessity fo finding the physical source of the force that causes thae acceleration toward the center of the circle.

African Virtual University 55

Activity 3: Gravitation



• 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.


According to Newton’s postulated law of universal gravitation, two bodies of

mass m1 and

m2 separated44 by a distance r . exert equal attractive forces on

each other of magnitude given by F = Gm1m

2r2 . , 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, propor-tional 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



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 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 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.

List of Relevant MM Resources (for the Learning Activity)

Resource #3;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 para-meters of the projectile.


List of Relevant Useful Links (for the Learning Activity)


Useful Link #3 Universal Gravitation from Wikipedia

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.

Date Consulted: - April 2007


Useful Link #4 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 the physics class room.

Rationale: Reach in discussion topics and interactive problems.

Useful Link #5 Wikipedia

Title: Gravitational Field URL: http://en.wikipedia.org/wiki/Gravitational_field

Screen Capture:


Description: Gravitational 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...


Detailed Description of the Activity (Main Theoretical Ele-ments)

Introduction

In Newoton’s day the force of gravity was thought to be a local force that influence objects near the earth. Prior to Newton, scientists believed that physics on the earth diffeered from physics in the heavens. The motions of falling apples, for example, was never connected to the motion of an orbiting planet. Newton’s realization that gravity is universal had important implications. He recognized that the laws of motion must also be universal in character. i.e. the same laws of motion govern both celestial and terrestrial motion.

This was indeed the beginning of understanding nature that paved the way to present day space exploration and a number of artificial satellites that orbit the earth.

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

a⊥=

v2

R=

(2πR / T )2

R=

4π 2 RT 2

= 2.72 ×10−3 m/ s2

g = 9.8m/s2

a⊥

g=

1

3606

• From the knowledge:

radius of earth (RE)

radius of moon's orbit (R )=

1

60, he observed


a1

g=

RE

R

⎛

⎝⎜⎞

⎠⎟

2

⇒ gRE

2 = a⊥

R 2

⇒ 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 F g : m1m

2 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


Note:

i. Gravitational forces between two particles are an acriton- and- reaction pair

ii. G is universal.iii. 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. 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.

v. The gravitation force of a sphere of radius R is

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 = F dr

⇒ S = F dr∞

r

∫ =GmM

r 2

∞

r

∫ dr = GmM∞

r 1

r 2dr

= GmM −1

r∞

r

=−GmM

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 r

2


W = ΔU = GmM1

r1

−1

r2

⎛

⎝⎜

⎞

⎠⎟

⇒ If r1< r

2 W will be positive

r1> r

2W 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 =

GMr

Example 1: Gravitational field due to a non-homogenous spherical mass

A non-homogenous spher of radius R has the following density variation:

p = ρ0 for :r<

R

3

=ρ

0

2 for :

R

3< r ≤

3R

4

=ρ

0

8 for :

R

3< r ≤ R

What is the gravitational field due to the sphere at

r =

R4

,R2

,5R6

and 2R


Solution

g =Gmr 2

=GpV

r 2

∴g R 4( ) =Gρ

0

43π R 4( )3

R 4( )2=π

3Gρ

0R

gR2

⎛

⎝⎜⎞

⎠⎟= Gρ

0

4

3π

R 3( )3

+12

R 2( )3

− R 2( )3⎡⎣⎢

⎤⎦⎥

(R / 2)2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

= Gρ0

4 3( )πR1

21 3( )2

+1

21 2( )3⎡

⎣⎢

⎤

⎦⎥ × 4

= Gρ0πR

4

3× 2 1 27 +1 8⎡⎣ ⎤⎦ = 0.432πGρ

0R

g5R6

⎛

⎝⎜⎞

⎠⎟= Gρ

0

4

3π

R 3( )3

+12

3R 4( )3

− R 3( )3⎡⎣⎢

⎤⎦⎥+

18

5R 6( )3

− 3R 4( )3⎡⎣⎢

⎤⎦⎥

(R / 2)2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 0.48πGρ0R

Similarly g(2R)=Gρ0

4 3( )πR 1 3( )3

+1

23 4( )3

− 1 3( )3⎡

⎣⎢

⎤

⎦⎥ +

1

81− 3 4( )3⎡⎣⎢

⎤⎦⎥÷ 4

= 0.1πGρ0R


Example 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

ρ = 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 .ω S ' = r '2 .ω

S cosα = r 2ω S cosα = r '2 w

mass of S = r 2ωρ / cosα

mass of S = r '2 ωρ / cosα

∴ intensity at p due to S=r2ωρ

cosαr 2G =

Gωρ

cosαand

intensity at P due to S' =r'2ωρ

cosαr '2G =

Gωρ

cosα

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: 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

Thus we find that the torque gets as if the entire mass M of the body were conc at its cm.


Motion of Planets and Satellites

All planets move in elliptical orbits, the sun being at one focus. We may for sim-plicity 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,

GMr 2

= mω 2r or GMr 2

= ω 2r

GMr 2

= ω 2 =4π 2

T 2

From this, we may deduce the period of the planet or the satellite.

T 2 =

4π 2

GMr 2

Kepler’s Law

Johannes Kepler culminated his 30 years of research with the publication of the last of his three laws of planetary motion. The three laws of planetary motion can be stated as:

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,


GM(R

E+ h)2

=mv

0

2

(RE+ h)

where RE

is the radius of the earth

v0=

GM(R

E+ h)

If h < RE,v

0= gR

E

The amount of work required to move a body from the earth to infinity is given by the formula

GE

MmR

E

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 criti-

cal initial speed v0 called escape velocity is given by the formula

1

2mv

0

2 =GM

E m

RE

v0=

2GME

RE

= 11.2 km/second

Synchronous Satellite (Geostationary Orbit)

Most communication satellites, and many meteorological satellites, circle the Earth in orbits above the equator. From the surface of the Earth, they appear to stand still due to the fact that they orbit the Earth once every 23 hours and 56 minutes. That also is the amount of time required for the Earth to complete one rotation with respect to distant stars.

The amount of time required for one revolution around the Earth, is determined by the satellite’s distance from the Earth’s center. In order to revolve once every 23 hr and 56 minutes, these geostationary satellites must be at the correct dis-tance from the Earth. One method of calculating this ‘magic’ distance is to use Newton’s law of gravity and his second law of motion. i.e. putting period to be 24hours and using the equation


T = 2πr

(RE+ h)3

GM= 86400seconds.

‘h’ may be calculated. It is about 22,700 miles above the earth’s surface.

Example 4: Acceleration due to gravity high above the surface of Earth.

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,

GMEm

r2= mg ⇒ g =

GME

r2

At the earth's surface, wher g=go= 9.8m/s2 and r = r

E= 6400km

From these fromulas for g and go we find that, at the distance r from

the earth's center, the acceleration of gravity is

g=r

E

r

⎛

⎝⎜

⎞

⎠⎟

2

go

using r=rE+h=6400km + 1000km =7400km

Thus g=7.3m/s2




1. How would the weight of a body vary while being carried to the centre of the earth?

2. What is the meaning of weightlessness as applied to objects in space? Is there any place where weight is zero? If so why?


1) Why do a gram of weight and a quintal of weight, released simultaneously from the top of a tower reach the ground at the same time?

2) If you jump off a chair, you accelerate towards the earth. Does the earth also accelerate towards you? Explain.

3) What would happen if the force of gravity were to disappear suddnely?

4) The earth is constantly acted upon by the gravitational attraction of the sun. Then why does not the earth fall into the earth.?

5) A body is taken from Nairobi (Kenya) (very close to the equator of the earth) to South Africa. What will be the effect on its weight? On its mass?

6) In uninformed discussions of satellites, one hears questions such as “what keeps the satellite moving in its orbit?” and “What keeps the satellite up?” How do you answer these questions?

7) Are your answers to the above question applicable to the moon? Explain.

8) 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?

9) A rocket has a mass of 2.0 × 104 kg of which half is fuel. Assume that the fuel is consumed at a constant rate as the rocket is fired and that there is a

constant thrust of 5.0 × 106 Newtons. Neglecting air resistance and any possible variation of g , compute

a) the initial acceleration andb) the acceleration just as the last fuel is used

10) The mass of the moon is

1

81 and its radius is about a quarter of the earth’s

radius. What is the accelreation due to gravity on the surface of the moon?


11) Find the accelration the moon towards the earth assuming that the moon is situated at a distance which is 60 times earth’s raduius being measured from the centre of the earth.

12) If R

1 and

R

2 be the radii of two planets and

ρ

1 and

ρ

2 their mean densities

show that values of acceleration due to gravity on the two planets will be in

the ratio R1

ρ1: R

2ρ

2

Optional Formative Evaluation 3


Gravitation in schools is introduced to show how Newton used the work of Kepler and Brahe in combination with his own work (the equation of centripetal accele-ration) to identify the law governing the gravitational force and its universality.

Physical laws are more meaningful to students if their origins are understood. Derivation of the law of universal gravitation is simple enough for beginners to follow. Therefore leading students to students to an understanding of what quantities can be determined from the application of the law of gravitation to the motion of planets.

A field is difficult concept for school students, but attempts to help students by introducing analogy with electric and magnetic fields will help.


Activity 4: Relativity of Motion



• Describe the relativity of motion • Use Galilean transformation to solve problems in gravitation


The origin of a coordinate system is needed to describe the velocity of a body just like its position. Ordinarily this origin is taken to be fixed in some other body, but this second body may be in motion relative to a third and so on. Thus when we speak of the “the velocity’’ of an object we usually mean its velocity relative to the earth. But the earth itself is in motion relative to the Sun and the Sun is in motion relative to some other star, and so on.

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 inertial reference frames; In particular, Einstein extended the Galilean principle of relativity to electromagne-tism and optics which describe the theory of light

Galilean transformation also called Newtonian Transformation, is a set of equa-tions 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 (for the Learning Activity).

Copyright free readings should also be given in electronic form (to be provided on a CD with the module)

Reading 9: Velocity and Relative Motion .

Complete reference : From html version of Simple Nature, by Benjamin Crowell. URL : http://www.lightandmatter.com/html_books/0sn/ch02/ Accessed on the 20th April 2007

Abstract :


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.


Detailed Description of the Activity (Main Theoretical Ele-ments)

Introduction

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 par-ticular, Einstein extended the Galilean principle ofrelativity to electromagnetism and optics which describethe theory of light

Relative Velocity

Have you observed out of the window of a fast moving bus or train? The countryside appears to flash past the window. Although you already know that the countryside is not moving, it appears to be moving relative to you inside a moving bus or train.

Consider two cars A and B travelling with a velocity of v

A and

v

B as shown

in figure 4.1a. Car A overtakes B. The dirver in A sees car B apparently moving towards west. However, a passanger in the back of car B sees car A is catching up with her car

Figure 4.1a

The velocity of A relative to B is the velocity which A appears to have to an ob-server who is moving with B. Thus the velocityh of A relative to B is effectively the resultant velocity of A when when B is made stationary and the same retarding force is applied to A.

Suppose a long train of flatcars is moving to the east along a straight level track, and that the two automobiles are moving on the flat car as shown in figure 4.1b.

A BVBF VFE

Figure 4.1b


In figure 4.1, V

FErepresents the velocityof the flatcar relative to the earth E and

V

AF the velocity of the automobile A relative to the flat car. The Velocity of the

flat car relative to earth is the sum of the velocities of automobile A relative to the flatcar and that of the flatcar relative to the earth. Thus:

o

r

VAE

=r

VAF

+r

VFE

This same equation holds for the relative velocity of automobile B relative to the flatcar and that of the flatcar relative to the earth

o

r

VBE

=r

VBF

+r

VFE

Notice that the speed of automobile A relative to the earth is the sum of the two velocities where as the speed of automobile B relative to the earth is the difference of the two velocities.

Example 1: An automobile driver A, travelling relative to the earth at 75 km/hr on a straight level road is ahead of a motor cyclist. B travelling in the same di-

rection at 90 km/hr . What is the velocity of B relative to A.

Solution

We h a v e

r

VAE

= 75 km/hr, r

VBE

= 90 km/hr a n d w e w i s h t o

find

r

VBA

. From the above rule of combination of velocities we have:

r

VBA

=r

VBE

+r

VEA

But r

VEA

= −r

VAE

r

VBA

=r

VBE

−r

VAE

= 90 km/hr − 75 km/hr=15 km/hr

o

The cyclist is overtaking the driver at 15km/hr.

Example 2: Verify that the relative posions of the two do not matter when it come to the relative velocity. I.e.the relative velocity would not be altered if B were ahead of A.

Solution

The relative positions of the bodies do not matter. The velocity of B relative to


A remains 15km/hr but now pulling ahead of A at 15km/hr.

Example 3: 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= V

1−V

2, the two

objects will collide if

1. Vrel> 0

2. Vrel< 0

3. Vrel= 0

Solution

For object 1 to overtake object 2 and thus collide, V

1must be greater than

V

2 this gives Vrel

> 0 . Since both V1 and V2

are positive.

Example 4.4: 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

Galilean Transformation

Relativty concerns the laws of physics as they are formulated by observers in relative motion. For example, consider two observers: a man standing on the ground and looking into a passing train and a passanger standing in a train. The passanger in the train dropped a coin. The man standing outside sees the con fall along a parabolic trajectory. The passannger sees it drop straight down as he continues along at constant velocity. The outsider and the passanger observe different positions and diffenrnet positon and different velocities, but both trajec-tories hold relative to each individulal reference frame, as predicted by Newton’s law of motion. Even thouth the observers disagree on the actulal trajectory, the do agree that Newton’s laws are applicable.

Experiments show that if Newtonian mechanics is valid in one reference frame, it is valid in all other reference frames moving at constant velocity relative to the first. This is a principle of relativeity. The key requirement here is the absence fo any significant acceleration of the reference system. A convenient interpretation


is no significant acceleration (including rotation) with respect to the fixedc stars or with respect to distant galaxies. These constant velocity reference frames are called inertial systems because Newton’s first law holds in these systems. In modern terminology we say that Newtonian mechanics is invariant to the choice of inertial system.

The invariance of Newtonian mechanics to the choice of inertial reference fram is called Galilean relativity.

The term invariance means that observers in diffenetn inertial systems agree that Newtonian mechanics adequately describes the observed motions. Newtonian mechanics, then, is invariant to the choice of inertial reference frame. However, observers in different inertial systems disagree on the actual positons and vselo-cities in the observed trajectories.

Now 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. O and O’ frames are related as follows:


Here we’re following Newton’s assumption of one universal time. We will refer to equations relating quantities in primed and unprimed references as the Galilean Transformations.



1. Give an example of an object that is at rest in one reference frame while in motion in another?

2. The path followed by a moving body may be straigh in one frame of reference while being parabolic in the other. Give an example.



1) A package is dropped out of an airplane in level flight. If air resistance could be neglected, how would the motion of the package look to the pilot? To an observer on the ground?

2) If an artificial earth satellite has a period of exactly one day, how does its motion look to an observer on the rotating earth?

3) A passenger on a ship travelling due east with a speed of 18miles observes

that the stream of smoke from the ship’s funnels makes an angle of 20o with the ship’s wake. The wind is blowing from south to north. Assume that the smoke acquires a velocity (with respect to the earth) equal to the velocity of the wind, as soon as it leaves the funnels. Find the velocity of the wind.

4) A river flows due north with a velocity of 2m/s. A man rows a boat across the river, his velocity relative to the water being 3m/s due east .

a) What is his velocity relative to the earth?b) If the rivere is 1000m wide, how far north of his starting point will he

lreach the opposite bank?c) How long a time is required to cross the river

5)

a) In what diretion should the rowboat in the aboe problem be headed in order to reach a point on the opposite bank directly east from the start?

b) What will be the velocity of the boat relative to the earth?c) How long a time is required to cross the river?


This activity introduces the concepts of relativity of motion. Daily life experiances like apparent motion of road side objects when students are in a moving van may be cited as the case of relativtiy of motion. Why a bus moving in opposite direction apeears faster than a bus taking over the vehicle in which we are moving is also a good point to illustrate the relative velocity concept at school level.


XI. KeyConcepts (Glossary)

1. Momentum: The momentum of a particle, a vector quantity, is the product of the particle’s mass and velocity.

Jur= P

ur2 −P

ur1

In terms of momentum, Newton’s second law for a particle may be expressed as

rF∑ =

d rpdt

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.

rj =

rF(t

2− t

1)∑ =

rFΔt∑

if the net force varies with time, impulse is

rJ = F

urdt∑

t1

t 2

∫ 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 rJ =

rP

2−

rP

1

The momentum of a body equals the impulse that accelerated it from rest to its present speed.

3. InternalForce: 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. TotalMomentum: The total momentum of a system of particles A , B , C , K is the vector sum of the momenta of the individual particles:

o rP =

rP

A+

rP

B+

rP

C+L = m

A

rvA+m

B

rvB+m

C

rvC+L

5. ConservationofMomentum: 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. ElasticCollision: 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. CenterofMass: The center of mass of a system is the average position of the mass of the system. Its motion under given forces are the same as though all the mass were concentrated at the center of mass.

8. RigidBody: A rigid body is a body with a definite and unchanging shape and size.

9. AngularDisplacement(θ ):-is the measure of change in angular position of a rotating object.

10.AngularDisplacement(ω ): is the time rate of change of angular displace-ment:

ω =

dθdt

11. AngularAcceleration α( ) : is the time rate of change of angular velocity

α =

dωdt

12.MomentofInertia

I( ) : Is the rotational analog of mass. the greater the mo-ment 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 τ( ) : The torque exerted by a force on a body is a measure of its ef-fectiveness 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 τ( ) excreted by the force about O

has the magnitude: τ = Force( ) × Moment arm( )

14.KineticEnergyofRotation.: The kinetic energy of a body of moment of inertia I and angular velocity ω (in rad/s) is

K E =

1

2Iω 2

15.RotationalWork(W):- The work done by a constant torque τ( ) that acts

on a bodyh while it experiences the angular displacement θ in rad is

W = τθ

16.AngularMomentum: 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( )= Iω

17.LawofUniversalGravitation: States every body in the universe attracts every other body with a force that is directly proportional to each of their mas-ses and inversely proportional to the square of the distance between them

Graviataional Force = F = Gm

1m

2

r 2

G = 6.67 ×10−11 Nm2 /kg2

18.GravitationalField: The Newtonian theory of gravity is based on 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 gravi-tational 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.GravitationalPotentialEnergy: - Potential energy arises in systems with parts that exert forces on each other of a magnitude dependent on the confi-guration, 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 separa-ting 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.GravitationalfieldStrength: - 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:

g =

GMEarth

r2

21.GalileanTransformation: 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 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; and that the speed of light depends upon the relative motion of the observer


XII.CompulsoryReadings


Completereference:ConservationofMomentumFrom 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:


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.

Completereference:MomentumConservationandTransferFrom 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 mo-mentum 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.

Completereference:AngularAccelerationincircularmotionFrom 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 acce-leration, 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.


Completereference:ConservationofMomentumFrom 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.

Completereference:TorqueandAngularMomentumincircularmotionFrom 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 acce-leration, producing angular acceleration and constant torque case rotational kinematics are treated.

Rationale:This article covers topics on torque and angular momentum, sys-tem of particles, conservation of angular momentum, nonplanar rigid bodies. The problem supplement and the model exam at the end makes this site popu-lar.


Reading 6: Universal Gravitation:.

Completereference:Newton’slawofUniversalGravitationFrom 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:.

Completereference:Orbitalmotioninaninverse-squarelawforcefieldFrom 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 plane-tary 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 9: Velocity and Relative Motion .

Completereference:FromhtmlversionofSimpleNature,byBenjaminCrowell.URL : http://www.lightandmatter.com/html_books/0sn/ch02/ 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.


XIII. CompulsryResources

Resource #1

Title:MotionofCentreofMass

URL: http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html

Screen Capture:

Description:Applet shows the motion of the centre of mass of a dumbbell sha-ped object. The red and blue dots represent two masses and they are connected by a mass less rod. The dumbbell’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 Rotating Stool


URL:- http://hyperphysics.phy-astr.gsu.edu/hbase/rstoo.html#sm CompleteReference:- 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.

Resource #3;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 para-meters of the projectile.


XIV. UsefulLinks

Useful Link #1 Classical Mechanics

Title:ClassicalMechanicsURL: 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 Tutorial on torque from university of Guelph

Title:TorqueURL: 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 tor-que...

Useful Link #3 Universal Gravitation from Wikipedia

Title:UniversalGravitationURL: 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 #4 From The physics Class room

Title:UniversalGravitationandPlanetaryMotionURL: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/circles/u6l3c.html

Screen Capture:

Description: Lecture notes and discussion forum from the physics class room.

Rationale: Reach in discussion topics and interactive problems.


Useful Link #5 Wikipedia

Title:GravitationalFieldURL: http://en.wikipedia.org/wiki/Gravitational_field

Screen Capture:

Description: Gravitational 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:GeostationaryorbitURL: 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. SynthesisOftheModule

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 discussed 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.

The fourth activity has illustrated that motion is a relative concept. Quantities of motion like position, displacement and velocity are not universal and yet Newton’s laws of motion hold in all inertial reference frames. The quantities of motion in different frames of reference are related by Galilean Transformation.


XVI. SummativeEvaluation

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 velo-cities. 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) hap-pens 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 centre?

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 systemd. May or may not change under the influence of a net external torque, de-

pending upon the direction of the torque

2. Which of the following is not a unit of impulse?

a. N.Sb. lb.sc. lb.hd. N.m

3. If the momentum of a body increases by 20%, the percentage increase in its K.E. is equal to:

a. 44b. 88c. 66d. 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. 2sb. 0.02 sc. 0.2 sd. 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 inertiab. Acceleratec. Move rapidlyd. 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/sc. 4m/sd. 2m/s

8. A bomb dropped from an aeroplane explodes in air. Its total

a. Momentum decreasesb. Momentum increasesc. Kinetic energy increasesd. 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 forceb. 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 velocityω . 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. ω M ( M + m)

b. ω M ( M = 2m)( M + 2m)

c. ω M ( M + 2m)

d. ω ( 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 -1c. A positive number which varies from 0 to 1d. A positive number

12. If the polar ice caps melts, the length of the day will

a. Increaseb. Decreasec. Remain the samed. 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 forced. Centripetal force

14. The centripetal acceleration of a particle in circular motion is

a. Equal to its tangential acceleration b. Less than its tangential accelerationc. More than its tangential accelerationd. May be more or less than its tangential acceleration

15. If the frequency is 2 rev/s, the angular speed is

a. 4π rad / s

b. π 8 rad / sc. 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 directionc. Direction onlyd. 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/ s2

b. 40m/ s2

c. 4m/ s2

d. 20m/ s2

19. In an equilibrium problem, the axis about which torques are computed

a. Must pass through the centre of gravity of the bodyb. Must pass through one end of the bodyc. Must intersect the line of action of at least one force acting on the bodyd. 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 hoopd. 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/sb. 20 m/sc. 3 m/sd. 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 mb. 4 mc. 16 md. 4 rad

23. A quantity not directly involved in the rotational motion body is

a. Massb. Torque c. Moment of inertiad. Angular velocity

24. A full circle contains

a. π /2 radiansb. π /4 radiansc. π radians d. 2π radians

25. Wheel with angular momentum of 10 kg m2 / s has a moment of inertia equal

to 6.5 kg m2 . Its angular speed is

a. 5 rad/s b. 20 rad/s c. 40 rad/sd. 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 ab. 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 m2

c. kg /m2

d. kg m

29. An object is travelling in a circle with a constant speed. Its acceleration is constant in

a. Magnitude onlyb. Both magnitude and direction c. Direction onlyd. 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 l about an axis through

its centre and perpendicular to its length is I = M l 2 / 12b. The M.I. of a uniform circular disc of Mass M and radius r about per-

pendicular to the plane of the disc is I = 1 / 2 Mr 2

c. The M.I. in case of (b) about the axis through one end and perpendicular

to its length is I = M l 2 / 4d. 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 ω be its moment of inertia and the angular speed respectively, its angular momentum L about the given axis is given by

a. L = Iω

b. L = ω / I

c. L = I /ω

d. L = Iω


32. If the mass of a rotating body moves towards the axis of rotation, in moment of intertia

a. Remains the sameb. Decreasesc. Increasesd. May increases of decreases

33. When milk is churned, the cream separates from it due to

a. Gravitational forces b. Centrifugal forces c. Cohesive forcesd. 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 longerb. Speed of the all is increasedc. 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 degreesd. 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. Accelerationb. Angular velocityc. Velocityd. Angular acceleration

37. When the angular position of the swing is 450 , its linear velocity is

a. Negative and increasingb. Positive and decreasing c. Positive and increasingd. 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 cylinderb. Hollow lead cylinder c. Solid lead cylinderd. Solid lead sphere

39. Due to redistribution of a mass of rotating body if its moment of inertia in-creases then its angular velocity

a. Decreases b. Remains unchangedc. Increasesd. 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

a.

2E 2

2 Ib. 2E 2 I

c. 2E Id. 2E I

41. Physical quantity not directly involved in rotatory motion is

a. Moment of inertia b. Mass c. Angular velocityd. Torque

42. Vibration of a body is given by the differential equation

d2 xdt2

+ω 2 x = 0 The amplitude and time period are

a. 6cm and 10 πb. 8cm and 10πc. 10cm and 0.2 secd. 14cm and 2 sec.


43. The acceleration of a body rolling down an inclined plane is given by

a.

gsinθ

1+k 2

R 2

b.

gsinθ

1+R 2

K 2

c.

gsinθ

R 2 +k 2

R 2

d.

gsinθ

K 2 +k 2

R 2

44. If the earth stopped rotating, the weight of objects at the equator would

a. Be greaterb. Be the same as beforec. Be lessd. 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 earthc. Is more than the force with which the moon attracts the earthd. Varies with the phase of the moon

48. If the earth stopped rotating, the weight of objects at either pole would

a. Be lessb. Be greaterc. Be the same as befored. Vary with latitude

49. The gravitational constant is 6.67 ×10−11 N .m2 / kg2 . What is the gravi-tational 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 kg2

b. 2.81× 20−11 N .m2 kg2

c. 5.68 ×10−11 N .m2 kg2

d. 3.00 ×10−1

1N .m2 kg2


51. The gravitational force between bodies does not depend upon

a. The product of their n assesb. Their separationc. The sum of their massesd. 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/ sec2

b. 19.6m/ sec2

c. 4.0m/ sec2

d. 2.45m/ sec2

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. Ellipsesb. Circles c. Combinations of circles forming looped orbitsd. 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. Increased. Depend upon the weather.

57. If the earth had another satellite at twice the distance of the moon , its period would be

a. 28days22 / 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/ sec2

b. 19.5m/ sec2

c. 9.8m/ sec2

d. 39.2m/ sec2

59. If the earth were three times farther from the sun than it is now, the gravita-tional 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?

a.

v = R

g(R − h)

⎡

⎣⎢

⎤

⎦⎥

1/ 2

b.

v = R

g(R − h)

⎡

⎣⎢

⎤

⎦⎥

c.

v = R

g(R − h)

⎡

⎣⎢

⎤

⎦⎥

2

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 weightc. Might be so great as to break the roped. 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 emptyb. The period will go on increasing till the sphere is empty c. The period will remain unchanged throughoutd. 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 velocityb. Escape velocity c. Angular velocityd. Terminal velocity

65. The square of a satellite time of revolution round the earth is

a. Directly proportional to the cube of radius of orbitb. Inversely proportional to the cube of radius or orbit c. Directly proportional to radius of orbitd. Inversely proportional to radius of orbit

66. The earth retains its atmosphere because

a. The earth is sphereb. 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 goingc. Either of these two d. None of the above

68. To hit the target, one has to point his rifle

a. Higher than the targetb. Lower than the targetc. 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. Altitudec. Longituded. Depth

71. A rocket can go vertically upwards earths atmosphere because

a. Of gravitational pull of the sum b. It is highter than airc. 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

b. 11.2 ×

1

2km/ s

c. 11.2 × 2km/ s

d. 11.2 km/ s


XVII. ReferencesThis 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

Finn, C. B.P (1993). Thermal Physics, Chapman & Hall, London.

Raymond A. Serway (1992). PHYSICS for Scientists & Engineers. Updated Version.

Kleppner & Kolenkow An introduction to mechanics.

Douglas D. C. Giancoli Physics for scientists and engineers. Vol. 2. Prentice Hall.

Sears, Zemansky and Young, College Physics, 5th Ed.

Sena L.A. (1988) Collection of Questions and Problems in physics, Mir Publishers Moscow.

Nelkon & Parker (1995) Advanced Level Physics, 7th Ed, CBS Publishers & Di-tributer, 11, Daryaganji New Delhi (110002) India. ISBN 81-239-0400-2.

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.

Beiser A., (2004) Applied Physics, 4th ed., Tata McGraw-Hill edition, New Delhi, India

Halliday D., Resnick R., and Walker J. (1997), Fundamentals of Physics, 5th ed., John Wiley and Sons


Xviii.MainAuthorOfTheModule

About the author of this module

Dr. Tilahun Tesfaye

Department of physics, Addis Ababa University,

Ethiopia, East Africa.P.O.Box 80359 (personal), 1176 (Institutional)E-mail: [email protected]; [email protected]: +251-91-1418364 Fax : +251-11-1223931

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.


XIX. FileStructure

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 compi-led 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).

abc

Documents

23rd april 2007abstract

24th april 2007abstract

20th april 2007abstract

main theoretical elements

ang ular momentu

conservationofmomentumfrom html version

relevant mm resources

producing angular acceleration