REGAN
PHY34210
PHYS 34210 PHYSICS I
Notre Dame,
London Programme,
Fall 2013
Prof. Paddy Regan
Dept. of Physics, University of
Surrey, Guildford, GU2 7XH, UK
E-Mail: p.regan@surrey.ac.uk
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•
•
•
Course & General Information
REGAN
PHY34210
Lectures, usually, Tuesdays 2.15-5.00
– first lecture Tues 27th August 2010
– One ‘makeup’ lecture Mon. 30th Sept. 5.15 – 8pm (no class on Tues. 29th Oct)
Grading
– 3 x 2 hour class examinations
• Exam 1 : Tues. 24th September (30%);
• Exam 2: Tues 5th November (35%),
• Exam 3: Tues. 26th November (35%)
Some information about Prof. Paddy Regan FInstP CPhys:
–
National Physical Lab. & University of Surrey Chair Professor in Radionuclide Metrology, ( staff since 1994).
– BSc University of Liverpool (1988); DPhil University of York (1991).
– Adjunct Assoc. Prof. at ND London 2002-7; Full Professor from 2007 - present
– Held post-doctoral research positions at:
• University of Pennsylvania, Philadelphia, USA (1991-2)
• Australian National University, Canberra, Australia (1992-4);
• Yale University (sabbatical researcher 2002 ; Flint Visiting Research Fellow 2004 – 2013)
– Co-author of >200 papers in nuclear physics; supervised 25 PhD students so far + 100 Masters.
– Led RISING and PreSPEC projects (major nuclear physics research project at GSI, Germany).
– Married (to a nurse), 4 kids.
– Understands gridiron, baseball, (ice) hockey etc., regular visitor to US (and other countries)
– Still plays squash and golf (poor, 27); formerly football (soccer), cricket & a bit of rugby (union).
– Occasional half marathons for the mental health charity, MIND (see
– http://uk.virginmoneygiving.com/Paddy-James-Clare-Regan
– Have also done some (physics related) media work in the UK and USA, see e.g.,
– http://www.bbc.co.uk/news/world-asia-pacific-12744973
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REGAN
PHY34210
Course textbook,
Fundamentals of Physics,
Halliday, Resnick & Walker,
published by Wiley & Sons.
Now in 9th Edition.
http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001575.html
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Course Timetable (2013)
PART 1
Lect 1: 27 Aug (Cp 1,2)
Lect 2: 03 Sept (Cp 3,4)
Lect 3: 10 Sept (Cp 5,6)
Lect 4: 17 Sept (revision)
Lect 5: 24 Sept Exam 1
PART 2
* Lect 6: Mon. 30th Sept.
(Ch. 7,8) 5.15 - 8pm
* Lect 7: 01 Oct (Ch 9,10)
* Lect 8: 08 Oct. (Ch 11,12)
* Lect 9: 15 Oct. (revision)
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PHY34210
4
PART 3
Lect 11: 12 Nov. (13,14)
Lect 12: 19 Nov. (15,16)
Lect 13: Weds. 20 Nov
(17,18) 5.15-8pm.
Lect 14: 26 Nov Exam 3
Break, no lect. 22nd Oct.
No lect. 29th Oct (resched).
* Lect 10: 5th Nov Exam 2
Course notes and past papers/solns can be found at the following link:
http://personal.ph.surrey.ac.uk/~phs1pr/lecture_notes/notre_dame/
• 1: Measurement
1st Section:
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PHY34210
– Units, length, time, mass
• 2: Motion in 1 Dimension
– displacement, velocity, acceleration
• 3: Vectors
– adding vectors & scalars, components, dot and cross products
• 4: Motion in 2 & 3 Dimensions
– position, displacement, velocity, acceleration, projectiles,
motion in a circle, relative motion
• 5: Force and Motion: Part 1
– Newton’s laws, gravity, tension
• 6: Force and Motion: Part 2
– Friction, drag and terminal speed, motion in a circle
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2nd
Section:
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PHY34210
• 7: Kinetic Energy and Work
– Work & kinetic energy, gravitational work, Hooke’s law, power.
• 8: Potential Energy and Conservation of Energy
– Potential energy, paths, conservation of mechanical energy.
• 9: Systems of Particles
– Centre of mass, Newton’s 2nd law, rockets, impulse,
• 10: Collisions.
– Collisions in 1 and 2-D
• 11 : Rotation
– angular displacement, velocity & acceleration, linear and
angular relations, moment of inertia, torque.
• 12: Rolling, Torque and Angular Momentum
– KE, Torque, ang. mom., Newton’s 2nd law, rigid body rotation
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3rd section:
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PHY34210
• 13: Equilibrium and Elasticity
– equilibrium, centre of gravity, elasticity, stress and strain.
• 14: Gravitation
– Newton’s law, gravitational potential energy, Kepler’s laws.
• 15: Fluids
– density and pressure, Pascal’s principle, Bernoulli’s equation.
• 16 : Oscillations
– Simp. Harm. Mo. force and energy, pendulums, damped motion.
• 17 & 18 : Waves I and II
– Types of Waves, wavelength and frequency, interference,
standing waves, sound waves, beats, Doppler effect.
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Recommended Problems and Lecture Notes.
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PHY34210
Problems are provided at the end of each book chapter.
Previous years examinations papers will also be provided with
solutions (later) for students to work through at their leisure.
No marks will be give for these extra homework problems
Final grade will come from the three class exams.
Full lecture notes can be found on the web at
http://www.ph.surrey.ac.uk/~phs1pr/lecture_notes/phy34210_13.ppt
and
http://www.ph.surrey.ac.uk/~phs1pr/lecture_notes/phy34210_13.pdf
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1: Measurement
REGAN
PHY34210
Physical quantities are measured in specific UNITS, i.e., by
comparison to a reference STANDARD.
The definition of these standards should be practical for the
measurements they are to describe (i.e., you can’t use a ruler to
measure the radius of an atom!)
Most physical quantities are not independent of each other (e.g.
speed = distance / time). Thus, it often possible to define all other
quantities in terms of BASE STANDARDS including length
(metre), mass (kg) and time (second).
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SI Units
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PHY34210
10
The 14th General Conference of Weights and Measures (1971) chose
7 base quantities, to form the International System of Units
(Systeme Internationale = SI).
There are also DERIVED UNITS, defined in terms of BASE UNITS,
e.g. 1 Watt (W) = unit of Power = 1 Kg.m2/sec2 per sec = 1 Kg.m2/s3
Scientific Notation
In many areas of physics, the measurements correspond to very large
or small values of the base units (e.g. atomic radius ~0.0000000001 m).
This can be reduced in scientific notation to the ‘power of 10’ ( i.e.,
number of zeros before (+) or after (-) the decimal place).
e.g. 3,560,000,000m
= 3.56 x 109 m = 3.9 E+9m
&
0.000 000 492 s = 4.92x10-7 s = 4.92 E-7s
Prefixes
For convenience, sometimes,
when dealing with large or small
units, it is common to use a prefix
to describe a specific power of 10
with which to multiply the unit.
e.g.
1000 m = 103 m = 1E+3 m = 1 km
0.000 000 000 1 m = 10-10 m = 0.1 nm
•
•
•
•
•
•
•
•
•
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PHY34210
1012 = Tera = T
109 = Giga = G
106 = Mega = M
103 = Kilo = k
10-3 = milli = m
10-6 = micro = m
10-9 = nano = n
10-12 = pico = p
10-15 = femto = f
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Converting Units
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PHY34210
It is common to have to convert between different systems of units
(e.g., Miles per hour and metres per second). This can be done most
easily using the CHAIN LINK METHOD, where the original value
is multiplied by a CONVERSION FACTOR.
NB. When multiplying through using this method, make sure you
keep the ORIGINAL UNITS in the expression
e.g., 1 minute = 60 seconds, therefore (1 min / 60 secs) = 1
and
(60 secs / 1 min) = 1
Note that 60 does not equal 1 though!
Therefore, to convert 180 seconds into minutes,
180 secs = (180 secs) x (1 min/ 60 secs) = 3 x 1 min = 3 min.
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Length (Metres)
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PHY34210
13
Original (1792) definition of a metre (meter in USA!) was
1/10,000,000 of the distance between the north pole and the equator.
Later the standards was changed to the distance between two lines
on a particular standard Platinum-Iridium bar kept in Paris.
(1960) 1 m redefined as 1,650,763.73 wavelengths of the (orange/red)
light emitted from atoms of the isotope 86Kr.
(1983) 1 m finally defined as the length travelled by light in vacuum
during a time interval of 1/299,792,458 of a second.
• To Andromeda Galaxy
• Radius of earth
• Adult human height
• Radius of proton
~ 1022 m
~ 107 m
~2m
~ 10-15 m
Time (Seconds)
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PHY34210
Standard definitions of the second ?
Original definition
1/(3600 x 24) of a day, 24 hours = 1day, 3600 sec per hours, thus
86,400 sec / day, 3651/4 days per year and 31,557,600 sec per year.
But, a day does not have
a constant duration!
(1967) Use atomic clocks,
to define 1 second as
the time for
9,192,631,770 oscillations
of the light of a specific
wavelength (colour) emitted
from an atom of caesium (133Cs)
From HRW, p6
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Mass (Kg, AMU)
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PHY34210
1 kg defined by mass of Platinum-Iridium cylinder near to Paris.
Masses of atoms compared to each other for other standard.
Define 1 atomic mass unit = 1 u (also sometimes called 1 AMU) as
1/
12
the mass of a neutral carbon-12 atom.
1 u = 1.66054 x 10-27 kg
Orders of Magnitude
It is common for physicists to ESTIMATE the magnitude of
particular property, which is often expressed by rounding up (or
down) to the nearest power of 10, or ORDER OF MAGNITUDE,
e.g.. 140,000,000 m ~ 108m,
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Estimate Example 1:
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PHY34210
A ball of string is 10 cm in diameter, make an order of magnitude
estimate of the length, L , of the string in the ball.
4 3
Volume of string, V  d L  r
3
r  radius of ball  10cm/2 5cm  0.05m
2
r
assume cross- sectionof string~ 3mmsquare
d
d
4
3
2
2
V   0.05m   3m m L  0.003m  L
3
4
 0.05m 3 4  0.05 0.05 0.05m 3
 L 3

2
0.003 0.003m 2
0.003m 
 L  55.5m  60m
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REGAN
PHY34210
E.g., 2: Estimate Radius of Earth (from the beach.)
From P ythagoras
,
d 2  r 2  r  h   r 2  2rh  h 2
2
d

 d 2  2rh  h 2 , but h   r
 d 2  2rh
h
r
r
 is the angle through which the
sun moves around the earth during the
time between the ‘two’ sunsets (t ~ 10 sec).

t
t ( 10sec)  360(deg) 3600
o


(deg)




(deg)

0
.
04
360o 24 hours
24 60 60sec
86,400
2h
Now, from trigonometry,d  r tan thusr 2 tan2   2rh  r 
tan2 
2  2m
4m
6
if h  human height ~ 2m, thensubstituting, r 


8

10
m
2
o
7
tan (0.04 ) 4.9 10

(acceptedvalue for earthradius  6.4x108 m!)
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2: Motion in a Straight Line
REGAN
PHY34210
Position and Displacement.
To locate the position of an object we need to define this RELATIVE
to some fixed REFERENCE POINT, which is often called the
ORIGIN (x=0).
In the one dimensional case (i.e. a straight
line), the origin lies in the middle of an
AXIS (usually denoted as the ‘x’-axis)
which is marked in units of length.
x = -3 -2 -1 0 1 2 3
Note that we can also define
NEGATIVE co-ordinates too.
The DISPLACEMENT, Dx is the change from one position to
another, i.e., Dx= x2-x1 . Positive values of Dx represent motion in
the positive direction (increasing values of x, i.e. left to right looking
into the page), while negative values correspond to decreasing x.
Displacement is a VECTOR quantity. Both its size (or ‘magnitude’)
AND direction (i.e. whether positive or negative) are important.
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REGAN
PHY34210
19
Average Speed and Average Velocity
We can describe the position of an
object as it moves (i.e. as a function
of time) by plotting the x-position
of the object (Armadillo!) at different
time intervals on an (x , t) plot.
The average SPEED is simply the
total distance travelled (independent
of the direction or travel) divided by
the time taken. Note speed is a
SCALAR quantity, i.e., only its
magnitude is important (not its
direction).
From HRW
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PHY34210
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The average VELOCITY is defined
by the displacement (Dx) divided
by the time taken for this
displacement to occur (Dt).
x  x Dx
vav  2 1 
t2  t1 Dt
The SLOPE of the (x,t) plot gives
average VELOCITY.
Like displacement, velocity is a
VECTOR with the same sign as
the displacement.
The INSTANTANEOUS VELOCITY
is the velocity at a specific
moment in time, calculated by making
Dt infinitely small (i.e., calculus!)
Dx dx
v  lim Dt  0

Dt dt
Acceleration
Acceleration is a change in
velocity (Dv) in a given time (Dt).
The average acceleration, aav,
is given by
dx2 dx1

v2  v1
Dv
dt
dt
aav 


t 2  t1
t 2  t1
Dt
The instantaneous ACCELERATION
is given by a, where,
dv d  dx  d 2 x
a
   2
dt dt  dt  dt
SI unit of acceleration is
metres per second squared (m/s2)
REGAN
PHY34210
HRW
21
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PHY34210
22
Constant Acceleration and the Equations of Motion
For some types of motion (e.g., free
fall under gravity) the acceleration
is approximately constant, i.e., if
v0 is the velocity at time t=0, then
By making the assumption
that the acceleration is a
constant, we can derive a set
of equations in terms of the
following quantities
v  v0
a  aav 
t 0
x  x0
 thedisplacement
v0
 initialvelocity(at timet  0)
v
a
t
 velocityat timet
 acceleration (constant)
 timetakenfromt  0
Usually in a given problem, three of these quantities are given
and from these, one can calculate the other two from the following
equations of motion.
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PHY34210
Equations of Motion (for constant a).
x  x0
v  v0  at  (1) recallingthatvav 
, then
t 0

v0  v 
since by definition, vav 
, 2vav  v  v0
2
1
substituting into (1) for v0 gives vav  v0  at ,
2
d x  x0 
1 2
x  x0  v0t  at  (2) , note
 v  v0  at
2
dt
combining(1) and (2) to eliminatet , a and v0 gives
1
v  v  2a ( x  x0 )  (3) ; x  x0  v0  v t  (4) ; and
2
1 2
x  x0  vt  at  (5)
2
2
2
0
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Alternative Derivations (by Calculus)
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PHY34210
24
dv d 2 x
a
 2 by definit ion  dv  a  dt
dt dt
  dv   a dt and thusfor a  constant
v  at  C , evaluatedby knowing thatat
t  0, v  v0  v0  a  0  C thus, v  v0  at  (1)
dx
 v  dx  v  dt , v not constant,but v0 is, therefore
dt
by substitution,  dx   v0  atdt   dx  v0  dt  a  t dt
1 2
integrating gives, x  v0t  at  C , calculateC by
2
knowing x  x0 at t  t0  C  x0
1 2
and x  x0  v0t  at  (5)
2
Free-Fall Acceleration
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PHY34210
At the surface of the earth, neglecting any effect due to air resistance
on the velocity, all objects accelerate towards the centre of earth
with the same constant value of acceleration.
This is called FREE-FALL ACCELERATION, or ACCELERATION
DUE TO GRAVITY, g.
At the surface of the earth, the magnitude of g = 9.8 ms-2
Note that for free-fall, the equations of motion are in the y-direction
(i.e., up and down), rather than in the x direction (left to right).
Note that the acceleration due to gravity is always towards the centre
of the earth, i.e. in the negative direction, a= -g = -9.8 ms-2
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Example
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PHY34210
26
A man throws a ball upwards with an initial velocity of 12ms-1.
(a) how long does it take the ball to reach its maximum height ?
(a) since a= -g = -9.8ms-2,
Therefore, time to max height from
initial position is y0=0 and
v  v0 0  12m s1
at the max. height vm a x=0 v  v0  at  t  a   9.8m s2  1.2s
(b) what’s the ball’s maximum height ?
2
2
2
1


v

v

2
a
(
y

y
)

(
12
)

2
a
y

0
&
v

0
,
v


12
m
s
,
(b)
0
0
0
2
2
1 2
v

v
0

(
12
m
s
)
2
0
a   g  9.8m s  y 

 7.3m
2
2a
2  9.8m s
REGAN
PHY34210
( c) How long does the ball take to reach a point 5m above its initial
release point ?
v0  12m s1 , a  9.8m s2 , y  y0  5m
1 2
1
1
 from y  y0  v0t  at  5m  12m s t  9.8m s 2t 2
2
2
assuming SI units, we havea quadratic equation,4.9t 2  12t  5  0
recallingat 2  bt  c  0 solutionsare given by
 b  b 2  4ac
t
 t  0.5s AND  1.9s
2a
Note that there are TWO SOLUTIONS here (two different ‘roots’ to
the quadratic equation). This reflects that the ball passes the same
point on both the way up and again on the way back down.
27
3: Vectors
REGAN
PHY34210
• Quantities which can be fully described just by
their size are called SCALARS.
– Examples of scalars include temperature, speed,
distance, time, mass, charge etc.
– Scalar quantities can be combined using the standard
rules of algebra.
• A VECTOR quantity is one which need both a
magnitude (size) and direction to be complete.
– Examples of vectors displacement, velocity,
acceleration, linear and angular momentum.
– Vectors quantities can be combined using special
rules for combining vectors.
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
b
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PHY34210
Adding Vectors Geometrically
Any two vectors can be added using the
  
VECTOR EQUATION, where the sum of s  a  b
vectors can be worked out using a triangle.



      
r  a b c  a  b c


s

a

b
Note that two vectors can be
added together in either order to
get the same result. This is called
the COMMUTATIVE LAW.
Generally, if we have more than
2 vectors, the order of combination
does not affect the result. This is
called the ASSOCIATIVE LAW.
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
a

a

s
   
s  b a  a b

b

b

a

c

s

r

c

b
=

a
'
s 
r
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PHY34210
Subtracting Vectors, Negative Vectors


 b is thesame magnitudeas b
but in theoppositedirection.

b 

d
a

b

s
  
s  a b

b

a

s
as with usual algebra, we can
 
   

d  a b  a  b
re - arrangevectorequations,
  
  
e.g., d  a  b  d  b  a
Note that as with all quantities, we can only add / subtract vectors
of the same kind (e.g., two velocities or two displacements).
We can not add differing quantities e.g., apples and oranges!)
30
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PHY34210
Components of Vectors
A simple way of adding vectors can
y
be done using their COMPONENTS.
The component of a vector is the
ay 
projection of the vector onto the
x, y (and z in the 3-D case) axes in the a sin 
Cartesian co-ordinate system.
Obtaining the components is known as
RESOLVING the vector.
The components can be found using the rules
for a right-angle triangle. i.e.
a x  a cos and a y  a sin 
thiscan also be writtenin
MAGNIT UDE- ANGLE NOT AT IONas
a  a  a , tan 
2
x
2
y
ay
ax
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
a

ax 
a cos
x
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PHY34210
Unit Vectors
A UNIT VECTOR is one whose magnitude is exactly equal to 1.
It specifies a DIRECTION. The unit vectors for the Cartesian
co-ordinates x,y and z are given by,iˆ, ˆj and kˆ respectively.
The use of unit vectors can make the
addition/subtraction of vectors simple.
One can simply add/subtract together
the x,y and z components to obtain
the size of the resultant component in
that specific direction. E.g,
z1
ˆj
y
1
kˆ
iˆ
1


ˆ
a  a x iˆ  a y ˆj  a z k , b  bx iˆ  by ˆj  bz kˆ
  
thenif s  a  b using vectoraddition by components

s  (a x  bx )iˆ  (a y  by ) ˆj  (a z  bz )kˆ  s x iˆ  s y ˆj  s z kˆ
x
32
Vector Multiplication
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PHY34210
There are TWO TYPES of vector multiplication.
One results in a SCALAR QUANTITY (the scalar or ‘dot’ product).
The other results in a VECTOR called the vector or ‘cross’ product.
For the SCALAR or DOT PRODUCT,

   
a.b  a cos b   a b cos  also, a . b  b . a
In unit vector notation,

a.b  a x iˆ  a y ˆj  a z kˆ . bx iˆ  by ˆj  bz kˆ but since



cos0o  1 and cos90o  0 expandingthisreduces to

a.b  a x bx  a y by  a z bz
since iˆ.iˆ 
iˆ. ˆj  iˆ.kˆ 
ˆj. ˆj  kˆ.kˆ  1 (11cos0o  1) and
ˆj.kˆ  0 (1 1 cos90o  0)
33
Vector (‘Cross’) Product


The VECTOR PRODUCT of two vectors a and b
REGAN
PHY34210
34
produces a third vector whose magnitude is given by
c  absin 
The direction of the
resultant is perpendicular
to the plane created by the
initial two vectors, such that
 is the angle between
the two initial vectors


 
 
also a  b   b  a and


 
a  b  a xiˆ  a y ˆj  a z kˆ  bxiˆ  by ˆj  bz kˆ

but a x iˆ  bxiˆ  a xbx iˆ  iˆ  0 and
a xiˆ  by ˆj  a xby iˆ  ˆj  a xby kˆ, thus
 
a  b  a y bz  by a z iˆ  a z bx  bz a x  ˆj  a xby  bx a y kˆ
 

 b
c


a
REGAN
PHY34210
Example 1:
Add thefollowing threevectors

 
 ˆ
a  i  4 ˆj , b  3iˆ  2 j , c  iˆ  2 ˆj

    ˆ
r  a  b  c  i  4 ˆj  3iˆ  2 j   iˆ  2 ˆj

 r  3iˆ  4 ˆj  rx  3 , ry  4

y

r


b
 
 
4
r  3  4  5 , tan    53.1o
3
2

a

c
35
x
2

Example 2:
REGAN
PHY34210
Whatare the(a) scalar and (b) vect orproductsof the two vect ors
aˆ  2iˆ  3 ˆj  4kˆ and bˆ  4iˆ  20 ˆj  12kˆ
 
(a) Scalar P roduct: a . b  ab cos  2iˆ  3 ˆj  4kˆ . 4iˆ  20 ˆj  12kˆ
recallingonlyiˆ.iˆ  ˆj.ˆj  kˆ.kˆ  1, iˆ.ˆj  iˆ.kˆ  ˆj.kˆ  0



then a . b  a x bx  a y by  a z bz  (2.4)  (3.  20)  (4.12)
 a . b  8 - 60 - 48  -100
 
(b) Vector P roduct,a  b  2iˆ  3 ˆj  4kˆ  4iˆ  20 ˆj  12kˆ
 
recallinga  b  a y bz  by a z  iˆ  a z bx  bz a x  ˆj  a x by  bx a y  kˆ
 
 a b 



(3).(12)  (20).(4)  iˆ  (4).(4)  (12).(2)  ˆj  (2).(20)  (4)(3) kˆ
36  80 iˆ  (16  24) ˆj  (40  12) kˆ

 
 a  b  44 iˆ  40 ˆj  52 kˆ  4 . 11iˆ  10 ˆj  13kˆ

36
4: Motion in 2 and 3 Dimensions
REGAN
PHY34210
37
The use of vectors and their components is very useful for describing
motion of objects in both 2 and 3 dimensions.
Position and Displacement
If in general theposit ionof a particlecan be
descibed in Cartesianco - ordinatesby

r  xiˆ  yˆj  zkˆ , then theDISP LACEMENT is


  
ˆ
Dr  r2  r1  x2 i  y2 ˆj  z 2 k  x1i  y1 ˆj  z1kˆ

  x  x i   y  y  ˆj   z  z kˆ

 
2
1
2
1

2
1
 Dxiˆ  Dyˆj  Dzkˆ


ˆ
ˆ
ˆ
e.g., if r1  3i  2 j  5k and r2  9iˆ  2 ˆj  8kˆ
  
thenDr  r2  r1  (9  (3))iˆ  (2  2) ˆj  (8  5)kˆ  12iˆ  3kˆ
Velocity and Acceleration
REGAN
PHY34210
The average velocity is given by


Dr Dxiˆ  Dyˆj  Dzkˆ Dx ˆ Dy ˆ Dz ˆ
vav 


i
j
k
Dt
Dt
Dt
Dt
Dt
While the instantaneous velocity is given by making Dt tend to 0, i.e.

 dr d ( xiˆ  yˆj  zkˆ) dx ˆ dy ˆ dz ˆ
v

 i
j k
dt
dt
dt
dt
dt

 v  v x iˆ  v y ˆj  v z kˆ
Similarly, the average acceleration is given by,
 


v2  v1 Dv Dvxiˆ  Dv y ˆj  Dvz kˆ Dvx ˆ Dv y ˆ Dvz ˆ
aav 



i
j
k
Dt
Dt
Dt
Dt
Dt
Dt
While the instantaneous acceleration is given by

 dv d (vxiˆ  v y ˆj  vz kˆ) dvx ˆ dvy ˆ dvz ˆ
a


i
j
k
dt
dt
dt
dt
dt
38
Projectile Motion
REGAN
PHY34210
39
The specialist
 case where a projectile is ‘launched’ with an initial
velocity, v0 and a constant free-fall acceleration, g .
Examples of projectile motion are golf balls, baseballs, cannon balls.
(Note, aeroplanes, birds have extra acceleration see later).
We can use the equations of motion for constant acceleration and
what we have recently learned about vectors and their components
to analyse this type of motion in detail.
T heinitialproject ilevelocity

(at t  0) is v0  v0 x iˆ  v0 y ˆj
where v0 x  v0 cos
and
v0 y  v0 sin 
More generally,

v  vxiˆ  v y ˆj

v0
v0 sin 
v0 cos
where vx  v cos and v y  v sin 
REGAN
PHY34210
Horizontal Motion
40
In the projectile problem, there is NO ACCELERATION in the
horizontal direction (neglecting any effect due to air resistance).
Thus the velocity
x  x0  v0 xt (equationof motion 1)
component in the
x (horizontal) direction and v0 x  v0 cos 0 , thus x  x0  v0 cos 0  t
remains constant throughout the flight, i.e.,
Vertical Motion
1 2
1 2
y  y0  v0 y t  gt  v0 y sin  0  t  gt
2
2
2
2
v y  v0 sin 0  gt and v y  v0 sin 0   2 g  y  y0 

v0
Max.height occurs when
v y  0,
i.e., v0 sin 0  gt
v0 sin 
v0 cos
vy
REGAN
PHY34210
41
The Equation of Path for Projectile Motion
Given that x  x0  v0 cos 0  t and y  y0  v0 sin  0  t 
1 2
gt
2
substituting for thetimebetween the two upper equns.


x  x0   1   x  x0  
 y  y0  v0 sin  0 
  g





v
cos

2
v
cos

0
0 
0 

 0
2
1  x  x0  
 y  y0  tan 0  x  x0   g 

2  v0 cos 0   Note that this is an
equation of the form
y=ax+bx2 i.e., a parabola
(also, often y0=x0=0.)
2
REGAN
PHY34210
The Horizontal Range
T herange, R  x  x0 is defined when theproject ilehits theground
1 2
i.e., when y  y0 , thenR  v0 cos 0  t and 0  v0 sin  0  t  gt
2

R
1 
R

 0  v0 sin  0 
 g 
v0 cos 0  2  v0 cos 0  
2
R sin  0
2v02 cos 0 sin  0
gR2

 2
 R
2
cos 0
2v0 cos  0
g

v0
since in general
2 cos 0 sin  0  sin 2 0 , then
v02 sin 2 0
R
g
(noteassumes y  y0 )
v0 sin 
(y0,x0)
vy=0
Max
height
v0 cos
Range
42
Example
REGAN
PHY34210
43
At what angle must a baseball be hit to make a home run if the
fence is 150 m away ? Assume that the fence is at ground level, air
resistance is negligible and the initial velocity of the baseball is 50 m/s.
Recalling that therange  150mis given
v02 sin 2 0
by R 
, v0  50m s1 , g  9.8m s 2
g
gR 9.8m s- 2 150m 1470
 sin2 0  2 

1
1
v0 50m s  50m s
2500
 2 0 sin 1 0.588  36o OR 143o
  0 18o or 71.5o
R
How far
must the fence be moved back for no homers to be possible ?
2
R
v0 sin 2 0
is a maximumwhen sin2 0  max  1, i.e., when 2 0  900 ,
g
1
1
50
m
s

50
m
s
thus  0max  45o  Rmax 
 255m  840 feet!
2
9.8m s
Uniform Circular Motion
REGAN
PHY34210

v
44

A particle undergoes UNIFORM CIRCULAR
v

MOTION is it travels around in a circular arc at a
a
CONSTANT SPEED. Note that although the


speed does not change, the particle is in fact
a
r
ACCELERATING since the DIRECTION OF THE

a
VELOCITY IS CHANGING with time.

The velocity vector is tangential to the instantaneous
v
direction of motion of the particle.
The (centripetal) acceleration is directed towards the centre of the circle
Radial vector (r) and the velocity vector (v) are always perpendicular
T hePERIODOF REVOLUT ION timetakenfor theparticleto go
around thecircle.If thespeed (i.e., themagnitudeof the velocit yfor
UCM) of theparticle v, the timetakenis, by definition
Circum ference 2r
T

velocity
v
Proof for Uniform Circular Motion
REGAN
PHY34210
yp
xp

ˆ
ˆ
ˆ
ˆ
v  v x i  v y j   v sin   i  v cos  j , but sin  
and cos 
, thus
r
r

  vy p  ˆ  vx p  ˆ
 dv  v dyp  ˆ  v dxp  ˆ
i  
 j . Acceleration is a 
i   
 j
v   
  
dt  r dt   r dt 
 r   r 
but
dyp
 v yp  v cos and
dxp
 v xp  v sin 

yp r
45

v
dt
dt

  v dyp  ˆ  v dxp  ˆ  v 2 cos  ˆ  v 2 sin   ˆ
i   
 j
i   
 j   
 a   
xp
r
r 
 r dt   r dt 

 
v2
v2
2
2
2
2
 cos    sin   , thus, a 
m agnitude, a  a x  a y 
r
r
 v 2 sin  


a y 
r  sin 
ACCELERAT ON
I DIRECT ION from tan 


 tan ,
2
a x  v cos  cos
 

r


i.e.,   ,  acceleration is along theradius....towards centreof circle
Relative Motion
If we want to make measurements
of velocity, position, acceleration etc.
these must all be defined RELATIVE to
a specific origin. Often in physical
situations, the motion can be broken down
into two frames of reference, depending
on who is the OBSERVER.
A
( someone who tosses a ball up in a
moving car will see a different motion
to someone from the pavement).
REGAN
PHY34210

rAp
p

vAB  const

rBp
 B
rAB
If we assume that different FRAMES OF REFERENCE always move
at a constant velocity relative to each other, then using vector addition,



i.e., acceleration
d rpA  rpB  rBA 





rpA  rpB  rBA , v pA 
 v pB  vBA is the SAME for
dt
both frames of




d v pA  v pB  vBA
d v pB


reference!
a pA 

 0  a pB
(if VAB=const)!
dt
dt




 
46
5: Force and Motion (Part 1)
REGAN
PHY34210
47
If either the magnitude or direction of a particle’s velocity changes
(i.e. it ACCELERATES), there must have been some form of
interaction between this body and it surroundings. Any interaction
which causes an acceleration (or deceleration) is called a FORCE.
The description of how such forces act on bodies can be described by
Newtonian Mechanics first devised by Sir Isaac Newton (1642-1712)..
Note that Newtonian mechanics breaks down for (1) very fast speeds,
i.e. those greater than about 1/10 the speed of light c, c=3x108ms-1
where it is replaced by Einstein’s theory of RELATIVITY and (b)
if the scale of the particles is very small (~size of atoms~10-10m),
where QUANTUM MECHANICS is used instead.
Newton’s Laws are limiting cases for both quantum mechanics and
relativity, which are applicable for specific velocity and size regimes
Newton’s First Law
REGAN
PHY34210
Newton’s 1st law states
‘ If no force acts on a body, then the body’s velocity can not change,
i.e., the body can not accelerate’
This means that
(a) if a body is at rest, it will remain at rest unless acted
upon by an external force, it; and
(b) if a body is moving, it will continue to move at that velocity
and in the same direction unless acted upon by an external force.
So for example,
(1) A hockey puck pushed across a ‘frictionless’ rink will move
in a straight line at a constant velocity until it hits the side of the rink.
(2) A spaceship shot into space will continue to move in the direction
and speed unless acted upon by some (gravitational) force.
48
Force
REGAN
PHY34210
49
The units of force are defined by the acceleration which that force
will cause to a body of a given mass.
The unit of force is the NEWTON (N) and this is defined by the
force which will cause an acceleration of 1 m/s2 on a mass of 1 kg.
If two or more forces act on a body we can find their resultant value
by adding them as vectors. This is known as the principle of
SUPERPOSITION. This means that the more correct version of
Newton’s 1st law is
‘ If no NET force acts on a body, then the body’s velocity can not
change, i.e., the body can not accelerate’
Mass: we can define the mass of a body as the characteristic
which relates the applied force to the resulting acceleration.
Newton’s 2nd Law
REGAN
PHY34210
50
Newton’s 2nd law states that
‘ The net force on a body is equal to the product of the body’s mass
and the acceleration of the body’
We can write the 2nd law in the form of an equation:


Fnet  ma
As with other vector equations, we can make three equivalent equations
for the x,y and z components of the force. i.e.,
Fnet, x  max , Fnet, y  may and Fnet, z  maz
The acceleration component on each axis is caused ONLY by the force
components along that axis.
REGAN
PHY34210
51
If the net force on a body equals zero and thus it has no acceleration,
the forces balance out each other and the body is in EQUILIBRIUM.
We can often describe multiple forces acting on the same body using
a FREE-BODY DIAGRAM, which shows all the forces on the body.
FA  220 N, FB  ?, FC  170 N
  

 

FA  FB  FC  ma  0  FB   FA  FC

FA  220 N
components along the x and y axes cancel
 
F   F  F  0   F cos 133   F cos
137
 F cos133   220  0.682
 cos 

 0.883
y
FBy   FAy  FCy   FA sin 47  FC sin     FB
o
o
Bx
Ax
Cx
A
C
o
43o
47o

Fc  170N

x
0
A
FC
 
170
 
   28 and FA sin 47  FC sin 28  FB
0
o
o
FB  220 0.731  170 0.47  241 N

FB  ?? N
HRW
p79
The Gravitational Force
REGAN
PHY34210
52
The gravitational force on a body is the pulling force directed towards
a second body. In most cases, this second body refers to the earth (or
occasionally another planet).
From Newton’s 2nd law, the force is related to the acceleration by


Fg  ma  takingcomponentsin theverticaldirection
( y positivecorresponds to upward) then  Fgy  m ay  m g 


ˆ
ˆ
In vectorform,we have Fg   Fg j  m g j  mg
A body’s WEIGHT equals the magnitude of the gravitational force on the
body, i.e, W = mg. This is equal to the size of the net
force to stop a body falling to freely as measured by someone at
ground level. Note also that the WEIGHT MUST BE MEASURED
WHEN THE BODY IS NOT ACCELERATING RELATIVE TO THE
GROUND and that WEIGHT DOES NOT EQUAL MASS.
Mass on moon and earth equal but weights not ge=9.8ms-2, gm=1.7ms-2
The Normal Force
REGAN
PHY34210
53
The normal force is the effective ‘push’ a body feels from a body
to stop the downward acceleration due to gravity, for example the
upward force which the floor apparently outs on a body to keep it
stationary against gravity.
General equation for block on a table is

  
Fnet  ma  N  Fg

NormalForce, N
y  component,m ay  N  Fg  N  m g
 N  m a y  g  i.e., if block is at rest then
N  m 0  g   m g i.e. same magnitudeas
gravitational forcebut in oppositedirection.
Note the NORMAL FORCE is ‘normal’
(i.e. perpendicular) to the surface.


Gravitational Force, Fg  mg
REGAN
PHY34210
Example
A person stands on a weighing scales in a lift (elevator!) What is the
general solution for the persons measured weight on the scales ?

a


 
Fnet  ma  Fg  N

N
Fy ,net  m ay   Fg  N  m g  N
 N  m a y  g 
Fgy  mg
So, if lift accelerates upwards (or the
downward speed decreases!) the persons
weight INCREASES, if the lift accelerates
downwards (or decelerates upwards)
the persons weight DECREASES
compared to the stationary (or constant
velocity) situation.
54
REGAN
PHY34210
Tension
55
Tension is the ‘pulling force’ associated with a rope/string pulling a
body in a specific direction. This assumes that the string/rope is taught
(and usually also massless).
For a frictionless surface and a massless, frictionless

pulley, what are the accelerations of the sliding and
N
hanging blocks and the tension in the cord ?
x components, Fnet , x  MaMx  T


y components( a downward thus  amy )
M
T
Fnet , y  m amy  T  m g


FgM  Mg

T
m
magnitudesmust be equal, aMx  amy  a
mg
M m
mg
Mm g
M

T
M m M m
  m amy  MaMx  m g  a 

 Fnet , x  T  MaMx
Fgm  mg
Newton’s Third Law
REGAN
PHY34210
56
Two bodies interact when they push or pull on each other. This leads
to Newton’s third law which states,
‘ When two bodies interact, the forces on the bodies from each other
are always equal in magnitude and opposite in direction ’
Sometimes this is differently stated as

NormalForce, N
‘ for every action there is an equal
but opposite reaction ’
The forces between two interacting
bodies are called a ‘third-law pair
forces’.
e.g., Table pushes up block with
force N, block pushes down table
with force Fg, where Fg=N


Gravitational Force, Fg  mg
REGAN
PHY34210
Example
N
N T cos50o
T
y
57
T
x
40 o
40 o
mg =
g x 15kg
Question? Whatis the tensionin thestring?
  

T  N  Fg  ma  0
x component,T  0  m g sin   Fx ,net  0
 T  m g sin   15kg  9.8m s 2 sin 400  94.5 N
y component,0  N  m g cos  0
 N  112.6 N
50o
40 o
mg
mg sin 40o
6: Force and Motion (Part II)
REGAN
PHY34210
58
Friction: When two bodies are in contact, the resistance to movement
between their surfaces is known as FRICTION. The properties of
frictional forces are that if a force, F, pushes an object along a surface
(e.g., a block along a surface),
1) If the body does not move, the STATIC FRICTIONAL FORCE,
fs is equal in magnitude and opposite in direction to the
component of the pushing force, F, along the surface.
2) The magnitude of the frictional force, fs, has a maximum value,
f s,max, which is given by f s,max=msN where ms is the coefficient
of static friction.
3) If the body begins to move along the surface, the magnitude of the
frictional force reduces to fk=mkN, where mk is the coefficient of
kinetic friction.
Drag Force and Terminal Speed
REGAN
PHY34210
59
When a body passes through a fluid (i.e., gas or a liquid) such as a ball
falling through air, if there is a relative velocity between the body and
the fluid, the body experiences a DRAG FORCE which opposes this
relative motion and is in the opposite direction to the motion of the
body (i.e., in the direction which the fluid flows relative to the body).
The magnitude of this drag force is related
1
2
to the relative speed of the body in the fluid
D  C r Av
by a DRAG COEFFICIENT, C, which is
2
experimentally determined. The magnitude
of the drag force is given by the expression for D, which depends
on the fluid density (i.e. mass per unit volume, r), the effective
cross-sectional area, A (i.e. the cross-sectional area perpendicular to the
direction of the velocity vector), and the relative speed, v.
REGAN
PHY34210
60
Note that the drag coefficient. C, is not really a constant, but rather
a quantity associated with a body which can varies with the speed, v.
(for the purposes of this course, however, assume C = constant).
The direction of the drag force is opposed to the motion of the object
through the fluid. If a body falls through air, the drag force due to the
air resistance will start at zero (due to zero velocity) at the start of the
fall, increasing as the downward velocity of the falling body increases.
Ultimately, thedrag force will be cancelthedownward acceleration.
In general, Fnet , y  m ay  D  Fg
For ' terminalspeed', a y  0, thus
2 Fg
1
2m g
2
CrAvt  Fg  0  vt 

2
CrA
CrA
Forces in Uniform Circular Motion
REGAN
PHY34210
61
Recalling that for a body moving in a circular arc or radius, r, with
constant speed, v, the MAGNITUDE of the ACCELERATION, a, is
given by a = v2/r, where a is called the centripetal acceleration.
We can say that a centripetal force accelerates a body by changing the
direction of that body’s velocity without changing its speed.
Note that this centripetal force is not a ‘new’ force, but rather a
consequence of another external force, such as friction, gravity or
tension in a string.
Examples of circular motion are
(1) Sliding across your seat when your car rounds a bend:
The centripetal force (which here is the frictional force between
the car wheels and the road) is enough to cause the car to accelerate
inwards in the arc. However, often the frictional force between you
and your seat is not strong enough to make the passenger go in this
arc too. Thus, the passenger slides to the edge of the car, when its push
(or normal force) is strong enough to make you go around the arc.
REGAN
PHY34210
(2) the (apparent) weightlessness of astronauts on the space shuttle.
Here the centripetal force (which causes the space shuttle to orbit the
earth in a circular orbits) is caused by the gravitational force of the
earth on all parts of the space shuttle (including the astronauts).The
centripetal force is equal on all areas of the astronauts body so he/she
feels no relative extra pull etc. on any specific area, giving rise to a
sensation of weightlessness.
Note that the magnitude of the centripetal FORCE is given, (from
Newton’s second law) by : F = ma = m v2/r
Note that since the speed, radius and mass are all CONSTANTS so
is the MAGNTIUDE OF THE CENTRIPETAL FORCE. However,
DIRECTION IS NOT CONSTANT, varying continuously so as to
point towards the centre of a circle.
62
REGAN
PHY34210
Example:
r
At what constant speed does the roller
coasters have to go to ‘loop the loop’
of radius r ?
At the top of the loop, the free body
forces on the roller coaster are gravity
(downwards) and the normal
force (also inwards). The total acceleration is
also inwards (i.e., in the downwards direction).
Fg
N
Fy ,net   N  Fg  m a y  , limit at N  0 (no contact!)
v2
v2
thus,  Fg  m a y   m.   m g  
 g  v  gr
r
r
i.e., independent of mass!
63
7: Kinetic Energy and Work
REGAN
PHY34210
One way to describe the motion of objects is by the use of Newton’s
Laws and Forces. However, an alternative way is describe the motion
in terms of the ENERGY of the object.
The KINETIC ENERGY (K) is the energy associated with the
MOTION of an object. It is related to the mass and velocity of a body
by K= 1/2 mv2 , where m and v are the mass and velocity of the body.
The SI unit of energy is the Joule (J) where 1 Joule = 1kg.m2s-2.
Work:
`Work is the energy transferred to or from an object by means of a
force acting on it. Energy transferred to the object is positive work,
while energy transferred from the object is negative work.’
For example, if an object is accelerated such that it increases its
velocity, the force has ‘done work’ on the object.
64
Work and Kinetic Energy
REGAN
PHY34210
65
The work done (W) on an object by a force, F, causing a displacement,
d, is given by the SCALAR PRODUCT, W = F.d =dFcos where
Fcos is the component of the force along the object’s displacement.
This expression assumes a CONSTANT FORCE (one that does not
change in magnitude or direction) and that the object is RIGID (all
parts of the object move together).
Example: If an object moves in a straight line with initial velocity, v0
and is acted on by a force along a distance d during which the velocity
increases to v due to an acceleration, a, from Newton’s 2nd Law the
magnitude of the force is given by F = max . From the equations of
motion v2=vo2+2axd . By substituting for the acceleration, ax, we have,
Fx
1 2 2
1 2 1 2
d
v  v0 ,
 ax  Fx d  m v  m v0  DK  work done
2ax
m
2
2


1 2 1 2
mv  mv 0  DK  work done is the Work-Kinetic Energy Theorem
2
2
Work Done by a Gravitational Force.
REGAN
PHY34210
66
If an object is moved upwards against gravity, work must be done.
Since the gravitational force acts DOWNWARDS, and equals Fgr=mg ,
the work done in moving the object upwards in the presence of this
force is W=F.d = mg . d where d is the (vector) displacement in the
upward direction, (which we assume is the positive y-axis).

 

Wgr  mg.d  m gdcos , mg and d are in oppositedirections,
   180o , W  m gd.  sign shows gravity transfersKINET IC
ENERGY to GRAVIT AT IONAL POT ENT IALENERGY.
When theobject falls back down,   0 and W   m gd
 sign impliesgravitational force transfersenergyT O the
object from potentialenergy tokinetic.
REGAN
PHY34210
Work Done Lifting and Lowering an Object.
67
If we lift an object by applying a vertical (pushing) force, F, during
the upward displacement, work (Wa) is done on the object by this
applied force. The APPLIED FORCE TRANSFERS ENERGY TO
the object, while the GRAVITY TRANSFERS ENERGY FROM it.
From the work - kineticenergy theorem,
DK  K f  K i  net workdone  Wa  Wg .
If theobject is stationarybeforeand after thelift
(v  0 at start and finish) then DK  0  Wa  Wg
 Wa  Wg and Wa  m gdcos where is theangle


between Fg (i.e.' downwards' ) and thedisplacement,d .
If theobject is liftedup,   1800 and the work done by the
appliedforce, Wa   m gd. If theobject falls,  0o , Wa  m gd
Spring Forces and Hooke’s Law
REGAN
PHY34210
68
The spring force is an example of a VARIABLE FORCE.
For a PERFECT SPRING, stretching or compressing gives rise to
RESTORING FORCE which is proportional to the displacement
of the spring from its relaxed state. This is written by Hooke’s Law
(after
17th century British scientist) as

 Robert Hooke,
Frestoring  kd , where k  springconstant,stiffnessof thespring.
In the1 - d case, we can simply use the x - direction F  kx
The work done by a perfect spring can not be obtained from F.d, as the
force is not constant with d. Instead, the work done over the course of
the extension/compression must be summed incrementally.
xf
1


x
x
Ws   F j Dx , as Dx  0 then, Ws   xif Fdx   xif (kx)dx   kx2 
 2
 xi


1 2  1 2 1
1 2
2
2
 Ws   kx f    kxi   k xi  x f  if xi  0, Ws   kx f
2
2
 2
 2
Work Done by an Applied Force
REGAN
PHY34210
69
During the displacement of the spring, the applied force, Fa, does
work, Wa on the block and the spring restoring force, Fs does work Ws.
T hechangein kineticenergy (of theblock attachedto thespring)
due to thesetwo energy transfersis given by
DK  K f  K i  Wa  Ws
T hus, if
DK  0 , Wa  Ws
If the block attached to a spring is stationary before and after its
displacement, then the work done on the spring by the applied force
is the negative of the work done on it by the spring restoring force.
Work Done by a General Variable Force.
REGAN
PHY34210
T he work done by a forceaveragedover a distance,Dx, is
DW j  F j ,ave Dx. T otalwork done equals sum of all j th increments,
W   DW j  F j ,ave Dx. As Dx  0 , W 
j
j
f
F
D
x


 j ,ave
xi F  x dx
x
j , Dx 0
Workdone by 1 - D force AREA UNDERT HE CURVE
of F ( x) against x.

In 3 - D, F  F iˆ  F ˆj  F kˆ. If F only depends on x, F on y and
x
y
z
x
y
Fz on z , theby SEP ARAT INGT HE VARIABLES if theparticle

movesthroughan incremental displacement,dr  dxiˆ  dyˆj  dzkˆ ,
theincrementof work in dr, dW, is given by
 
dW  F.dr  Fx dx  Fy dy  Fz dz , then thetotalwork is
rf
xf
yf
zf
ri
xi
yi
zi
W   dW   Fx dx   Fy dy   Fz dz
70
Work-Kinetic Energy Theorem
with a General,Variable Force
xf
xf
REGAN
PHY34210
71
xf
dv
W   F x dx   m ax dx   m dx
dt
xi
xi
xi
dv dv dx
dv
using theCHAIN RULE,  .  v
dt dx dt
dx
dv
dv
 we can CHANGE T HE VARIABLE, m dx  m v . dx  m vdv
dt
dx
xf
vf
vf
dv
W   m dx   m vdv m  vdv
dt
xi
vi
vi
1 2 1 2
W  m vf  m vi  K f  K i  DK ,
2
2
which is the WORK- KINET ICENERGY T HEOREM
Power
REGAN
PHY34210
72
POWER is the RATE AT WHICH WORK IS DONE. The AVERAGE
POWER done due to a force responsible for doing work, W in a time
period, Dt is given by Pave = W/D t .
The INSTANTANEOUS POWER is given by
dW
P
dt
The SI unit of power = Watt (W), where 1 W= 1 J per sec=1 kg.m2/s3
Note that the imperial unit of horsepower (hp) is still used, for
example for cars. 1hp = 746 W
The amount of work done is sometimes expressed as the product of the
power output multiplied by time taken for this. A common unit for this
is the kilowatt-hour, where 1kWh = 1000x3600 J = 3.6 x106J = 3.6MJ.
We can also describe the instantaneous power in terms of rate at
which a force does work on a particle,

dW F cos  dx
dx
P

 F cos 
 F cos v  F .v
dt
dt
dt
REGAN
PHY34210
Example 1:
What is the total energy associated with a collision between two
locomotives, at opposite ends of a 6.4km track accelerating towards
each other with a constant acceleration of 0.26 m/s2 if the mass of
each train was 122 tonnes (1 tonne =103kg) ?
Using, v 2  v02  2ax  x0 
x  x0  3.2 103 m, v0  0, a  0.26m s- 2
T he velocit yof the trainsat collisionis then
v  2  0.26m s 2  3.2 103 m  40.8m s1
T hekineticenergyof each locomotiveis given by


1 2 1
5
1 2
K  m v  1.2210 kg  40.8m s  108 J
2
2
T hus totalenergyof collisionis 2 K  200MJ
73
Example 2:
REGAN
PHY34210
74
If a block slide across a frictionless

floor through a displacement of
d  3iˆm
-3m in the direction, while at the same
time a steady (i.e. constant) force of
F=(2i-6j) Newtons pushes against the
crate,

(a) How much work does the wind force do
F  2iˆ  6 ˆj N
on the crate
during
this
displacement
?


  


W  F .d  2iˆ  6 ˆj N .  3iˆ m  6 J
T hus, the' wind' forcedoes 6 J of NEGAT IVE WORK on thecrate
i.e. it transfers 6 J of kineticenergy FROM T HECRAT E

(b) If the crate had a kinetic energy of 10J at the start of the
displacement, how much kinetic energy did it have at the end of the
-3m ? Work-kineticenergy theorem,W  ΔK   6 J  K f  10J
 K f  6  10J  4 J
i.e., block is slowed down by wind force.
REGAN
PHY34210
Example 3:
If a block of mass, m, slides across a
frictionless floor with a constant speed
of v until it hits and compresses a perfect
spring, with a spring constant, k.
At the point where the spring is compressed
such that the block is momentarily stopped,
by what distance, x, is the spring compressed ?
Using the work - energy theorem,the work
done on theblock by thespring forceis
1
Ws   kx2 . T he work is also relatedto thechange
2
in kineticenergy of theblock,i.e.,W  ΔK  K f  K i

1 2
1 2
m
 kx  0  m v  x  v
2
2
k
v
k
x
v=0
m
75
REGAN
PHY34210
8: Potential Energy & Conservation of Energy
Potential energy (U) is the energy which can be associated with
configuration of a systems of objects.
One example is GRAVITATIONAL POTENTIAL ENERGY,
associated with the separation between two objects attracted to each
other by the gravitational force. By increasing the distance between
two objects (e.g. by lifting an object higher) the work done on the
gravitational force increases the gravitational potential energy of the
system.
Another example is ELASTIC POTENTIAL ENERGY which is
associated with compression or extension of an elastic object (such as
a perfect spring). By compressing or extending such a spring, work
is done against the restoring force which in turn increases the elastic
potential energy in the spring.
76
Work and Potential Energy
REGAN
PHY34210
77
In general, the change in potential energy, DU is equal to the negative
of the work done (W) by the force on the object (e.g., gravitational
force on a falling object or the restoring force on a block pushed by a
perfect spring), i.e., DU=-W
Conservative and Non-Conservative Forces
If work, W1, is done, if the configuration by which the work is done is
reversed, the force reverses the energy transfer, doing work, W2.
If W1=-W2, whereby kinetic energy is always transferred to potential
energy, the force is said to be a CONSERVATIVE FORCE.
The net work done by a conservative force in a closed path is zero.
The work done by a conservative force on a particle moving between
2 points does not depend on the path taken by the particle.
NON-CONSERVATIVE FORCES include friction, which causes
transfer from kinetic to thermal energy. This can not be transferred
back (100%) to the original mechanical energy of the system.
REGAN
PHY34210
Determining Potential Energy Values
xf
xf
xi
xi
W   F ( x)dx , DU    F ( x)dx . For GRAVIT .P OT .ENERGY,
yf
yf
yi
yi
DU    F ( y )dy   

yf
 m gdy  m g  dy  m gy f
 yi

yi

 DU grav  m g y f  yi  m gDy
Only CHANGESin gravit at ional P ot .energy are meaningful,
i.e., it is usual to defineU i  0 at yi , then U  y   m gy
For theELAST ICP OT ENT IALENERGY,
xf
xf
xf
DU elas    F ( x)dx     kxdx  k  xdx 
xi
xi


xi
 
1
k x2
2
xf
xi
1
k x 2f  xi2 . P ot energyis relative,thus we chose
2
1 2
U  0 at xi  0 T hen, U  x   kx , x is extension/compression.
2

DU elas 
78
Conservation of Mechanical Energy
REGAN
PHY34210
79
T hemechanicalenergyis thesum of kineticand pot entialenergies,
Emech  K  U . If thesystemis isolatedfromits environment and
no externalforcecauses any internalenergy changes,
DK  W & DU  W ,  DK  DU  K f  K i  U f  U i 
 K f  U f  K i  U i i.e, T hesum of thekineticand pot entialenergies
( themechanicalenergy)is thesame for all statesof an isolatedsystem,
i.e. theMECHANICALENERGY of an ISOLAT ED SYST EM where
thereare only conservative forcesis CONST ANT .
T hisis theP RINCIP LEOF CONSERVAT ION OF MECHANICAL
ENERGY (note,conservation is due to CONSERVAT IVE FORCES ).
T hiscan also be writtenas DEmech  DK  DU  0
The Potential Energy Curve
REGAN
PHY34210
For the1 - D case, the work done,W , by a force,F , movingan
object through a displacement,Dx equals, FDx , therefore, the
potentialenergy can be writtenas
DU  x 
dU  x 
DU  x   W   FDx  F  

Dx
dx
e.g ., Hooke's Law, if theelastic potentialis given by,
1 2
U  x   kx thendifferentiatinggives, F   kx
2
also, in thegravitational case, U  x   m gh  F   m g
In the general, the force at position x,
can be calculated by differentiating
the potential curve with respect to x
(remembering the -ve sign). F(x) is minus
the SLOPE of U(x) as a function of x
80
REGAN
PHY34210
Turning Points
For conservative forces, the
mechanical energy of the system
is conserved and given by,
U(x) + K(x) = Emec
where U(x) is the potential energy
and K(x) is the kinetic energy.
Therefore, K(x) = Emec-U(x).
Since K(x) must be positive ( K=1/2mv2),
the max. value of x which the particle
has is at Emec=U(x) (i.e., when K(x)=0).
Note since F(x) = - ( dU(x)/dx ) ,
the force is negative.
Thus the particle is ‘pushed back.
i.e., it turns around at a boundary.
81
K  0 at ymax , Emec  mgymax
F ( y)  
dU ( y)
 m g
dy
Emec  K ( y )  U ( y )
1 2
 m v  m gy
2
Equilibrium Points
REGAN
PHY34210
82
Equilibrium Points: refer to points where, dU/dx=-F(x)=0.
Neutral Equilibrium: is when a particle’s total mechanical energy is
equal to its potential energy (i.e., kinetic energy equals zero). If no
force acts on the particle, then dU/dx=0 (i.e. U(x) is constant) and
the particle does not move. (For example, a marble on a flat table top.)
Unstable Equilibrium: is a point where the kinetic energy is
zero at precisely that point, but even a small displacement from this
point will result in the particle being pushed further away (e.g., a
ball at the very top of a hill or a marble on an upturned dish).
Stable Equilibrium: is when the kinetic energy is zero, but any
displacement results in a restoring force which pushes the particle
back towards the stable equilibrium point. An example would be a
marble at the bottom of a bowl, or a car at the bottom of a valley.
REGAN
PHY34210
U(x)
x
D
B
C
A
Particles at A,B, C and D are in at equilibrium points
where dU/dx = 0
A,C are both in stable equilibrium ( d 2U/dx2 = +ve )
B is an unstable equilibrium ( d 2U/dx2 = -ve )
D is a neutral equilibrium ( d 2U/dx2 = 0 )
83
Work Done by an External Force
REGAN
PHY34210
Previously we have looked at the work done to/from an object.
We can extend this to a system of more than one object.
Work is the energy transferred to or from a system by means of an
external force acting on that system.
No friction (conservative forces)
W  DK  DU  DEmec
Including friction
From Newt ons2 nd law, F  f k  m a ,
t heforce(t husacceleraton)
i is const ant ,
t hereforewe can use v 2  v02  2ad
 v 2  v02 
  f k and
By substit ution, F  m
 2d 
1 2 1 2
Fd  m v  m v0  f k d  DK  f k d
2
2
84
Conservation of Energy
REGAN
PHY34210
85
This states that
‘ The total energy of a system, E, can only change by amounts of
energy that are transferred to or from the system. ’
Work done can be considered as energy transfer, so we can write,
W  DE  DEmec  DEth  DEin
DEmec is thechangein mechanicalenergy,DEth is thechangein thermal
energy(i.e.,heat)and DEin is thechangein internalenergyof thesystem.
If a system is ISOLATED from it surroundings, no energy can be
transferred to or from it. Thus for an isolated system, the total energy
of the system can not change, i.e., DE  DE  DE  DE  0
mec
th
in
Emec , 2  Emec ,1  DEth  DEin
Another way of writing this is,
which means that for an isolated system, the total energies can be
related at different instants, WITHOUT CONSIDERING THE
ENERGIES AT INTERMEDIATE TIMES.
REGAN
PHY34210
Example 1:
A child of mass m slides down a helter
skelter of height, h. Assuming the
slide is frictionless, what is the speed of
the child at the bottom of the slide ?
h=10m
From theCONSERVAT ION OF MECHANICALENERGY,
Emec ,i  Emec , f  U i  K i  U f  K f
1 2
U i  m gh , U f  0, K i  0, K f  m v
2
1 2
 m gh 0  0  m v  v  2 gh
2
Note that thisis thesame speed that thechild would have
if it fell directlyfroma height h.
86
Example 2:
REGAN
PHY34210
A man of mass, m, jumps from a
ledge of height, h above the ground,
attached by a bungee cord of length
h
L. Assuming that the cord obeys
Hooke’s law and has a spring constant,
k, what is the general solution for the
maximum extension, x, of the cord ?
By CONSERVAT ION OF MECHANICALENERGY,
DK  DU  0 , if v  0 at topand bottom,K i  K f  0
 DK  0
87
L
x
m
  1 2 


also, DU  DU grav  DU elas  m g Dy   0   kx  

2

1 2
1 2
 kx  m gL  x   m gL m gx 
kx  m gx m gL  0 ,
2
2
solving thisquadratic equation, x 
mg
m g2  2km gL
k
, x  ve root
9: Systems of Particles
REGAN
PHY34210
88
Centre of Mass (COM): The COM is the point that moves as though
all the mass of a body were concentrated there.
For 2 particlesof mass, m1 and m2 separatedby d , if theorgin of x - axis
coincides with theparticleof mass m1 , thecent reof mass of thesystemis
m2
xcom 
d . Moregenerally,if m1 is at x1 and m2 is at x2 , theCOM
m1  m2
m1 x1  m2 x2 m1 x1  m2 x2
is defined by xcom 

where M is the totalmass
m1  m2
M
T hegeneralformfor a n-particlesystem is given by
m1 x1  m2 x2  m3 x3  m4 x4   1 n
xcom 

mi xi , similarly,for 3 - D

M
M i 1

1 n
1 n
ycom 
mi yi and zcom 
mi zi . In vectorform,if r  xiˆ  yˆj  zkˆ


M i 1
M i 1
n



1
then rcom  xcomiˆ  ycom ˆj  zcom kˆ and rcom
m
r
 ii
M i 1
Centre of Mass for Solid Bodies
REGAN
PHY34210
89
Solid objectshaveso manyparticles(atoms)that theycan be considered
to be made up of manyinfinitessimallysmall MASS ELEMENT S,dm.
1
1
1
T hen, xcom 
xdm , ycom 
ydm , zcom 
zdm ,



M
M
M
Often, theintegralsare simplifiedassuming a UNIFORMDENSIT Y( r )
M dm
where r 

, where dV is the volumeoccupiedby mass, dm.
V dV
1
1
1
substituting, xcom 
xdm 
xrdV   xdV and similarly,


M
rV
V
1
1
ycom   ydV , zcom   zdV ,
V
V
Note that thecentreof mass need not necessarily lie in thevolumeof the
object (for examplea doughnut or an igloo).
Newton’s
2nd
Law for a System of Particles.





Mrcom  m1r1  m2 r2  m3 r3   mn rn
differentiating with respect totime,


dmn rn
 mvn we get
since,
dt






drcom
 Mvcom  m1v1  m2 v2  m3v3    mn vn
M
dt

dvn 
 an
differentiatingonceagain, and recalling
dt
and Newton's 2 nd law,






dvcom
 Macom  m1a1  m2 a2  m3 a3   mn an
M
dt

  

 Fcom  F1  F2  F3   Fn
REGAN
PHY34210
90
Linear Momentum
REGAN
PHY34210
91


T heLINEARMOMENT UMis defined by p  mv
 dp d mv  m dv

F


 ma (for m is constant ).
dt
dt
dt
Thus we can re-write Newton’s 2nd law as
‘ The rate of change of the linear momentum with respect to time is
equal to the net force acting on the particle and is in the direction of
the force.’

For a systemof particles,thesystemhas a totallinear momentum,P which
is the vectorsum of theindividual particle linear momenta,i.e.,
  








P  p1  p2  p3    pn  m1v1  m2 v2  m3v3    mn vn  P  M vcom
The linear momentum of a system of particles is equal to the product of
the total mass of the system, M, and the velocity of the centre of mass,
Conservation of Linear Momentum
REGAN
PHY34210
92


 dP

dvcom
Since, F 
m
 macom , in a closed system,if the
dt
dt
net externalforceis zero,and no particlesenteror leave the



 
dP
system,then, Fnet 
 0  P  constant i.e., Pi  Pf
dt
This is the law of CONSERVATION OF LINEAR MOMENTUM
which we can write in words as
‘In no net external force acts on a system of particles, the total linear
momentum, P , of the system can not change.’
also, leading on from this,
‘ If the component of the net external force on a system is zero along
a specific axis, the components of the linear momentum along that
axis can not change.’
Varying Mass: The Rocket Equation
REGAN
PHY34210
93
For rockets, the mass of the rocket is is not constant, (the rocket fuel is
burnt as the rocket flies in space). For no gravitational/drag forces,
By conservaton
i of momentum,Pi  Pf

T heinitialP of therocketplus theexhaust

fuel equals the P of theexhaust product splus

the P of therocketafter timeinterval,dt.
Mv  dMU  M  dM v  dv
a) time = t
M
v
if vrel is therelativespeed between the rocketand
theexhaust product s(and dM - ve) , then,v  dv  vrel  U
 Mv  dMU  M  dM vrel  U 
 Mv  dM v  dv  vrel   M  dM v  dv
 Mv  dMv  dMdv dMvrel  Mv  Mdv  dMv  dMdv
 0  dMvrel  Mdv  
dM
dv
vrel  M

dt
dt
b) time = t+dt
-dmM+dm
U
dv
if  R is therateof mass loss, then,we obt ain Rvrel  M
 Ma
dt
v+dv
1st rocket
equation
Rvrel is called theT HRUST (T ) of therocketengine.
REGAN
PHY34210
94
M is themass at timet and a is theacceleration, T  Ma,
which is Newt on's 2 nd law. T o find the velocityas themass changes,
dM
Mdv  vrel dM  dv  vrel
M
vf
Mf
vi
Mi
Integrat ing, we obt ain,  dv  -vrel

dM
where vi and v f
M
are theinitialand final rocket vel
ocit ies,corresponding to rocket
masses of M i and M f respectively.
1
Since, in general, dx  ln x , then
x
 M i  2nd rocket
Mf 

  v rel ln
v f  vi  vrel ln M f  ln M i   vrel ln
 M  equation
 Mi 
 f 
thus increasein velocity greatestfor small M f (use of mult i- st age rockets!)
Internal Energy Changes and External Forces
REGAN
PHY34210
Energy can be transferred ‘inside a system’ between internal and
mechanical energy via a force, F. (Note that up to now each part of an
object has been rigid). In this case, the energy is transferred internally,
from one part of the body to another by an external force.
T hechangein internalenergyof thesystemis given by,
ΔEint   Fd cos

where d is thedisplacement of theCENT REOF MASS and is the


angle between the directionsof theforce F and displacement d .
 

T heassociatedchangein theMECHANICALENERGY is then
ΔE mec  ΔK  ΔU  Fd cos
95
10: Collisions
REGAN
PHY34210
96
‘A collision is an isolated event in which two or more colliding
bodies exert forces on each other for a short time.’
Impulse
For a head on collision between two bodies, the3rd forcepair, -F(t) F(t)
F(t) and -F(t)acts between the two at time,t.
F(t)is a T IME- VARYING FORCE.
From Newton's 2 nd law, theseforceswill change thelinear momenta

of both bodies. T heamountby which p changesdepends on thetime
interval,Δt , during which these forcesact.

From Newton's 2 law dp  F t dt 
nd

pf
tf

pi
ti


 dp   F t dt  IMP ULSE,J
T heIMP ULSEis theCHANGE IN LINEARMOMENT UMof thebody acted
on by F(t) (right handside). T hisis also equal to theproduct of thestrengthand
durationof theappliedforce,and theAREA UNDERT HECURVE of F t  versust.
REGAN
T heIMP ULSE-LINEARMOMENT UMT HEOREMstates
PHY34210
that thechangein thelinear momentumof each body in a
collisionis equal to theIMP ULSE thatactson thatbody,



i.e., p f  pi  Dp  J
Since, impulse is a VECT OR, we can also write thisin
componentform,









p f x  pix  Dp x  J x , p f y  pi y  Dp y  J y , p f z  piz  Dp z  J z
If Fave is the timeaveragedforceover a period, Δt ,
the magnitudeof theimpulse is given by J  Fave Dt
E.g A 140g is pitched with a horizontal speed of vi=39m/s. If it is hit
back in the opposite direction with the same magnitude of speed
what is the impulse, J, which acts on the ball ?
J  p f  pi  mv f  vi  takingtheinitialvelocitydirectionas the
NEGAT IVEdirection,J  0.1439   39kgm s1  10.9kgm s1
97
REGAN
PHY34210
98
Momentum and Kinetic Energy in Collisions
In any collision, at least one of the bodies must be moving prior to
the collision, meaning that there must be some amount of kinetic
energy in the system prior to the collision. During the collision, the
kinetic energy and linear momentum are changed by the impulse from
the other colliding body.
If the total kinetic energy of the system is equal before and after
collision, it is said to be an ELASTIC COLLISION.
However, in most everyday cases, some of this kinetic energy is
transferred into another form of energy such as heat or sound.
Collisions where the kinetic energies are NOT CONSERVED
are known as INELASTIC COLLISIONS.
In a closed system, the total linear momentum, P of the system can
not change, even though the linear momentum of each of the
colliding bodies may change.
REGAN

PHY34210
By CONSERVAT ION OF LINEARMOMENT UM,Pi  Pf
99
totalmomentumbeforecollision  totalmomentumaftercollision
For a 2 BODY COLLISION,








p1,i  p2,i  p1, f  p2, f  m1v1,i  m2 v2,i  m1v1, f  m2 v2, f
For a COMP LET ELYINELAST ICCOLLISION,the two particles


stick aftercollision(e.g., a rugby t ackle!) , then m1v1,i  m1  m2 V
For an isolat edsystem,the velocit yof thecentreof mass can not change
in a collisionas thesystemis isolat edand thereis no net ext ernalforce.





Recalling P  Mvcom  m1  m2 vcom  m1v1  m2 v2

 

P
p1  p2
 vcom 

m1  m2 m1  m2
Elastic Collisions in 1-D
REGAN
PHY34210
100
In an elastic collision, the total energy before the collision is equal to
the total kinetic energy after the collision. Note that the kinetic energy
of each body may change, but the total kinetic energy remains constant.
after elastic collision
before elastic collision
m1, v1,f m2, v2,f
m1, v1,i m2, v2,i=0
For a head - on collision between two billiard balls, with mass, m2 at rest.



By conservation of linear momentum,m1v1,i  m1v1, f  m2 v2, f
 m1 v1,i  v1, f   m2 v2, f (1- D case, magnitudesalongsame axis).
In an elasticcollision the totalkineticenergyis conserved
1
1
1
 m1v12,i  m1v12, f  m2 v22, f  m2 v22, f  m1 v1,i  v1, f v1,i  v1, f 
2
2
2
m1  m2
2m1
which leads to, v1, f 
v1,i and v2, f 
v1,i
m1  m2
m1  m2
Note thatv2 ,f is always positive(i.e.m2 is always pushed forward).
REGAN
PHY34210
For 1 - D elasticcollisions, v1, f 
101
m1  m2
2m1
v1,i & v2, f 
v1,i
m1  m2
m1  m2
T heselead to thefollowinglimitingcases.
1) Equal masses, m1  m2 (e.g. poolballs) : v1, f  0 , v2, f  v1,i
i.e., for a head - on collision between equal masses, theprojectilestops
followingcollisionand the targetmovesoff with theprojectile' s velocity.
2) Massive target,m2  m1 (e.g., golf ball on a cannonball) :
 2m1 
v1,i i.e., light projectilebounces back with
v1, f  v1,i , v2 ,f  
 m2 
similar velocity(but oppositedirection)to incomingprojectile.
Heavy target movesforwards with small velocity.
3) Massive projectile m1  m2 (e.g., cannonball on golf ball) :
v1,f  v1,i , v2 ,f  2v1,i i.e. heavyprojectilecontinuesforwards at approx.
unchangedvelocity,light target movesoff with twice theprojectilevelocity
Example 1:
REGAN
PHY34210
102
Nuclear reactors require that the energies of neutrons be reduced by
nuclear collisions with a MODERATOR MATERIAL to low energies
(where they are much more likely to take part in chain reactions). If the
mass of a neutron is 1u~1.66x10-27kg, what is the more efficient
moderator material, hydrogen (mass = 1u) or lead (mass~208u)?
Assume the neutron-moderator collision is head-on and elastic.
We want theMAXIMUMtransferof kineticenergy FROM T HE NEUT RONfor
a single collisionas a functionof moderatormass. T heinitialand final kinetic
1
1
mn vn2,i and K f  mn vn2, f
2
2
K i  K f vn2,i  vn2, f
 T hefractionalenergyloss per collisionis F 

Ki
vn2,i
energiesof theorginaland scatteredneutronare K i 
For a closed neutron- nucleus collision& themoderatingnucleus initiallyat rest,
fromcons.of lin. mom.
vn , f
vn , i
mn  mMOD
4mn mMOD

, thusF 
therefore,
2
mn  mMOD
mn  mMOD 
F  4 / 4  1 for hydrogen proton(NB. water  H 2 O) and ~ 4/208~ 1/50 for Pb!
Example 2: The Ballistic Pendulum
A ballastic pendulum uses the transfer
of energy to measure the speed of
bullets fired into a wooden block
suspended by string.
vbul
REGAN
PHY34210
Mblock
By conservation of linear momentum,mbul vbul  mbul  M block vblock
Also know thatis theblock systemis closed, we can assume
a conservation of mechanicalenergy,then
1
2
mbul  M block vblock
 mbul  M block gh
2
where h is theincreasein height of theblock as it swings upwards.
 mbul vbul  
1
 mbul  M block 
  mbul  M block gh
2
 mbul  M block  
2
2
 vbul

mbul  M block
2 gh
2


m

M

v

bul
block
bul
2
mbul
mbul
2 gh
h
103
1-D Collisions with a Moving Target
REGAN
PHY34210
104
By conservation of linear momentum,








m1v1,i  m2 v2,i  m1v1, f  m2 v2, f  m1 v1,i  v1, f   m2 v2,i  v2, f 
For an elasticcollision,kineticenergyis conserved,thus
1
1
1
1
m1v12,i  m2 v22,i  m1v12, f  m2 v22, f
2
2
2
2
 m1 v12,i  v12, f  m2 v22, f  v22,i 




m1 v1,i  v1, f v1,i  v1, f   m2 v2, f  v2,i v2, f  v2,i 
before elastic collision
m1, v1,i
m2, v2,i
solving thesesimultaneous equations,we obtain thegeneralrelations,
v1, f
m1  m2
2m2
2m1
m2  m1

v1,i 
v2 ,i & v2 , f 
v1,i 
v2 ,i
m1  m2
m1  m2
m1  m2
m1  m2
T hesubscripts1 and 2 are arbitrary.Note,if we set v2,i  0 (stationary
target)we obtain thepreviousresults of
v1, f 
m1  m2
v1,i
m1  m2
& v2 , f 
2m1
v1,i
m1  m2
Collisions in Two Dimensions
REGAN
PHY34210
105
When two bodies collide, the
m2, v2,f
y
impulses of each body on the other
m2, v2,i
determine the final directions following
2
x
the collision. If the collision is not
1
head-on (i.e. not the simplest 1-D case)
in a closed system, momentum remains m1, v1,i
conserved, thus, for an elastic collision
m1, v1,f
where Ktot,I=Ktot,f , we can write,




1
1
1
1
2
2
2
P1,i  P2,i  P1, f  P2, f and m1v1,i  m2 v2,i  m1v1, f  m2 v22, f
2
2
2
2
For a 2-D glancing collision, the collision can be described in terms of
momentum components. For the limiting case where the body of m2 is
initially at rest, if the initial direction of mass, m1 is the x-axis, then,
x axis, m1v1,i  m1v1, f cos1  m2 v2, f cos 2
y axis,
0  m1v1, f sin 1  m2v2, f sin  2
For an elastic collision,m1v12,i  m1v12, f  m2 v22, f
11: Rotation
REGAN
PHY34210
Most motion we have discussed thus far refers to translation.
Now we discuss the mechanics of ROTATION, describing
motion in a circle.
First, we must define the standard rotational properties.
A RIGID BODY refers to one where all the parts rotate
about a given axis without changing its shape.
(Note that in pure translation, each point moves the same
linear distance during a particular time interval).
A fixed axis, known as the AXIS OF ROTATION is
defined by one that does not change position under rotation.
Each point on the body moves in a circular path described by an
angular displacement D. The origin of this circular path is centred at
the axis of rotation.
106
Summary of Rotational Variables
REGAN
PHY34210
107
All rotational variables are defined relative to motion about a
fixed axis of rotation.
The ANGULAR POSITION, , of a body is then the angle between
a REFERENCE LINE, which is fixed in the body and perpendicular
to the rotation axis relative to a fixed direction (e.g., the x-axis).
If  is in radians, we know that =s/r where s is the length of arc
swept out by a radius r moving through an angle . (Note
counterclockwise represent increase in positive .
axis of
Radians are defined by s/r and are thus
rotation
pure, dimensionless numbers without
units. The circumference of a circle
(i.e., a full arc) s=2r, thus in radians,
the angle swept out by a single, full
reference
revolution is 360o = 2r/r=2. Thus,
line
r 
1 radian = 360 / 2 = 57.3o
s
x
= 0.159 of a complete revolution.
The angular displacement, D represents the change in
PHY34210
the angular position due to rotational motion.
In analogy with the translational motion variables, other angular
motion variables can be defined in terms of the change (D), rate of
change ( ) and rate of rate of change ( ) of the angular position.
REGAN
Angular position(radians),
Angular displacement (radians),
108
s

r
D   2  1
D  2  1
AverageAngular Velocity(radian per second), av 

Dt
t 2  t1
Instantane
ous Angular Velocity(rad/s),
AverageAngular Acceleration, (radiansper s 2 ),
Instantane
ous Angular Acceleration, (rad/s2 ),
d

dt
D
 av 
Dt
d
 av 
dt
Relating Linear and Angular Variables
REGAN
PHY34210
109
For the rotation of a rigid body, all of the particles in the body take the
same time to complete one revolution, which means that they all have
the same angular velocity,, i.e., they sweep out the same measure of
arc, d in a given time. However, the distance travelled by each of
the particles, s, differs dramatically depending on the distance, r, from
the axis of rotation, with the particles with the furthest from the axis
of rotation having the greatest speed, v.
at and ar are the tangential and radial accelerations respectively.
We can relate the rotational and linear variables using the following
(NB.: RADIANS MUST BE USED FOR ANGULAR VARIABLES!)
ds
d
dv d r 
d
s  r ; v 
r
 r ; at 

r
 r
dt
dt
dt
dt
dt
v 2 r 
Radial componentof theacceleration is ar 

 r 2
r
r
2r 2
P eriodof revolution, T 

v

2
Rotation with Constant Acceleration
REGAN
PHY34210
For translational motion we have seen that for the case of a
constant acceleration, we can derive a series of equations of motion.
By analogy, for CONSTANT ANGULAR ACCELERATION, there
is a corresponding set of equations which can be derived by
substituting the translational variable with its rotational analogue.
TRANSLATIONAL
v  v0  at
1 2
x  x 0  v0t  at
2
v 2  v02  2ax  x0 
 v  v0 
x  x0  
t
 2 
1 2
x  x0  vt  at
2

ROTATIONAL
  0   t
1 2
   0   t   t
2
  2  02  2    0 


   0 
  0  
t
 2 
1 2
   0  t   t
2
110
Example 1:
REGAN
PHY34210
A grindstone rotates at a constant
angular acceleration of =0.35rad/s2.
At time t=0 it has an angular velocity
of 0=-4.6rad/s and a reference line on
its horizontal at the angular position, 0=0.
111
ref. line
for 0=0
axis of
rotation
(a) at what time after t=0 is the reference line at =5 revs ?
1
2
   0  0t  t 2 :   5rev  10 rad;  0  0 ; 0  4.6rad / s ;   0.35rad / s 2
  4.6 
1

10π-0  -4.6t    0.35 t 2   t 
2

 4.62  4  0.175  10 
2  0.175

4.6  6.56
 32s
0.35
Note that while 0 is negative,  is positive. Thus the grindstone starts
rotating in one direction, then slows with constant deceleration before
changing direction and accelerating in the positive direction.
At what time does the grindstone momentarily stop to reverse direction?
t
  0
a

0   4.6rad / s 
 13s
2
0.35rad / s
Kinetic Energy of Rotation
REGAN
PHY34210
112
For a compositebody which we can treatas a collectionof
masses, mn , movingat differentspeeds, vn , thekineticenergyis
1
1
1
1
2
2
2
K  m1v1  m2 v2  m3v3    mn vn2
2
2
2
n 2
1
1
2
2  2
 K   mn rnn  . BUT  is constant,thus K    mn rn  .
2 n
n 2

Now we can define I   mn rn2 where




n
I is theMOMENTOF INERT IAor ROT AT IONALINERT IA
1 2
 Kineticenergy of rotationis given by K  I
2
T husin general,a smaller momentof inertiameansless work
is needed to be done (i.e.less K ) for rotation ot takeplace.
Calculating to the Rotational Moment of Inertia

For a rigid body, I   mn rn2

REGAN
PHY34210
113
where r is theperpendicular distanceof then th
n


part iclefrom therotationaxis. For a continuousbody, I   mn rn2   r 2 dm.
n
The Parallel-Axis Theorem
To calculate I if the moment of inertia about a parallel axis passing
through the body’s centre of mass is known, we can use I=Icom+Mh2,
where, M= the total mass of the body, h is the perpendicular distance
between the parallel centre of mass axis and the axis of rotation and
Icom is the moment of inertia about the centre of mass axis.


If h 2  a 2  b 2 , I   r 2 dm   x  a    y  b  dm

2
2



 I   x 2  y 2 dm  2a  xdm  2b  ydm   a 2  b 2 dm
1
Now, since x  y  R and since xcom 
x dm, assuming we

M
take thecentreof mass as theorgin, thenby definition,
2
2
2


2b  xdm  2b  ydm  0  I   R 2 dm   a 2  b 2 dm  I com  h 2 M
Example 2:
REGAN
PHY34210
114
The HCl molecule consists of a hydrogen atom (mass 1u) and a
chlorine atom (mass 35u). The centres of the two atoms are separated
by 127pm (=1.27x10-10m). What is the moment of inertia, I, about an
axis perpendicular to the line joining the two atoms which passes
through the centre of mass of the HCl molecule ?
We can locatethecent reof mass of the2 - particle
systemusing xcom
m x  m2 x2
 1 1
M  m1  m2 
a
Cl
d-a
com
H
If the x co - ordinatefor thecent reof mass x  0 then,
 mCl a  mH d  a 
mH
0
a
d . Now
mCl  mH
mCl  mH
I com   mi ri 2  mH d  a   mCl a 2 
2
d
rotation axis
i
2
2


 mH

mH mCl 2
mH
 1 35 
2
mH  d 
d   mCl 
d  
d  I 
u  127pm
mCl  mH 
mCl  mH
 1  35 

 mCl  mH 
 I  15,250u. pm2 (not eunits for rotationalmomentsof inertiafor molecules).
Torque and Newton’s
2nd
Law
REGAN
PHY34210
The ability of a force, F, to rotate an object
depends not just on the magnitude of its
tangential component, Ft but also on how
far the applied force is from the axis of
rotation, r. The product of Ft r =Frsin
is called the TORQUE (latin for twist!)  .
115
F
Ft

Frad
r
O
r
T ORQUE,   r F sin    rFt AND   r sin  F   r F . r is the

perpendicular distancebetween O and a line running throughF .

r is theMOMENTARM OF T HE FORCE F. SI unit of T orqueis Nm,
 
which are equivalent to theunit of work W  F.d . W in Joules, in Nm.


Relatingthe tangential force to the tangential acceleration, Ft  m at .
T orqueactingon theparticleis   Ft r  m at r , since, at  r ,
  m rr   m r2  I  τ net  Iα  Newton's 2nd law for rotation.
Work and Rotational Kinetic Energy
From the WORK- KINET ICENERGY T HEOREM,
1 2 1 2
m vf  m vi  W
2
2
1
1
2
2
since v  rω , then mr  f  mr i  W
2
2
Recallingfor a single - particlebody, I  m r2 ,
DK  K f  K i 
1 2 1 2
then W  DK  I f  Ii
2
2
Workdone, W  Fs  Ft rD  D
Workdone in an angular displacement 1 to 2
2
is given by W   d
1
dW
d
P OWERis given by P 

 
dt
dt
REGAN
PHY34210
116
REGAN
PHY34210
117
12: Rolling, Torque and Angular Momentum
Rolling: Rolling motion (such as
a bicycle wheel on the ground) is
a combination of translational
and rotational motion.
O
COM
motion.
R
 O
P
P
S
A wheelrollingat a CONST ANTSP EED, means that t hespeed of the
centreof mass, vcom is constant.In a timeintervaldt, thecentreof mass
travelsthesame distanceas thedistance theoutside of the wheel moves
throughan arc of length,s  R , where R is the wheel radius and is
its angular displacement.
d
T heangular speed of the wheel about its centreis  
, while thespeed
dt
ds d R 
d
of thecentreof mass is given by vcom 

R
 R
dt
dt
dt
REGAN
PHY34210
118
The kinetic energy of rolling.
R
A rolling object has two types of
COM
kinetic energy, a rotational
O
 O
motion.
kinetic energy due to the
rotation about the centre of
P
P
S
mass of the body and
translational kinetic energy due to the translation of its centre of mass.
We can viewthesituationas pure rotationabout an axis through he
t point,P.
1
T hekineticenergyof thisrotationis given by K  I P 2 where I p is the
2
momentof inertia through he
t pointof contact with theground, P.
From thePARALLELAXIST HEOREM,we can write
1
I P  I com  MR 2 and thus, K  I com  MR 2  2
2
1
1
1
1
2
2
2
2
 K  I com  M R   I com  Mvcom
2
2
2
2



N
REGAN
119
Rolling Down a Ramp
PHY34210
If a wheel rolls at a constant speed, it
R

has no tendency to slide. However, if this
Fg sin 
P Fg cos
wheel is acted upon by a net force (such

as gravity) this has the effect of speeding
Fg

up (or slowing down) the rotation, causing
an acceleration of the centre of mass of the system, acom along the
direction of travel. It also causes the wheel to rotate faster. These
accelerations tend to make the wheel SLIDE at the point, P, that it
touches the ground. If the wheel does not slide, it is because the
FRICTIONAL FORCE between the wheel and the slide opposes the
motion. Note that if the wheel does not slide, the force is the STATIC
FRICTIONAL FORCE ( fs ).
Since therotationalfrequencyis given by R  vcom , then
d R  d vcom 
by differentiatingboth sides,

 acom  R
dt
dt

N
Rolling down a ramp (cont.)
For a uniform body of mass, M and radius, R,
rolling smoothly (i.e. not sliding) down a ramp Fg sin 
tilted at angle,  (which we define as the x-axis
in this problem), the translational acceleration

down the ramp can be calculated, from
REGAN
PHY34210
120
R
P

Fg
Fg cos
theforcecomponentsalong theslope, Fx ,net  Macom, x  f s  Mg sin 
where f s  m s N  m s Mg cos . Rot.formof Newton's 2 nd law is   Fr  I .
T heonly forcecausing a rollingmotionin thefigure is theFRICT IONat point P .
T hegravitational and Normalforcesall act through theCOM and thus haveR  0.

 acom 
  net  I com  Fr  f s R. For smoothrolling, 
(notesign)
R
I com acom, x
I com I com .  acom, x 
 f s  Macom, x  Mg sin  


R
R
R
R2
I com acom, x
g sin 

 Macom, x  Mg sin   acom  
2
R
 I com 

1 
 MR 2 
The Yo-Yo: If a yo - yo rolls down a distanceh it loses gravitational
potentialenergy,m gh. T hisis transferred into kinetic
REGAN
PHY34210
T
1 2

energyin both translational K trans  m v  and rotational
R0 R
2



1

2
 K rot  I  forms.As the yo - yo climbs back up thestring,
2


it loses thiskineticenergy and transfersit back topotentialenergy.
Mg
T heexpressionfor thevalue of theacceleration of the yo - yo rolling
down thestringcan be calculatedassuming Newton's 2 nd law (as for
a body rollingdown a hill) with thefollowingassumptions.
(1) the yo - yo rolls directlydown thestring(i.e.  900 ).
(2) the yo - yo rolls around theaxle with radius R0 , not theouter radius, R.
(3) the yo - yo is slowed by thetensionin thestringrather tha
n friction.
g
T hisanalysisleads to theexpression, acom  
 I

1   com2 
 MRO 
121
Example 1:
A uniform ball of mass M=6 kg and radius R rolls
smoothly from rest down a ramp inclined at
30o to the horizontal.
(a) If the ball descends a vertical height of
1.2m to reach the bottom of the ramp, what is the
speed of the ball at the bottom ?
REGAN
PHY34210
122
1.2m
By conservation of mechanicalenergy, K i  U i  K f  U f
 Mgh  0  0  K rot  K trans 
1
1
2
Mvcom
 I com 2 .
2
2
2
MR 2 for a sphere,
5
2
1
1
1
12
1
1
2
2
2
2  vcom 
2
2


Mgh  Mvcom  I com  Mvcom   MR  2   Mvcom  Mvcom
2
2
2
25
5
 R  2
For smoothrolling,vcom  R and subsit uting, I com 
v
2
com

gh
7
10
 vcom
10gh

 4.1m s1 (note,Mass independent, marble
7
and bowling ball reach bottomat same time!)
REGAN
PHY34210
Example 1 (cont):
123
(b) A uniform ball, hoop and disk, all of mass M=6 kg and
radius R roll smoothly from rest down a ramp inclined
at 30o to the horizontal. Which of the three objects reaches
the bottom of the slope first ?
1.2m
2
1
2
T hemomentsof inertiafor a sphere MR ; disk  MR 2 ; and hoop MR 2 .
5
2
T hefractionof kineticenergy which goes intoT RANSLAT IONAL MOT ION,
2
Mvcom
vcom
2
f 1
. In general,I com  MR , wit h   a constantand  
2
1
 2 I com
R
2 Mv
1
2
2
com
 f 
1
2
2
Mvcom
2


v
2
2
com
1
1
 2 
2 Mvcom  2   MR 
 R 

1
1 
1
2
 For hoop,  1, f  0.5 ; For disk,   , f  0.66* ; For sphere,  , f  0.71.
2
5
Sphere rolls fast est,followed by thedisk. Any size marble will beat disk.
REGAN
PHY34210
124
Torque was defined previously for a rotating rigid body as =rFsin.
More generally, torque can be defined for a particle moving along
ANY PATH relative to a fixed point. i.e. the path need not be circular.
z
z
rxF

= 
F redrawn
at origin
O 
O
x
x
 F


r
r
F
r
y
F
y


 
T he torqueis defined by   r  F . T hedirectionof the torqueis found using


the vectorcross product right - hand rule, (i.e.perpendicular toboth r and F ).
T heMAGNIT UDEOF T HET ORQUE is given by   rF sin   r F  rF 

where r  r sin  and F  F sin  .
REGAN
PHY34210
125
Angular Momentum



A particleof mass m, with velocity v (i.e. with linear momentum,p  mv ),
  
 
has an ANGULAR MOMENT UM given by l  r  p  mr  v  .
z
z
rxp
= l
l
O
y

r

p redrawn
at origin

p
p
O
x
y
r
 p
r

x
p
T heangular moment umdirect ionis given by t hevect orcross product


 
(t heright - hand rule shows t hatl is  t o bot h r and p, v ).
T hemagnit udeof t heangular moment um(in unit s of kg.m2 /s  Js ) is
given by l  rp sin   r p  p  r where r  r sin  and p   p sin  .
Newton’s
2nd
Law in Angular Form.
REGAN
PHY34210
126


dp
Newton's 2 nd law in translationalform can be writtenas Fnet 
dt
  
 
If theangular momentumof a particleis given by l  r  p  mr  v 
Differentiatingboth sides with respect totimegives,

 


   dv    dr  
   
dl d mr  v 

 m r    v     mr  a   v  v 
dt
dt
dt   dt  

 
v  v  0 since these vectorsare parallel(sin  0)

  
  
 

dl

 mr  a   r  ma  r  Fnet   ri  Fi   net
dt
i
i.e. therateof changeof angular momentumwith respect totime

 dl 

  is equal to the vectorsum of torquesactingon theparticle net .
 dt 
 


REGAN
127
For a SYST EM OF P ART ICLES,the totalangular momentum,
PHY34210


L is theVECT OR SUM of theangular momenta,l of theindividual particles,


n 
n
n

   

dli
dL
i.e., L  l1  l2  l3    ln   li 

  net ,i
dt i 1 dt i 1
i 1

Only EXT ERNALtorqueschange theT OT ALANGULAR MOMENT UM( L )
i.e., thosedue to forceson theparticlesfromexternalbodies.
If  net is the NET EXT ERNALT ORQUE, i.e. the vectorsum of all

dL

externaltorques, τ net 
, we obtain a formfor Newton's 2 nd law :
dt
The net external torque, net acting on a system is equal to the rate
of change of the total angular momentum of the system ( L ) with time.
REGAN
128
For a given part iclein a rigid body rot at ingabout a fixed
PHY34210
axis, t he magnitudeof t heangular moment umof a mass element Δmi , is
l  ri pi sin 900  ri Dmi vi .
T heangular moment umcomponentparallelt o t herot at ionDm
(z) axis is liz  li sin   ri sin  Dmi vi   ri Dmi vi
T hecomponentfor t heENT IREBODY is t hesum
of t heseelement alcont ribut ons
i
z
r

r


pi

y
x
 n

i.e., L z   liz  Dmi vi ri   Dmi r i ri     Dmi r2i 
i 1
i 1
i 1
 i 1

n
n
n
n
 is a CONST ANTfor all point son t herot at ingbody and  Dmi r2i  I ,
i 1
t he momentof inert iaof t hebody about a fixed axis, we can writ e,
Lz  I
Usually t he ' z ' is dropped,assuming t hat L is about therot at ionaxis.
REGAN

dPHY34210
L 
Since thenet torqueis relatedto thechangein angular momentumby,
 τ net ,
dt

dL
if NO NET T ORQUEactson thesystem,then
 0 and thus
dt
T HE ANGULAR MOMENT UMOF T HESYST EM IS CONSERVED.
Conservation of Angular Momentum
129
T hismeans that thenet angular momentumat timeti , is equal to thenet angular
momentumof thesystemat some other time
,tf .
We can thussay thatif thenet externaltorqueactingon a systemis zero, the

angular momentumof thesystem,L remainsconstant,no matter what changes
takeplace WIT HINthesystem.
Similarly,if theCOMP ONENT of thenet externaltorqueon a systemalonga
fixed axis is zero, then thecomponentof angular momentumalong thataxis
can not change,no matter what takesplace within the system.
T heconservation law can be writtenin algebraic formas I ii  I f  f .
T hismeans thatif themomentof inertiaof a systemdecreases,its rotational
speed increasesto compensate, (e.g., pirouetting skaters,neutronstarsand nuclei!)
Example1: Pulsars (Rotating Neutron Stars)
Crab nebula, SN remnant
observed by chinese in 11th century
before
REGAN
PHY34210
after!
SN1987A
Pulsars have similar
periodicities ~0.1-1s.
Vela supernova
remnant, pulsar
period ~0.7 secs
130
REGAN
PHY34210
131
Rotational period of crab nebula (supernova remnant) =1.337secs
Lighthouse
effect
Star
quakes
optical




I ii  k .MRi2 i  I f  f  k .MR 2f  f , k  constant
Tf
i
 R f  Ri
 Ri
, T  periodof rotation
f
Ti
Ri ( sun) ~ 7 108 m, Ti ( sun) ~ 2.5 106 s, T f ( pulsar) ~ 1s
1s
 R f ~ 7 10 m
~ 400Km.
6
2.5 10 s
8
x-ray
PULSAR = PULSAting Radio Star (neutron-star)
REGAN
PHY34210
TRANSLATIONAL
ROTATIONAL



 dp

dl 
Force,
F  ma 
 Torque,   I 
 r F
dt
dt


 

Mom entum Linear
,
p  mv
 Ang. Mom. l  I  r  p


 dp

dL
nd
Newtons 2 Law, F 
  net 
dt
dt
Conservation, LinearMom.
 Angular mom.




dp
dL
Fnet  0 
 net  0 
dt
dt
132
13: Equilibrium and Elasticity
REGAN
PHY34210
133
An object is in ‘equilibrium’ if p=Mvcom and L about an any axis are
constants (i.e. no net forces or torques acts on the body).
If both equal to zero, the object is in STATIC EQUILIBRIUM.
If a body returns to static equilibrium after being moved (by a restoring
force, e.g., a marble in a bowl) it is in STABLE EQUILIBRIUM.
If by contrast a small external force causes a loss of equilibrium, it has
UNSTABLE EQUILIBRIUM (e.g., balancing pennies edge on).


dp
F net
 0 (i.e.balanceof forces)for
dt
T RANSLAT IONAL EQUILIBRIUM
and


dL
 net 
 0 (i.e.balanceof torques)for
dt
ROT AT IONALEQUILIBRIUM
The Centre of Gravity
REGAN
PHY34210
The gravitational force acts on all the individual atoms
in an object. In principle these should all be added together
vectorially.
However, the situation is usually simplified by the concept
of the CENTRE OF GRAVITY (cog), which is the point
in the body which acts as though all of the gravitational force
acts through that point.
If the acceleration due to gravity, g, is equal at all points
of the body, the centre of gravity and the centre of mass
are at the same place.
134
Elasticity
REGAN
PHY34210
135
A solid is formed when the atoms which make up the solid take up
regular spacings known as a LATTICE. In a lattice, the atoms take up
a repetitive arrangement whereby they are separated by a fixed, well
defined EQUILIBRIUM DISTANCE (of ~10-9->10-10m) from their
NEAREST NEIGHBOUR ATOMS.
The lattice is held together by INTERATOMIC FORCES which can
be modelled as ‘inter-atomic springs’. This lattice is usually extremely
rigid (i.e., the springs are stiff).
Note that all rigid bodies are however, to some extent ELASTIC.
This means that their dimensions can be changes by pulling, pushing,
twisting and/or compressing them. STRESS is defined as the
DEFORMING FORCE PER UNIT AREA= F/A,
which produced a STRAIN, which refers to a unit deformation.
The 3 STANDARD type of STRESS are (1) tensile stress ->DL/L
(stretching) ; (2) shearing stress -> Dx/L (shearing) ; and (3) hydraulic
stress -> DV/V (3-D compression).
REGAN
PHY34210
136
STRESS and STRAIN are PROPORTIONAL TO EACH OTHER.
The constant of proportionality which links these two quantities is
know as the MODULUS OF ELASTICITY, where
STRESS = MODULUS x STRAIN
F
T heSTRESSon an object for simple tensionor compression is given by ,
A
where F is themagnitudeof theforceappliedperpendicularly tothearea A
(T hisalso defines thepressure at thatpoint).
T heSTRAIN is theunit deformation. For tensile stress, this is a dimensionless
ΔL
corresponding to the fractionalchange inthe length
quantitydefined by
L
of theobject ( L is theorginallength,ΔL is theextension).
T heYOUNG'S MODULUS ( E ) for tensile or
F
compressive stressis defined by
E
A
F
For SHEARING, thestressis still , but
A
F
ΔL L+
DL
L

F is parallel
Δx
t o theplaneof thearea. T hestrain is now
, leading
l
F
Dx L
t o theSHEAR MODULUS,(G) where
G
A
l
HYDRAULICSTRESSis defined as thefluid pressure P,
REGAN
PHY34210
L
F
Dx F
V
ΔV
(i.e.forceper unit area).T hestrain is defined as
,
V
V-DV
whereV is theinitial volumeand ΔV is the volumechange.
DV
T heBULK MODULUS( B ) is defined by P  B
V
137
F
DV
REGAN
PHY34210
If we plots stress as a function of strain,
for an object, over a wide range,
there is a linear relationship. This means
that the sample would regain its original
dimensions once the stress was removed
(i.e., it is ‘elastic’).
However, if the stress is increases BEYOND
THE YIELD STRENGTH, Sy,of the specimen,
it will become PERMANENTLY DEFORMED.
If the stress is increased further, it will ultimately
reach its ULTIMATE STRENGTH, Su, where the
specimen breaks/ruptures.
138
Su (rupture)
Sy (perm. deformed)
Strain (Dl/l)
Example 1:
REGAN
139
PHY34210
F=62kN
A cylindrical stainless steel rod has a radius r = 9.5mm and
length, L = 81cm. A force of 62 kN stretches along its length.
(a) what is the stress on the rod ?
F
F
6.2 104 N
8
2
stress   2 

2
.
2

10
Nm
2
3
A r
  9.5 10 m

A
l=
81cm

F=62kN
(b) If the Young’s modulus for steel is 2.2 x
are the elongation and strain on the cylinder ?
Dl F
l
From thedefinitionof Young's modulus, E

 Δl   stress
l
A
E
0.81m  2.2 108 Nm  2
4
 Δl 

8
.
9

10
m
11
2
2.2 10 Nm
Dl 8.9 10 4 m
strain 

 1.110 4  0.11%
l
0.81m
1011


Nm-2, what
14: Gravitation
REGAN
PHY34210
140
Isaac Newton (1665) proposed a FORCE LAW which described the
mutual attraction of all bodies with mass to each other. He proposed
m1m2
that each particle attracts any other particle via the
F G 2
GRAVITATIONAL FORCE with magnitude given by
r
G=6.67x10-11N.m2/kg2=6.67x10-11m3kg-1s-2 is the gravitational constant
‘Big G’ (as opposed to ‘little g’ the acceleration due to gravity).
m1
The two particles m1 and m2 mutually attract with a force
of magnitude, F. m1 attracts m2 with equal magnitude
F
but opposite sign to the attraction of m2 to m1. Thus,
r

F and -F form a third force pair, which only depends on
F
the separation of the particles, r, not their specific positions.
m2
F is NOT AFFECTED by other bodies between m1 and m2.
THE SHELL THEOREM:
While the law described PARTICLES, if the distances between the
masses are large, the objects can be estimated to be point particles.
Also, ‘a uniform, spherical shell of matter attracts a particle outside
the shell as if all the shell’s mass were concentrated at its centre’.
Gravitation Near the Earth’s Surface
REGAN
PHY34210
141
The earth can be thought of a nest of shells, and thus all its mass
can be thought of as being positioned at it centre as far as bodies
which lie outside the earth’s surface are concerned.
Assuming theearthis a UNIFORM SP HERE of mass M , themagnitudeof
thegravit ational forcefrom theearthon a particleof mass m, at a distancer ,
Mm
from theearth's centreis given by : Fgrav  G 2
r
If theparticleis released,it will accelerateto theearth's centreunder gravit y

with a GRAVIT AT IONAL ACCELERAT ON,
I a g , whose magnitudeis given
Mm
GM
by Fgrav  m ag  G 2  a g  2 . T hus theacceleration due to gravit y
r
r
depends on the' height'at whichan object is droppedfrom.
average ag at earth’s surface = 9.83 ms-2 altitude = 0 km
ag at top of Mt. Everest
= 9.80ms-2 altitude = 8.8 km
ag for space shuttle orbit
= 8.70 ms-2 altitude = 400km
REGAN
PHY34210
142
We have assumed the free fall acceleration g equal the
gravitational acceleration, ag, and that g=9.8ms-2 at the earth’s surface,
In fact, the measured values for g differ. This is because
• The earth is not uniform. The density of the earth’s crust varies. Thus
g varies with position at the earth’s surface.
• The earth is not a sphere. The earth is an ellipsoid, flattened at the
poles and extended at the equator. (rpolar is ~21km smaller than requator).
Thus g is larger at poles since the distance to the core is less.
• The earth is rotating. The rotation axis passes through a line joining
the north and south poles. Objects on the earth surface anywhere apart
these poles must therefore also rotate in a circle about this axis of
rotation (joining the poles), and thus have a centripetal acceleration
directed towards the centre of the circle mapped out by this rotation.
Centripetal Acceleration at Earth’s Surface
T henormalforceon a surface object is

from Fnet  m ar  N  m ag  m   2 R
N

T henormalforce,N is equal to the weight, m g


 m g  m ag  m ω 2 R  g  a g   2 R
REGAN
PHY34210

R
m
mag
R is theradius which theobject rot at esaround
and  is therot at ionalvelocity.
R is max.at theequator, R  Rearth  6.37106 m.
Δθ
2 rads
 can be estimatedfrom

Δt 24 3600s
 acentr   2 R  0.034m s 2 (cf.m ag  9.8m s- 2 )
i.e. verysmall comparedto m ag . Assuming
the weight equals thegravitational acceleration
is usually (on earthat least!) well justified.
S
‘above’ view,
looking from pole,
RN
m

143
Gravitation Inside the Earth
REGAN
PHY34210
144
‘A uniform shell of matter exerts no NET force on a particle located
inside it.’
Therefore, a particle inside a sphere only feels a
net gravitational attraction from the portion of
the sphere inside the radius at which it is at.
No
net
Force
m
r
Net
force
In the example on the left, for r = M/V = constant
R
a planet of radius, R and total mass M.
An object of mass m, which burrows downwards such that it is now at
a distance r from the centre of the planet (with r < R ).
The object will experience a gravitational attraction from the mass of
the planet inside the ‘shell’ of radius r and none from the portion of
the planet between radii r and the outer radius R.
M ins  rV  r  43 r 3 and since theforceexperienced by theparticledue to the
GM ins m Gr  43 r 3 4Grr  m
mass inside theshell is F 


i.e. Fnet  kr
2
2
r
r
3
Gravitational Potential Energy
REGAN
PHY34210
145
T hegravitational potentialenergyis defined by theexpression,
Mm
U  G
and defined to be zero at infiniteseparation(r  ).
r
PROOF





In general,work done is W   F r .dr , F r .dr  F r dr cos ,
R
 



Mm
cos  cos1800  -1  F r .dr  G 2 dr  W   F r .dr
r
R

W    G
R


Mm
1
GMm
GMm
 GMm 
dr


GMm
dr


0



R r 2
 r 
r2
R
R
R
W  WORK REQUIRED to movea mass, m froma distance R out to.
Since potentialenergyand work done are relatedby thegeneralexpression,
GMm
DU  W  U   U R ,  U R  W  
R
Potential Energy and Force
REGAN
PHY34210
146
Gravityis a conservative forceand changesin thegrav.potentialenergy
onlydependon theinitialand final positions,NOT T HE PAT HT AKEN.
Since we can derive thegravitational potentialenergyfrom theexpressionfor
theforce,theconverseis also true.F  
dU
d 
Mm 
Mm
  G


G

dr
dr 
r 
r2
Escape Speed (Velocity)
A mass m projectileleavinga mass M planetof radius R has an
ESCAP E SP EED,v. T hiscauses theobject t omoveup with
const antspeed, v against gravity,unt ilit slows down tov  0 at infinitedistance.
1 2
GMm
m v ; T hegravitational potentialenergy, U  
.
2
R
At infinitedistance,K  U  0 (i.e.zero velocit yand at thezero potentialenergy
T hekineticenergy, K 
configuration for r  ). T husfrom theprincipleof conservaton
i of energy
K U 
1 2  GMm 
2GM
earth
1
m vesc   

0

v

;
v

11
.
2
km
s

esc
esc
2
R 
R

Johannes Kepler’s (1571-1630) Laws
• THE LAW OF ORBITS:
All planets move in elliptical
orbits with the sun at one focus.
• THE LAW OF AREAS:
A line that connects a planet
to the sun sweeps out equal
areas in the plane of the planet’s
orbits in equal times.
i.e., dA/dt=constant.
• THE LAW OF PERIODS:
The square of the period of
any planet around the sun is
proportional to the cube of
the semi-major axis of the orbits.
REGAN
PHY34210
a
b a
b
147
The Law of Orbits
REGAN
PHY34210
148
If M >> m,the centre of
Rp
Ra
mass of the planet-sun
system is approximately
m
at the centre of the sun.
r
The orbit is described
by the length of the
M

semi-major axis, a
f
f’
and the eccentricity
ea
ea
parameter, e.
The eccentricity is defined
by the fact that the each
a
focus f and f’ are distance
ea from the centre of the ellipse. A value of e=0 corresponds to a
perfectly circular orbit.
Note that in general, the eccentricities of the planetary orbits are small
(for the earth, e=0.0167). Rp is called the PERIHELION (closest
distance to the sun); Ra is the APHELION (further distance).
Ra
r
Rp
y
ea
149
r

f
REGAN
PHY34210
ea

f’
x
f
a
In generalfor an ellipse, theeccentricity is defined by ,  
rmax  rmin
rmax  rmin

f’
Ra  R p
Ra  R p
If we taketheorigin of theco - ordinatesas thefocus, f ,
r0
r
r
r
 rmax  0 , rmin  0
1   cos
1 
1 
In Cart esianco - ordinates r  x 2  y 2 , x  r cosθ, y  r sin θ


Ellipseis defined by theequation, 1   2 x 2  2r0 x  y 2  r02
1
 1
A  Lengt hof majoraxis  2  a  rmax  rmin  RP  Ra  r0 

1  1 
r
A
 T helength of theSEMI - MAJOR AXIS  0 2
2 1 
2r0



2
 1 
REGAN
PHY34210
The Law of Areas
150
If theDA is thearea swept out in timeDt , ΔA can be EST IMAT EDassuming
the wedge of area swept out is a T RIANGLEof height,r , and base s  rD .
 p
T hearea swept out is approximately
p
1
1
DA   base  height  .rDθ.r.
2
2
T hisexpressionbecomesmoreexact
for smaller values of ΔA .
As D  0 and ΔA  0,
r
D
m
DA
M
ΔA dA 1 2 d 1 2

 r
 r
Δt
dt 2 dt 2
where  is theangular speed of therotatingline connectingthesun and planet
(i.e., rotationalvelocityof planetaround thesun).
T heang. mom.of theplanetaround thestar is, L  rp  r m v   r m r   m r2
dA 1 2
L

 r 
dt 2
2m
dA
. T hus,if L is conserved,
 constant.
dt
REGAN
PHY34210
The Law of Periods
For a circular orbit,using Newton's 2 nd law,

Mm
v2
rω
 G 2  mg  m  m
 m rω 2
r
r
r
m
2
F  m agrav
151
r
M
Mm
 G 2  m rω 2  GM  r 3ω 2
r
2π
T heperiodof revolution, T 
, substit uting in we get
ω
2
2

 3
4
π
4

3
2
r . For ellipse, r  a  semi - majoraxis.
GM  r
 T  
2
T
 GM 

T 2  4π 2

Exact version of law predicts 3  
a
 G M  m  
T2
2
i.e., 3  constantfor M  m, ( Ta 3  3.010-34 yr2 m -3for solar system).
a
REGAN
PHY34210
Satellites, Orbits and Energies
152
T hepotentialenergy of systemis given by
GMm
U 
, U  0 for infiniteseparation
r
T heKINET ICENERGY OF A CICRULARLY
ORBIT INGSAT ELLIT E,via Newt on's 2 nd law is
K(r)
r
Etot(r)
GMm
v2
1
GMm
F  2  m  K  m v2 
r
r
2
2r
=-K(r)
U
T herefore,K  
for a sat ellit ein a circular orbit.
2
T he totalmechanicalenergyis given by E  U  K
GMm GMm
GMm
 E 


 K
r
2r
2r
i.e. thetotalenergyis equal to the NEGAT IVEOF T HE KINET ICENERGY
GMm
For an ellipt icalorbit substitutea (semi- majoraxis length)for r , ie. E  
2a
Example 1:
REGAN
PHY34210
153
A satellite in a circular orbit at an altitude of 230km above the earth’s
surface as a period of 89 minutes. From this information, calculate
the mass of the earth ?
2

 3
4

rd
2
r assuming M  m
From Kepler's 3 law : T  
 GM 
r  R  h whereR  6.37106 m  theearth's radius
and G  6.6710-11 m3kg 1s  2


 4π  R  h  
 6.6 10 m
4




 M earth  

2
2
11 3
1  2 

G
T
6
.
67

10
m
kg
s


89

60
s




 M earth  6 1024 kg
2
3
2
6
3
15: Fluids
REGAN
PHY34210
154
Fluids (liquids and gases), by contrast with solids, have the ability to
FLOW. Fluids push to the boundary of the object which holds them.
DENSIT Y( r ) : is theratioof themass Δm and thesize of thevolume
elementΔV , i.e., r  DDMV . For a uniformdensity ρ  MV .
Density has SI units of kg/m3 . In general, the density of liquids does
not vary (they are incompressible); gases are readily compressible.
Pressure: The pressure at any point in a fluid is defined by the limit of
the expression, p = DF /DA as DA is made as small as possible.
If the force is UNIFORM over a FLAT AREA, A, we can write p=F/A
The pressure in a fluid has the same value no matter what direction
the pressure WITHIN the fluid is measured. Pressure is a SCALAR
quantity (i.e.,independent of direction).
The SI unit of pressure is the PASCAL (Pa) where 1 Pa=1Nm-2. Other
units of pressure include ‘atmospheres’ (atm), torr (mmHg) and lbs/in2
where 1 atm = 1.01x105Pa=760 torr = 760 mm Hg =14.7 lb/in2
REGAN
PHY34210
Fluids at Rest
For a tank of water open to air. The water pressure
increases with depth below the air-water interface,
while air pressure decreases with height above the
water. If the water and air are at rest, their pressures
are called HYDROSTATIC PRESSURES.
AIR
WATER
155
y=0
A F1 y
F2 mg
1
y2
For wat erin ST AT ICEQUILIBRIUM, (stationary and theforceson it balance)
F1 and F2 are theforcesat thetopand bottomof a cylinderof water with t op
and bottomat depths y1 and y2 respectively and cross- sectionalarea, A.
T hemass of water in cylinder  m  rV  ρA y1  y2  , V  volumeof cylinder.
Balancingtheforcesgives F2  F1  m g where F1  p1 A and F2  p2 A.
 F2  F1  m g  p2 A  p1 A  rA y1  y2   g   p2  p1  r  y1  y2   g 
If y1  surface level, y2  h  depth, p1  p0  atmospheric pressure
 ph  p0  rgh
(P rinciplebehind theMERCURY BAROMET ER)
The pressure at a point in a fluid in static equilibrium depends on the
depth of that point but NOT on any horizontal dimension of the fluid.
Pascal’s Principle
REGAN
PHY34210
156
‘ a change in the pressure applied to an enclosed incompressible fluid
is transmitted undiminished to every portion of the fluid and to the
walls of its container ’
i.e. squeezing a tube of toothpaste at one end pushes it out the other.
We can write Pascal’s principle as Dp=Dpext, i.e. the change in pressure
in the liquid equals the change in the applied external pressure.
This is the basis behind the concept
of the HYDRAULIC LEVER.
Fi
Fo
A downward force on one platform (the
load
Ao
‘input piston’) causes a change in
Ai
pressure of the INCOMPRESSIBLE
do
LIQUID, resulting in the movement
of a second platform (the ‘output piston’).
di
Fo
For equilibrium, there must be a
downward force due to a load on the
output piston which balances the upward
force, Fo.
The Hydraulic Lever
T heappliedforce Fi , and thedownward
Fi
REGAN
PHY34210
157
Ao
do
load
Fo
force Fo , from theload on theright hand piston
Ai
producesa changein thepressure of theliquid
Fi Fo
Ao
Dp 

 Fo  Fi
Ai Ao
Ai
di
Fo
If theinput pistonmovesdown a distanced i ,
theoutput pistonmovesupwards a distanced o such thatthesame volumeV ,
of theincompressible fluid is displaced at both pistons,then,
Ai
V  Ai d i  Ao d o  d o  d i
 if Ao  Ai , d i  d o
Ao
i.e., theoutput pistonmovesa smaller distance than theinput piston.
 Ao   Ai 
   d i
  Fi d i
T heOUT P UT WORK is given by W  Fo d o   Fi
 Ai   Ao 
With a hydraulic lever, a given force applied over a given distance can
be transformed to a greater force over a smaller distance.
Archimedes’ Principle
REGAN
PHY34210
158
‘when a body is fully or partially submerged in a fluid, a BUOYANT
FORCE, Fb from the surrounding fluid acts on the body. The force is
directed upwards and has a magnitude equal to the weight, mfg of the
fluid that has been displaced by the body.
This net upward bouyant force exists because the water pressure around
the submerged body increases with depth below the surface (Dp=rgh).
Thus the pressure at the bottom of the object is larger than at the top.
If a body submerged in a fluid has a greater density that then fluid, there
is a net force downwards (Fg>Fb), while if the density is less than the
fluid, there will be net force upwards (since Fb<Fg).
For a body to float in a fluid, the magnitude of the bouyant force Fb,
equals to the magnitude of the gravitational force Fg or,
the magnitude of the gravitational force on the body is equal to the
weight, mfg of the fluid which has been displaced by the body.
The weight of a body in fluid is the APPARENT WEIGHT (Wapp)
where Wapp= actual weight - magnitude of bouyant force.
Since floating bodies have Fb=mg, their apparent weight is ZERO!
Flow of Ideal Fluids In Motion
REGAN
PHY34210
159
The flow of real fluids is very complicated mathematically. Often
matters are simplified by assuming an IDEAL FLUID.
This requires 4 basic assumptions:
A) STEADY FLOW: in steady flow, at a fixed point, the velocity of the
moving fluid does not change in magnitude or direction.
B) INCOMPRESSIBLE FLOW: This assumes the fluid has constant
and fixed density (i.e. it is incompressible).
C) NONVISCOUS FLOW: Viscosity is a measure of how resistive a
fluid is to flow and is analogous to friction in solids. For example,
honey has a higher viscosity than water). An object moving through
an ideal, non-viscous fluid experiences NO VISCOUS DRAG force
(i.e. no resistive force due to the viscosity of the fluid).
D) IRROTATIONAL FLOW: In irrotational flow, a body can not
rotate about its own centre of mass as it flows in the fluid. (Note that
this does not mean that it can not move in a circular path).
The Equation of Continuity
REGAN
PHY34210
160
Everyday experience tells us that the velocity of a fluid emerging from
a tube depends on the cross-sectional areas of the tube. (For example,
you can speed up the water exiting a hose by squeezing the end).
If we have a tube of cross-sectional area, A1,
v1
which narrows to area A2. In a time interval,
v2
A2
Dt, a volume DV of fluid enters the tube, with A1
velocity v1. Since the fluid is ideal, and thus
incompressible, the same volume of fluid must exit the smaller end of
the tube with velocity, v2, some time interval later. The volume of fluid
element at both ends (DV) is given by the product of the cross-sectional
area (A) and the length it flows (Dx). Also, by definition, v=Dx/Dt thus,
Applyingt o bot h ends of t he t ube segment ,DV  A1v1Dt  A2 v2 Dt 
A1v1  A2 v2  EQUAT IONOF CONT INUIT Y
. T hiscan be rewrit t enas
RV  Av  const ant where RV is t heVOLUME FLOW RAT E.
If t hedensit y of t hefluid is const ant ,
t he MASS FLOW RAT E,Rm  rRV  rAv  const ant
Bernoulli’s Equation
REGAN
PHY34210
161
If an ideal (incompressible) fluid flows through a
p2 ,v2
tube at a steady rate. If in time Dt , a volume of
fluid, DV enters the tube and an identical
p1 ,v1
y2
volume emerges from the other end. If y1, v1
and p1 are elevation, speed and pressure of the
y1
fluid entering the tube and y2, v2 and p2 are the same quantities for the
fluid emerging from the other end of the tube. These quantities are
related by the BERNOULLI’S EQUATION, which states,
p1  12 rv12  rgy1  p2  12 rv22  rgy2 which can be re - writtenas
p  12 rv 2  rgy  constant.If thefluid does not changeelevationin its flow,
then y1  y2  y  0, Bernoulli's equn. reduces to p1  12 rv12  p2  12 rv22 .
For fluids at rest v1  v2  0, Bernoulli's equn becomes p2  p1  rg  y1  y2 
If the speed of a fluid element increases as it travels along a horizontal
streamline, the pressure of the fluid must decrease and vice versa.
Proof of Bernoulli’s Equation
REGAN
PHY34210
T he work - kineticenergy theoremstatesW  DK .
i.e., thenet workdone on thesystemequals thechange
in kineticenergy ΔK , resultingfrom thechangein fluid
162
p2 ,v2
p1 ,v1
y2
y1

speed between the ends of the tube. ΔK  12 Δm v22  12 Δm v12  12 ρΔV v22  v12

Δm  ρΔV is thefluid mass enteringand leaving theoutput in timeinterval,Δt .
T he work done on thesystemis due to the work done by thegrav.force
on thefluid elementof mass Dm, during its verticallift from theinput to
theoutput level. i.e. Wg  Dm g y2  y1    rDVg  y2  y1  . T he- sign
arises since themotionand gravitational forceare in oppositedirections.
Workis also done ON thesystemby pushing thefluid through he
t tube at
theentranceand BY thesystem to push forwardfluid at theemergingend.
Since, in general,the work done is given by W  FDx   pAΔx  pDV
T he NET work done is W p   p2 ΔV  p1 ΔV   p2  p1 ΔV . T he work - kinetic

theoremW  Wg  W p  ΔK   rgDV  y2  y1   ΔV  p2  p1   12 rDV v22  v12


which gives,  rg  y2  y1    p2  p1   12 r v22  v12  Bernoulli' s equation

16: Oscillations
REGAN
PHY34210
163
Oscillations describe motions which are repetitive. An important
property of oscillatory motion is its FREQUENCY, f, which describes
the number of oscillations per second. The SI unit for frequency is
the Hertz (Hz), where 1 Hz = 1 oscillation per second =1 s-1.
Motion which regularly repeats is called periodic or harmonic motion.
T heperiod,T , is the timeto completeone oscillation, where T  1f .
For SIMPLEHARMONICMOT ION,the timedependenceof the
displacement of a particlex, is given by xt   xm cost    where
xm is theAMP LIT UDE(maximumvalue of x).
  ANGULAR FREQUENCYdefined by xm cost   xm cos t  T  .
Since cos  cos  2  , thisleads to therelation hat
t   2Tπ  2f .
 is theP HASE ANGLE of themotion,which is determinedby the
displacement and velocityof theparticleat timet  0.
t    is called theP HASE of themotion.
REGAN
164
The Velocity of Simple Harmonic Motion
PHY34210
The velocity of a particle undergoing simple harmonic motion can
be found by differentiating the displacement, x(t) with respect to time.
dxt  d xm cost   
vt  

  xm sin t     vm sin t   
dt
dt
where  xm  vm is called theVELOCIT YAMPLIT UDE.
Note, in SHM, the magnitude of the velocity is greatest when the
displacement is smallest and vice versa, since cos( )=sin(+/2)
The Acceleration of Simple Harmonic Motion
The acceleration can be found by differentiating the velocity in SHM,
dvt  d   xm sin t   
at  

  2 xm cost   
dt
dt
 2 xm  am is known as theacceleration amplitude.
 at    2 xt  , which is thesignatureequation for SHM
‘ In SIMPLE HARMONIC MOTION, the acceleration a(t), is
proportional to the displacement x(t), but opposite in sign, and the
two quantities are related by the square of the angular frequency 2 ’
The Force Law for Simple Harmonic Motion


REGAN
PHY34210
From Newton's 2 nd law, F  m a  F  m a  m -ω2 x  kx
which is HOOKE'S LAW! T hisis CONSIST ENT with theidea
of a REST ORINGFORCE.Substituting for theSPRING CONST ANT ,
thenfor a Simple HarmonicMotion, k  m 2
Re - arrangingthisgives that theangular frequencyfor a simple harmonic
k
oscillatoris relatedto thestrengthof thespringconstantby  
m
2π
m
 2
ω
k
‘ Simple Harmonic Motion is the motion which is described by a
particle of mass m subject to a force which is proportional to the
displacement of the particle but opposite in sign’
T hus thePERIODof oscillation for a linear oscillatoris T 
A LINEARHARMONICOSCILLAT ORdescribes a system where the
forceF x , is proportion
al to x (rather th
an someotherpower of x).
165
Energy in Simple Harmonic Motion
REGAN
PHY34210
166
T heP OT ENT IALENERGY of a linear oscillatoris given by


1 2 1
U t   kx  k xm2 cos2 t   
2
2
T heKINET ICENERGY of thesystemis given by
1 2 1


K t  m v  m 2 xm2 sin 2 t   
2
2
T heMECHANICALENERGY, E, is defined as




1
1
k xm2 cos2 t     m  2 xm2 sin 2 t   
2
2
recallingthatfor SHM, k  mω 2 , thensince cos2  sin 2  1,
E U  K 


1 2
1 2
2
2
E  U  K  kxm cos t     sin t     kxm
2
2
Therefore, the mechanical energy of a linear oscillator is constant and
time independent.
Angular Simple Harmonic Motion
REGAN
PHY34210
167
An ANGULAR SIMPLE HARMONIC
fixed end
PENDULUM (also known as a
torsion wire
TORSION PENDULUM) is an angular
version of the linear simple harmonic
+m
oscillator. The disk oscillates in the
reference
horizontal plane, with the reference line
line, =0
- m
oscillating with an oscillation amplitude m.
The torsion wire twists, thereby storing potential energy in the same
way that a compressed spring does in the linear SHO case. The torsion
wire also gives rise to the RESTORING TORQUE, .
For angular simple harmonicmotion,rot at ingthedisk through an angle
 fromits rest posit ion(at   0) causes a REST ORINGT ORQUE
given by τ  -θ , where is called theT ORSIONCONST ANT .
By analogy with t hesimple harmonicoscillatorcase, theP ERIOD
of an angular simple harmonicoscillatoris given by T  2π
I
κ
Simple Pendulums
REGAN
PHY34210
A SIMPLE PENDULUM has a bob of
mass m hanging from a massless string
of constant length l, fixed at the other
end to that which the bob is attached.
τ  r F  rF  l Fg sin    lm g sin 
 
From Newton's 2 nd law,   I  lm g sin 
I  mom.of inertia,   angular acceleration
of thependulum at angular displacement, .
lm g
I
 T  2
I
lm g
, I  m l2

l
T
s=r
Fgsin

Fgcos
Fg
For small angular displacement,in radians,
 2 
l
m
T he sign indicatesthat thetorque , acts
to REDUCE theangular displacement, .
 lm g 
sin    I  lm g    

 I 
168
simple pendulums have SHM
ONLY for small values of .
 T  2
l
(no m dependence)
g
Real (‘Physical’) Pendulums
REGAN
PHY34210
O
Real pendulums exhibit similar behaviour
to simple pendulums, but the restoring
component of the gravitational force, Fgsin,
has a moment arm of distance h from the
pivot point. h is the distance from the pivot
point to the centre of mass of the object.

h
s=r
169
cog

Fgsin Fgcos
Fg
For a physicalpendulum,for small amplitudes,
I
the periodis given by T  2
. (For thesimple pendulum, I  m l2 ).
m gh
For real pendulums I differs for each case. For a uniformrod of lengt hl ,
I through the cent re(of mass) is 121 m l2 . T heP ARALLELAXIST HEOREM
gives thatfor a pivotat one end of therod, I 0 
1
12
m l  m
2

l 2
2
1 2
 ml
3
I
m l2
2l
8π 2l i.e. can measure
 since T  2
 2
 2
g
l
m gh
3m g 2 
3g
3T 2 g directly using
Focault’s pendulum
Damped Simple Harmonic Motion
REGAN
PHY34210
170
If the motion of an oscillator is reduced as a result of an external force,
the oscillator and its motion are described as DAMPED.
If thedampingforceis proportion
al to the velocity
of theof theoscillating system,then Fd  bv ,
where,b is a DAMPINGCONST ANT .T heminus
sign indicatesthat thisforceopposesthemotion.
From Newton's 2 nd law, Fnet  m a  kx  bv
spring,
constant, k
Fk=-kx
m
Fb=-bv
water
tank
d 2x
dx
 m 2  kx  b  0 . T hesolutionfor this2 nd order differential equation
dt
dt
k b2
where ω' 

is theangular
m 2m
is given by xt   xm e bt 2 m cos ' t   
k
frequencyfor thedampedoscillator. For b  0 (no damping)ω' 
as in SHM.
m
If thedampingconstantis small,i.e., b  km , then ω'  .

T heamplitudefor a dampedoscillator xm e
bt 2 m
bt 2 m 2
1


  Emec t  2 k xme   const.
Forced Oscillations and Resonances
REGAN
PHY34210
171
If a body oscillates without an external force on the body, the body
is said to undergo FREE OSCILLATION.
However, if there is an external force periodically pushing the system
(such as someone pushing a swing), this is known as FORCED or
DRIVEN OSCILLATION.
2 angular frequencies are associated with a system undergoing forced
oscillations, namely the
(i) NATURAL ANGULAR FREQUENCY ( ) of the system, which is the frequency
at which the system would oscillate if it was disturbed and left to oscillate freely;
and the
(ii) ANGULAR FREQUENCY OF THE EXTERNAL DRIVING FORCE (d )
which is the angular frequency of the force causing the driven oscillations.
If =d, the system is said to be ‘in resonance’. If this condition is achieved, the
velocity amplitude, vm is maximised (and so approximately is the displacement, xm).
17: Waves - Part 1
REGAN
PHY34210
172
Waves describe situations where the energy of the system is spread
out over the space through which it passes. This is in contrast to
particles which imply a tiny concentration of matter which is capable
of transmitting energy by moving from one place to another.
There are THREE main types of waves:
a) Mechanical Waves: These are governed by Newtons’s Law and
can only exist within a medium (such as a taut string, water, air, etc.)
b) Electromagnetic Waves: These are massless objects which require
no medium to travel in. All EM-waves travel through vacuum at
the same, constant speed (‘the speed of light, c=3x108ms-1). Examples
of EM waves are visible light, UV and IR radiation, radio-waves,
x-rays and gamma-rays. (the only difference between these waves
is their wavelength and their mode of origin, (atomic, nuclear etc.)
c) Matter Waves: These are quantum descriptions of subatomic
particles such as electrons, protons etc. They are described by the
‘de Broglie’ wavelength, dependent on the particle’s (lin.) momentum.
Transverse and Longitudinal Waves
REGAN
PHY34210
173
One way to investigate wave motion is to look at the WAVEFORM,
which describes the shape of the wave (i.e. y=f(x)).
Alternatively, one can monitor the motion of a particular element of the
wave medium (e.g., a string) as function of time (i.e., y=f(t)).
In cases where them displacement of (for example) each element in an
oscillating string is perpendicular to the direction of travel of the wave,
the wave said to be TRANSVERSE (i.e. a transverse wave, such as
waves on a string.)
By contrast, if the displacement is parallel to the direction of motion
of the wave (as in sound waves), the motion is described at
LONGITUDINAL (i.e., transmitted via a longitudinal wave such as
sound).
Wavelength and Frequency
REGAN
PHY34210
174
To completely describe a wave on a string (and the motion of any
element along its length) a function which
describes the shape of the wave as a function
of time t, is required. This means we need a
function of the form, y = f (x,t) , where y is the
displacement in the ‘up-down’ direction and x
is the position along the string.
For a sinusoidal wave, thedisplacement y, as a
functionof timet , for an elementat positionx
along thestringis y( x, t )  ym sin kx  t 
HRW p374
The amplitude (ym) is the magnitude of
the maximum displacement.
The phase is the argument of ( kx-t ).
As the wave passes through a string
element at a position, x, its phase
changes linearly with time.
Wavelength and Angular Wave Number
REGAN
PHY34210
175
The wavelength l , of a wave is the distance
(parallel to the waves direction of travel)
between repetitions of the shape of the wave.
T heshape of he
t wavecan be seen for t  0 as
y  x, t  0   ym sin kx  t   ym sin kx
By definition, thedisplacement, y is thesame
at both ends of a single wavelength, thus
y  x,0   y  x  l ,0   ym sin kx  ym sin k  x  l 
Since thefunctionsin repeatsitself every2
radians, thusif kx  kx  kl  kl  2
 T heANGULAR WAVENUMBER, k 
HRW p374
2
l
Period, Angular Frequency and Frequency
REGAN
PHY34210
176
We can monitor the time dependence of the displacement of a fixed
position on a vibrating string. This can be done by taking x=0.
HRW p375
T hedisplacement of thestringat
x  0 as a functionof timeis then
y 0, t   ym sin 0  t   ym sin  t 
sin      sin    y 0, t    ym sin t 
The PERIOD OF OSCILLATION T, is defined as the time for any
string element to move through one oscillation. The displacement at
both end of the period of oscillation are, by definition, equal. Thus,
 ym sin t1   ym sin  t1  T    ym sin t1  T 
2
T hiscan only be trueif ωT  2π  Angular Frequency,  
T
1 ω
T hefrequencyis defined as f  
T 2π
Speed of a Travelling Wave
REGAN
PHY34210
If the wave travelsin thex - direction,we can define the
Δx
dx
WAVE SP EED v as theratioof
 . If we takea
Δt
dt
fixed pointon themovingwaveformwhich is defined by
havingthesame displacement y, then thephase of the
waveformmust remainconstantsince y  ym sin kx  t .
 T hephase,kx  ωt  const . differentiating with respect
dx
dx

to timegives k
-ω0 
v
dt
dt
k
2
2
2π
l
T
Re calling k 
and ω 
 v  2   l f
l
T
T
l
177
Wave Speed on a Stretched String
REGAN
PHY34210
178
For a wave to pass through a medium, the particles in the medium
must oscillate as the waves passes through. The medium must have
mass (so the particles have kinetic energy = 1/2mv2) and elasticity (for
potential energy = 1/2 kx2). The mass and elasticity of the medium
determine how fast the wave can travel through the medium.

A small stringelementof length Δl , formsa circular arc of 
Dl


F
 
radius R subtending an angle 2θ. If a force with a
R
v
magnitudeequal to the tensionin thestringpulls
O
tangentially tothe two ends, thehorizontalcomponents

cancel.T he verticalcomponentssum, providingtherestoringforceF of
magnitudeof thisforceis, F  2τ sin θ   2τ θ  τ
Δl
R
. T hemass of theelement
Δm  μDl , where m is thelinear densityof thestring. The speed of a wave
T hestringelementmovesin a circle, thus a 
v2
R
.
 v2 

F  m a leads to
 m a   μDl      v 
R
m
R
Dl
on an ideal stretched
string only depends
on the string’s tension
and linear density.
Energy and Power of a Travelling Wave
REGAN
PHY34210
2
179
T hekineticenergy of a stringelementof mass dm is dK  12 dm  u
whereu
is theT RANSVERSE SP EED of theoscillating stringelement.u 
dy
dt
 u  ym coskx  t .
energyis given by dK 
1
2
Using thelinear density, dm  μ.dx, thekinetic
 μ.dx ym 2 cos2 kx  t 
.
T herateat which he
t kineticenergy of a stringelementis thus
dK 1  dx 
2
2
 2  μ.  ym  cos2 kx  t   12  μv ym  cos2 kx  t 
dt
 dt 
 theAVERAGE rateat whichkineticenergy transportedis given by




dK 1
2
 2  μv ym  cos2 kx  t  ave  14 mv 2 ym2
averageof cos2 X  12
dt
Recallingfor an oscillating system,theaveragekineticand potentialenergies
are equal, theAVERAGE P OWERis theaveragerate which thetotal
 dK 
2 2
mechanicalenergy transmittedby thewave i.e., P  2
  12 mv ym
 dt  ave
The Principle of Superposition
REGAN
PHY34210
180
The principle of superposition states that when several effects occur
simultaneously, their net effect is the sum of the individual effects.
y' x, t   y1 x, t   y1 x, t 
Mathematically, this means,
Overlapping waves add
algebraically to produce
a RESULTANT or NET WAVE.
Note however, that overlapping
HRW p383
waves do not in any wave affect
each others travel.
Interference of Waves
If 2 sinusoidal waves of the same wavelength and amplitude overlap,
the resultant wave depends on the relative PHASES of the waves.
If they are perfectly ‘in phase’ they will add coherently, doubling the
displacement observed for individual waves. By contrast, if they are
completely out of phase (peaks of one wave matched by troughs of the
other), they will completely cancel out resulting in a ‘flat’ string.
REGAN
181
If y1 x, t   ym sinkx  t  and y2 x, t   ym sinkx  t    PHY34210
These waves have the same frequency determined by  , wavelength
from l and amplitude ym . They differ only by the phase constant .
HRW p384
From the principle of superposition,
y '  x, t   ym sin kx  t   ym sin kx  t   
Since, sin   sin   2 sin
 cos 
 
 
2
2
ym sin kx  t   ym sin kx  t   

  
y '  x, t   2 ym cos 2 sin kx  t  2


 amplitudeof resultantwave  2 ym cos 2
and phaseangle  2 . If   0, the two initial waves are ' in phase'the
waves interferefully ' constructively'.If    radians, the waves are
completelyout of phaseand interferecompletely, DEST RUCT IVELY.
If 2 sinusoidal waves of the same amplitude and wavelength travel in
the same direction along a stretched string, they interfere to produce a
resultant sinusoidal wave travelling in that direction.
Phasors
REGAN
PHY34210
182
y’=y1+y2
Waves can be represented in vector form using the idea of PHASORS.
This is a vector whose amplitude is represented by the length which is
equal to the magnitude of the wave and which rotates around the origin
of a set of Cartesian co-ordinates. The angular speed of the phasor
about the origin is equal to the angular frequency,  of the wave.
As the phasor rotates about the origin,

y’m
its projection, y1 onto the vertical axis
y2
y2
y
m,2
varies sinusoidally between +ym and -ym.

2 waves which travel along the same
y1 
ym,1
y1
string in the same direction can be

added using a PHASOR DIAGRAM.
If y1 x, t   ym,1 sin kx  t  and y2 x, t   ym, 2 sin kx  t    , the
resultantis of theform y ' x, t   ym' sin kx  t    where ym' and  can
be found using thePHASOR DIAGRAM.Adding vectorially thephasors
y1 x, t  and y2 x, t  at any instant,themagnitudeof theresultantequals
y ' x, t  and  is theangle between the resultantand thephasorfor y1 x, t .
Standing Waves
REGAN
PHY34210
183
If two sinusoidal waves travel in opposite directions along a string,
their sum can be found using the principle of superposition. There are
specific places along the resultant wave which DO NOT MOVE,
known as NODES. Halfway between neighbouring nodes (the
‘anti-nodes’) the amplitude of the resultant wave is maximised. Such
wave patterns are called STANDING WAVES since the wave patterns
do not move in the x-direction (i.e. they are stationary left to right).
HRW
If two sinusoidal waves of the same amplitude and wavelength travel in
opposite directions along a stretched string, their intereference with
each other produces a standing wave.
Analysis of Standing Waves
REGAN
PHY34210
184
T he t wo combiningwaves which makeup t hest anding wave are
y1  x, t   ym sin kx  t  and y2  x, t   ym sin kx  t  .
From t heprincipleof superposition, y '  x, t   y1  x, t   y2  x, t 
 y '  x, t   ym sin kx  t   ym sin kx  t  .
       
Recallingsin  sin   2 sin 
 cos

 2   2 
y '  x, t   2 ym sin kxcost  Note, this does not describe a travelling
wave, but rather a standing wave.
• The absolute value of [2ymsin(kx)] is the amplitude of oscillation at x.
• The amplitude varies with position for a standing wave.
T heamplitudeis zero if sin kx  0. i.e., for integern where kx  nπ 
2π
λ
x
 amplitudeis zero (i.e., nodes) occur at x  n 2λ . For a standing waves,
adjacent nodes are separatedby a distanceof half the wavelength  l2 .
Similarly,themax.amp.is 2 y for sin kx  1  kx  n  12   x  n  12  l2
Standing Waves and Resonance
REGAN
PHY34210
185
A standing wave can be set up by allowing a wave to be reflected at
a boundary of a string. The interference of the original (incident) and
reflected wave can interfere to give rise to a standing wave. (Note
that for ‘hard’ reflection, the reflection point must be a fixed node.)
HRW
If a taught string is fixed at both ends (such as in a guitar) and a
continual sinusoidal wave is sent down from one end, it will be
subsequently reflected at the other end. The reflected wave and the
next transmitted wave will interfere. If more waves are continually
sent from the generator, many such waves can add coherently.
REGAN
PHY34210
186
At certain frequencies, such behaviour results in STANDING WAVE
PATTERNS on the string. Such standing waves ‘RESONATE’ at
fixed ‘RESONANT FREQUENCIES’. (Note that if the string is
oscillated at a non-resonant frequency, a standing wave is NOT set up.)
2L
A standing wave can be set up on a stringof length L, by a wave if λ 
n
v
v
where
T hiscorresponds to resonancefrequencies given by f   n
2L
l
v is the wavespeedalong thestring.
n  1 is called thefundamental mode or ' first harmonic'.
n  2 is called thesecond harmonic,n  3 the thirdharmonicand so on.
T hefrequencies associatedwith these modes are oftengiven the
symbols, f1,f2 , f 3, , f n
Example 1:
REGAN
PHY34210
187
Two identical sinusoidal waves moving in the same direction along a
stretched string interfere with each other. The amplitude of each
wave is 9.8 mm and the phase difference between them is 100o.
(a) What is the amplitude of the resultant wave due to the interefence
between these two waves ?

 
ym'  2 ym cos 2  2 9.8 mm. cos 500  13 mm
(b) What phase difference (in both radians and in fractions of
wavelength) will give a resultant wave amplitude of 4.9 mm ?


Since ym'  2 ym cos 2  4.9mm 2  9.8mm  cos 2

4.9 mm
 cos 2 
 0.25    2.6 radians
2  9.8 mm
single wavelength corresponds to  2 , thus,in wavelengths, the

phase differenceis given by
  rads 
2  rads / wavelength 
 2.6
2
 0.42 wavelengths
Example 2:
REGAN
PHY34210
188
Two sinusoidal waves y1(x,t) and y2(x,t) have the same wavelength
and travel together in the same direction along a string. Their
amplitudes are y1,m=4.0 mm and y2,m=3.0mm and their phase constants
are 0 and /3 respectively.
What are the amplitude, y’ and phase constant  of the resulting wave ?
ym2
/
Adding thehorizontalcomponents
y'mh  ym1 cos0  ym 2 cos3   4  3 cos3   5.5m m
Adding the verticalcomponents
3
ym1
y’m

ym1
y'mv  ym1 sin 0  ym 2 sin 3   0  3 sin 3   2.6m m
 From P ythagorastheorem,theresultantwave has
an amplitudeof y'm 
5.52  2.62 m m  6.1m m
ym2
-1  2.6 
T hephaseconstantis   tan 
  0.44 rads
 5.5 
 y ' x, t   6.1m msin kx  t  0.44rads
18: Waves - Part 2
REGAN
PHY34210
189
ray
Sound Waves: can be generally defined
as longitudinal waves whose oscillations
are parallel to the direction of travel
through some medium (such as air).
P
wavefronts
planes
ray
If a point source, P, emits sound
waves, wavefronts and rays describe the
direction of travel of the waves. Wavefronts correspond
to surfaces over which the wave has the same displacement value.
Rays are lines drawn perpendicular to wavefronts which indicate the
direction of travel of the waves. Note that in real bodies, wavefronts
spread out in 3 dimensions in a spherical pattern.
Far from the point source the wavefronts can appear as planes or
straight lines to an observer.
Speed of Sound
REGAN
PHY34210
190
The speed of any mechanical wave depends on the physical properties
of the medium through which it travels.
As a sound wave passes through air, we can associate a potential energy
with periodic compressions and expansions of small volume elements.
The BULK MODULUS is the property which determines the volume
change in a material when exposed to an external pressure (p=F/A).
Dp
Recalling that t heBULK MODULUSis defined by B  
DV V
where ΔV/V is thefract ionalchangein volumeproducedby a
changein pressure Δp. Since thesigns of Δp and ΔV are always
oppositea minus sign is included to make B a positivequanity.
T hespeed of sound for a longitudinal wave in a medium is give by
v
B
r
where r is thedensity of
Vair(20oC)=343 ms-1
themedium. Vwater(20oC)=1482 ms-1
Vsteel = 5941 ms-1
The Doppler Effect
REGAN
PHY34210
191
The Doppler effect describes how sound waves from a point source
(such as a car or train or star or galaxy!)) are apparently shifted in
frequency for an observer which is moving relative to that source.
T hegeneralexpressionfor theDOPP LERFORMULAfor sound waves is
 v  vD 
 where f is theemittedfrequencyof thesource and
f '  f 
 v  vS 
f ' is thedetectedfrequencyby theobserver.v is thespeed of sound through
theair, vD is therelativespeed of thedetector(or ' observer') relativeto an
air - fixed frameand vS is thesource speed relativeto thesame air - fixed
referenceframe.(Notein most cases, either thesource or thedetectoris
stationary, i.e., vD or vS  0).
When the motion of the detector or source are towards each other,
the sign on its speed gives an UPWARD SHIFT IN FREQUENCY.
When the motion of the detector or source are away from each other
the sign on its speed gives a DOWNWARD SHIFT IN FREQUENCY.
Example 1
REGAN
PHY34210
192
A rocket moves at a speed of 242 m/s through stationary air directly
towards a stationary pole while emitting sound waves at a source
frequency of f =1250Hz.
(a) What frequency is measured by a detector attached to the pole ?
 v  vD 


343m s1  0



  4245Hz. T he- sign on bottom
f ' f 
 1250
1
1 

 343m s  242m s 
 v  vS 
gives an INCREASE in observedfrequencyfor relativemotion tow
ards source.
(b) If the some of the sound waves reflect from the pole back to the
rocket, what frequency f ’’does the rocket detect for the echo ?
 v  vD 
 343m s1  242m s1 
  4245
  7240Hz. T he sign on topgives
f '  f 
1
343m s  0


 v  vS 
an INCREASE in observedfrequencyfor relativemotion tow
ards source (pole).