PHYSICS 2
2020
Learner Guide
ONLINE
Faculty:
APPLIED AND COMPUTER
SCIENCES
Department:
NON - DESTRUCTIVE
TESTING AND PHYSICS
Subject:
PHYSICS 2
Subject Code:
APHYT2A / APPHT2A
Compiled and Edited By:
By:
Year & Semester:
F.P. NEMALILI
2020 – SEMESTER 2
TABLE OF CONTENTS
WORD OF WELCOME .................................................................................................................. iii
CONTACTS PERSONS ................................................................................................................ iii
SEMESTER PLANNING ............................................................................................................... iv
LEARNING MATERIAL ................................................................................................................. iv
TEACHING STRATEGY................................................................................................................. v
ASSESSMENTS ............................................................................................................................. v
HOW TO ACCESS PHYSICS COURSE ON VUTELA.................................................................. vi
ASSESSMENT 1 LEARNING UNITS
LEARNING UNIT 1 ......................................................................................................................... 1
LEARNING UNIT 2 ......................................................................................................................... 9
LEARNING UNIT 3 ....................................................................................................................... 15
LEARNING UNIT 4 ....................................................................................................................... 20
ASSESSMENT 2 LEARNING UNITS
LEARNING UNIT 5 ....................................................................................................................... 23
LEARNING UNIT 6 ....................................................................................................................... 27
LEARNING UNIT 7 ....................................................................................................................... 31
LEARNING UNIT 8 ....................................................................................................................... 34
LEARNING UNIT 9 ....................................................................................................................... 41
ASSESSMENT 3 LEARNING UNITS
LEARNING UNIT 10 ..................................................................................................................... 45
LEARNING UNIT 11 ..................................................................................................................... 48
LEARNING UNIT 12 ..................................................................................................................... 52
LEARNING UNIT 13 ..................................................................................................................... 60
LEARNING UNIT 14 ..................................................................................................................... 62
INFORMATION SHEET ................................................................................................................ 65
PERIODIC TABLE........................................................................................................................ 66
ii
WORD OF WELCOME
The Department of Non-Destructive Testing & Physics welcomes you as a learner to the Faculty of Applied and
Computer Sciences at the Vaal University of Technology.
The department strives towards integration of existing knowledge with new knowledge to afford learner the ability to:


Think logically and visualise concepts.
Gain knowledge of Physics in order to make a positive contribution in your chosen career once you have
completed your studies.
DEPARTMENTAL STAFF CONTACTS
TITLE AND SURNAME
OFFICE NUMBER
DR. I.Q. SIKAKANA
F112-B
MR. F.P. NEMALILI
F210-B
MR. G.S. NKOSI
F210-G
DR. V. NOLTING
E016-4
MR. M.L MBANDEZI
F210-E
MR. F. MMETHI
F210-F
MR. L.T. MELATO
F210-D
MISS. M.N RANTHO
F210-C
MISS. P. SITHOLE
F312-I
MR. L.N. MBULI
E016-2
MR. T. SITHOLE
E016-2
DR. L. LODYA
D 10 - SECUNDA
CAMPUS
Telephone number and
E- Mail address
016 - 950 9761
Ike@vut.ac.za
016 – 950 6655
patrickn@vut.ac.za
016 – 950 9814
sibusiso@vut.ac.za
016 – 950 9608
volkmarn@vut.ac.za
016 - 950 9824
mxolisi@vut.ac.za
016 - 950 7774
fikim@vut.ac.za
016 - 950 7858
lucasm3@vut.ac.za
016 - 950 9338
mologadir@vut.ac.za
016-950 7509
pulanem2@vut.ac.za
016 - 950 9322
nicholasm@vut.ac.za
016 - 950 9322
sithokozanemoses@gmail.com
017 - 631 1971
lonzechel@vut.ac.za
TECHNICIANS (PHYSICS LABORATORY ASSISTANCE)
MR. M.J. THEBE
C202-1
MRS. L. MALINGA
C201-2
MISS. K.H. MODISAKENG
C201-1
MISS. T.G. MALATSI
C201-1
016-950 9904
mohapit@vut.ac.za
016-950 7848
livhuhanis@vut.ac.za
016-950 9365
kgomotsom2@vut.ac.za
016-950 9365
tshegofatsom3@vut.ac.za
iii
2020 SEMESTER 2 PLANNING: PHYSICS 2 (APHYT2A/APPHT2A)
WEEK
1
2
3
4
6
7
8
9
11
12
13
14
DATE OF FIRST
DAY OF WEEK
LEARNING
UNIT
21/09/2021
28/09/2021
05/10/2021
05/10/2021
12/10/2021
1
2
3
4
19/10/2021
26/10/2021
26/10/2021
02/11/2021
09/11/2021
5
6&7
8
9
23/11/2021
30/11/2021
07/12/2021
14/11/2021
07/01/2021
10
11
12
13 & 14
TEXT BOOK SEMESTER TEST DATES
CHAPTERS (ONLINE)
CH. 20
CH. 21
CH. 22
CH. 23
TEST 1: 16 OCTOBER 2020
(TEST SCOPE: LU 1 – 4)
CH. 3
CH. 5 & 8
CH. 9
CH. 10
TEST 2: 13 NOVEMBER 2020
(TEST SCOPE: LU 5 – 9)
CH. 11
CH. 14
CH. 15
CH. 30 & 32
TEST 3: 07 DECEMBER 2021
(TEST SCOPE: LU 10 – 14)
ALL TEST WILL BE WRITTEN FROM 17:00 UNTIL 18:30
REASSESSMENT 1
13 JANUARY 2021
REASSESSMENT 2
14 JANUARY 2021
REASSESSMENT 3
15 JANUARY 2021
*Times and dates might change.
LEARNING MATERIAL
1.
Prescribed textbook
John D Cutnell and Kenneth W Johnson, Physics, 10th Edition, John Wiley & Sons, Inc.
All references in this study guide are given for the 10th Edition.
2.
Learner guide.
In this learner guide the following words are used extensively in the outcomes: Explain and Define.
The following meaning has been attributed to these words in this guide:
Explain: Clarify or give reasons for something in your own words. You must prove that you understand the
contents. It may be useful to use examples or illustrations where possible.
Define:
Give the mathematical formula of the law or concept if possible, the meaning of all variables must
be stated clearly. Otherwise, state a definition in words.
3.
Additional learning material
Additional material related to this module will be handed out as class notes and or tutorial questions in due
course.
4.
Website:
BLACKBOARD LMS:
https://vut.blackboard.com/
iv
PROBLEM SOLVING IN PHYSICS 2
Solving problems is a big stumbling block for many learners of physics. You only learn to solve problems by doing
them. Even if you think you “understand” how a problem is done, you must do other similar problems to see if you
really understand how to solve a problem.
“Word problems” can most effectively be solved by following the following steps:
Step 1: Review the problem to get an overview of the question. Don’t get bogged down by details. Don’t worry
about numbers of formulas that you must use. Read the question without analysing it in detail.
Step 2: Identify what it is you must calculate. Look for words such as “Find…” or “How many…” or “What is…”.
They let you know where you are heading.
Step 3: Look for the information provided in the question. Don’t worry about the numbers yet. Simply examine
the nature of the data. Tabulating the given data is useful so that it is not cluttered with words.
Step 4: Consider the kinds of calculations that must be performed on the data. Still don’t worry about the
numbers. You must now write the necessary formulas to do the calculation.
Step 5: You have compiled the necessary information and decided how the problem will be solved. Insert the
numbers into the formulas and solve.
Step 6: Look whether your answer is reasonable. Are the units correct? If so, you are finished.
CONSULTATION HOURS
The timetable and consultation hours will be pasted on the office door of the lecturer for this module; students are
thereby requested to adhere to this.
Consultation hours will only be used to clarify concepts in class or further assistance from tutorials and class exercises,
not for the presentation of a lecture.
ASSESSMENT (ONLINE)

Summative Assessment events
THREE (1½ Hour) summative assessment events, which take the form of ONLINE exams, will be conducted during
the duration of the module. The summative assessment events (exams) will represent a formal proof of the
student’s knowledge acquired over a particular time frame. To pass a summative assessment event, a minimum of
50% must be obtained. The summative assessment events will be weighed to contribute 100% of the final mark.
Physics 2: Theory (APHYT2A/APPHT2A)
Activity
Summative Assessment Event 1*
Summative Assessment Event 2*
Summative Assessment Event 3*
TOTAL (* - moderated)

Weight
33.33%
33.34%
33.33%
100%
LU
1-4
5-9
10-14
Dishonesty
Any dishonesty during the writing of a summative assessment event will be regarded as a serious offence and
can result in a nil mark. It is your own responsibility to ensure that your answer sheets are submitted at the end
of the assessment period. No late submission of answers will be accepted.

Re-assessment
Only the outcomes of assessment not achieved by the student will be re-assessed (1½ Hour) for each
summative event.
The Final mark will be an average of three Assessments, if the average is 50% or more then that student Pass
the module, but if the average is less than 50% then the student Fail the module. However, those who Fail the
assessments will be given a second chance to write re-assessments for the modules they failed to upgrade their
failed assessments results.
v
TIPS ON HOW TO ACCESS PHYSICS MATERIALS USING VUTELA
1. On you internet browser type: https://vut.blackboard.com and enter your login details (student
number and password).
2. Click on the Physics 2 to access, learning materials, past test and exam papers,
3. The tutorials for all learning units will be posted and marked on the blackboard, and the marks will be
used on your final year mark
4. All the announcement and changes during the course of the semester will be posted on the
blackboard, so visit your VETELA site all the time.
Figure 1: Opening Blackboard on Internet
Figure 2: Login menu
Please enter your student
number and your password.
User must put this website:
vut.blackboard.com in internet
explorer textbox.
Figure 3: Main Menu
User can click the link to
view Physics 2 Home page
vi
LEARNING UNIT 1
CHAPTER 20: ELECTRIC CIRCUITS
After completion of this learning unit you should be able to:
Explain Alternating Current
Define Alternating Voltage, Alternating Current
Define Power and Average Power.
Define rms Voltage and Current.
Define Kirchoff’s Loop Rule and Junction Rule.
Explain the different connections of capacitors.
Explain what an RC circuit is.
Explain what happens to the charge in an RC circuit when the capacitor is charging and discharging.
Define the charge on the capacitor plates when the capacitor charges and discharges.
Define the time constant.
Apply the reasoning strategy in solving Kirchoff problems.
Solve Alternating Current, Kirchoff, equivalent capacitance and RC circuit problems.
Study the following paragraphs in the textbook:
20.5
Alternating current ............................................................... p 611 (p 585)
20.10
Kirchoff’s Rules.................................................................... p 623 (p 596)
20.12
Capacitors in Series and Parallel ......................................... p 628 (p 601)
20.13
RC Circuits .......................................................................... p 630 (p 603)
Know the following concepts:








Sinusoidal voltage
RMS (root mean square) values
Peak values
Average Power
Junction
Closed Loop
Potential rise/drop
Capacitors in series/parallel
Use the following hints and information when solving problems:





When calculating the angle of the alternating voltage, the calculator should be set to Radian mode.
The rms values are not the same as peak values. RMS values are smaller by a factor of 2.
Outside a battery, conventional current is always directed from a higher potential (+) toward a lower
potential (-).
In Kirchoff calculations, it is always helpful to mark the resistors with plus and minus signs to keep track of
the potential rises and drops in the circuit.
All capacitors in series, regardless of their capacitances, contain charges of the same magnitude, +q
and –q, on their plates.
1|Page
Summary
Alternating current
More electric circuit uses battery and involved Direct Current (dc).However, there are more circuits that operate with
Alternating Current (ac). In an AC circuit, the charge flow reverses direction periodically. The ac generators serve the
same purpose as the battery serves in a dc circuit that is they give energy to the moving charges. Electric outlets in a
house provide ac current, we all use ac circuit regularly.
Alternating current and Voltage
Io
V  Vo sin(2ft )
I rms 
I  I o sin(2ft )
P  I rmsV rms
2
Vrms 
Vo
2
(Average power)
Application of Kirchhoff’s rules
1.
Name currents and choose direction of currents. The current can be in any direction and not necessary the
direction of the actual current in the circuit.
2.
Go along each current in each branch. The head of the resistor will be “+” and the tail of the resistor “ “.
3.
4.
5.
Outside a battery, conventional current is always directed from a higher potential (+) toward a lower
potential (-).
In Kirchoff calculations, it is always helpful to mark the resistors with plus and minus signs to keep track of the
potential rises and drops in the circuit.
Set up equations (junction rule and loop rule).
Junction rule
The sum of the magnitude of the currents going into the junction equals the sum of the magnitude of the currents
going out of the junction: ∑Iinto = ∑Iout.
2|Page
Loop rule
Around any closed circuit-loop, the sum of the potential drops equals the sum of the potential rises: ∑Vrise = ∑Vdrop.
Reasoning and Strategy
Concept Question and Answers for Kirchhoff’s Rule Application:
1. Draw the current in each branch of the circuit. Choose any direction.
Question: The currents through the three resistors are labeled I1, I2, and I3. Does it matter which direction has
been chosen for each current?
Answer: No, it doesn’t matter. If we initially select the wrong direction for a current, there is no problem. The value
obtained for that particular current will turn out to be negative number, indicating that the actual current is in the
opposite direction.
2. Mark each resistor with + (positive) at one end and – (negative) at the other end in a way that is consistent with
your choice for current direction in step 1. Outside a battery, conventional current is always directed from a higher
potential (the end marked +) to a lower potential (the end marked –).
Question: When place the + and  signs at the ends of each resistor, does it matter which end is + and which  ?
Answer: Yes, it does matter. Once the direction of the current has been selected, the + and  signs must be
chosen so that the current goes from + end towards the  end of the resistor. Notice that this is the case for
resistors in the figure.
3. Notice that in one of the three loops in the circuit does not have battery on it.
Question: Does it matter that there is no battery in the loo (i.e. loop 1), but only resistors?
Answer: No it doesn’t matter. A loop can have any number of batteries, including none at all.
4. When we evaluate the potential rises and drops around a closed loop,
Question: Does it matter which direction, clockwise and counter-clockwise, is chosen for the evaluation?
Answer: No it doesn’t matter, the direction is arbitrary. If we choose a clockwise direction, for example, we will
have a certain number of potential drops and rises. If we choose a counter-clockwise direction, all the potential
drops becomes potential rises, and vice versa.
5.
Apply the junction rule and the loop rule to the circuit, obtaining in the process as many independent equations as
there are unknown variables. Solve these equations simultaneously for the unknown variables.
Now use the circuit above to calculate the currents I1, I2, and I3.
3|Page
Capacitance
Capacitanc e (C ) 
Chaerge (q)
q
 C  , S.I. Unit is Farad (F)
Voltage (V )
V
Capacitors in series and parallel
A capacitor is a device for storing charge, made up of two parallel plates with a space between them. The plates have
an equal and opposite charge on them, creating a potential difference between the plates. A capacitor can be made of
conductors of any shape, but the parallel-plate capacitor is the most common kind. In circuit diagrams, a capacitor is
represented by two equal parallel lines.
Series
Parallel
1
1
1
1



CT C1 C 2 C 3
Ct = C1 + C2 + C3
VT  V1  V2  V3
VT  V1  V2  V3
qT  q1  q2  q3
qT  q1  q2  q3
Energy of a capacitor
E
1
1
1 q2
qV  CV 2 
2
2
2 C
Charging a Capacitor
Discharging a Capacitor
40
40
35
35
30
30
25
25
20
20
15
15
10
10
5
5
0
0
0
0.5
1
1.5
q  q0 (1  e
V  V0 (1  e
I  I 0e
t
t
t
2
2.5
RC )
RC
)
RC
3
3.5
4
0
0.5
q  q0 e
1
t
1.5
2
2.5
3
3.5
4
RC
𝑉 = 𝑉𝑜 𝑒 −𝑡/𝑅𝐶
I  I 0e
t
RC
Time constant of capacitor
Time taken for the capacitor to charge or discharge to 63.2% of its maximum charge.
  RC
4|Page
TUTORIAL QUESTIONS:
1.1
A light bulb is connected to a 120.0-V wall socket. The current in the bulb depends on the time t according to
the relation I = (0.707 A) sin [(314 Hz)t ].
a.
b.
c.
1.2
The figure shows variation of the current through the heating element with time in an iron when it is plugged
into a standard 120 V, 60 Hz outlet.
a.
b.
1.3
1.4.
What is the frequency of the alternating current? Ans.: 50 Hz
Determine the resistance of the bulb’s filament. Ans.: 240 Ω
What is the average power delivered to the light bulb? Ans.: 60 W
What is the peak voltage? Ans.: 169,71 V
What is the rms value of the current in this circuit? Ans.: 7,07 A
A portion of a circuit is shown, with the values of the currents given for some branches. What is the direction
and magnitude of the current I?
Find the currents I1, I2, and I3 in the circuit shown in Figure below. Ans.: 2 A, –3 A, –1A
5|Page
1.5.
A dead battery is charged by connecting it to the live battery of another car with jumper cables (Figure below).
Determine:
a.
the current in the starter, in the live battery and in the dead battery, Ans.: 0,85 A down the starter;
0,46 A down the dead battery; 1,31 A up the live battery.
b.
the power delivered to starter (8,00 Ω resistor). Ans.: 5,78 W
1.6.
Under steady state conditions (situation in which no current present in any branch of the circuit containing a
capacitor). use below circuit to calculate the following:
a.
the unknown currents in the multi loop circuit shown in the figure below. (Use the labels as given and
the currents as indicated in the figure) Ans.: 1,39 A; 0,37 A; 1,02 A
b.
the charge on the capacitor. (Hint: Use the loop rule). Ans.: 66 µC
1.7
Determine the voltage across the 0,06 Ω resistor in the drawing, and which end of the 0,06 Ω resistor is at the
higher potential (x or y)? Ans.: 10,26 V
6|Page
1.8
1.9.
Three resistors and two batteries are connected as shown in the circuit diagram. What are the magnitudes of
the currents I1, I2, I3? [Use loop ABCDEFA and CDEFC and the directions of the currents as indicated.]
Ans.: 0,14 A; 0, 53 A; 0,67 A.
In the following circuit diagram calculate the following:
a.
b.
c.
1.10
the resistance, R2
the current, I5
the resistance, R3.
Ans.: 6,5 Ω
Ans.: 15 A
Ans.: 3,5 Ω
For the circuit shown in the drawing, find the current I through the 2,00 Ω resistor and the voltage V of the
battery to the left of this resistor. Ans.: 5,00 A; 46,0 V.
1.11. Find the equivalent capacitance between points A and B for the group of capacitors connected as shown in
Figure below. Take C1 = 5,00 µF, C2 = 10,0 µF, and C3 = 2,00 µF. Ans.: 6,05 µF
7|Page
1.12. For the system of capacitors shown in Figure below, find the following:
a.
b.
c.
d.
the equivalent capacitance of the system,
the charge on each capacitor,
the potential across each capacitor,
the total energy stored by the group.
Ans.: 3,33 µF
Ans.: 120 µC; 120 µC; 180 µC; 180 µC
Ans.: 60 V; 30 V; 60V; 30 V
Ans.: 0,0135 J
1.13. Four identical capacitors are connected with a resistor in two different ways. When they are connected as in
part (a) of the drawing, the time constant to charge up this circuit is 0,72 s. What is the time constant when
they are connected with the same resistor as in part (b)? Ans.: 0,29 s
C
C
C
R
R
C
+
(a)

C
C
C
C
+

(b)
1.14. An RC circuit consists of a resistor, R with resistance 1,0 k, a 120 V battery, and two capacitors, C1 and C2,
with capacitances of 20 µF and 60 µF, respectively. Initially, the capacitors are uncharged; and the switch is
closed at t = 0 s.
a.
b.
c.
d.
What is the current through the resistor a long time after the switch is closed? Explain your reasoning.
Ans.: 0A; Capacitors are fully charged
What is the time constant of the circuit? Ans.: 0,08 s
How much charge will be stored in each capacitor after a long time has elapsed? Ans.: 2,4 mC; 7,2 mC
Determine the total charge on both capacitors two time constants after the switch is closed.
Ans.: 8,3 mC (2,1 mC; 6,2 mC)
1.15 The figure shows a simple RC circuit consisting of a 100,0 V battery in series with a 10,0 µF capacitor and a
resistor. Initially, the switch S is open and the capacitor is uncharged. Two seconds after the switch is closed,
the voltage across the resistor is 37 V.
a.
b.
c.
What is the voltage across the capacitor after 2 s? 63 V
Determine the numerical value of the resistance R. Ans.: 2,01 x 105 
How much charge is on each plate of the capacitor 2,0 s after the switch is closed? Ans.: 6,3 x 104 C
8|Page
LEARNING UNIT 2
CHAPTER 21: MAGNETIC FORCES AND MAGNETIC FIELDS
After completion of this learning unit you should be able to:
Explain what a magnetic field is.
Explain how a magnetic field exert a force on a charge.
Define the magnetic field.
Explain the motion of a charged particle in a magnetic field.
Define the radius of the circular trajectory motion.
Explain the working of the mass spectrometer.
Explain the force acting on a current carrying wire in a magnetic field.
Explain the torque on a current carrying coil.
Explain the working of a dc electric motor.
Explain how currents produce magnetic fields.
Define the magnetic field produced by a:
 long straight wire
 loop of wire
 solenoid
Explain Ampere’s Law.
Solve Magnetic Force and Magnetic Field problems.
Study the following paragraphs in the textbook:
21.1
Magnetic Fields ................................................................... …. p 647 (p 621)
21.2
The Force that a Magnetic Field exerts on a Moving Charge ..... p 649 (p 622)
21.3
The Motion of a Charged Particle in a Magnetic Field ................ p 652 (p 626)
21.4
The Mass Spectrometer ............................................................ p 656 (p 630)
21.5
The Force on a Current in a Magnetic Field............................... p 656 (p 630)
21.6
The Torque on a Current-Carrying Coil ..................................... p 660 (p 634)
21.7
Magnetic Fields Produced by Currents...................................... P 662 (p 636)
21.8
Ampere’s Law ........................................................................... p 670 (p 642)
Use the following hints and information when solving problems:







The direction of the magnetic field at any point in space is the direction indicated by the north pole of a small
compass needle placed at that point.
A charged particle will only experience a magnetic force if the charge is moving and it has a velocity
component perpendicular to the magnetic field.
The magnetic force, velocity of the charge and the magnetic field are always perpendicular with each other.
If the charge is negative, then the magnetic force acting on the charge is opposite to the direction of the force
acting on a positively charged particle.
When a current carrying loop is placed in a magnetic field, the loop tends to rotate such that its normal
becomes aligned with the magnetic field.
Do not confuse the formula for the magnetic field produced at the centre of a circular loop with that of a very
long straight wire. These equations are similar, but they are not the same.
NB: When describing direction – use words like:
Out of the page
Into the page
To the top of the page
To the bottom of the page
To the left, Right of the page.
DO NOT USE UP or DOWN – UP OR DOWN DOES NOT REALLY SAY WHETHER YOU MEAN
9|Page
Summary
There is a single force—the electromagnetic force—that governs the behavior of both magnets and electric charges.
The Earth’s Magnetic Field
The Earth itself acts like a huge bar magnet. The presence of a magnetic field around the Earth allows us to use
compasses that point northward, and creates a spectacular aurora over the northern and southern skies.
Definition of the Magnetic Field
The magnitude of the magnetic field at any point in space is defined as
B
F
qo v sin  
where the angle (0<θ<180o) is the angle between the velocity of the charge and the direction of the magnetic field.
SI Unit of Magnetic Field: Tesla
Magnetic Force on Charges
Magnetic field strength is measured in teslas (T).
Force on a positive charge in a magnetic field,  - angle between B and v F  qvB sin 
Force on conventional current magnetic field,  - angle between B and L F  BIL sin 
Right Hand Rule Number. 1
Extend the right hand so the fingers point along the direction of the
magnetic field and the thumb points along the velocity of the charge.
The palm of the hand then faces in the direction of the magnetic force
that acts on a positive charge.
If the moving charge is negative, the direction of the force is opposite to that
predicted by RHR-1.
Exercise1:
Complete the following table (from (a) to (d)) for charged particles in a magnetic field:
Magnetic field
Magnetic force
Charge
Velocity of Charged Particle
Into the page
To the bottom of the page
(a)
Negative
(b)
To the right of the page
To the top of the page
Positive
To the left of the page
(c)
Out of the page
Positive
To the top of the page
To the left of the page
(d)
Into the page
10 | P a g e
As long as the particle’s velocity vector is perpendicular to the magnetic field lines, the force vector will be perpendicular
to both the velocity vector and the magnetic field. As we saw in the chapter on circular motion and gravitation, a force
that always acts perpendicular to the velocity of an object causes that object to move in circular motion with radius r.
2
v
r
Fc  m
v2
qvB  m
r
r
mv
qB
(radius of charge in constant B)
The mass spectrometer
Used for identify unknown molecules, determining the relative masses and
abundance of isotopes.
Atoms or molecules are vaporized and then ionized by the ion source by removing
an electron form the particle giving it a charge of +e.
The ions are the accelerated by through potential difference V.
The ions with speed v, are then deflected into a semi-circle path by a constant
magnetic field B.
Only specific ions will strike the detector. The mass of the ions that do strike the
detector can be determined by the following equation.
qr 2 2
m
B
2V
Exercise 2: Two charged particles are traveling in circular orbits with the same speed in a region of uniform
magnetic field that is directed into the page, as shown. The magnitude of the charge on each particle is
identical, but the signs of the charges are opposite. Which one of the entries in the table below is correct?
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
R1 x
x
x
x
x
x
x
x
x
x
x
x
x
x
A.
B.
C.
D.
E.
Q2
R2
Q1
Mass Relationship
m1 = m2
m1 < m2
m1 > m2
m1 > m2
m1 < m2
Sign of charge Q1
–
–
–
+
+
Sign of charge Q2
+
+
+
–
–
11 | P a g e
Torque on current-carrying coil
magnetic
moment

  NIA B sin 
Number of turns of wire
(Torque on N loops with area A in magnetic field B,
 - angle between perpendicular to area (normal) and B.
Right-Hand Rule No. 2. Curl the fingers of the right hand into the shape of a half-circle. Point
the thumb in the direction of the conventional current, and the tips of the fingers will point in the
direction of the magnetic field.
Magnetic fields produced by currents
Straight conductor
B
Loop of wires
0 I
2r
BN
Solenoid
0 I
B  0nI n 
2R
Force between 2 current carrying wires
N
L
F  BIL sin 
F
0 I
IL sin 
2r
Ampere's Law:
Ampere's Law states that for any closed loop path, the sum of the length elements times the
magnetic field in the direction of the length element is equal to the permeability times the electric
current enclosed in the loop.
 B    I
||
o
B      o I
B2 r  o I
B
o I
2 r
12 | P a g e
TUTORIAL QUESTIONS:
2.1
Two long, straight, parallel wires separated by a distance d carry currents as shown in the figure. The right wire
carries a current of 6 A. Point C is at the midpoint between the wires and point O is a distance 0.5d from the
6A wire as suggested in the figure. The total magnetic field at point O is zero tesla.
a.
b.
c.
2.2
What must the direction of the unknown current in wire I must be?
Determine the value of the current, I, in the left wire. Ans.: 18 A
Determine the magnitude and direction of the magnetic field at point C if d = 0.10 m. Ans.: 9.6 x 10-5 T
The figure shows two concentric metal loops, each carrying a current. The larger loop carries a current of
8,0 A and has a radius of 0,06 m. The smaller loop has a radius of 0,04 m. What is the value of a current in
the smaller loop that will result in zero total magnetic field at the centre of the system? Ans.: 5,33 A
8,0A
0,04m
0,06m
2.3
A mass spectrometer is used to separate two isotopes of uranium with masses m1 and m2 where m2 > m1. Two
types of uranium atom exit an ion source S with the same charge of +e and are accelerated through a potential
difference V. The charged atoms then enter a constant, uniform magnetic field B as shown. If r1 = 0,5049 m
and r2 = 0,5081 m, what is the value of the ratio m1/m2? Ans.: 0.9874
B
r1 r2
V
+
S
2.4
A single circular loop of wire with radius 0,020 m carries a current of 8,0 A. It is placed at the centre of a solenoid
that has length 0,65 m, and 1400 turns, as in the diagram below. Calculate the value of the current in the solenoid
so that the magnetic field at the centre of the loop is zero tesla. Ans.: 9,29 x 10-2 A
2.5
The radius of a coil of wire with N turns is r = 0,22 m. A current Icoil = 2 A flows clockwise in the coil, as shown.
A long, straight wire carrying a current Iwire = 31 A toward the left is located 0,05 m from the edge of the coil.
The magnetic field at the center of the coil is zero tesla. Determine N, the number of turns. Ans.: 4.02 turns
13 | P a g e
2.6
The drawing shows an end-on view of three wires. They are long, straight and perpendicular to the plane of the
paper. Their cross sections lie at the corners of a square. The currents in wires 1 and 2 are equal (I1 = I2), and
are directed into the paper. The net magnetic field at the empty corner (top right) is zero.
a.
b.
What is the direction of the current in wire 3? Ans.: Out of page
What are the ratio I3/I, such that? Ans.:2
2.7
A rectangular loop has sides of length 0,06 m and 0,08 m. The wire carries a current of 10 A in the direction
shown. The loop is in a uniform magnetic field of magnitude 0,2 T and directed in the positive x direction. What
is the magnitude of the torque on the loop? Ans.:8,31 x 103 N.m
2.8
Four long, straight wires are parallel to each other; and their cross-section forms a square. Each side of the
square is 0,02 m as shown in the figure. If each wire carries a current of 8,0 A in the direction shown in the
figure, determine the magnitude of the total magnetic field at P, the center of the square. Ans.: 2,26 x 104 T
2.9
A long, straight wire carries a 10,0 A current in the +y direction as shown in the figure.
Next to the wire is a square copper loop that carries a 2,0 A current as shown. The length of each side of the
square is 1,0 m.
a.
b.
2.10
What is the magnitude of the net magnetic force that acts on the loop? Ans.: 1.67 x 105 N
What is the direction of the net magnetic force that acts on the loop? Ans.: Towards the wire
A circular coil consists of 5 loops, each of diameter 1,0 m. The coil is placed in an external magnetic field of
0,5 T. When the coil carries a current of 4,0 A, a torque of magnitude 3,93 N.m acts on it. Determine the angle
between the normal to the plane of the coil and the direction of the magnetic field. Ans.: 30,02°
14 | P a g e
LEARNING UNIT 3
CHAPTER 22: ELECTROMAGNETIC INDUCTION
After completion of this learning unit you should be able to:
Explain what electromagnetic induction is.
Explain what Induced EMF and Induced current is.
Explain the 3 ways how induced current can be produced.
Explain/Define what motional EMF is.
Explain what magnetic flux is.
Explain/Define Lenz’s Law.
Explain the working of an electric generator.
Explain what counter torque and Back EMF is.
Explain what a transformer is.
Solve Motional EMF, Faraday’s Law, electric generator, Back EMF and transformer problems.
Study the following paragraphs in the textbook:
22.1
Induced EMF and Induced Current .................................... p 686 (p 660)
22.2
Motional EMF ..................................................................... p 688 (p 661)
22.3
Magnetic Flux ..................................................................... p 692 (p 664)
22.4
Faraday’s Law of Electromagnetic Induction ....................... p 695 (p 667)
22.5
Lenz’s Law ......................................................................... p 698 (p 670)
22.7
The Electric Generator ........................................................ p 702 (p 674)
22.9
Transformers ...................................................................... p 711 (p 681)
Know the following concepts:














Induced current/EMF
Changing magnetic field
Electromagnetic induction
Motional EMF
Magnetic flux
Faraday’s Law of electromagnetic induction
Induced magnetic field
Lenz’s Law
Counter torque
Back EMF
Two sources of emf for an electric motor.
Transformer equation
Step up/Step down transformer
Power of transformer
Use the following hints and information when solving problems:






The direction of the current induced is such that a magnetic force acts on the rod to oppose its motion,
thereby slowing it down.
The minus in Faraday’s Law is a reminder that the polarity of the induced emf sends the induced current in
the proper direction so as to give rise to the opposing magnetic force.
The magnetic flux is determined by the magnitude of B, A as well as the angle of the normal (of the coil) to
the magnetic field:   BA cos
In the equation E0 = NAB, the angular frequency  must be in rad.s-1.
The current in an electric motor depends on both the applied EMF, V, and any back emf, E, developed
because the coil of the motor is rotating.
A transformer that steps up the voltage simultaneously steps down the current and vice versa.
15 | P a g e
Summary
Faraday’s law for electromagnetic induction
  N
  N
A cos ( B2  B1 )
t

( BA cos )
 N
t
t
  N
  N
BA(cos 2  cos 1 )
t
B cos ( A2  A1 )
t
Lenz’s law
The induced emf resulting from a changing magnetic flux () has a polarity that leads to an induced current whose
direction is such that the induced magnetic field opposes the original flux change.
1. Direction of Bexisting (NS or RHR2)
2. decrease
or
increase
3. Binduced same direction as Bexisting or opposite direction
4. RHR2 to determine the direction of induced current.
Motional emf

t
 ( BA cos  )
  N
t
B (l  s )
B A
 

t
t
Bls
  
 vBL
t
  N
Where does the electrical energy comes from?
Even if it is a frictionless track on which the bar moves, the hand must still move it because of the opposing magnetic
force. Lenz’s law: Existing magnetic field into the book, flux increase, induced magnetic field out of the book, induced
current CC. RHR1: Current to top of bar (top of page), existing magnetic field into book, F magnetic to the left.
16 | P a g e
Electric motor (21.6)
Electric generator (22.7)
TRANSFORMERS
TRANSFORMERS EQUATIONS
I s Vp N p


I p Vs N s
Electrical energy
converted to
mechanical energy
  NIAB sin 
I
V 
(Ohm’s law)
R
Mechanical energy converted to
electrical energy
  NAB sin t
  2ft
A transformer that steps up the voltage simultaneously
steps down the current, and a transformer that steps
down the voltage steps up the current.
Average Power P, delivered to the Primary is always
the same as that delivered to the Secondary.
P  I sVs  I pVp
Step-Up and Step-Down Transformers
17 | P a g e
TUTORIAL QUESTIONS:
3.1
The area of a 333-turn conducting coil is 7,85 x 103 m2. The resistance of the coil is 10,4 . If the coil is
oriented as shown in a magnetic field B, at what rate in T/s should the magnitude of B change to induce a
current of 2,50 x 10-3 A in the coil? Ans.: 0.0155 T/s
3.2
A circular coil of 100 turns has a radius of 0,115 m. The flux density through the coil is changing at the rate of
0,14 T/s. The coil is connected in series to a 100  resistor.
a.
What is the magnitude of the induced emf in the circuit? Ans.: 0,58 V
b.
What current flows in this circuit? (The resistance of the coil is negligible) Ans.: 5,8 x 103 A
3.3
An emf is induced in a conducting loop of wire 1.12 m long and resistance 3 , as its shape is changed from a
square to a circle. The change in shape occurs in 4.25 s and the local 0.105 T magnetic field is perpendicular
to the plane of the loop. What is the average magnitude of the induced emf? Ans.: 0.000523 V
3.4
A circular coil of wire has 25 turns and has a radius of 0,075 m. The coil is located in a variable magnetic field
whose behavior is shown on the graph. At all times, the magnetic field is directed at an angle of 75° relative to
the normal to the plane of a loop. What is the average emf induced in the coil in the time interval from t = 5,00
s to 7,50 s? Ans.: 0.018 V
3.5
A uniform magnetic field passes through two areas, A1 and A2. The angles between the magnetic field and the
normals of areas A1and A2 are 30,0° and 60,0°, respectively. If the magnetic flux through the two areas is the
same, what is the ratio A1/A2? Ans.: 0,58
3.6
A flexible, circular conducting loop of radius 0,15 m and resistance 4,0  lies in a uniform magnetic field of
0,25 T. The loop is pulled on opposite sides by equal forces and stretched until its enclosed area is essentially
zero m2, as suggested in the drawings. It takes 0,30 s to close the loop.
a.
b.
c.
Determine the magnitude of the emf induced in the loop. Ans.: 5,89 x 102 V
At what rate is heat generated in the loop? Ans.: 8.67 x 104 W
What is the direction of the induced current and induced magnetic field while the loop is being stretched?
Ans.: B-Into page, I-Clockwise
18 | P a g e
3.7
A permanent magnet is moved away from a loop of wire as indicated in the diagram. The external circuit attached
to the loop consists of a resistance R.
a.
b.
c.
d.
3.8
What is the direction of the magnetic field lines through the loop due to the magnet? Ans.: To the right
What is the direction of the induced magnetic field lines (through the loop of wire)? Ans.: To the right
In which direction does the induced current flow through the resistor R, (A to B, or B to A)? Ans.: B to A
Which, A or B, is the positive of the induced emf? Ans.: B
A circuit is pulled with a 16 N force toward the right to maintain a constant speed v. At the instant shown, the
loop is partially in and partially out of a uniform magnetic field that is directed into the paper. As the circuit
moves, a 6.0 A current flows through a 4.0 Ω resistor.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
R
3.9
x
v
a. In which direction is the induced current flows around the circuit?
b. Calculate the magnitude of speed v? Ans: 9.0 m/s
In two situations below, a metal ring is dropped from rest below a bar magnet that is fixed in position as
suggested in the figure A and B. In figure A an observer views the ring from above, while in figure B an observer
views the ring from below. In each of the situations below, determine which direction does the induced current
flow through the loop of the metal ring?
Figure A
Figure A
19 | P a g e
3.10
A five-sided object, whose dimensions are show in the drawing, is placed in a uniform magnetic field. The
magnetic field has a magnitude of 0.25 T and points along the positive y –direction. Calculate the flux through
each of the five sides. Ans: 0 Wb, 0 Wb, 0 Wb, 0.090 Wb, 0.090Wb.
3.11
A power plant produces a voltage of 6.0 kV and 150 A. The voltage is stepped up to 240 kV by a transformer
before it is transmitted to a substation. The resistance of the transmission line between the power plant and the
substation is 75 . What is the current in the transmission line from the plant to substation? Ans: 3,75 A
3.12
A transformer is used to increase the line voltage of 120 V to the 24 000 V accelerating potential required by a
colour television picture tube. The primary coil has 100 turns. How many turns does the secondary coil have?
Ans.:20 000 turns
3.13
A transformer has 10 turns and 100 turns, respectively, in its primary and secondary coils.
a.
When 5,0 A flows in the secondary circuit, what minimum current must exist in the primary coil?
Ans.: 50 A
b.
Is this a step up or a step down transformer? Give a reason for your answer. Ans.: Step up; Ns > Np
3.14 The coil of an ac motor has a resistance of 4.1 ohms. The motor is plugged into an outlet where the voltage is
120.0 volts (rms), and the coil develops a back emf of 118.0 volts (rms) when rotating at normal speed. The
motor is turning a wheel. Find (a) the current when the motor first starts up and (b) the current when the motor
is operating at normal speed. Ans: 29 A, 0.49 A
3.15 The angular speed of a motor is 262 rad/s. The back emf generated by the motor is 89.4 V. Assuming all other
factors remain the same, determine the back emf if the angular speed of the motor is reduced to 154 rad/s.
Ans: 52.4 A
3.16 A motor is designed to operate on 117 V and draws a current of 12.2 A when it first starts up. At its normal
operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back
emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
3.17
Ans: 9.59 A, 95 V, 8.90 A
20 | P a g e
LEARNING UNIT 4
CHAPTER 23: ALTERNATING CURRENT CIRCUITS
After completion of this learning unit you should be able to:
Explain what capacitive reactance is.
Define capacitive reactance.
Explain what inductive reactance is.
Define inductive reactance.
Explain what an LCR circuit is.
Explain what impedance is.
Define impedance.
Define phase angle.
Define power factor.
Solve LCR circuit problems.
Study the following paragraphs in the textbook:
23.1
Capacitors and Capacitive Reactance.................................. p 726 (p 697)
23.2
Inductors and Inductive Reactance ...................................... p 729 (p 699)
23.3
Circuits Containing Resistance, Capacitance, and Inductance...... (p 701)
23.4
Resonance in Electric Circuits ..................................................... (p 736)
Know the following concepts:









Capacitive reactance
Capacitive reactance against frequency graph
Phasor diagrams
Inductive reactance
Inductive reactance against frequency graph
LCR circuit
RMS voltage
Impedance
Power factor
Use the following hints and information when solving problems:





The average power used by a capacitor in an ac circuit is zero.
The average power used by an inductor in an ac circuit is zero.
On average, only the resistor uses power in an LCR circuit.
In an LCR circuit, the rms voltages across the resistor, capacitor, and inductor do not add up to equal the
rms voltage across the power supply.
The impedance against frequency graph, is a combination of the capacitive reactance and inductive
reactance graphs.
Summary
21 | P a g e
Resistor
Resistance constant
Capacitor
Inductor
X L  2fL
1
Xc 
2fC
Inductive reactance indirectly
proportional to frequency
Inductive reactance indirectly
proportional to frequency
Current leads the voltage by 90o
Energy stored in electric field
Current lags the voltage by 90o
Energy stored in magnetic field
(Φ=-90o)
(Φ=90o)
In alternating current circuit:
Current and voltage are in phase
Energy conversion
LRC Phasor Diagrams
These Phasors rotate anticlockwise at a frequency f.
In a series RLC circuit, the total opposition to the flow is called
the impedance.
tan  
Vrms  I rms Z
VL  VC X L  X C

VR
R
 = 0,  XL = XC,
 > 0,  XL > XC,
Impedance
 < 0,  XL < XC
Resistors in series
RT = R1 + R2 + R3
RCL circuit
VT = V1 + V2 + V3
VT  V R  (V X L  V X C ) 2
IT = I1 = I2 = I3
IT = I1 = I2 = I3
Z  R2  ( X L  X C )2
2
In an inductor and a capacitor energy is stored and then released. Therefore, on average, the power is zero. It uses
or convert no energy in a circuit. The only energy conversion takes place in the resistor.
P  IVresistor  IVT cos . Where cos is called the power factor.
22 | P a g e
TUTORIAL QUESTIONS:
4.1
An ac generator is connected across the terminals of a 3,25 µF capacitor. Determine the frequency at which
the capacitive reactance is 495 . Ans.: 98,93 Hz
4.2
A series RCL circuit operating at 60.0 Hz contains a 35  resistor and an 8.2 µF capacitor. If the power factor
of the circuit is +1.00, what is the inductance of the inductor in this circuit? Ans.: 0,86 H
4.3
The following table gives the reactance and rms voltage across the elements of a series RCL circuit:
Circuit element
Reactance Voltage across element
resistor
200 
86 V
capacitor
663 
285 V
inductor
377 
162 V
a.
b.
c.
d.
e.
What is the rms current in the circuit? Ans.:0,43 A
What is the impedance of the circuit? Ans.: 348,99 
Determine the peak (not rms) voltage of the ac generator. Ans.: 212,22 V
What is the power factor for this circuit? Ans.:0,57
What is the average power consumed by the circuit?
Ans.:36,78 W
4.4
A capacitor of 2,65 F is connected to a 110 V 60 Hz power supply.
a.
What is the capacitive reactance? Ans.: 1 000,97 
b.
When the capacitor is placed in the circuit, with a 175  resistor, what is the impedance of the circuit?
Ans.: 1 016,15 
c.
What effective current flows? Ans.: 0,11 A
d.
What is the phase angle? Ans.: -80,08°
e.
Does the voltage lags the current or does the current lags the voltage? Ans.: Voltage lags current
4.5
An ac generator supplies a rms (not peak) voltage of 180 V at 60 Hz. It is connected in series with a 0,5 H
inductor, a 6,0 µF capacitor and an 300  resistor.
a.
Determine the peak voltage of the generator. Ans.: 254,56 V
b.
What is the impedance of the circuit? Ans.: 392,83 
c.
What is the peak (not rms) current through the resistor? Ans.: 0,65 A
d.
What is the phase angle for this circuit? Ans.: -40,21°
e.
What is the average power supplied to the circuit? Ans.: 63,18 W
4.6
A series RCL circuit operating at 55.0 Hz contains a 35Ω resistor and an 820 mH inductor. If the power factor
of the circuit is + 0.707, determine the capacitance of the capacitor in this circuit? Ans.: 12.67 F
4.7
In a series RLC circuit, the applied voltage has a maximum value of 120 V and oscillates with a frequency of
60 Hz. The circuit contains an inductor whose inductance can be varied, R = 800 , and C = 4 F.
Determine the value of inductance, L such that the voltage across the inductor is out of phase with the applied
voltage by 300, with Vmax leading VR, as in the phasor diagram below. Ans.: 5,43 H
4.8
The graph shows the voltage across and the current through a single circuit element connected to an ac
generator.
a.
b.
c.
d.
Determine the frequency of the generator. Ans.: 12.5 Hz
Determine the rms voltage across this element. Ans.: 70.004 V
Determine the rms current through this element Ans.: 1.98 A
Identify the circuit element. Explain your reasoning.
23 | P a g e
LEARNING UNIT 5
CHAPTER 3: KINEMATICS IN TWO DIMENSIONS
After completion of this learning unit you should be able to:
Define displacement, velocity and acceleration in two dimensions.
Explain Projectile Motion.
Apply the reasoning strategies in solving problems of Kinematics in Two Dimensions
Solve problems of Kinematics in Two Dimensions
Study the following paragraphs in the textbook:
3.1
Displacement, Velocity and Acceleration ................................... p59 (p 56)
3.2
Equations of Kinematics in Two Dimensions ............................ p 60 (p 57)
3.3
Projectile Motion ...................................................................... p 65 (p 60)
Know the following concepts:


Displacement, Velocity and Acceleration
Projectile motion, both X-motion and Y-motion
Use the following hints and information when solving problems:

In two-dimensional kinematics, it is important to realise that the x part of the motion occurs exactly as it
would if the y part did not occur at all, and vice versa.

The velocity of a projectile at any location along its path is: v 

(because velocity is a vector), Both the horizontal and vertical components contribute to the speed.
When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero. The
horizontal component of its velocity is not zero.
v x2  v 2y , and direction tan  
vy
vx
Summary
The motion of any object moving in two dimensions can be broken into x- and y-components. The most common
problems of this kind involve projectile motion: the motion of an object that is shot, thrown, or in some other way
launched into the air. The type of problems you will get can be divided into 4 types. Read the problem and decide the
type first by looking at the path of the object. The velocity in the x direction is ALWAYS constant (ignore friction), and
the velocity in the y-direction is free falling motion.
24 | P a g e
Under the influence of gravity alone, an object near the surface of the Earth will accelerate
2
downwards at 9.80m/s .
a y  9.80 m s 2
Type 1
&
vx  vox  constant
ax  0
Type 2
x-component
ax = 0
y-component
ay=9.8 m/s2
y=0
tx = ty = t
Type 3
x-component
ax = 0
y-component
ay=9.8 m/s2
vy=0
tx = ty = t
Type 4
?
x-component
ax = 0
y-component
ay=9.8 m/s2
voy=0
tx = ty = t
Not one of the first 3.
x-component
ax = 0
y-component
ay=9.8 m/s2
tx = ty = t
Equations of Kinematics In Two Dimension
x – components
y – components
v x  vo  a x t
v y  voy  a yt
v x  vo2  2a x x
v y2  voy2  2a y y
x  vo t  12 a x t 2
y  voyt  12 a y t 2
vo  vx  t
y  12 voy  v y  t
2
x
1
2
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and negative ( ).
3. Write down the values that are given for any of the five kinematic variables associated with
each direction.
4. Verify that the information contains values for at least three of the kinematic variables. Do
this for x and y. Select the appropriate equation.
5. When the motion is divided into segments, remember that the final velocity of one segment is
the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a kinematics problem.
25 | P a g e
TUTORIAL QUESTIONS:
5.1
This time, William Tell is shooting at an apple that hangs on a tree (see the figure below). The apple is a
horizontal distance of 20.0m away and at a height of 4.00m above the ground, if the arrow is released from a
height of 1.00m above the ground and hits the apple 0.700s later, what is the arrow’s initial velocity.
Ans.: 29.59 m/s, 15.12o N of E
5.2.
A package of supplies is to be dropped from an airplane so that it hits the ground at a designated spot near
some campers. The airplane, moving horizontally at a constant velocity of 140 km/h, approaches the spot at an
altitude of 0.500 km above level ground. Having the designated point in sight, the pilot prepares to drop the
package.
a.
b.
5.3.
Calculate the horizontal distance between point of release and the point of landing? Ans: 392.69 m
What should the angle be between the horizontal and the pilot’s line of sight when the package is
released? Ans.: 51.85o below horizontal
A stone is thrown upward from the top of a building at an angle of 30 o to the horizontal and with an initial speed
of 30 m/s, as shown in figure below. If the height of the building is 45 m, determine:
a.
b.
c.
d.
how long is the stone “in flight”?
Ans.: 4,93 s
what is the speed of the stone just before it strikes the ground? Ans.: 42,25 m/s
where does the stone strikes the ground (find x)?
Ans.: 128,09 m
what is the total displacement of the stone after it strikes the ground? Ans.: 135,76 m;19,36°below +x
5.4
A soccer player kicks a stationary ball, giving it a speed of 15.0 m/s at an angle of 15.0º to the horizontal.
a.
What is the maximum height reached by the ball? Ans.: 0.77 m
b.
What is the ball’s range? Ans.: 11.45 m
c.
How could the range be increased?
5.5.
A basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. The ball enters the
basket 0.96 s after it is launched. Calculate the magnitude and direction of the total displacement travelled by
the basketball until it enters the basket? Note: The drawing is not to scale. Ans.: 5.48 m, 9.57o N of E
v0
y
45°
x
26 | P a g e
5.6.
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell
strikes the ground 1330 m from the base of the cliff as sown in the figure below.
a.
b.
c.
Determine the initial speed of the shell. Ans.: 329,21 m/s
What is the speed of the shell as it hits the ground? Ans.: 331,58 m/s
What is the magnitude of the acceleration of the shell just before it strikes the ground?
Ans.: 9,8 m/s2 downwards
5.7.
An arrow has an initial launch speed of 18 m/s. If it must strike a target at 31 m away at the same elevation,
what should be the project angle? (Remember that 2sinθcosθ = sin2θ) Ans.: 34.83o
5.8.
A rugby player kicks a ball in projectile motion from ground level at an angle of θ = 12.0° above the horizontal.
It returns to ground level. To what value should the launch angle be adjusted, without changing the launch
speed, so that the range doubles? Ans.: 27.2o
5.9.
A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed,
how much height will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
Ans.: 0.844 m
5.10. The figure below shows a boy throwing a stone at an initial speed of 18 m/s at an angle of 57o with horizontal.
The rock lands on the roof of a building which is 15 m far away from the boy in the x-direction as shown in the
figure below (ignore air-resistance).
Determine
a. How long the rock is in the air [time of flight]. Ans.: 1.53 s
b. The height (H) of the building. Ans.: 13.13 m
27 | P a g e
LEARNING UNIT 6
CHAPTER 5: DYNAMICS OF UNIFORM CIRCULAR MOTION
After completion of this learning unit you should be able to:
Define Uniform Circular Motion.
Explain and Define Centripetal Acceleration.
Explain and Define what Centripetal Force is.
Study the following paragraphs in the textbook:
5.1
Uniform Circular Motion ....................................................... p 135 (p 125)
5.2
Centripetal Acceleration ....................................................... p 136 (p 125)
5.3
Centripetal Force ................................................................. p 139 (p 128)
5.4
Banked Curves…………………………………………………..p 143
5.7
Vertical Circular Motion………………………………………….p 151
Know the following concepts:


Period, Speed
Centripetal acceleration, Centripetal Force
Use the following hints and information when solving problems:

Centripetal acceleration and Centripetal force are always directed to the centre of the circle.
Summary
Uniform circular motion occurs when a body moves in a circular path with constant speed.
v
2 r
T
Centripetal acceleration
An object moves in a circle with constant speed. Show the direction of the centripetal acceleration in each of the five
points of the diagram. “Centripetal comes from a Latin word meaning “center-seeking.”
v2
ac 
r
Total acceleration
An object, can have a tangential acceleration (not moving with constant speed but accelerate in a circle).
To calculate the magnitude and direction of the total acceleration:
  arctan
a total  a c 2  a tan gental 2
atan

atotal
ac
a vertical
a horizontal
An object accelerates clockwise around a circle. Show the
direction of the centripetal acceleration, tangential
acceleration and total acceleration in each of the five points
of the diagram. Also show how you would calculate  in
each of the cases.
28 | P a g e
Centripetal Force
Wherever you find acceleration, you will also find force. For a body to experience centripetal acceleration, a
centripetal force must be applied to it. The vector for this force is similar to the acceleration vector: it is of constant
magnitude, and always points radially inward to the center of the circle, perpendicular to the velocity vector.
We can use Newton’s Second Law and the equation for centripetal acceleration to write an equation for the
centripetal force that maintains an object’s circular motion.
v2
Fc  mac  m
r
Example: On an unbanked curve, the static
frictional force provides the centripetal force.
Remember Physics 1 (Mechanics) about forces!
Reasoning Strategy for the problem Solving:
1. Draw a free body diagram. Choose one point on the circular path and draw the forces (weight, normal force,
friction force, Tension in the rope) working on the object at that point on the circular path. Always choose positive
direction in the direction of acceleration – to the center of the circle if it is centripetal acceleration. Remember:
Centripetal force is NOT one of the forces you should draw in the free body diagram because centripetal force is
the resultant force.
2. Set up equations to solve for the unknowns:
3. If the mass or weight is not given, rather use mg in the equations (not W), because the mass might cancel out in
some equations.
For objects moving in a horizontal circle with
constant speed:
For objects moving in a vertical with constant
speed:
Case 1:
If a car is moving around an Unbanked horizontal circle
v2
Fx  mac  m
r
Fy  0
Fx  0
Case 2:
If a car is moving around a Banked curve, it is moving
in a horizontal circle!
v2
Fy  mac  m
r
v2
Fc  FN sin   m
r
FN cos  mg
tan  
v2
rg
29 | P a g e
Vertical Circular Motion
v12
FN 1  mg  m
r
FN 2
v22
m
r
v32
FN 3  mg  m
r
FN 4
v42
m
r
TUTORIAL QUESTIONS:
6.1
A child swings a yo-yo of weight mg in a horizontal circle so that the cord makes an angle of 30,0° with the
vertical, as in the figure. Find the centripetal acceleration of the yo-yo.Ans.: 5,67 m/s2
6.2
A child is swinging a 0,0120 kg ball on a string in a horizontal circle whose radius is 0,100 m. The ball travels
twice around the circle in 1,00 s.
a.
Determine the centripetal force acting on the ball. Ans.: 0,19 N
b.
If the speed is doubled, does the centripetal force double? If not, show by what factor does the
centripetal force increase? Ans.: No, factor of 4
6.3
The earth rotates once per day about an axis that passing through the north and south poles, an axis that is
perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6,38 × 106 m,
determine the speed and centripetal acceleration of a person situated:
a.
at the equator. Ans.: 463,97 m/s; 0,033 m/s2
b.
at a latitude of 30,0° north of the equator. Ans.: 401,81 m/s; 0,029 m/s2
30 | P a g e
6.4
A car with a constant speed of 83.0 km/h enters a circular flat curve with a radius of curvature of 0.400 km. If
the friction between the road and the car’s tires can supply a centripetal acceleration of 1.25 m/s2, does the car
negotiate the curve safely? Justify your answer. Ans.: No, because ac = 1.33 m/s2 > 1,25 m/s2
6.5
A student is to swing a bucket of water in a vertical circle without spilling any? If the distance from his shoulder
to the center of mass of the bucket of water is 1.0 m.
a.
b.
what is the minimum speed required to keep the water from coming out of the bucket at the top of the
swing? Ans.: 3,1m/s
If the combined mass of water and the bucket is 7.5 kg, what is the net force supplied to the swing in order
to keep the system in uniform circular motion? Ans.: 73,50 N
6.6
An object of mass m is on a horizontal rotating platform. The mass is located 0.22 m from the center and makes
one revolution every 0.74 s. The friction force needed to keep the mass from sliding is 13 N. What is the
object's mass? Ans.: 0,82 kg
6.7
How large must the coefficient be between the tires and the road if a car is to round a level curve of radius 95
m at a speed of 90 km/h? At what angle should it be banked so that it can go around the circle 90 km/h without
depending on friction? Ans.: 0.67; 33,870
6.8
A ball attached to a string swings in a circle of radius 1.20 m at 28o angle with the vertical. The tension in the
string is 1.80 N.
28o
1.20 m
a.
b.
What is the mass of the ball? Ans.: 0,162 kg
How long does the ball take to complete one orbit? Ans.: 2,99 s
6.9
A 60 kg glider pilot traveling in a glider at 40 m/s wishes to turn an inside vertical loop such that he exerts a
350 N force on the seat when the glider is at the top of the loop. What must be the radius of the loop under
these conditions? Ans.: 102 m
6.10
A stone rests in a pail that is moved in a vertical circle of radius 60 cm. What is the least speed the stone must
have as it rounds the top of the circle if it is to remain in contact with the pail? Ans.: 2,4 m/s
6.11
A box rests at a point 2 m from the axis of a horizontal platform.  = 0.25. The angular speed of the platform is
slowly increased from 0 rad/s. At what angular speed will the box starts to slide? Ans.: 1,1 rad/s
6.12
A 4.0 kg mass is attached to one end of a rope 2 m long. If the mass is swung in a vertical circle from the free
end of the rope, what is the tension in the rope when the mass is at its highest point if it is moving with a speed
of 5 m/s? Ans.: 10,8 N
6.13
A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger
feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the
dip. If r = 20.0 m, how fast is the roller coaster traveling at the bottom of the dip? Ans.: 14,0 m/s
31 | P a g e
LEARNING UNIT 7
CHAPTER 8: ROTATIONAL KINEMATICS
Study the following paragraphs in the textbook:
8.1
Rotational Motion and Angular Displacement ...................... p 223 (p 207)
8.2
Angular Velocity and Angular Acceleration .......................... p 226 (p 209)
8.3
The Equations of Rotational Kinematics .............................. p 228 (p 212)
8.4
Angular Variables and Tangential Variables ........................ p 231 (p 215)
8.5
Centripetal Acceleration and Tangential Acceleration .......... p 232 (p 216)
8.6
Rolling Motion.................................................................... p 235 (p 218)
Summary
Axis of Rotation
The rotational motion of a rigid body occurs when every point in the body moves in a circular path around a line
called the axis of rotation, which cuts through the center of mass.
Angular Displacement
When a rigid body rotates about an fixed axis, the angular displacement is the angle Δθ swept out by a line passing
through any point on the body and intersecting the axis of rotation perpendicular.
The S.I unit of angular displacement is radians: 1 revolution
 360 degree  2π radians
     o
 (in radians) 
Arc length s

Radius
r
Angular Velocity and Angular Acceleration
Angular velocity, , is defined as the change in the angular displacement per unit time.

Angular acceleration, , is defined as the rate of change of angular velocity per unit time.

t

, units of rad/s.

. units of rad/s2.
t
32 | P a g e
Rolling Motion
Linear motion
v f  vo  at
Angular motion
 f  o  t
v f  v o  2as
 f   o  2
2
2
s  vo t 
2
1
at
2
2
  ot 
1
t
2
1
1
vo  v f t
   o   f t
2
2
Relation between linear and angular quantities
s

s
r

v
r

a
r
ac  rw 2
If 2 objects rotate together with different radius:
Connected at their sides:
s1  s 2
 1 r1   2 r2
and
v1  v 2
1 r1   2 r2
and
Rotate around the same axis:
a1  a 2
1   2
 1 r1   2 r2
s1
s and v1 v 2 and a1 a 2
 2


r1
r2
r1
r2
r1
r2
1   2
1   2
33 | P a g e
TUTORIAL QUESTIONS:
7.1
During the spin-dry cycle of washing machine, the motor slows from 95 rad/s to 30 rad/s while turning the drum
through 4320 degrees. What is the magnitude of the angular acceleration of the motor? Ans.: 53,88 rad/s2
7.2
A roulette wheel with a diameter of 2.4 m reaches a maximum angular speed of 21.8 m/s before it begins
decelerating. After reaching this maximum angular speed, it turns through 35 revolutions before it stops. How
long did it take the wheel to stop after reaching its maximum angular speed? Ans.: 24,44 s
7.3
A motorcycle accelerates uniformly from rest and its wheels reach an angular speed of 78 rad.s-1 in a distance
of 100 m. The radius of each wheel of the motorcycle is 0.28 m. Calculate the angular as well as the linear
acceleration of each wheel on the motorcycle. Ans.: 8.25 rad/s; 2.38 m/s
7.4
Determine which has the greater angular speed: particle A, which travels 160º in 2.00 s, or particle B, which
travel 4π rad in 8.00 s or particle C travel with 13,50 rpm. Ans.: particle B
7.5
The hour, minute and second hands on a clock are 0.25 m, 0.30 m and 0.35 long, respectively. What are the
distances travelled by the tips of the hand in a 30 min interval? Ans.: 65,97 m, 0,94 m, 0,065 m
7.6
At the end of her routine, an ice skater spins through 7.50 revolutions with her arms always fully outstretched at
right angles to her body. If her arms are 60.0 cm long, through what linear arc length distance do the tips of her
fingers move during her finish? Ans.: 28,27 m
7.7
A jogger on a circular track that has a radius of 0.250 km runs a distance of 1.00 km. What angular distance
does the jogger cover in (a) radians and (b) degrees? Ans.: 4 rad; 229,30o
7.8
What is the period of revolution for (a) 9500 rpm centrifuge and (b) a 7000 rpm computer hard-disk drive?
Ans.: 6.32 x103 s, 8.57 x103 s
7.9
You measure the length of a distant car to be subtended by an angular distance of 1.5º. If the car is actually
5.0 m long, approximately how far away is the car? Ans.: 190,99 m
7.10
An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. How
many revolutions does the propeller undergo during this time? Ans.: 5 rev
7.11
A father gently places his son on a rotating merry-go-round. The merry-go-round is essentially a disk with a
mass of 300 kg and a radius of 3.00 m and initially (before the boy placed) rotating at one revolution every
5,00 seconds. Assume the boy has a mass of 42 kg and is placed (without slipping) near the edge of the
merry-go-round. Determine the final angular speed of the merry-go-round + boy combination. Ans.: 0,98 rad/s
7.12
If a particle is rotating with an angular speed of 3,5 rad/s, how long does the particle take to go through one
revolution? Ans.: 1,80 s
7.13
An exercise bike that you peddle in place has a bicycle chain connecting the wheel you peddle to the large
wheel in front. For the wheel diameters shown, calculate how many rpm must you produced to turn the large
wheel at 75 rpm? Ans.: 30 rpm
7.14 A space station consists of two donut-shaped living chambers, A and B that have the radii shown in the drawing.
As the station rotates, an astronaut in chamber A is moved 240 m along a circular arc. How far along a circular
arc is an astronaut in chamber B moved during the same time? Ans.: 825 m
34 | P a g e
LEARNING UNIT 8
CHAPTER 9: ROTATIONAL DYNAMICS
Study the following paragraphs in the textbook:
9.1
The Action of Forces and Torques on Rigid Objects ..................................
p 248 (p 232)
9.2
Rigid Objects in Equilibrium ......................................................................
p 251 (p 234)
9.4
Newton’s Second Law for Rotational Motion About a Fixed Axis ...............
p 261 (p 242)
9.5
Rotational Work and Energy .....................................................................
p 269 (p 248)
9.6
Angular Momentum ..................................................................................
p 271 (p 251)
Use the following hints and information when solving problems:



Lever arm and the force must always be perpendicular to produce torque.
Draw a free body diagram for the object under consideration.
Choose a direction for the unknown forces. If the answer is negative, it only means that the direction of the
force was chosen incorrectly.

Choose any point for the axis of rotation. It is best to choose the axis at such a point that one or more of
the unknown forces acts through the axis (thus the lever arm is zero and torque produced is also zero).

Use the Moments of Inertia as in Table 9.1 p 244. Note that the Moment of Inertia is axis dependent.
Summary
Leverarm
Shortest (or perpendicular) distance between the axis of rotation and the work line of a force.
Torque
Definition of Torque = (Magnitude of the Force) x (Lever arm)
  F
SI Unit of Torque: newton x meter (N·m)
The torque is positive when the force tends to produce a counterclockwise rotation about the axis.
A force of 10 N is working on a 1 m beam, with an axis of ration
each of the following sketches:
 Show the work line of the force.
 Show the leverarm.
 Determine the lenth of the leverarm
as shown in each of the following sketches. In
35 | P a g e

Determine the torque.
90o
30o
Leverarm
Torque
Leverarm
Torque
120o
Leverarm
Torque
Leverarm
Torque
Newton’s First Law and Equilibrium
If the net torque acting on a rigid object is zero, it will rotate with a constant angular velocity.
The most significant application of Newton’s First Law in this context is with regard to the concept of equilibrium.
When the net torque acting on a rigid object is zero, the object is not rotating.
Rigid Objects in Equilibrium: If a rigid body is in equilibrium, neither its linear motion nor its
rotational motion changes.
 0
ax  a y  0
 Fx  0
 Fy  0
  0
Newton’s Second Law
The angular acceleration of a body is proportional to the torque applied to it. Of course, force is also proportional to
mass, and there is also a rotational equivalent for mass: the moment of inertia, I, which represents an object’s
resistance to being rotated. Using the three variables, , I, and , we can arrive at a rotational equivalent for Newton’s
 Moment of
Net external torque  
 inertia
  Angular

  

  accelerati on 
  I 
Rotational Kinetic Energy
There is a certain amount of energy associated with the rotational motion of a body, so that a ball rolling down a hill
does not accelerate in quite the same way as a block sliding down a frictionless slope. The kinetic energy of a
rotating rigid body is: Ekr=½Iω2.
Moment of Inertia
The moment of inertia of a particle spinning a distance r away from its mass is I=mr 2. The axis of rotation is not going
through the object and is not tangential to the object.
36 | P a g e
r
m
Example: The Moment of Inertial Depends on Where the Axis Is.
Two particles each have mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the
moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at
(a) one end and (b) the center.
The moment of inertia for a few common shapes – axis of rotation is going through the object or
is tangential to the object.
Thin-walled hollow cylinder or hoop, I = MR2
Solid cylinder or disk, I =
1
MR2
2
1
ML2
12
Thin rod, axis through one edge, I =
2
MR2
5
Thin-walled spherical shell, axis through center, I =
Thin rod, axis perpendicular to rod passing through center,
I=
Solid sphere, axis through center , I=
Thin rectangular sheet, axis parallel to one edge and passing
through center of the other edge, I =
1
ML2
3
2
MR2
3
1
MR2
12
Thin rectangular sheet, axis through one edge, I =
1
MR2
3
37 | P a g e
Angular Momentum
The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia and its
angular velocity with respect to that axis. The angular momentum vector always points in the same direction as the
angular velocity vector.
L  I
SI Unit of Angular Momentum: kg·m2/s
Requirement: The angular speed must be expressed in rad/s.
Conservation of Angular Momentum
If the net torque action on a rigid body is zero, then the angular momentum of the body is constant or conserved.
I A A  I BB
How would you distinguish between an angular momentum and conservation of energy problem?
Subject
Mass/Moment of Inertial
Force/Torque
Newton Second Law
Work Done
Power
Momentum
Impulse
Kinetic Energy
Total Mechanical Energy
Linear Motion
Rotational Motion
m
F
I  kmr 2
F  ma
W  Fs
P  Fv
P  mv
Impulse  Ft
1
KE  mv 2
2
KE 
1
mv 2  mgh
2
  Fl
  I
W  
P  w
L  Iw
Impulse  t
KE 
1 2
Iw
2
KE 
1
1
mv 2  Iw 2  mgh
2
2
TUTORIAL QUESTIONS:
8.1
A string is wrapped around a pulley of radius 0,10 m and moment of inertia 0,15 kg • m 2. The string is pulled
with a force of 12 N. What is the magnitude of the resulting angular acceleration of the pulley? Ans.: 8 rad.s-2
8.2
A wheel rolls five revolutions on a horizontal surface without slipping. If the center of the wheel moves 3.2 m,
what is the radius of the wheel? Ans.: 0,102 m
8.3
Two children are sitting on opposite ends of a uniform seesaw of negligible mass.
a.
Can the seesaw be balanced if the masses of the children are different? How?
b.
If a 35 kg child is 2,0 m from the pivot point (or fulcrum), how far from the pivot point will her 30 kg
playmate have to sit on the other side for the seesaw to be in equilibrium? Ans.: 2,33 m
8.4
A fixed 0.15 kg solid disk pulley with a radius of 0.075 m is acted on by a net torque of 6.4 N.m. What is the
38 | P a g e
angular acceleration of the pulley? Ans.: 1,52 x 104 rad/s2
8.5
A 15 kg uniform sphere with a radius of 15 cm rotates about an axis tangent to its surface at 3,0 rad/s. A
constant torque of 10 N.m ten increases the rotational speed to 7,5 rad/s. Through what angle does the
sphere rotate while accelerating? Ans.: 1,12 rad
8.6
A person opens a door by applying a 15 N force perpendicular to it at a distance 0,90 m from the hinges. The
door is pushed wide open (to 120º) in 2.0 s. (a) How much work was done? (b) What was the average power
delivered? Ans.: 28,22 J, 14,1 W
8.7
A constant torque of 10 N.m is applied to a 10 kg uniform disk of radius 0,20 m. What is the angular speed of
the disk about an axis through its center after it rotates 2,0 revolutions from rest? Ans.: 35,45 rad/s
8.8
A thin 1.0-m-long rod pivoted at one end falls (rotates) from a horizontal position, starting from rest and with
no friction. What is the angular speed of the rod when it is vertical? [Hint: Consider the center of mass and use
the conservation of mechanical energy. Ans.: 5.42 rad/s
8.9
In a tumbling clothes dryer, the cylindrical drum (radius 50,0 cm and mass 35.0 kg) rotates once every second
(a) Determine its rotational kinetic energy about its central axis. (b) If it started from rest and reached that
speed in 2.50 s, determine the average net torque on the dryer drum. Ans.: 172.18 J, 21,96 N.m
8.10
An industrial flywheel with a moment of inertia of 4.25 x 102 kg·m2 rotates with a speed of 7500 rpm. (a) How
much work is required to bring the flywheel to rest? If this work is done uniformly in 1,50 min, how much power
is expended? Ans.: 1,31 x 108 J; 1,46 x 106 W
8.11
A father gently places his small son on a rotating merry-go-round. The merry-go-round is essentially a disk
with a mass of 250 kg and a radius of 2.50 m initially rotating at one revolution every 5.00 seconds. Assume
the boy has a mass of 15.0 kg and is placed (without slipping) near the edge of the merry-go-round.
Determine the final angular speed of the merry-go-round/boy combination. Ans.: 1,12 rad/s
8.12
A turntable with moment of inertia 5,4 x 103 kg.m2 rotates freely with angular speed of 33.33 rpm. Suddenly, a
cute 36 g mouse falls on the rim of the turntable, 15 cm from the center. Calculate the angular speed of the
turntable after the mouse falls on it. Ans.: 28,98 rpm.
8.13
A skater has a moment of inertia of 100 kg·m2 when his arms are outstretched and a moment of inertia of
75 kg·m2 when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 2,0 rps
(revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?
Ans.: 16.57 rad/s
8.14
An ice skater spinning with outstretched arms has an angular speed of 4,0 rad/s. She tucks in her arms,
decreasing her moment of inertia by 7,5%.
a.
What is the resulting angular speed? Ans.: 4,33 rad/s
b.
By what factor does the skater’s kinetic energy change? (Neglect any frictional effects.) 1,084
c.
Where does the extra kinetic energy come from?
8.15
A 2,0 kg solid cylinder of radius 0.5 m rotates at a rate of 40 rad/s about its cylindrical axis. What power is
required to bring the cylinder to rest in 10 s? Ans.: 20 W
8.16
A block of mass 0,05 kg is attached to a cord passing through a hole in a horizontal frictionless surface, as in
the drawing. The block is originally revolving at a distance of 0,2 m from the hole, with an angular velocity of
3 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to
0,1 m. The block may be considered a point mass.
a.
b.
8.17
What is the new angular velocity? Ans.: 12 rad/s
What is the change in kinetic energy of the block? Ans.: 0,027 J
The sign below is 4,5 m long, weighs 1350 N, and is made of uniform material. A weight of 225 N hangs 1 m
from the end. Find the tension in each support cable.
Ans.: 1125N; 450 N
39 | P a g e
1m
2m
1,5 m
Larry’s Barber and Styling
1m
8.18
A 3,0-kg ball and a 1,0 kg ball are placed at opposite ends of a massless beam so that the system is in
equilibrium as shown. Note: The drawing is not drawn to scale. What is the ratio of the lengths, b/a? Ans.: 4
8.19
The span of a certain bridge is 32,0 m long and has a mass of 5,38 x 104 kg. A truck of mass 11 700 kg is
stopped 11,4 m from one end. Find the upward force that must be exerted by each pier. Ans.: 3,04 x 105 N;
3,38 x 105 N
11,4 m
20,6 m
16 m
Pier P
Pier Q
FBridge
8.20
A solid sphere of radius 0,35 m is released from rest from a height of 1,8 m and rolls down the incline as
shown. What is the angular speed of the cylinder when it reaches the horizontal surface? Ans.: 14,34 rad/s
1,8 m
8.21
A spinning skater draws in her outstretched arms thereby reducing her moment of inertia by a factor of 2.
Determine the ratio of her final kinetic energy to her initial kinetic energy. Ans.: 2
8.22
A thin-walled hollow cylinder and a solid cylinder of the same mass, m and same radius, r start from rest at the
same time of an incline of height ho = 65 cm. Show with the necessary calculation that solid cylinder will reach
the bottom of the incline first with the greatest translational speed . Ans.: 2,90 m/s; 2,52 m/s
8.23
In this drawing, the large wheel has a radius of 8,5 m. A rope is wrapped around the edge of the wheel and a
7,6 kg-box hangs from the rope. A smaller disk of radius 1,9 m is attached to the wheel. A rope is wrapped
40 | P a g e
around the edge of the disk as shown. An axis of rotation passes through the centre of the wheel disk system.
What is the value of the mass M that will prevent the wheel from rotating? Ans.: 34 kg
8.24
Uniform 20 kN beam of length L is hinged at point P as shown in the sketch. Calculate the tension in the rope
3
attached L from P. Ans.: 22.402 kN
4
8.25
A solid cylindrical 5,00 kg reel with a radius of 0,600 m and a frictionless axle, starts from rest and speeds up
uniformly as a 3,00 kg bucket falls into a well, making a light rope unwind from the reel (Fig. below). The
bucket starts from rest and falls for 4,00 s. (a) What is the linear acceleration of the falling bucket? (b) How far
does it drop? (c) What is the angular acceleration of the reel? Ans.: 5,35 m/s2; 42,8 m; 8,91 rad/s2
8.26
Consider the following diagram. The 4420 N crate is being lifted by a motor through the duel pulley system.
The pulleys are fastened together, have radii r1 = 0,6 m and r2 = 0,2 m and turn as a single unit with moment
of inertia I = 50 kg.m 2. If a tension of magnitude T1 = 2150 N is maintained in the cable connected to the
motor, calculate:
a. The angular acceleration of the pulley. Ans.: 5,967 rad/s2
b. The linear acceleration of the crate. Ans.: 1,1934 m/s2
c. The tension in the cable attached to the crate. Ans.: 4958,25 N
41 | P a g e
r1
T1
r2
motor
crate
42 | P a g e
LEARNING UNIT 9
CHAPTER 10: SIMPLE HARMONIC MOTION AND ELASTICITY
Study the following paragraphs in the textbook:
10.1
The Ideal Spring and Simple Harmonic Motion ..................... p 286 (p 267)
10.2
Simple Harmonic Motion and the Reference Circle............... p 290 (p 270)
10.3
Energy and Simple Harmonic Motion ................................... p 295 (p 275)
10.4
The Pendulum ..................................................................... p 299 (p 279)
10.5
Damped Harmonic Motion ................................................... p 301 (p 280)
10.6
Driven Harmonic Motion and Resonance ............................. p 302 (p 281)
10.7
Elastic Deformation.............................................................. p 303 (p 282)
10.8
Stress, Strain, and Hooke’s Law .......................................... p 308 (p 286)
Summary
Angular Frequency
Angular frequency, f, is defined as the number of circular revolutions in a given time interval. It is commonly
measured in units of Hertz (Hz), where 1 Hz = 1 s–1. For example, the second hand on a clock completes one
revolution every 60 seconds and therefore has an angular frequency of 1 /60 Hz.
The relationship between frequency and angular velocity is:   2f .
Angular Period
Angular period, T, is defined as the time required to complete one revolution and is related to frequency by the
equation: T 
1
.
f
Restoring force of spring
F= - kx
F- restoring force of spring (The force applied by spring to pull/push it back to equilibrium position)
k- spring constant
x- displacement of the spring
43 | P a g e
Simple harmonic motion
The vibration motion in which a system obeys Hooke’s law. Oscillates with a constant amplitude.
If friction (or any other mechanism that dissipate energy) is present, the amplitude decreases – damped harmonic
motion.
Critical damping: The smallest degree of damping that will eliminate the oscillation.
If some force is applied at all times, the amplitude will increases – driven harmonic motion.
Resonance
Resonance is the condition in which a time-dependent force can transmit large amounts of energy to an oscillating
object, leading to large-amplitude motion. Resonance occurs when the frequency of the force matches the natural
frequency at which the object will oscillate.
Displacement, velocity and acceleration
Displacement of SHM: x = A cos t with  = 2f. Because  is directly proportional to f, it is often called angular
frequency.
Velocity of SHM: v = A sin t. (vmax = A)
Acceleration of SHM: a = A2 cos t. (amax = A2)
30
x = 5 cos (2t)
v = 10 sin (2t)
a = 20 cos (2t)
20
10
x
0
v
0
0.785
1.57 2.355 3.14
3.925 4.71 5.495
6.28 7.065
a
-10
-20
-30
The angular frequency with which a mass m on a spring will vibrates (frequency of vibration):
  2f 
k
m
Elastic potential energy
E = ½kx2 . The energy stored in a stretched or compressed spring.
Conservation of energy
½kx12 + ½mv12 + mgh1 = ½kx22 + ½mv22 + mgh2
For a simple pendulum:
  2f 
g
L
44 | P a g e
Elasticity
Elasticity is the property by which a body returns to its original size and shape when the forces that deformed it are
removed.
Elastic deformation:
F
L
 Y ( ) Y - Young’s modulus. (F is perpendicular to the area A)
A
Lo
Hookes law: Stress  strain,
Shear deformation:
F
L
( )
A
Lo
F
X
 S(
)
A
Lo
S - Shear modulus (F is parallel to the area A)
Bulk deformation:
F
V
( )  P  B(
)
A
Vo
B- Bulk modulus
TUTORIAL QUESTIONS:
9.1
A 0,5 kg mass attached to a spring with a spring constant of 8 N/m vibrates in simple harmonic motion with an
amplitude of 10 cm. Calculate the:
a.
maximum value of its speed and acceleration Ans.: 0,4 m/s; 1,6 m/s2
b.
speed and acceleration when the mass is 6 cm from the equilibrium position. Ans.: 0,32 m/s; 0,97 m/s2
9.2
A 1,0 kg object is suspended from a spring with k = 16 N/m. The mass is pulled 0,25 m downward from its
equilibrium position and allowed to oscillate. What is the maximum kinetic energy of the object? Ans.: 0,5 J
9.3
A certain spring compressed 0,20 m has 10 J of elastic potential energy. The spring is then cut into two halves
and one of the halves is compressed by 0,20 m. How much potential energy is stored in the compressed half
of the spring? Ans.: 20 J
9.4
A pendulum is transported from sea level, where the acceleration due to gravity g = 9,80 m/s2, to the bottom of
Death Valley. At this location, the period of the pendulum is decreased by 3,00%. What is the value of g in
Death Valley? Ans.: 10,42 m/s2
9.5
A spring with constant k = 40,0 N.m-1 is at the base of a frictionless, 30,0° inclined plane. A 0,50 kg block is
pressed against the spring, compressing it 0,20 m from its equilibrium position. The block is then released. If
the block is not attached to the spring, how far along the incline will it travel before it stops? Ans.: 0,326 m
0.5 kg
30.0°
9.6
Gina's favourite exercise equipment at the gym consists of various springs. In one exercise, she pulls a handle
grip attached to the free end of a spring to 0,80 m from its unstrained position. The other end of the spring
(spring constant = 53 N/m) is held in place by the equipment frame. What is the magnitude of the force that
Gina is applying to the handle grip? Ans.: 42,4 N
9.7
A 0,750 kg object hanging from a vertical spring is observed to oscillate with a period of 2,00 s. When the 0,750
kg object is removed and replaced by a 1,25 kg object, what will be the period of oscillation? Ans.: 2,58 s
45 | P a g e
9.8
A ping-pong ball weighs 0,025 N. The ball is placed inside a cup that sits on top of a vertical spring. If the
spring is compressed 0,055 m and released from rest, the maximum height above the compressed position that
the ball reaches is 2,84 m. Neglect air resistance and determine the spring constant. Ans.: 47 N/m
compressed
position
9.9
A spring is mounted as in the figure. By attaching a spring balance to the free end and pulling sideways, it is
determined that the force is proportional to the displacement, a force of 4,00 N causing a displacement of
0,02 m. A 2,00 kg body is attached to the end, pulled aside a distance of 0,04 m, and released.
X=0
a.
b.
c.
d.
e.
f.
What is the spring constant of the spring? Ans.: 200 N/m
What is the frequency of vibration, and the period? Ans.: 10 rad/s; 0,63 s
What is the maximum velocity attained by the vibrating body? Ans.: 0,4 m/s
What is the maximum acceleration of the body? Ans.: 4 m/s
Calculate the velocity when the body moved halfway in towards from its initial position? Ans.: 0,35 m/s
How much time is required for the body to move halfway in toward from its initial position? Ans.: 0.105 s
9.10 When a 19.6 N block is suspended from a vertically hanging spring, it stretches the spring from its original
length of 5 cm to 6 cm. The same block is attached to the same spring and placed on a horizontal, frictionless
surface. The block is then pulled so that the spring stretches to a total length of 0,10 m. The block is released
at time t = 0s and undergoes simple harmonic motion, as in the sketch.
a.
b.
c.
9.11
What is the frequency of the motion? Ans.: 4,98 Hz
What is the speed of the block each time the spring is 5,0 cm long? Ans.: 2,71 m/s
What is the maximum acceleration of the block? Ans.: 97,91 m/s
The acceleration of a block attached to a spring is given by: a = – 0.302cos (2,41t) m/s2.
a. What is the frequency of the block’s motion? Ans.: 0,384 Hz
b. What is the Amplitude of the block’s motion? Ans.: 0,0519 m
9.12
A vertical spring with spring constant 450 N/m is mounted on the floor. From directly above the spring, which
is unstrained, a 0.3 kg block is dropped from rest. It collides and sticks to the spring, which is compressed by
2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible; from what height above
the compressed spring was the block dropped? Ans.: 0,048 m
9.13
A 500 N object is hung from the end of a wire of cross-sectional area 0,010 cm2. The wire stretches from its
original length of 200,00 cm to 200,50 cm. Determine the Young's modulus of the wire. Ans.: 2.0 x 1011 N/m2
9.15
The brick shown in the drawing is glued to the floor. A 3500 N force is applied to the top surface of the brick
as shown. If the brick has a shear modulus of 5,4  109 N/m2, how far to the right does the top face move
relative to the stationary bottom face? Ans.: 2,6  10–6 m
3500 N
0.10 m
0.10 m
0.25 m
46 | P a g e
LEARNING UNIT 10
CHAPTER 11: FLUIDS
After completion of this learning unit you should be able to:
Explain Archimedes’ Principle.
Define Buoyant Force.
Explain why solid objects can float.
Explain what viscosity is.
Define Pouiselle’s Law
Solve Buoyant Force and Viscous Flow problems.
Study the following paragraphs in the textbook:
11.6
Archimedes’ Principle ......................................................... p 332 (p 309)
11.11
Viscous Flow ..................................................................... p 345 (p 323)
Summary
Archimedes principle
Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the
buoyant force equals the weight of the fluid that the object displaces:
F
B
Magnitudeof
buoyantforce
 Wfluid

Weight of
displacedfluid
If the object is floating (partially or completely) then the magnitude of the buoyant force is equal to the magnitude of
its weight.
One day, while getting into his bath he noticed water spilling over the sides. In a
flash, Archimedes realized the relation between the water that had fallen out and
the weight of his body - in other words he discovered why some objects float and
some sink! Archimedes was so excited with his discovery that he hopped out of
the bath, and rushed naked into the street yelling triumphantly, 'Eureka!' 'Eureka!'
(Greek word for 'I have found it!).
47 | P a g e
Reasoning Strategy For Problem Solving:
1. Draw a free body diagram - the object and all the forces (weight, normal force, friction force, Tension in the
rope, Buoyant Force, FB) working on the object.
2. FB = m x g (weight of fluid displaced)
FB = ρ x V x g (ρ - of fluid displaced, V – of fluid displaced = V object underneath the water)
3. Set up equations to solve for the unknowns.
FORCE NEEDED TO MOVE A LAYER OF VISCOUS FLUID WITH CONSTANT VELOCITY
The magnitude of the tangential force, F, required to move a fluid layer at a constant speed, v. The layer has an area
A, and y is the distance from the immobile surface is given by:
F
vA
y
.
Coefficient of viscosity
SI Unit of Viscosity: Pa·s
POISEUILLE’S LAW
A fluid with viscosity, , flowing through a pipe radius, R and length L, has a volume flow rate. Q given by:
R 4 P
Q
8L
TUTORIAL QUESTIONS:
10.1
Normally, a Goodyear airship, contains about 5400 m3 of helium (He) whose density is 0.179 kg/m3. Find the
weight of the load that the airship can carry in equilibrium at an altitude where the density of air is 1.20 kg/m3.
Ans.: 5.40 x 104 N
10.2
A 2 kg block displaces 10 kg of water when it is held fully immersed. The object is then tied down as shown in
the figure; and it displaces 5 kg of water. What is the tension in the string? Ans.: 29.4 N
10.3
An object is solid throughout. When the object is completely submerged in ethyl alcohol (806 kg/m3), its
apparent weight is 15.2 N. When completely submerged in water (1000 kg/m3), its apparent weight is 13.7 N.
What is the volume of the object? Ans.: 7.9 x 10-4 m3
48 | P a g e
10.4
A balloon inflated with helium gas (density = 0.2 kg/m3) has a volume of 6 x 103 m3. If the density of air is
1.3 kg/m3, what is the buoyant force exerted on the balloon? Ans.: 0.08 N
10.5
A piece of steel wire, which is 3 m long, and of 1mm diameter hangs vertically from the ceiling. A 5 kg mass,
made from iron, (density = 7,9 x 103 kg.m-3) is attached to the free end. What is the tension in the cord if the
mass is totally immersed in water? Ans.: 42.8 N
10.6
A piece of wood (=600 kg.m-3) is 0.6 m long, 0.25 m wide, and 0.08 m thick. What volume of lead
(=11 300 kg/m3) must be fastened underneath it to sink the wood in calm water so that its top is just even with
the water level? Ans.: 4.66 x 10-4 m3.
10.7
When a block of volume 1,00 x 10-3 m3 is hung from a spring scale as shown in Figure A, the scale reads 10,0
N. When the same block is then placed in an unknown liquid, it floats with 2/3 of its volume submerged as
suggested in Figure B.
a.
b.
Determine the mass of the block. Ans.: 2,02 kg
Determine the density of the unknown liquid. Ans.: 3,03 x 103 kg/m3
10.8
The density of ice is 0,92 g/cm3, and the density of seawater is 1,03 g/cm3. A large iceberg floats in Arctic
waters. What fraction of the volume of the iceberg is exposed? Ans.: 0,107
10.9
A flat-bottom river barge is 12,0 m wide; 36,0 m long; and 5,00 m deep.
a.
How many m3 of water will it displace if the top stays 1,00 m above the water?
Ans.: 1728 m3
b.
What load (in N) will the barge contain under these conditions if the barge weighs 4,50 x 106 N in
dry dock? Ans.: 1,24 x 107 N
10.10
A wood plaque contains some embedded pure silver handicraft work. Its total mass is 2,98 kg. When immersed
in water, it displaces 2500 cm 3 (Density of wood =550 kg/m 3, Density of silver = 10500 kg/m3).
a.
What is the mass (in grams) of silver contained in the plaque? Ans.: 1,69 kg
b.
What is its apparent weight while submerged? Ans.: 4,7 N
10.11
To verify her suspicion that a rock specimen is hollow, a geologist weighs it in water. She finds it weighs twice
as much in air than in water. The solid part of the specimen has a density of 5 x 103 kg.m-3. What fraction of
the specimen’s apparent volume is hollow? Ans.: 0,6
10.12
A foam plastic (0,58 g/cm 3) is used as a life preserver. What volume foam plastic must be used if it should keep
20% of an 80 kg man’s volume above water in a lake? The average density of a man is 1,04 g/cm 3.
Ans.: 4,4 x 10–2 m3
10.13
A syringe is filled with a solution whose viscosity is 1,5x10–3 Pa·s. The internal radius of the needle is
4,0x10–4m. The gauge pressure in the vein is 1900 Pa. What force must be applied to the plunger, so that
1,0x10–6m3 of fluid can be injected in 3.0 s? Ans.: 0,25 N
49 | P a g e
LEARNING UNIT 11
CHAPTER 14: THE IDEAL GAS LAW AND KINETIC THEORY
Study the following paragraphs in the textbook:
14.1
Molecular Mass, the Mole, and Avogadro’s Number ............. p 417 (p 394)
14.2
The Ideal Gas Law............................................................... p 419 (p 396)
14.3
Kinetic Theory of Gases ....................................................... p 424 (p 399)
14.4
Diffusion .............................................................................. p 429 (p 403)
Use the following hints and information when solving problems:



The pressure in the Ideal Gas Law is the absolute pressure and not the gauge pressure.
The temperature used in the Ideal Gas Law must be in Kelvin.
Guy-Lussa’c Law is not in the textbook. It can also be derived from the Ideal Gas Law, like Boyle’s Law.
For Guy-Lussac’s Law, n and V must be constant then:
P1 P2

T1 V2
.
Summary
Molecular Mass, the Mole, and Avogadro’s Number
One mole of a substance contains as many particles as there are atoms in 12 grams of the isotope cabron-12.
The number of atoms per mole is known as Avogadro’s number, NA.
N A  6.022 1023 mol 1
n
number of
moles
N
NA
number of Atoms
(or molecules) in a sample
Avogadro’s number
n
mparticleN
mparticleN A
mparticle 

mass of the sample m

Mass per mole
Mr
Mass per mole
NA
mparticle 
mass of the sample
N
Example 1 (Text Book)
The Hope diamond (44.5 carats) is almost pure carbon. The Rosser Reeves ruby (138 carats) is primarily aluminum
oxide (Al2O3). One carat is equivalent to a mass of 0.200 g. Determine (a) the number of carbon atoms in the Hope
diamond and (b) the number of Al2O3 molecules in the ruby.
50 | P a g e
The Ideal Gas Law
The absolute pressure of an ideal gas is directly proportional to the Kelvin temperature and the number of moles of
the gas and is inversely proportional to the volume of the gas.
R  8.31J mol  K  - Universal gas constant
PV  nRT
 R 
T  NkT
PV  nRT  N 
N
 A
Boltzman _ cons tan t , k 
R
8.31 J mol  K 

 1.38  10 23 J K
23
1
N A 6.022  10 mol
In the equation everything MUST be in SI units
Universal gas law
P1V1 P2V2

T1
T2
In the above equation Temperature must be in the SI unit (Kelvin). Pressure and volume can be in any unit though
pressure must be absolute pressure and not gauge pressure.
Boyle’law
Pressure is indirectly proportional to
volume if temperature and mass is
kept constant.
Gay lussac’s law
Pressure is directly proportional to
temperature if volume and mass is
kept constant.
Charle’s law
Volume is directly proportional to
temperature if pressure and mass is
kept constant.
P1V1  P2V2
P1 P2

T1 T2
V1 V2

T1 T2
Kinetic theory of gases
Temperature is a measure of the kinetic energy of the molecules in a material. Because individual molecules are so
small, and because there are so many molecules in most substances, it would be impossible to study their behaviour
individually. However, if we know the basic rules that govern the behaviour of individual molecules, we can make
statistical calculations that tell us roughly how a collection of millions of molecules would behave. This, essentially, is
what thermal physics is: the study of the macroscopic effects (temp. ,volume and pressure) of the microscopic
molecules that make up the world of everyday things.
The distribution of particle speeds in an ideal gas at constant temperature is the Maxwell speed distribution. The
kinetic theory of gases indicates that the Kelvin temperature T of an ideal gas is related to the average translational
kinetic energy of a particle.
PV 
 
2
2 1
2 
N E k  N  mvrms

3
3 2

vrms – root-mean square speed.
The average particle speed squared. The kinetic energy in the equation is the average kinetic energy of the gas as a
whole, a property of the gas and not a property of a single gas particle individually.
PV  NkT  NkT 
2 1 2
1 2
3
N ( mv rms )  mv rms
 kT
3 2
2
2
2
KE  12 mvrms
 32 kT
51 | P a g e
Internal energy of a monatomic gas
1 2
3
3
3
U  N ( mv rms
)  U  N ( kT )  NkT  U  nRT
2
2
2
2
Diffusion
The process in which molecules move from a region of higher concentration to one of lower concentration is called
Diffusion.
Fick’s law of diffusion
The mass m of solute that diffuses in a time t through a solvent contained in a channel of length L and cross sectional
area A is
Diffusion constant
m
DAC t
L
concentration gradient
between ends
SI Units for the Diffusion Constant: m2/s
TUTORIAL QUESTIONS:
11.1
A runner weighs 580 N, and 71% of this weight is water. (a) How many moles of water are in the runner’s body?
(b) How many water molecules (H2O) are there? Ans.: 2,3 x 103 mol, 1,4 x 1027
11.2
A mass of 135 g of a certain element is known to contain 30,1 x 10
Ans.: Aluminum
11.3
The artificial sweetener NutraSweet is a chemical called aspartame (C14H18N2O5). What is (a) its molecular
mass (in atomic mass units), (b) the mass (in kg) of an aspartame molecule? Ans.: 294,31 u, 4,89 x 10–25 kg
11.4
In a diesel engine, the piston compresses air at 305 K to a volume that is one-sixteenth of the original volume
and a pressure that is 48,5 times the original pressure. What is the temperature of the air after the compression?
Ans.: 925 K
11.5
What is the density (in kg/m3) of nitrogen gas (N2) at a pressure of 2,0 atmospheres and a temperature of
310 K? Ans.: 2,2 kg/m3
11.6
Atmosphere is composed primarily of Nitrogen N2 (78%, M = 28 g/mol) and Oxygen O2 (21%, M = 32 g/mol)
23
atoms. What is the element?
a.
Is the kinetic energy of Nitrogen greater than, less than, or the same as the kinetic energy of Oxygen?
Explain your answer with the aid of equations.
b.
Is the rms speed of Nitrogen greater than, less than, or the same as the rms speed of Oxygen. Explain
your answer with the aid of equations.
52 | P a g e
11.7
What is the root-mean-square speed of a hydrogen molecule (H2) at 300 K? Ans.:1,92 x 103 m/s
11.8
If the translational rms speed of the water vapor molecules (H2O) in air is 648 m/s, what is the translational rms
speed of the carbon dioxide molecules (CO2) in the same air? Both gases are at the same temperature.
Ans.: 414 m/s
11.9
Very fine smoke particles are suspended in air. The translational rms speed of a smoke particle is
2.8 x 10 – 3 m/s, and the temperature is 301 K. Find the mass of a particle. Ans.: 1,6 x 10-15 kg
11.10
An oxygen molecule is moving near the earth’s surface. Another oxygen molecule is moving in the ionosphere
(the uppermost part of the earth’s atmosphere) where the Kelvin temperature is three times greater. Determine
the ratio of the translational rms speed in the ionosphere to that near the earth’s surface. Ans.: 1,73
11.11
Near the surface of Venus, the rms speed of carbon dioxide molecules (CO 2) is 650 m/s. What is the
temperature (in kelvins) of the atmosphere at that point? Ans.: 750 K
11.12
Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of
tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface
of the insect’s body and penetrate into the interior. Suppose that a trachea is 1,9 mm long with a cross-sectional
area of 2,1 x 10– 9 m2. The concentration of oxygen in the air outside the insect is 0,28 kg/m3, and the diffusion
constant is 1,1 x 10 – 5 m2/s. If the mass per second of oxygen diffusing through a trachea is 1,7 x 10 – 12 kg/s,
find the oxygen concentration at the interior end of the tube. Ans.: 0,14 kg/m3
11.13
Ammonia, which has a strong smell, is diffusing through air contained in a tube of length L and cross-sectional
area 4,0 x 10 – 4 m2. The diffusion constant for ammonia in air is 4,2 x 10 – 5 m2/s. In a certain time, 8.4 x 10 – 8
kg of ammonia diffuses through the air when the difference in ammonia concentration between the ends of the
tube is 3,5 x 10 – 2 kg/m3. Find the speed at which ammonia diffuses through the air—that is, the length of the
tube divided by the time to travel this length. Ans.: 7,0 x 10-3 m/s
11.14
It is found that the amino acid glycine diffuses through water at a mass rate of m/t = 6,00 x 104 kg/s. The
diffusion constant is 10,6 x 10-10 m2/s. A tube of water has a radius of 1,4 cm. What difference in concentration
per unit length of the tube C/L must be maintained to give this flow? Ans.: 9,19 x 1016 kg/m
11.15
A frictionless gas-filled cylinder is fitted with a movable piston, as the drawing shows. The block resting on the
top of the piston determines the constant pressure that the gas has. The height h is 0,120 m when the
temperature is 273 K and increases as the temperature increases. What is the value of h when the temperature
reaches 318 K? Ans.: 0,14 m
11.16
Suppose that a tank contains 680 m 3 of neon at an absolute pressure of 1,01 x 10 5 Pa. The temperature is
changed from 293,2 to 294,3 K. What is the increase in the internal energy of the neon? Ans.: 3,9 x 105 J
11.17
Consider two ideal gases, A and B, at the same temperature. The rms speed of the molecules of gas A is
twice that of gas B. How does the molecular mass of A compare to that of B? Ans.: 4
11.18
An ideal gas at 0 °C is contained within a rigid vessel. The temperature of the gas is increased by 1 C°. What
is Pf/Pi the ratio of the final to initial pressure? Ans.: 1,004
53 | P a g e
LEARNING UNIT 12
CHAPTER 15: THERMODYNAMICS
Study the following paragraphs in the textbook:
15.1
Thermodynamic Systems and Their Surroundings .............. p 442 (p 417)
15.2
The Zeroth Law of thermodynamics .................................... p 443 (p 417)
15.3
The First Law of thermodynamics ....................................... p 443 (p 418)
15.4
Thermal Processes ............................................................. p 446 (p 420)
15.5
Thermal Processes Using an Ideal Gas .............................. p 449 (p 423)
15.6
Specific Heat Capacities ..................................................... p 451 (p 424)
15.7
The Second Law of Thermodynamics ................................. p 452 (p 426)
15.8
Heat Engines ...................................................................... p 453 (p 427)
15.9
Carnot’s Principle and the Carnot Engine ............................ p 455 (p 428)
15.10
Refrigerators, Air Conditioners, and Heat Pumps ................ p 457 (p 430)
15.11
Entropy ............................................................................... p 460 (p 433)
Use the following hints and information when solving problems:






Heat is positive when the system gains heat, and negative if the system looses heat.
Work is positive when the system is doing the work, and it is negative if the work is done on the system.
The area under a pressure-volume graph is the work for any kind of process.
When determining the efficiency of a Carnot Engine, be sure that the temperature of the cold an hot
reservoirs are in Kelvin.
No heat engine (even a perfect heat engine) can have an efficiency of 100%.
An air conditioner and a heat pump are the reverse of each other.
Summary
Dynamics is the study of why things move the way they do. The prefix thermo denotes heat, so thermodynamics is
the study of what compels heat to move in the way that it does. The laws of thermodynamics are a bit strange. There
are four of them, but they are ordered zero to three, and not one to four. They weren’t discovered in the order in
which they’re numbered, and some—particularly the Second Law—have many different formulations, which seem to
have nothing to do with one another.
Zeroth Law
If system A is at thermal equilibrium with system C, and B is at thermal equilibrium with system C, then A is at thermal
equilibrium with B.
Two systems are at thermal equilibrium if they have the same
temperature. If A and C have the same temperature, and B and C
have the same temperature, then A and C have the same
temperature. The significant consequence of the Zeroth Law is that,
when a hotter object and a colder object are placed in contact with
one another, heat will flow from the hotter object to the colder object
until they are in thermal equilibrium.
A
C
B
First Law
U  U f  U i  Q  W
Heat (Q) is positive when the system gains heat and negative when the system loses heat.
Work (W) is positive when it is done by the system and negative when it is done on the system.
U – INCREASE/DECREASE in internal energy, U = nRT
W- Work done BY/ON the system, W = PV
54 | P a g e
Q
W=PV
U
The area under a pressure-volume graph is the work for any kind of process.
The First Law is just another way of stating the law of conservation of energy. Both heat and work are forms of
energy, so any heat or work that goes into or out of a system must affect the internal energy of that system.
Thermal processes
Quasi-static process
A thermal process that occurs so slowly, that uniform pressure and temperature exist throughout the system at all
times.
ISOBARIC PROCESS
A process taking place at constant pressure. Ex: process in container with movable lid/piston.
W = PV = nRΔT
Q = U + W =
5
nRT
2
V1 V2

T1 T2
ISOCHORIC PROCESS
A process taking place at constant VOLUME. No work done!
W = PV = 0
Q = U =
3
nRT
2
Pi Pf

Ti T f
55 | P a g e
ISOTHERMAL PROCESS
Heat bath
No temperature change
U =
3
nRT = 0
2
Q = W = PV =
Vf
Vf
Vi
Vi
Vf 

i 
1
 PdV = nRT  V dV = nRT ln  V
PiVi  Pf V f
ADIABATIC PROCESS
Insulator
No heat transfer (Q=0)
W = –U = –
3
nRT
2
3
nR(Tf – Ti)
2
3
W=
nR(Ti – Tf)
2
W=–

PiVi  Pf V f

Derive equation for the relations to show relation between P and T, and the relation between V and T in an
adiabatic process from the equation P1V1 = P2V2
Thermal
processes
(∆U = Q - W)
Gas law
relation
Change in
internal
energy
Work done by
the system
Heat added to
the system
Isobaric
(∆P = 0: Pi = Pf)
Vi V f

Ti T f
∆U = 3 nR∆T
W = P∆V,
W= nR∆T
Q = ∆U + W
Isochoric
(∆V = 0: Vi = Vf)
Pi Pf

Ti T f
∆U = 3 nR∆T
2
W=0J
Isothermal
(∆T = 0: Ti = Tf)
PiVi  Pf V f
∆U = 0 J
W= P ʃ dV
Adiabatic
(Q = 0J )
PiVi  Pf V f
∆U = 3 nR∆T
Vf
W = nRT ln 
 Vi
W = – ∆U,

γ=C /C
P
V
2
Q = 5 nR∆T
2

2
Q = ∆U = 3 nR∆T
2
W=–



Vf
Q = W= nRT ln 
 Vi



Q=0J
3
nR∆T
2
56 | P a g e
Specific heat capacities
Q  mcT 
m
cMT  nCT
M

Qconstant pressure  U  W  32 nRT f  Ti   nRT f  Ti   52 nRT





Qconstant volume  U  W  32 nR T f  Ti   0  32 nRT



For a monatomic gas
3
nRT
2
5
QV  nCV T  nRT
2
5
C
5
3
C P  R and CV  R , with   P  and CP  CV  R
CV 3
2
2
Q P  nC P T 
Heat Engines
What is a heat engine?
A heat engine is a machine that converts heat into work. A large number of the machines we use—most notably our
cars—employ heat engines.
A heat engine is any device that uses heat to perform work. It has three essential features.
1. Heat is supplied to the engine at a relatively high temperature from a place called the hot reservoir.
2. Part of the input heat is used to perform work by the working substance of the engine.
3. The remainder of the input heat is rejected to a place called the cold reservoir.
The engine of a car generates heat by combusting petrol. Some of that heat drives pistons that make the car do work
on the road, and some of that heat is dumped out the exhaust pipe.
The steam engine is an example of a heat engine. In the boiler the water becomes steam, which pushes a piston,
and external work is done. In the condenser the steam becomes a liquid again. This water is returned to the boiler by
pumping. The water is thermally cycled.
Efficiency or coefficient of performance = Output/input
Efficiency is usually expressed as a percentage rather than in decimal form. That the efficiency of a heat engine can
never be 100% is a consequence of the Second Law of Thermodynamics. If there were a 100% efficient machine, it
would be possible to create perpetual motion: a machine could do work upon itself without ever slowing down.
57 | P a g e
Carnot’s principle
No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than
a reversible engine operating between the same 2 temperatures. Furthermore, all reversible engines operating
between the same temperatures have the same efficiency.
QC TC

QH TH
Heat engine
QH  W  QC
QH  magnitude of input heat
QC  magnitude of rejected heat
W  magnitude of the work done
Use
e
QC TC

QH TH
to derive an equation for e for a Carnot engine in terms of TH and Tc:
output W
 ,
input QH
eCarnot 
QH  QC
Q
T
1 C 1 C
QH
QH
TH
Refrigerator / Aircon during Summer
QH = W + Qc
Use
QC TC

QH TH
COP = Qc/W
to derive an equation for COP for a Carnot refrigerator in terms of T H
and Tc:
COP 
output QC

input W
1
COPCarnot
1
 Q  QC 
 QH 
T

QC

 H
 1   H  1
 
QH  QC  QC 
 QC 
 TC 
1
58 | P a g e
Heat Pump / Air conditioner during Winter
QH = W + QC
COP = QH/W
QC TC
Use
 to derive an equation for cop for a Carnot heat pump in terms
QH TH
of TH and Tc:
COP 
output QH

input
W
1
COPCarnot
1
 Q  QC 
 QC 
 TC 
QH

 H
  1 
  1  
QH  QC  QH 
 QH 
 TH 
1
What is the relation between COPheat pump and eheat engine?_________________________________________
Second Law
The Second Law in Terms of Heat Flow
Heat flows spontaneously from a hotter object to a colder one, but not spontaneously in the opposite direction.
The Second Law in Terms of Heat Engines
One consequence of this law, which we will explore a bit more in the section on heat engines, is that no machine can
work at 100% efficiency: all machines generate some heat, and some of that heat is always lost to the machine’s
surroundings.
Entropy
The word entropy was coined in the 19th century as a technical term for talking about disorder.
Q
S 
Q – Heat transfer TO the system. T – Temperature in Kelvin.
T
The Second Law in Order-disorder statement
System goes from order to disorder spontaneously, but it can never goes from disorder to order spontaneously.
The Second Law in entropy statement
Suniverse>0 for a irreversible process.
Suniverse=0 for a reversible process.
59 | P a g e
TUTORIAL QUESTIONS:
12.1
A quantity of carbon monoxide gas is slowly adiabatically compressed in an insulated container to one-half of
its initial volume. The ratio of the specific heat capacities at constant pressure and constant volume, cP/cV, for
carbon dioxide is approximately 1,3. Determine the final pressure of the gas if the initial pressure is
2.0 x 105 Pa. Ans.: 4,93 x 105 Pa
12.2
The pressure and volume of a gas are changed along the path ABCA. Using the data shown in the graph,
determine the work done (including the algebraic sign) in each segment of the path: (a) A to B, (b) B to C, and
(c) C to A. Ans.: 0 J; +2,1 x 103 J; –1,5 x 103 J
12.3
The pressure of a monatomic ideal gas (γ = 1,67) doubles during an adiabatic compression. What is the ratio
of the final volume to the initial volume? Ans.: 0,66
12.4
A monatomic ideal gas (γ = 1,67) is contained within a perfectly insulated cylinder that is fitted with a movable
piston. The initial pressure of the gas is 1,5 x 105 Pa. The piston is pushed so as to compress the gas, with the
result that the Kelvin temperature doubles. What is the final pressure of the gas? Ans.: 8,49 x 105 Pa
12.5
The work done by one mole of a monatomic ideal gas (γ = 1,67) in expanding adiabatically is 825 J. The initial
temperature and volume of the gas are 393 K and 0,100 m3. Obtain (a) the final temperature and (b) the final
volume of the gas. Ans.: 327 K; 0,132 m3
12.6
An engine is run in reverse as a heat pump. An identical engine (with the same values of QH, QC and W as the
first engine) is run in reverse as a refrigerator. The coefficient of performance of the heat pump is three times
greater than the coefficient of performance of the refrigerator. Obtain the coefficient of performance of the
refrigerator, coefficient of performance of the heat pump, and the efficiency of the engine. Ans.: 0,5; 1,5; 66%
12.7
Engine 1 has an efficiency of 0,18 and requires 5500 J of input heat to perform a certain amount of work.
Engine 2 has an efficiency of 0,26 and performs the same amount of work. How much input heat does the
second engine require? Ans.: 3800 J
12.8
Engine A receives three times more input heat, produces five times more work, and rejects two times more
heat than engine B. Find the efficiency of (a) engine A and (b) engine B. Ans.: 0,55; 0,33
12.9
A Carnot engine has an efficiency of 0,700, and the temperature of its cold reservoir is 378 K. (a) Determine
the temperature of its hot reservoir. (b) If 5230 J of heat is rejected to the cold reservoir, what amount of heat
is put into the engine? Ans.: 1260 K, 1,74 x 104 J
12.10
A refrigerator, Heat pump, and Air-conditioner operates between temperatures of 296 and 275 K. What would
be their maximum coefficient of performance? Ans.: 13
12.11
The temperatures indoors and outdoors are 299 and 312 K, respectively. A Carnot air conditioner deposits
6.12 x 105 J of heat outdoors. How much heat is removed from the house? Ans.: 5,86 x 105 J
12.12
To keep a room at comfortable 21oC, a Carnot heat pump does 345J of work and supplies it with 3240J of heat.
a.
How much heat is removed from the outside air by the heat pump? Ans.: 2895 J
b.
What is the temperature of the outside air? Ans.: 262,7 K
c.
What is the coefficient of performance of the heat pump? Ans.: 9,39
12.13
On a cold day, 24 500 J of heat leaks out of a house. The inside temperature is 21 °C, and the outside
temperature is –15 °C. What is the increase in the entropy of the universe that this heat loss produces?
12.14
In an isothermal process 2,33 moles of an ideal gas is compressed to one-fifth of its initial volume at 285 K.
What quantity of heat is added to, or removed from, the system during this process? (State also whether this
heat is removed or added.) Ans.: –8,89 x 103 J, removed
12.15
The temperatures indoors and outdoors are 299 and 312 K respectively. A Carnot air conditioner deposits
6,12 x 105 J of heat outdoors. How much heat is removed from the house? Ans.: 5,86 x 105 J
60 | P a g e
12.16
The compression ratio of a certain diesel engine is 15. This means that air in the cylinders is compressed to
1/15 of its initial volume. The initial pressure is 1,0 x 10 5 Pa and the initial temperature is 300 K. The final
temperature is 886 K. Air is mostly a mixture of oxygen and nitrogen, and  = 1,40.
a.
Find the final pressure after compression. Ans.: 4,43 x 106 Pa
b.
How much work does the gas do during the compression if the initial volume of the cylinder is
1,0 x 10–3 m3? Ans.: –292,3 J
c.
Is this work done on the system or by the system? Give a reason for your answer. Ans.: Negative; gas
is compressed, then work is done on the system
12.17
From a hot reservoir at a temperature of T1, Carnot engine A takes an input heat of 5550 J, delivers 1750 J of
work, and rejects heat to a cold reservoir that has a temperature of 503 K. This cold reservoir at 503 K also
serves as the hot reservoir for Carnot engine B, which uses the rejected heat of the first engine as input heat.
Engine B also delivers 1750 J of work, while rejecting heat to an even colder reservoir that has a temperature
of T2. Find the temperatures T1 and T2. Ans.: 734,65K; 271,36 K
12.18
An engine is used to lift a 2700 kg truck to a height of 3,0 m at constant speed. In the lifting process, the engine
received 3,3 x 105 J of heat from the fuel burned in its interior. What is the efficiency of the engine?
Ans.: 24,05%
12.19
The engine of a Ferrari F1 sports car takes air in at 20 oC and 1 atm, and then compresses it adiabatically to
0,09 times its original volume. The air may be treated as an ideal gas with  = 1,4. Calculate the final
temperature and pressure of the air in the engine. Ans.: 767,66 K; 2,95 x 106 Pa
12.20
An ideal monatomic gas expands isothermally from state A to state B. It then cools at constant volume to state
C. The gas is then compressed isobarically to D where it is then heated until it returns to state A.
a.
b.
c.
d.
e.
f.
What is the internal energy of the gas at point B?
Ans.: 3,03 x 103 J
What is the pressure of the gas when it is in state B? Ans.: 5,05 x 105 Pa
What is the temperature of the gas when it is in state C (in °C)? Ans.: –33,04°C
What is the ratio of the internal energy of the gas in state C to that in state A? Ans.: 0,4
How much work is done on the gas as it is compressed isobarically from state C to D? Ans.: –404 J
What is the net amount of work done after one complete cycle? Ans.: 996,37 J
12.24 A cylinder with a piston contains 0,25 mole of ideal oxygen at pressure 2,4 x 105 Pa and temperature 355 K.
The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its
original volume and finally it is cooled isochorically to its original pressure.
a. Show the series of the process on a PV diagram.
b. Calculate the initial volume of the gas. Ans.: 3,07 x 10–3 m3
c. Calculate the temperature during the isothermal process. Ans.: 708,9 K
d. Calculate the maximum pressure. Ans.: 4,8 x 105 Pa
e. Calculate each of the following:
Initial isobaric expansion
Isothermal compression
Isochoric cooling
Work done by oxygen
(i).
(iv).
(vii).
Change in internal energy
(ii).
(v).
(viii).
Heat transfer to oxygen
(iii).
(vi).
(ix).
12.25 A refrigerator has a coefficient of performance of 2,1. In each cycle it absorbs 3,4 x 104 J of heat from the cold
reservoir. (a) How much work is done in each cycle? (b) In each cycle, how much heat is discarded to the high
temperature reservoir? Ans.: 16190.48 J; 50190 J
12.26 A first year student with nothing better to do add heat to 0,35 kg ice at 0 oC until it’s all melted. The source of
the heat is a very massive body at a temperature of 25 oC. Lf=33,5x104 J/kg.
a. What is the change in entropy of the water? Ans.: 429,5 J/K
b. What is the change in entropy of the hot object? Ans.: –393,5 J/K
c. Does the above (negative) answer contradict the 2nd law – entropy statement?
d. Is this process reversible or irreversible? Explain your answer.
61 | P a g e
LEARNING UNIT 13
CHAPTER 30: THE NATURE OF THE ATOM
After completion of this learning unit you should be able to:
Explain the different types of X-rays.
Explain how X-rays are produced.
Explain what the LASER is.
Explain how a LASER works.
Study the following paragraphs in the textbook:
30.7
X-Rays ................................................................................ p 955 (p 919)
30.8
The Laser ........................................................................... p 959 (p 923)
Know the following concepts:







A
B.
Characteristic X-rays
Bremsstrahlung
Fig 30.19 and Fig 30.20
Spontaneous emission
Stimulated emission
Population inversion
Metastable
X-rays
Draw a named sketch of a x-ray tube and explain how X rays are formed.
X rays are formed when electrons from a heated filament is accelerated through a large potential difference.
The electrons collide with a target metal contained in a vacuum. If the electron has enough energy, it knocks
out an electron from the K shell of the target metal.
An electron from a higher energy level fills the empty space and releases an X ray photon
Name three types of X-rays that are formed and explain how each one is formed
K X rays:
X rays formed when an electron from the n = 2 energy level fill the empty space.
K X rays:
X rays formed when an electron from the n = 3 energy level fill the empty space.
Bremsstrahlung: X rays formed when, an electron, decelerates when hitting the target metal.
62 | P a g e
The laser
A.
What does the word “laser” mean?
Light amplification by stimulated emission of radiation
B.
When does emission occur?
When an electron makes a transition from a higher energy state to a lower energy state and emits a photon
C.
Name the two types of emission processes and explain how each happens.
Spontaneous emission: A photon is emitted spontaneously in random direction.
Stimulated emission:
An incoming photon stimulates the electron to change energy levels.
D.
Name a condition for stimulated emission to occur.
The incoming photon must have an energy that exactly matches the difference in energy between the two
energy levels.
E.
State the three important features of stimulated emission
One photon goes in and two photons come out.
The emitted photon travels in the same direction as the incoming photon.
The emitted photon has the same phase as the incoming photon.
F.
Explain population inversion
When more electrons are exited to a higher energy state than remain in the lower energy state.
G.
Define a metastable state
A higher energy state in which electrons remain for a longer period than in an ordinary exited state.
H.
Draw a named sketch and explain the working of the helium/neon laser
A high voltage is discharged across a low pressure mixture of 15% helium and 85% neon in a cylindrical tube.
An atom spontaneously emits a photon parallel to the axis of the tube.
This photon stimulates another atom to emit two photons parallel to the tube axis.
These two stimulate the emission of four others and so on.
The ends of the tube are silvered to form mirror that reflects the photons back and forth through the mixture to
ensure continuous stimulation.
One end is only partially silvered so that some photons escape to form the laser beam.
63 | P a g e
LEARNING UNIT 14
CHAPTER 32: IONISING RADIATION, NUCLEAR ENERGY, AND
ELEMENTARY PARTICLES
After completion of this learning unit you should be able to:
Explain what Ionising Radiation is.
Explain the effects of Ionising Radiation on humans.
Explain what Induced Nuclear Reactions is.
Explain what Nuclear Fission is.
Explain the different types of Chain Reactions.
Explain what Nuclear Reactors are.
Explain the 3 types of Nuclear Reactors.
Explain the working of a Nuclear Power station (with the aid of a sketch).
Explain what Nuclear Fusion is.
Explain the difficulties of building a Nuclear Fusion reactor.
Solve LCR circuit problems.
Study the following paragraphs in the textbook:
32.1
Biological Effects of Ionizing Radiation .............................. p 1004 (p 967)
32.2
Induced Nuclear Reactions ............................................... p 1008 (p 969)
32.3
Nuclear Fission ................................................................. p 1010 (p 971)
32.4
Nuclear Reactors .............................................................. p 1012 (p 973)
32.5
Nuclear Fusion ................................................................. p 1014 (p 975)
Know the following concepts:



















Ionising radiation
Exposure
Absorbed Dose
Relative Biological effectiveness
Biological Equivalent Dose
Short and Long term effects
Radiation sickness
Induced nuclear transmutation
Thermal neutrons
Nuclear fission
Chain reaction (2 types)
Fuel elements
Reactor core
Moderator
Control rods
Reactor (3 types)
Nuclear Power Station (Fig 32.8) (Fig 32.7)
Nuclear Fusion
Thermonuclear reactions
64 | P a g e
Biological effects of ionising radiation
A
Define each of the following:
i
Ionising radiation:
Radiation that consists of photons that have sufficient energy to knock an electron from an atom to form
an ion.
ii
Exposure:
Measure of ionisation produced in air. Defined as the total charge per unit mass of air:
Exposure 
iii
q
in C/kg
m
Absorbed dose
Energy absorbed from the radiation per mass of absorbed material:
Absorbed dose 
Energy absorbed
in J/kg or Gray (Gy)
Mass of absorbing material
Another unit is rad and 1 rad=0.01 gray
iv
RBE (Relative biological effectiveness)
The dose of 200 keV X rays that
Relative biological effectiven ess 
produces a certain biological effect
The dose of radiation that produces
the same biological effect
v
Biological equivalent dose:
Biological equivalent dose (in rem) = Absorbed dose (rad) x RBE
B
Name the two categories that the effects of radiation can be divided into and explain each.
Short term effects:
effect that appear within minutes, day or weeks
Long term effects:
effect that appear years, decades or centuries later
C
What is radiation sickness and give at least three examples.
The acute effects of radiation. Examples: nausea, vomiting, fever, diarrhoea, loss of hair
A
A
Induced nuclear reactions
Define each of the following:
i
Nuclear reaction:
When an incident particle induces a change in the nucleus of the target material
ii
Induced nuclear transmutation:
When the incident particle induces a transmutation from one element into another element
iii
Thermal neutrons:
Neutrons with a kinetic energy of 0,04 eV or less
Nuclear fission
Define each of the following:
i.
Nuclear fission and give one example:
The splitting of a massive nucleus into two smaller nuclei.
Example:
0 n  92 U 56 Ba  36 Kr 30 n
Chain reaction:
A series of nuclear fissions whereby the neutrons produced causes additional fissions.
1
ii
iii
235
141
92
1
Controlled chain reaction
By limiting the number of neutrons in the environment the fissions can be controlled so that only one
neutron fissions another nucleus. By doing this the chain reaction and the rate of production are
controlled.
65 | P a g e
Nuclear reactors
A
B
What is a nuclear reactor?
A type of furnace in which energy is generated by a controlled fission chain reaction.
Name and discuss the three basic components of a nuclear reactor
Fuel elements: Contain the fissile, fuel such as Uranium
May be in the shape of thin rods about 1 cm in diameter.
In a large power reactor there may be thousands of fuel elements placed together- it is then
called the reactor core
Moderator
Material that slows down or moderates the speed of the neutrons so that they can fission
additional nuclei.
Commonly used moderator is water.
Energetic neutrons collide with the water molecules and losses energy with each collision.
Control rods
Can be moved into or out of the reactor core.
Contains an element such as boron or cadmium that absorbs neutrons without fissioning.
C
Define the following:
i.
Critical reactor
When each fission leads to one additional fission and the reactor produces, a steady output of energy it
is called a critical reactor.
ii
Sub critical reactor
When on average each fission leads to less than one additional fission and the reactor eventually dies
out it is called a sub critical reactor
iii
Super critical reactor
When each fission leads to more than one additional fission and the energy released by the reactor
increases, it is called a super critical reactor.
D
Discuss (with the aid of a sketch) the working of a pressurized water reactor
The heat generated in the fuel rods is given to the water around the rods.
The temperature of the water is allowed to rise to about 300°C to remove as much energy from the rods as
possible.
To prevent vaporization the -water is pressurized to more than 150- atmospheres.
The hot water is pumped through a heat exchanger where the heat is transferred to water flowing in a second
closed system.
The heat transferred to the second system produces steam that drives a turbine coupled to an electric
generator producing electrical power.
After leaving the turbine the steam is condensed and returned to the heat exchanger.
Nuclear fusion
A
Define nuclear fusion
Two very low mass nuclei with small binding energy combine to form a larger nucleus
B
Why is it difficult to build a fusion reactor?
The two nuclei must be brought very close to each other so that the strong nuclear force can pull them
together, leading to fusion.
The nuclei on the other hand repel each other since they are both positively charged.
A large kinetic energy is therefore needed for die charges to come sufficiently close. This is obtained by
working at a very high temperature
When these high temperatures are used the atoms are ionized to form a plasma.
This plasma can not easily be confined for a long time so that collision among the ions can lead to fusion.
C
Define thermonuclear reaction
Nuclear reaction that occur at very high temperatures, (about 4 x 108 K)
66 | P a g e
EXAMINATION FORMULAE SHEET AND CONSTANTS
g = 9.80 m/s2
water = 1 000 kg/m3
k = 1.38 x 10-23 J/K1
R = 8.314 J./(K.mol)
0 = 4 x 10-7 T/(m.A)
Acceleration due to gravity
Density of water
Boltzmann’s constant
Universal gas constant
Permeability of free space
I  I o sin wt
V RMS 
V0
2
; I RMS 
I0
2
V2
R
t



q  q 0 1  e RC 


q  q0 e
t
RC
F  IBL sin 
 er 2 
m   B 2
 2V 
  NIAB sin 
r
mv
qB
0 I
I
B 0
2R
2R
B  n 0 I
E  vBL
E  NAB  sin ωt
V E
I
R
N S VS I P
 
N P VP I S
1
XC 
X L  2fL
2fC
B
Z  R 2  X L  X c 
XL  XC
R
P  IV cos 
tan  
w f  wi  t
v 2  u 2  2as
w f  wi  2
2
2
1
s  ut  at
  wi t  t 2
2
1
1
s  u  v t
  wi  w f t
2
2
s  r v  rw a  rw
2r
v
T
Fc  ma c
v2
ac 
r
   I
  Fl
L  Iw
  2f
f 
k
m


W R  
Fspring  kx
1
T
g
L
1
KE R  Iw 2
2
1
PE elasti  kx 2
2
x  a cos t
v   A sin t
a   A cos t
2
 L 
F  Y   A
 Lo 
 x 
F  S   A
 Lo 
 V 
P   B 
 Vo 
n
2
2
1
2
P  IV  I 2 R 

v  u  at
N
NA
m
Mr
vA
F
y
n
R 4 P
Q
8L
1
2
KE  mv rms
2
KE  32 kT
U  32 NkT  32 nRT
PV  nRT
DACt
m
L
U  W  Q
Vf 
W  nRT ln  
 Vi 
W  PV  nRT
Q  nCT
Q P  52 nRT
QV  32 nRT
PiVi   Pf V f
QH
W
Q H TH

QC TC
e
COP 
QH
W
COP 
QC
W
67 | P a g e
1A
1
1
H
1,00
8
0
PERIODIC TABLE
2A
3A
4A
5A
6A
7A
2
He
4,00
3
6
C
12,0
1
14
Si
28,0
9
32
Ge
72,5
9
50
Sn
118,
7
82
Pb
207,
2
7
N
14,0
1
15
P
30,9
7
33
As
74,5
9
51
Sb
121,
8
83
Bi
209,
0
8
O
16,0
0
16
S
32,0
6
34
Se
78,9
6
52
Te
127,
6
84
Po
209,
0
9
F
19,0
0
17
Cl
35,4
5
35
Br
79,9
0
53
I
126,
9
85
At
210,
0
10
Ne
20,1
8
18
Ar
39,9
5
36
Kr
83,8
0
54
Xe
131,
3
86
Rn
222,
0
KEY / SLEUTEL
2
3
4
5
6
7
3
Li
6,94
1
11
Na
22,9
9
19
K
39,1
0
37
Rb
85,4
7
55
Cs
132,
9
87
Fr
223,
0
4
Be
9,01
2
12
Mg
24,3
1
20
Ca
40,0
8
38
Sr
87,6
2
56
Ba
137,
3
88
Ra
226,
0
17 Atomic number
Cl Symbol
35,4 Atomic mass
5
21
Sc
44,9
6
39
Y
88,9
1
57
La
138,
9
89
Ac
227,
0
22
Ti
47,9
0
40
Zr
91,2
2
72
Hf
178,
5
104
Unq
261,
1
23
V
50,9
4
41
Nb
92,9
1
73
Ta
180,
9
105
Unp
262,
1
24
Cr
52,0
0
42
Mo
95,9
4
74
W
183,
9
106
Unh
263,
1
5
B
10,8
1
13
Al
26,9
8
25
26
27
28
29
30
31
Mn
Fe
Co
Ni
Cu
Zn
Ga
54,9 55,8 58,9 58,7 63,5 65,3 69,7
4
5
3
0
5
8
2
43
44
45
46
47
48
49
Tc
Ru
Rh
Pd
Ag
Cd
In
98,9 101, 102, 106, 107, 112, 114,
1
1
9
4
9
4
8
75
76
77
78
79
80
81
Re
Os
Ir
Pt
Au
Hg
Tl
186, 190, 192, 195, 197, 200, 204,
2
2
2
1
0
6
4
107
Uns
262,
1
LANTHANIDES /
LANTANIEDE
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140, 140, 144, 144, 150, 152, 157, 158, 162, 164, 187, 168, 173, 175,
1
9
2
9
4
0
3
9
5
9
3
9
0
0
ACTINIDES / 90
91
92
93
94
95
96
97
98
99
100 101 102 103
AKTINIEDE Th
Pa
U
Np
Pu
Am Cm
Bk
Cf
Es
Fm Md
No
Lr
232, 231, 238, 237, 244, 243, 247, 247, 242, 252, 257, 258, 259, 260,
0
0
0
0
1
1
1
1
1
1
1
1
1
1
68 | P a g e