phys 34210 physics i notre dame, london programme, fall 2013
DESCRIPTION
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: [email protected]. Course & General Information. Lectures, usually, Tuesdays 2.15-5.00 first lecture Tues 27 th August 2010 - PowerPoint PPT PresentationTRANSCRIPT
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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: [email protected]
REGAN PHY34210
2Course & General Information• 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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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)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
* 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)
Break, no lect. 22nd Oct.No lect. 29th Oct (resched).
* Lect 10: 5th Nov Exam 2
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
PART 1 PART 2 PART 3
Course notes and past papers/solns can be found at the following link:
http://personal.ph.surrey.ac.uk/~phs1pr/lecture_notes/notre_dame/
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1st Section:• 1: Measurement
– 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: • 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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• 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.
3rd section:
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Recommended Problems and Lecture Notes.
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.pptand http://www.ph.surrey.ac.uk/~phs1pr/lecture_notes/phy34210_13.pdf
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1: Measurement
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 UnitsThe 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 NotationIn many areas of physics, the measurements correspond to very largeor 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
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Prefixes
• 1012 = Tera = T• 109 = Giga = G• 106 = Mega = M • 103 = Kilo = k• 10-3 = milli = m• 10-6 = micro = • 10-9 = nano = n• 10-12 = pico = p• 10-15 = femto = f
For convenience, sometimes,when dealing with large or smallunits, it is common to use a prefixto describe a specific power of 10with 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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Converting UnitsIt is common to have to convert between different systems of units(e.g., Miles per hour and metres per second). This can be done mosteasily using the CHAIN LINK METHOD, where the original valueis 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)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 vacuumduring a time interval of 1/299,792,458 of a second.
• To Andromeda Galaxy ~ 1022 m • Radius of earth ~ 107 m • Adult human height ~ 2 m• Radius of proton ~ 10-15 m
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Time (Seconds)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.
From HRW, p6But, a day does not have a constant duration!
(1967) Use atomic clocks,to define 1 second asthe time for 9,192,631,770 oscillations of the light of a specificwavelength (colour) emittedfrom an atom of caesium (133Cs)
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Mass (Kg, AMU)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 kgOrders 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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mmL
m
m
m
mL
LmLmmmV
r
rLdV
605.55
003.0003.0
05.005.005.04
003.0
05.034
003.0305.03
4
square 3mm~ string ofsection -cross assume
0.05m 5cm 10cm/2 ball of radius3
4 string, of Volume
2
3
2
3
223
32
Estimate Example 1:A ball of string is 10 cm in diameter, make an order of magnitudeestimate of the length, L , of the string in the ball.
r
d
d
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rhd
rhhrhd
hrhrhrrd
2
but ,2
2
,Pythagoras From
2
22
22222
)m!6.4x10 radiusearth for valueaccepted(
108109.4
4
)04.0(tan
22 ng,substituti then 2m, ~height human h if
tan
2 2tan thustan try, trigonomefrom Now,
04.0(deg) 400,86
3600
sec606024
(deg)360sec)10((deg)
hours 24360
8
672
222
mmm
r
hrrhrrd
tt
o
oo
E.g., 2: Estimate Radius of Earth (from the beach.)
r
r
d h
is the angle through which the sun moves around the earth during the time between the ‘two’ sunsets (t ~ 10 sec).
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2: Motion in a Straight LinePosition 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. Note that we can also define NEGATIVE co-ordinates too.
-3 -2 -1 0 1 2 3x =
The DISPLACEMENT, x is the change from one position to another, i.e., x= x2-x1 . Positive values of x 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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Average Speed and Average Velocity
From HRW
We can describe the position of an object as it moves (i.e. as a functionof time) by plotting the x-position of the object (Armadillo!) at differenttime intervals on an (x , t) plot.
The average SPEED is simply thetotal distance travelled (independentof the direction or travel) divided bythe time taken. Note speed is a SCALAR quantity, i.e., only its magnitude is important (not its direction).
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dt
dx
t
xtv
0lim
t
x
tt
xxvav
12
12
The average VELOCITY is definedby the displacement (x) divided by the time taken for this displacement to occur (t).
The SLOPE of the (x,t) plot gives average VELOCITY.Like displacement, velocity is a VECTOR with the same sign asthe displacement.
The INSTANTANEOUS VELOCITY is the velocity at a specificmoment in time, calculated by making t infinitely small (i.e., calculus!)
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t
v
ttdtdx
dtdx
tt
vvaav
12
12
12
12
AccelerationAcceleration is a change in
velocity (v) in a given time (t). The average acceleration, aav, is given by
The instantaneous ACCELERATIONis given by a, where,
2
2
dt
xd
dt
dx
dt
d
dt
dva
SI unit of acceleration is metres per second squared (m/s2)
HRW
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00
t
vvaa av
Constant Acceleration and the Equations of MotionFor some types of motion (e.g., free fall under gravity) the accelerationis approximately constant, i.e., ifv0 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
0 from taken time
(constant)on accelerati
at time velocity
)0 (at time velocity initial
ntdisplaceme the
0
0
tt
a
tv
tv
xx
Usually in a given problem, three of these quantities are given and from these, one can calculate the other two from the followingequations of motion.
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(5) 2
1
and ; (4) 2
1 ; (3) )(2
gives and , eliminate to(2) and (1) combining
note , (2) 2
1
, 2
1 gives for (1) into ngsubstituti
2 ,2
,definitionby since
then, 0
that recalling (1)
20
00020
2
0
002
00
00
00
00
atvtxx
tvvxxxxavv
vat
atvvdt
xxdattvxx
atvvv
vvvvv
v
t
xxvatvv
av
avav
av
Equations of Motion (for constant a).
REGAN PHY34210
24Alternative Derivations (by Calculus)
(1) thus,0 ,0
at that knowingby evaluated ,
constantfor thusand
definitionby
000
2
2
atvvCavvvt
Catv
adtadv
dtadvdt
xd
dt
dva
(5) 2
1 and at knowing
by calculate , 2
1 gives, gintegratin
on,substitutiby
thereforeis, but constant,not ,
200000
20
00
0
attvxxxCttxx
CCattvx
dttadtvdxdtatvdx
vvdtvdxvdt
dx
REGAN PHY34210
25Free-Fall Acceleration
At the surface of the earth, neglecting any effect due to air resistanceon the velocity, all objects accelerate towards the centre of earth with the same constant value of acceleration.This is called FREE-FALL ACCELERATION, or ACCELERATIONDUE 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
REGAN PHY34210
26ExampleA 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 ?
(b) what’s the ball’s maximum height ?
sms
ms
a
vvtatvv 2.1
8.9
120
2
10
0
(a) since a= -g = -9.8ms-2, initial position is y0=0 and at the max. height vm a x=0
Therefore, time to max height from
mms
ms
a
vvymsga
msvvyayyavv
3.78.92
)12(0
2 8.9
, 12 ,0 & 02)12()(2
2
212202
10
20
20
2
(b)
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( c) How long does the ball take to reach a point 5m above its initial release point ?
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.
1.9s AND 5.0 2
4
by given are solutions 0 recalling
05124.9 equation, quadratic a have weunits, SI assuming
8.92
1125
2
1 from
5 ,8.9 ,12
2
2
2
221200
021
0
sta
acbbt
cbtat
tt
tmstmsmattvyy
myymsamsv
REGAN PHY34210
283: Vectors• 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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Adding Vectors GeometricallyAny two vectors can be added using the VECTOR EQUATION, where the sum of vectors can be worked out using a triangle.
bas
a
b
s
Note that two vectors can be added together in either order toget the same result. This is called the COMMUTATIVE LAW. baabs
a a
b
b
s
Generally, if we have more than 2 vectors, the order of combination does not affect the result. This is called the ASSOCIATIVE LAW.
cbacbar
a
b
s c
r a
b
c
r
= 's
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direction. opposite in thebut
as magnitude same theis bb
bas
a
b
s
babad
b
ab
d s
abdbad
e.g.,
equations, vector arrange-re
can wealgebra, usual with as
Subtracting Vectors, Negative Vectors
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!)
REGAN PHY34210
31Components of Vectors
x
y
a
sina
ay
cosa
ax
A simple way of adding vectors canbe done using their COMPONENTS.The component of a vector is the projection of the vector onto the x, y (and z in the 3-D case) axes in the Cartesian co-ordinate system.Obtaining the components is known asRESOLVING the vector. The components can be found using the rulesfor a right-angle triangle. i.e.
x
yyx
yx
a
aaaa
aaaa
tan ,
as NOTATION ANGLE-MAGNITUDE
in written be alsocan this
sin and cos
22
REGAN PHY34210
32Unit 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, ly.respective ˆ and ˆ,ˆ kji
ksjsiskbajbaibas
bas
kbjbibbkajaiaa
zyxzzyyxx
zyxzyx
ˆˆˆˆ)(ˆ)(ˆ)(
componentsby addition vector using if then
ˆˆˆ , ˆˆˆ
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,
x
y
z11
1
jk
i
REGAN PHY34210
33Vector Multiplication
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.
)090cos1(1 0ˆ.ˆˆ.ˆˆ.ˆ
and 1)1cos0(1 1ˆ.ˆˆ.ˆˆ.ˆ since
.
toreduces thisexpanding 090cos and 10cos
sincebut ˆˆˆ . ˆˆˆ.
notation,r unit vectoIn
. . also, coscos.
o
o
zzyyxx
oo
zyxzyx
kjkiji
kkjjii
babababa
kbjbibkajaiaba
abbabababa
For the SCALAR or DOT PRODUCT,
REGAN PHY34210
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Vector (‘Cross’) Product
kabbajabbaiabbaba
kbajibajbia
iibaibia
kbjbibkajaiaba
yxyxxzxzzyzy
yxyxyx
xxxx
zyxzyx
ˆˆˆ
thus,ˆˆˆˆˆ
and 0ˆˆˆˆbut
ˆˆˆˆˆˆ
and also abba
The VECTOR PRODUCT of two vectors and produces a third vector whose magnitude is given by
a
b
sinabc is the angle between the two initial vectors
ab
c
The direction of the resultant is perpendicular to the plane created by the initial two vectors, such that
REGAN PHY34210
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Example 1:
o
yx
r
rrjir
jijijicbar
jicjibjia
53.1 3
4 tan, 543
4 , 3 ˆ4ˆ3
ˆ2ˆ2ˆ3ˆ4ˆ
ˆ2ˆ , 2ˆ3 , ˆ4ˆ
vectors threefollowing theAdd
22
a
b
c
y
x
r
REGAN PHY34210
36Example 2:
kjikjiba
kji
kji
ba
kabbajabbaiabbaba
kjikjiba
ba
babababa
k.jk.ij.i, k.kj.ji.i
kjikjiabba
kjibkjia
yxyxxzxzzyzy
zzyyxx
ˆ13 ˆ 10ˆ 11 . 4ˆ 52ˆ 40ˆ 44
ˆ )1240(ˆ )2416(ˆ 8036
ˆ)3)(4()20).(2(ˆ)2).(12()4).(4(ˆ )4).(20()12).(3(
ˆ ˆ ˆ recalling
ˆ12ˆ20ˆ4 ˆ4ˆ3ˆ2 Product,Vector (b)
-100 48-60-8 .
)12.4()20.3()4.2( . then
0ˆˆˆˆˆˆ1ˆˆˆˆˆˆonly recalling
ˆ12ˆ20ˆ4 . ˆ4ˆ3ˆ2cos . :ProductScalar (a)
ˆ12ˆ20ˆ4ˆ and ˆ4ˆ3ˆ2ˆ
vectors two theof products vector (b) andscalar (a) theareWhat
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4: Motion in 2 and 3 Dimensions
The use of vectors and their components is very useful for describingmotion of objects in both 2 and 3 dimensions.
Position and Displacement
kikjirrr
kjirkjirge
kzjyix
kzzjyyixx
kzjyixkzjyixrrr
kzjyixr
ˆ3ˆ12ˆ)58(ˆ)22(ˆ))3(9( then
ˆ8ˆ2ˆ9 and ˆ5ˆ2ˆ3 if .,.
ˆˆˆ
ˆˆ
ˆˆˆˆ
is NTDISPLACEME then the, ˆˆˆ
by ordinates-coCartesian in descibed
becan particle a ofposition thegeneralin If
12
21
121212
11122212
REGAN PHY34210
38Velocity and AccelerationThe average velocity is given by
kt
zj
t
yi
t
x
t
kzjyix
t
rvav
ˆˆˆˆˆˆ
While the instantaneous velocity is given by making t tend to 0, i.e.
kvjvivv
kdt
dzj
dt
dyi
dt
dx
dt
kzjyixd
dt
rdv
zyxˆˆˆ
ˆˆˆ)ˆˆˆ(
Similarly, the average acceleration is given by,
kt
vj
t
vi
t
v
t
kvjviv
t
v
t
vva zyxzyx
avˆˆˆ
ˆˆˆ12
While the instantaneous acceleration is given by
kdt
dvj
dt
dvi
dt
dv
dt
kvjvivd
dt
vda zyxzyx ˆˆˆ)ˆˆˆ(
REGAN PHY34210
39Projectile MotionThe specialist case where a projectile is ‘launched’ with an initial velocity, and a constant free-fall acceleration, .Examples of projectile motion are golf balls, baseballs, cannon balls. (Note, aeroplanes, birds have extra acceleration see later).
0v
g
We can use the equations of motion for constant acceleration and what we have recently learned about vectors and their componentsto analyse this type of motion in detail.
0v
cos0v
sin0v
More generally, sin and cos whereˆˆ vvvvjvivv yxyx
sin and
cos where
ˆˆ is )0(at
velocity projectile initial The
00
00
000
vv
vv
jvivvt
y
x
yx
REGAN PHY34210
40Horizontal MotionIn the projectile problem, there is NO ACCELERATION in the horizontal direction (neglecting any effect due to air resistance). Thus the velocity component in the x (horizontal) direction remains constant throughout the flight, i.e.,
tvxxvv
tvxx
x
x
cos thus, cos and
1)motion of(equation
000000
00
Vertical Motion
200
200 2
1 sin
2
1gttvgttvyy yy
02
002
00 2sin and sin yygvvgtvv yy
gtvvy 00 sin i.e., ,0
whenoccursheight Max.
0v
cos0v
sin0v
vy
REGAN PHY34210
41
The Equation of Path for Projectile Motion
2
00
0000
2
00
0
00
0000
2000000
cos2
1tan
cos2
1
cossin
equns.upper twoebetween th timefor the ngsubstituti
2
1 sin and cos Given that
v
xxgxxyy
v
xxg
v
xxvyy
gttvyytvxx
Note that this is an equation of the form y=ax+bx2 i.e., a parabola(also, often y0=x0=0.)
REGAN PHY34210
42
) assumes (note
2sin
then, 2sinsincos2
generalin since
sincos2
cos2cos
sin
cos2
1
cossin0
2
1 sin0 and cos then , when i.e.,
ground thehits projectile when thedefined is range, The
0
020
000
0020
022
0
2
0
0
2
000000
200000
0
yy
g
vR
g
vR
v
gRR
v
Rg
v
Rv
gttvtvRyy
xxR
The Horizontal Range
0v
cos0v
sin0v
Range(y0,x0)
Maxheight
vy=0
REGAN PHY34210
43ExampleAt 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.
How far must the fence be moved back for no homers to be possible ?
feet! 840 2558.9
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REGAN PHY34210
44
Uniform Circular Motion
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nceCircumfereT
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A particle undergoes UNIFORM CIRCULARMOTION is it travels around in a circular arc at aCONSTANT SPEED. Note that although the speed does not change, the particle is in fact ACCELERATING since the DIRECTION OF THEVELOCITY IS CHANGING with time. The velocity vector is tangential to the instantaneousdirection of motion of the particle. The (centripetal) acceleration is directed towards the centre of the circleRadial vector (r) and the velocity vector (v) are always perpendicular
r
REGAN PHY34210
45
circle of centre towardsradius.... thealong ison accelerati , i.e.,
,tancos
sin
cos
sin
tan from DIRECTION ONACCELERATI
thus,, sincos ,
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Proof for Uniform Circular Motion
r
yp
xp
v
REGAN PHY34210
46Relative MotionIf we want to make measurementsof velocity, position, acceleration etc.these must all be defined RELATIVE to a specific origin. Often in physicalsituations, the motion can be broken downinto two frames of reference, dependingon who is the OBSERVER. ( someone who tosses a ball up in a moving car will see a different motion to someone from the pavement).
AB
ABr
p
Apr
Bpr constvAB
If we assume that different FRAMES OF REFERENCE always move at a constant velocity relative to each other, then using vector addition,
pB
pBBApBpApA
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pABApBpA
adt
vd
dt
vvvda
vvdt
rrrdvrrr
0
, i.e., accelerationis the SAME for both frames of reference! (if VAB=const)!
REGAN PHY34210
475: Force and Motion (Part 1)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 byNewtonian 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
REGAN PHY34210
48Newton’s First LawNewton’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 directionand speed unless acted upon by some (gravitational) force.
REGAN PHY34210
49
Force
If two or more forces act on a body we can find their resultant valueby 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’
The units of force are defined by the acceleration which that forcewill 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.
Mass: we can define the mass of a body as the characteristicwhich relates the applied force to the resulting acceleration.
REGAN PHY34210
50
Newton’s 2nd Law
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: amFnet
As with other vector equations, we can make three equivalent equations for the x,y and z components of the force. i.e.,
zznetyynetxxnet maFmaFmaF ,,, and ,
The acceleration component on each axis is caused ONLY by the forcecomponents 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.
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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
The Gravitational Force
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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 netforce 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
REGAN PHY34210
53The Normal Force
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 itstationary against gravity.
N
Force, Normal
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as magnitude same i.e. 0
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General equation for block on a table is
Note the NORMAL FORCE is ‘normal’ (i.e. perpendicular) to the surface.
REGAN PHY34210
54
gamN
NmgNFmaF
NFamF
y
gynety
gnet
,
Example
A person stands on a weighing scales in a lift (elevator!) What is thegeneral solution for the persons measured weight on the scales ?
mgFgy
Na
So, if lift accelerates upwards (or the downward speed decreases!) the personsweight INCREASES, if the lift acceleratesdownwards (or decelerates upwards) the persons weight DECREASES compared to the stationary (or constant velocity) situation.
REGAN PHY34210
55
TmM
Mmg
mM
mgMMaTF
mM
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) thusdownward ( components
, components
TensionTension 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 hanging blocks and the tension in the cord ?
M
mgMFgM
gmFgm
N
T
T
REGAN PHY34210
56
‘ When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction ’
Newton’s Third LawTwo bodies interact when they push or pull on each other. This leadsto Newton’s third law which states,
The forces between two interacting bodies are called a ‘third-law pairforces’. e.g., Table pushes up block withforce N, block pushes down tablewith force Fg, where Fg=N
Sometimes this is differently stated as
‘ for every action there is an equal but opposite reaction ’
N
Force, Normal
gmFg
Force, nalGravitatio
REGAN PHY34210
57Example
N
mg =g x 15kg
T
mg
N Ty x
50o
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amFNT
netx
g
REGAN PHY34210
58
6: Force and Motion (Part II)Friction: When two bodies are in contact, the resistance to movementbetween 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=sN where s 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=kN, where k is the coefficient of kinetic friction.
REGAN PHY34210
59Drag Force and Terminal SpeedWhen 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 andthe fluid, the body experiences a DRAG FORCE which opposes thisrelative 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 to the relative speed of the body in the fluid by a DRAG COEFFICIENT, C, which is experimentally determined. The magnitude of the drag force is given by the expression for D, which dependson the fluid density (i.e. mass per unit volume, ), the effective cross-sectional area, A (i.e. the cross-sectional area perpendicular to thedirection of the velocity vector), and the relative speed, v.
2 2
1vACD
REGAN PHY34210
60
Note that the drag coefficient. C, is not really a constant, but rathera 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.
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REGAN PHY34210
61Forces in Uniform Circular MotionRecalling that for a body moving in a circular arc or radius, r, with constant speed, v, the MAGNITUDE of the ACCELERATION, a, isgiven 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
62
(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 theearth on all parts of the space shuttle (including the astronauts).The centripetal force is equal on all areas of the astronauts body so he/shefeels 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.
REGAN PHY34210
63
Example: At what constant speed does the roller coasters have to go to ‘loop the loop’ of radius r ?
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 isalso inwards (i.e., in the downwards direction).
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REGAN PHY34210
64
7: Kinetic Energy and WorkOne way to describe the motion of objects is by the use of Newton’sLaws and Forces. However, an alternative way is describe the motionin 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.
REGAN PHY34210
65Work and Kinetic EnergyThe work done (W) on an object by a force, F, causing a displacement, d, is given by the SCALAR PRODUCT, W = F.d =dFcoswhere 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=vo
2+2axd . By substituting for the acceleration, ax, we have,
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REGAN PHY34210
66
Work Done by a Gravitational Force.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).
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REGAN PHY34210
67
Work Done Lifting and Lowering an Object.
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If we lift an object by applying a vertical (pushing) force, F, during the upward displacement, work (Wa) is done on the object by thisapplied force. The APPLIED FORCE TRANSFERS ENERGY TOthe object, while the GRAVITY TRANSFERS ENERGY FROM it.
REGAN PHY34210
68Spring Forces and Hooke’s LawThe 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 Robert Hooke, 17th century British scientist) as
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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.
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REGAN PHY34210
69Work Done by an Applied ForceDuring the displacement of the spring, the applied force, Fa, doeswork, Wa on the block and the spring restoring force, Fs does work Ws.
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If the block attached to a spring is stationary before and after itsdisplacement, then the work done on the spring by the applied forceis the negative of the work done on it by the spring restoring force.
REGAN PHY34210
70Work Done by a General Variable Force.
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REGAN PHY34210
71Work-Kinetic Energy Theorem
with a General,Variable Force
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REGAN PHY34210
72PowerPOWER is the RATE AT WHICH WORK IS DONE. The AVERAGEPOWER done due to a force responsible for doing work, W in a time
period, t is given by Pave = W/t .
The INSTANTANEOUS POWER is given by dt
dWP
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 thisis the kilowatt-hour, where 1kWh = 1000x3600 J = 3.6 x106J = 3.6MJ.
vFvFdt
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We can also describe the instantaneous power in terms of rate at which a force does work on a particle,
REGAN PHY34210
73
Example 1:What is the total energy associated with a collision between twolocomotives, at opposite ends of a 6.4km track accelerating towardseach other with a constant acceleration of 0.26 m/s2 if the mass of each train was 122 tonnes (1 tonne =103kg) ?
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REGAN PHY34210
74Example 2:
NjiF ˆ6ˆ2
mid ˆ3
If a block slide across a frictionlessfloor through a displacement of -3m in the direction, while at the sametime a steady (i.e. constant) force ofF=(2i-6j) Newtons pushes against the crate, (a) How much work does the wind force do on the crate during this displacement ?
(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 ?
CRATE THE FROMenergy kinetic of 6 rsit transfe i.e.
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6ˆ3.ˆ6ˆ2.
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REGAN PHY34210
75
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 perfectspring, 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 ?
mk v
x
v=0
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REGAN PHY34210
768: Potential Energy & Conservation of Energy
Potential energy (U) is the energy which can be associated withconfiguration 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 betweentwo objects (e.g. by lifting an object higher) the work done on the gravitational force increases the gravitational potential energy of thesystem.
Another example is ELASTIC POTENTIAL ENERGY which is associated with compression or extension of an elastic object (such asa perfect spring). By compressing or extending such a spring, workis done against the restoring force which in turn increases the elastic potential energy in the spring.
REGAN PHY34210
77
Conservative and Non-Conservative ForcesIf 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 potentialenergy, 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 between2 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.
Work and Potential EnergyIn general, the change in potential energy, U 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., U=-W
REGAN PHY34210
78
Determining Potential Energy Values
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REGAN PHY34210
79
Conservation of Mechanical Energy
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REGAN PHY34210
80The Potential Energy Curve
mgFmghxUalso
kxFkxxU
gedx
xdU
x
xUFxFWxU
xFx
FW
case, nalgravitatio in the ,
gives, atingdifferenti then 2
1
by,given is potential elastic theif Law, sHooke' .,.
as written becan energy potential
the, therefore, equals, nt,displaceme aough object thr
an moving , force, aby , done, work thecase, D-1 For the
2
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
REGAN PHY34210
81Turning PointsFor 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.
mgymv
yUyKEmec
2
2
1
)()(
mgdy
ydUyF
mgyEyK mec
)(
)(
,at 0 max max
REGAN PHY34210
82Equilibrium PointsEquilibrium 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 noforce 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 amarble at the bottom of a bowl, or a car at the bottom of a valley.
REGAN PHY34210
83
Particles at A,B, C and D are in at equilibrium pointswhere 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 )
A
B
C
Dx
U(x)
REGAN PHY34210
84
Work Done by an External Force
mecEUKW No friction (conservative forces)
Including friction
dfKdfmvmvFd
fd
vvmF
advv
mafF
kk
k
k
20
2
20
2
20
2
nd
2
1
2
1
and 2
on,substitutiBy
2 usecan wetherefore
constant, is on)accelerati (thus force the
, law, 2 Newtons From
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.
REGAN PHY34210
85Conservation of EnergyThis states that ‘ The total energy of a system, E, can only change by amounts of energy that are transferred to or from the system. ’
system. theofenergy internalin change theis and heat) (i.e.,energy
in thermal change theis energy, mechanicalin change theis
in
thmec
inthmec
E
EE
EEEEW
If a system is ISOLATED from it surroundings, no energy can be transferred to or from it. Thus for an isolated system, the total energyof the system can not change, i.e., 0 inthmec EEEE
Work done can be considered as energy transfer, so we can write,
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.
inthmecmec EEEE 1,2,
REGAN PHY34210
86Example 1:A child of mass m slides down a helter skelter of height, h. Assuming theslide is frictionless, what is the speed of the child at the bottom of the slide ?
h=10m
.height a fromdirectly fellit if
have wouldchild that thespeed same theis that thisNote
22
100
2
1 ,0 ,0 ,
ENERGY, MECHANICAL OF ONCONSERVATI theFrom
2
2
,,
h
ghvmvmgh
mvKKUmghU
KUKUEE
fifi
ffiifmecimec
REGAN PHY34210
87Example 2:A man of mass, m, jumps from a ledge of height, h above the ground,attached by a bungee cord of lengthL. Assuming that the cord obeysHooke’s law and has a spring constant,k, what is the general solution for the maximum extension, x, of the cord ?
root ve ,
2 equation, quadratic thissolving
, 02
1
2
1
2
10 also, 0
0 bottom, and at top 0 if , 0U
ENERGY, MECHANICAL OF ONCONSERVATIBy
2
22
2
x k
kmgLmgmgx
mgLmgxkxmgxmgLxLmgkx
kxymgUUUK
KKvK
elasgrav
fi
L
x
h
m
REGAN PHY34210
889: Systems of ParticlesCentre of Mass (COM): The COM is the point that moves as though all the mass of a body were concentrated there.
n
1
n
1
n
1
n
1
44332211
2211
21
2211
221121
2
1
21
1 and ˆˆˆthen
ˆˆˆ if form, In vector .1
and1
D-3for similarly, , 1
by given is afor form general The
mass total theis where by defined is
COM the,at is and at is if generally, More .
is system theof mass of centre the, mass of particle with thecoincides
axis- oforgin theif ,by separated and mass, of particles 2For
iiicomcomcomcomcom
iiicom
iiicom
iiicom
com
com
rmM
rkzjyixr
kzjyixrzmM
zymM
y
xmMM
xmxmxmxmx
systemn-particle
MM
xmxm
mm
xmxmx
x mxmdmm
mx
m
xdmm
REGAN PHY34210
89Centre of Mass for Solid Bodies
igloo).an or doughnut a example(for object
theof volumein the liey necessarilnot need mass of centre that theNote
,1
, 1
similarly, and 1
1
1
ng,substituti
. mass,by occupied volume theis where, where
)( DENSITY UNIFORMa assuming simplified are integrals theOften,
, 1
, 1
, 1
Then,
. ELEMENTS, MASS smallimally infinitessmany of up made be to
considered becan that they(atoms) particlesmany so have objects Solid
zdVV
zydVV
y
xdVV
dVxV
xdmM
x
dmdVdV
dm
V
M
zdmM
zydmM
yxdmM
x
dm
comcom
com
comcomcom
REGAN PHY34210
90
Newton’s 2nd Law for a System of Particles.
ncom
nncomcom
nn
nncomcom
nnn
nncom
FFFFF
amamamamaMdt
vdM
adt
vd
vmvmvmvmvMdt
rdM
vmdt
rdm
rmrmrmrmrM
321
332211
nd
332211
332211
law, 2 sNewton' and
recalling and again, once atingdifferenti
get we since,
time,respect to with atingdifferenti
REGAN PHY34210
91Linear Momentum
constant). is (for
by defined is MOMENTUM LINEAR The
mamdt
vmd
dt
vmd
dt
vmp
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.’
comnnn vMPvmvmvmvmppppP
P
i.e., momenta,linear particle individual theof sum vector theis
which momentum,linear totala has system theparticles, of system aFor
332211321
The linear momentum of a system of particles is equal to the product ofthe total mass of the system, M, and the velocity of the centre of mass,
REGAN PHY34210
92
Conservation of Linear Momentum
finet
comcom
PPPdt
amdt
vdm
dt
i.e.,constant 0 then,system,
theleaveor enter particles no and zero, is force externalnet
theif system, closed ain , Since,
This is the law of CONSERVATION OF LINEAR MOMENTUMwhich 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.’
REGAN PHY34210
93Varying Mass: The Rocket EquationFor 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,
Ma dt
dvMRvR
dt
dvMv
dt
dMMdvdMv
dMdvdMvMdvMvdMvdMdvdMvMv
dvvdMMvdvvdMMv
UvdMMdMUMv
UvdvvdM
v
dvvdMMdMUMv
dtP
P
P
PP
rel
relrel
rel
rel
rel
rel
rel
fi
obtain we then,loss, mass of rate theis if
0
then,, ve)- (and productsexhaust the
androcket ebetween th speed relative theis if
. interval, after timerocket theof the
plus productsexhaust theof theequals fuel
exhaust theplusrocket theof initial The
momentum, ofon conservatiBy
Mv
M+dmv+dv
-dm
U
a) time = t
b) time = t+dt
1st rocketequation
REGAN PHY34210
94
)rockets! stage-multi of (use smallfor greatest y in velocit increase thus
ln lnlnln
then, ln1
general,in Since,
ly.respective and of masses
rocket toingcorrespond ocities,rocket vel final and initial theare
and where obtain, weg,Integratin
changes, mass theas velocity thefind Tolaw. 2 sNewton' iswhich
, on,accelerati theis and at time mass theis
engine.rocket theof )( THRUST thecalled is
nd
f
f
irel
i
frelifrelif
fi
fi
v
v
M
M
rel
relrel
rel
M
M
Mv
M
MvMMvvv
xdxx
MM
vvM
dM-vdv
M
dMvdvdMvMdv
MaTatM
TRv
f
i
f
i
2nd rocketequation
REGAN PHY34210
95Internal Energy Changes and External Forces
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.
cos
thenis ENERGY MECHANICAL in the change associated The
.nt displaceme and force theof directions ebetween th angle
theis and MASS OF CENTRE theofnt displaceme theis where
cos
by,given is system theofenergy internalin change The
int
FdΔUΔKΔE
dF
d
FdΔE
mec
REGAN PHY34210
9610: Collisions‘A collision is an isolated event in which two or more colliding bodies exert forces on each other for a short time.’
Impulse
. versus of CURVE THE AREA UNDER theand force, applied theofduration
andstrength theofproduct the toequal also is This side). hand(right by on
actedbody theof MOMENTUM LINEAR IN CHANGE theis IMPULSE The
IMPULSE, law 2 sNewton' From
act. forces se which theduring , interval,
timeon the depends changes by which amount The bodies.both of
momentalinear thechange willforces theselaw, 2 sNewton' From
FORCE.VARYING -TIME a is
. at time, twoebetween th acts and
pair, force 3 thebodies, obetween twcollision on head aFor
nd
nd
rd
ttF
F(t)
JdttFpddttFpd
Δt
p
F(t)
t-F(t)F(t)
f
i
f
i
t
t
p
p
F(t)-F(t)
REGAN PHY34210
97
tFJ
ΔtF
JpppJpppJppp
Jppp
-
ave
ave
zzifyyifxxif
if
zzyyxx
by given is impulse theof magnitude the
, period, aover force averaged time theis If
, ,
form,component
in this writealsocan weVECTOR, a is impulse Since,
i.e.,
body,on that acts that IMPULSE the toequal iscollision
ain body each of momentumlinear in the change that the
states THEOREM MOMENTUM LINEARIMPULSE The
E.g A 140g is pitched with a horizontal speed of vi=39m/s. If it is hitback in the opposite direction with the same magnitude of speed what is the impulse, J, which acts on the ball ?
11 9.10393914.0 direction, NEGATIVE
theasdirection velocity initial the taking
kgmskgmsJ
vvmppJ ifif
REGAN PHY34210
98
Momentum and Kinetic Energy in CollisionsIn any collision, at least one of the bodies must be moving prior tothe 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 fromthe 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
99
21
21
21
221121
21,11
,22,11,22,11,2,1,2,1
Recalling
force. externalnet no is thereand isolated is system theascollision ain
changenot can mass of centre theof velocity thesystem, isolatedan For
then , )le!rugby tack a (e.g.,collision after stick
particles two theCOLLISION, INELASTIC COMPLETELY aFor
COLLISION, BODY 2 aFor
collisionafter momentum totalcollision before momentum total
MOMENTUM, LINEAR OF ONCONSERVATIBy
mm
pp
mm
Pv
vmvmvmmvMP
Vmmvm
vmvmvmvmpppp
P P
com
comcom
i
ffiiffii
fi
REGAN PHY34210
100Elastic Collisions in 1-DIn an elastic collision, the total energy before the collision is equal tothe total kinetic energy after the collision. Note that the kinetic energy of each body may change, but the total kinetic energy remains constant.
forward). pushed always is (i.e. positive always is that Note
2 and to,leadswhich
2
1
2
1
2
1
conserved isenergy kinetic totalthecollision elastican In
axis). same along magnitudes case, D-(1
momentum,linear ofon conservatiBy
rest.at mass, with balls, billiard obetween twcollision on -head aFor
22
,121
1,2,1
21
21,1
,1,1,1,112,22
2,22
2,11
2,11
,22,1,11
,22,11,11
2
mv
vmm
mvv
mm
mmv
vvvvmvmvmvmvm
vmvvm
vmvmvm
m
,f
ifif
fififffi
ffi
ffi
m1, v1,i m2, v2,i=0
before elastic collision
m1, v1,f m2, v2,f
after elastic collision
REGAN PHY34210
101
v , vvv
mm
vm
mvvv
mm
vvvmm
vmm
mvv
mm
mmv
,i,f,i,f
,i,fif
iff
ifif
velocity.projectile thee with twicoff moveset light targ velocity,unchanged
approx.at forwards continues projectileheavy i.e. 2
:ball) golfon ballcannon (e.g., projectile Massive 3)
velocity.small with forwards moveset Heavy targ
.projectile incoming todirection) opposite(but locity similar ve
back with bounces projectilelight i.e., 2
,
:ball)cannon aon ball golf (e.g., target,Massive 2)
velocity.s'projectile with theoff moves target theandcollision following
stops projectile themasses, equalbetween collision on -head afor i.e.,
, 0 :balls) pool (e.g. masses, Equal 1)
cases. limiting following the tolead These
2 & ,collisions elastic D-1For
1211
21
12
12,1,1
12
,1,2,121
,121
1,2,1
21
21,1
REGAN PHY34210
102Example 1: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 themass 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.
Pb!for 1/50~4/208~ and O)H water(NB.proton hydrogenfor 14/4
e, therefor4
thus, mom. lin. of cons. from
rest,at initially nucleus moderating the&collision nucleus-neutron closed aFor
iscollision per lossenergy fractional The
2
1 and
2
1 areneutron scattered and orginal theof energies
kinetic final and initial The mass.moderator offunction a ascollision single a
for NEUTRON THE FROMenergy kinetic of transfer MAXIMUM want theWe
2
2,
,
2,
2,
2,
2,
2,
F
mm
mmF
mm
mm
v
v
v
vv
K
KKF
vmKvmK
MODn
MODn
MODn
MODn
in
fn
in
fnin
i
fi
fnnfinni
REGAN PHY34210
103
Example 2: The Ballistic PendulumA ballastic pendulum uses the transfer of energy to measure the speed of bullets fired into a wooden block suspended by string.
ghm
MmvMm
m
ghv
ghMmMm
vmMm
h
ghMmvMm
vMmvm
bul
blockbulbulblockbul
bulbul
blockbulblockbul
bulbulblockbul
blockbulblockblockbul
blockblockbulbulbul
22
2
1
upwards. swingsit asblock theofheight in increase theis where2
1
thenenergy, mechanical ofon conservati a
assumecan weclosed, is systemblock theis that know Also
momentum,linear ofon conservatiBy
2
22
2
2
Mblock
vbul
h
REGAN PHY34210
1041-D Collisions with a Moving Target
ifif
i
iifiif
ififfifi
iffi
ffii
fififfii
vmm
mvv
mm
mmv
v
vmm
mmv
mm
mvv
mm
mv
mm
mmv
vvvvmvvvvm
vvmvvm
vmvmvmvm
vvmvvmvmvmvmvm
,121
1,2,1
21
21,1
,2
,221
12,1
21
1,2,2
21
2,1
21
21,1
,2,2,2,22,1,1,1,11
2,2
2,22
2,1
2,11
2,22
2,11
2,22
2,11
,2,22,1,11,22,11,22,11
2 &
of results previous obtain the wetarget)
y(stationar 0set weif Note, arbitrary. are 2 and 1 subscripts The
2 &
2
relations, general obtain the weequations, ussimultaneo thesesolving
2
1
2
1
2
1
2
1
thusconserved, isenergy kinetic collision, elastican For
momentum,linear ofon conservatiBy
m1, v1,i m2, v2,i
before elastic collision
REGAN PHY34210
105Collisions in Two Dimensions
m1, v1,i
m2, v2,i
m1, v1,f
m2, v2,f
When two bodies collide, the impulses of each body on the otherdetermine the final directions followingthe collision. If the collision is nothead-on (i.e. not the simplest 1-D case)in a closed system, momentum remainsconserved, thus, for an elastic collisionwhere Ktot,I=Ktot,f , we can write,
2,22
2,11
2,22
2,11,2,1,2,1 2
1
2
1
2
1
2
1 and ffiiffii vmvmvmvmPPPP
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,
2,22
2,11
2,11
2,221,11
2,221,11,11
collision, elastican For
sinsin0 axis,
coscos axis,
ffi
ff
ffi
vmvmvm
vmvmy
vmvmvmx
y
x
REGAN PHY34210
10611: RotationMost 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 . The origin of this circular path is centred at the axis of rotation.
REGAN PHY34210
107Summary of Rotational VariablesAll rotational variables are defined relative to motion about a fixed axis of rotation. The ANGULAR POSITION, , of a body is then the angle betweena 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 .
x
axis ofrotation
reference line
s
r
Radians are defined by s/r and are thus pure, dimensionless numbers without units. The circumference of a circle (i.e., a full arc) s=2r, thus in radians, the angle swept out by a single, full revolution is 360o = 2r/r=2. Thus, 1 radian = 360 / 2= 57.3o
= 0.159 of a complete revolution.
REGAN PHY34210
108
dt
dt
dt
d
ttt
r
s
av
av
av
),(rad/s on,AcceleratiAngular ousInstantane
),sper (radians on,AcceleratiAngular Average
(rad/s),Velocity Angular ousInstantane
second),per (radian Velocity Angular Average
(radians),nt displacemeAngular
(radians),position Angular
2
2
12
12
12
The angular displacement, represents the change in 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 (), rate of change () and rate of rate of change ( ) of the angular position.
REGAN PHY34210
109Relating Linear and Angular VariablesFor the rotation of a rigid body, all of the particles in the body take thesame 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!)
22 ,revolution of Period
ison accelerati theofcomponent Radial
; ;
222
v
rT
rr
r
r
va
rdt
dr
dt
rd
dt
dvar
dt
dr
dt
dsvrs
r
t
REGAN PHY34210
110Rotation with Constant AccelerationFor translational motion we have seen that for the case of aconstant 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.
20
20
00
00
020
20
20
2
20
200
00
2
1
2
1
2
2
2 2
2
1
2
1
ttatvtxx
ttvv
xx
xxavv
ttattvxx
tatvv
TRANSLATIONAL ROTATIONAL
REGAN PHY34210
111Example 1:A grindstone rotates at a constant angular acceleration of =0.35rad/s2.At time t=0 it has an angular velocityof 0=-4.6rad/s and a reference line onits horizontal at the angular position, 0=0. axis of
rotation
ref. linefor 0=0
(a) at what time after t=0 is the reference line at =5 revs ?
stt.t.-π-
sradsradradtt
3235.0
56.66.4
175.02
10175.046.46.4350
2
164010
/35.0 ; /6.4 ; 0 ; 105rev : 2
1
22
200
200
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?
s
srad
srad
at 13
/35.0
/6.402
0
REGAN PHY34210
112Kinetic Energy of Rotation
place. takeorotation tfor ) less (i.e. done be toneeded is
work less means inertia ofmoment smaller a general,in Thus2
1by given isrotation ofenergy Kinetic
INERTIA ROTATIONALor INERTIA OF MOMENT theis
where definecan weNow
. 2
1 thusconstant, is BUT .
2
1
2
1
2
1
2
1
2
1
isenergy kinetic the, speeds,different at moving , masses,
of collection a ascan treat webody which composite aFor
2
2
222
2233
222
211
K
IK
I
rmI
rmKrmK
vmvmvmvmK
vm
nnn
nnn
nnnn
nnn
nn
REGAN PHY34210
113
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.
MhIdmbadmRIydmbxdmb
dmxM
xRyx
dmbaydmbxdmadmyxI
dmbyaxdmrIbah
com
com
2222
222
2222
222222
022
,definitionby then orgin, theas mass of centre thetake
weassuming , 1
since and since Now,
22
, If
The Parallel-Axis Theorem
.body, continuous aFor axis.rotation thefrom particle
theof distancelar perpendicu theis where body, rigid aFor
22
2
dmrrmI
nrrmI
nnn
th
nnn
Calculating to the Rotational Moment of Inertia
REGAN PHY34210
114Example 2:The HCl molecule consists of a hydrogen atom (mass 1u) and a chlorine atom (mass 35u). The centres of the two atoms are separatedby 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 ?
com
d
a d-a
rotation axis
ClH
.molecules)for inertia of moments rotationalfor units (note .250,15
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351
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REGAN PHY34210
115Torque and Newton’s 2nd Law
rotation.for law2 sNewton'
, since,, is particle on the acting Torque
. on,accelerati l tangentia the toforce l tangentia theRelating
nd2
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rarmarF
maF
net
ttt
tt
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 =Frsinis called the TORQUE (latin for twist!) .
.in Joules,in . work ofunit the toequivalent arewhich
, is Torque ofunit SI .F FORCE THE OF ARM MOMENT theis
. through running line a and between distancelar perpendicu
theis .sin AND sin TORQUE,
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rFrFrrFFr t
r
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Or
REGAN PHY34210
116Work and Rotational Kinetic Energy
dt
d
dt
dWP
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rFFsW
IIKW
mrI
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t
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1
THEOREM, ENERGY KINETIC- WORK theFrom
2
1
21
22
2
22
22
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.
Rdt
dR
dt
Rd
dt
dsv
dt
d
RRs
dtv
com
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by given is mass of centre theof
speed the while, is centre itsabout wheel theof speedangular The
nt.displacemeangular its
is and radius wheel theis where, length, of arcan through
moves wheel theof outside thedistance theas distance same thetravels
mass of centre the, interval timeaIn constant. is mass, of centre
theof speed that themeans SPEED, CONSTANT aat rolling A wheel
COM motion.
P
O
P
O
R
S
REGAN PHY34210
118
The kinetic energy of rolling. A rolling object has two types ofkinetic energy, a rotational kinetic energy due to the rotation about the centre of mass of the body and translational kinetic energy due to the translation of its centre of mass.
COM motion.
P
O
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S
2222
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can write weTHEOREM, AXIS PARALLEL theFrom
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P. point, he through taxisan about rotation pure assituation thecan view We
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REGAN PHY34210
119Rolling Down a RampIf a wheel rolls at a constant speed, ithas no tendency to slide. However, if thiswheel is acted upon by a net force (suchas gravity) this has the effect of speedingup (or slowing down) the rotation, causingan acceleration of the centre of mass of the system, acom along the direction of travel. It also causes the wheel to rotate faster. Theseaccelerations 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 STATICFRICTIONAL FORCE ( fs ).
gF
N
singF
cosgF
R
P
Radt
vd
dt
Rd
vR
comcom
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sides,both atingdifferentiby
then ,by given isfrequency rotational theSince
REGAN PHY34210
120
2
,2,
2,,
,
nd
,,
1
sin sin
.
sin
sign) (note rolling,smooth For .
0.R have thusand COM h theact throug all forces Normal and nalgravitatio The
P.point at FRICTION theis figure in themotion rolling a causing forceonly The
. is law 2 sNewton' of form Rot. .cos where
sin slope, thealong components force the
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Rolling down a ramp (cont.)For a uniform body of mass, M and radius, R,rolling smoothly (i.e. not sliding) down a ramptilted at angle, (which we define as the x-axisin this problem), the translational acceleration down the ramp can be calculated, from
REGAN PHY34210
121
21
,expression the toleads analysis This
O
com
com
MRI
ga
R0 R
T
Mg
friction.n rather tha string in the tension by the slowed is yo-yo the)3(
radius,outer not the , radius with axle thearound rolls yo-yo the)2(
).90 (i.e. string down thedirectly rolls yo-yo the(1)
s.assumption following with thehill) adown rollingbody a
for (as law 2 sNewton' assuming calculated becan string down the
rolling yo-yo theofon accelerati theof valuefor the expression The
energy. potential back toit transfersandenergy kinetic thislosesit
string, theupback climbs yo-yo theAs forms. 2
1
rotational and 2
1 lationalboth transin energy
kinetic into ed transferris This . energy, potential
nalgravitatio losesit distance adown rolls yo-yo a If :
0
0
nd
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R.R
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mvK
mgh
hThe Yo-Yo
rot
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REGAN PHY34210
122Example 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 ?
) time!sameat bottomreach ball bowling and
marble t,independen Mass (note, 1.47
10
5
1
2
1
5
2
2
1
2
1
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sphere, afor 5
2 g,subsitutin and rolling,smooth For
.2
1
2
100
energy, mechanical ofon conservatiBy
1
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22
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ffii
1.2m
REGAN PHY34210
123Example 1 (cont):(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
disk.beat willmarble sizeAny disk. by the followed fastest, rolls Sphere
.71.0,5
2 sphere,For ;0.66 ,
2
1 disk,For ; 5.01, hoop,For
1
1
andconstant a with , general,In .
MOTION, NALTRANSLATIO into goesch energy whi kinetic offraction The
. hoop and ; 2
1disk ;
5
2 sphere afor inertia of moments The
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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.
.sin and sin where
sinby given is TORQUE THE OF MAGNITUDE The
.) and both lar toperpendicu (i.e. rule, hand-rightproduct cross vector the
using found is torque theofdirection The .by defined is torqueThe
FFrr
rFFrrF
Fr
Fr
z
x
y
O
r
F
F redrawn at origin
r x F=
z
x
y
Or
F
rF
REGAN PHY34210
125Angular Momentum
. by given MMOMENTU ANGULARan has
), momentum,linear with (i.e. city with velo, mass of particleA
vrmprl
vmpvm
z
x
y
O
r
p
p redrawn at origin
r x p= l
z
x
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Or
l
rp
p
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.sin and sin wheresinby given
is ) of units(in momentumangular theof magnitude The
.), and both to is that shows rule hand-right (the
product cross vector by thegiven isdirection momentumangular The
2
pprrrpprrpl
Js/s kg.m
vprl
REGAN PHY34210
126Newton’s 2nd Law in Angular Form.
. particle on the acting torquesof sum vector the toequal is
timerespect to with momentumangular of change of rate thei.e.
0)(sin parallel are vectors thesesince 0
gives, timerespect to with sidesboth atingDifferenti
by given is particle a of momentumangular theIf
as written becan form tionalin transla law 2 sNewton' nd
net
neti
iinet
net
dt
ld
FrFramrarmdt
ld
vv
vvarmdt
rdv
dt
vdrm
dt
vrmd
dt
ld
vrmprl
dt
REGAN PHY34210
127
:law 2 sNewton'for form aobtain we, torques,external
all of sum vector thei.e. TORQUE, EXTERNAL NET theis If
bodies. external from particles on the forces todue thosei.e.,
)( MOMENTUM ANGULAR TOTAL thechange torques EXTERNALOnly
d i.e.,
particles, individual theof momenta,angular theof SUM VECTOR theis
momentum,angular total thePARTICLES, OF SYSTEM aFor
nd
net
1,
11321
dt
Ldτ
L
dt
ld
dt
LlllllL
lL
net
n
iinet
n
i
in
iin
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 PHY34210
128
axis.rotation about the is that assuming dropped, is '' eUsually th
can write, weaxis, fixed aabout body theof inertia ofmoment the
, andbody rotating on the points allfor CONSTANT a is
i.e.,
onscontributi elemental theseof
sum theis BODY ENTIRE for the component The
sinsin is axis (z)
rotation the toparallelcomponent momentumangular The
.90sin
is ,element mass a of momentumangular theof magnitude theaxis,
fixed aabout rotatingbody rigid ain particlegiven aFor
1
2
1 1 1
2
1
0
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IL
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rmrrmrvmlL
vmrvmrll
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iiiiiiiiz
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i
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xy
r
r
ip
m
REGAN PHY34210
129Conservation of Angular Momentum
)nuclei! and starsneutron skaters, gpirouettin (e.g., ,compensate toincreases speed
rotational its decreases, system a of inertia ofmoment theif that means This
. as form algebraicin written becan lawon conservati The
system. e within thplace t takesmatter wha no change,not can
axis that along momentumangular ofcomponent then thezero, is axis fixed
a along system aon torqueexternalnet theof TCOMPONEN theif Similarly,
system. the WITHINplace take
changest matter wha no constant, remains system, theof momentumangular
thezero, is system aon acting torqueexternalnet theifsay that can thus We
. ,other time someat system theof momentum
angular net the toequal is , at time momentumangular net that themeans This
CONSERVED. IS SYSTEM THE OF MOMENTUM ANGULAR THE
thusand 0 then system, on the acts TORQUE NET ON if
, by, momentumangular in change the torelated is net torque theSince
ffii
f
i
net
II
L
t
t
dt
Ld
τdt
Ld
REGAN PHY34210
130Example1: Pulsars (Rotating Neutron Stars)
Vela supernovaremnant, pulsarperiod ~0.7 secs
Crab nebula, SN remnant observed by chinese in 11th century
Pulsars have similarperiodicities ~0.1-1s.
before after!
SN1987A
REGAN PHY34210
131
Rotational period of crab nebula (supernova remnant) =1.337secs
optical
x-ray
.400~105.2
1107~
1~)( ,105.2~)( ,107~)(
rotation of period ,
constant , ..
68
68
22
Kms
smR
spulsarTssunTmsunR
TT
TRRR
kMRkIMRkI
f
fii
i
fi
f
iif
ffffiiii
PULSAR = PULSAting Radio Star (neutron-star)
Lighthouseeffect
Starquakes
REGAN PHY34210
132
dt
Ld
dt
dt
Ld
dt
pdFNewtons
prIl MomAngvmpLinearMomentum,
Frdt
ldITorque
dt
pdamFForce
netnet
net
0 0
mom.Angular Mom.Linear on,Conservati
Law, 2
..
, ,
nd
TRANSLATIONAL ROTATIONAL
REGAN PHY34210
13313: Equilibrium and ElasticityAn 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).
MEQUILIBRIU ROTATIONAL
for torques)of balance (i.e. 0
and
MEQUILIBRIU NALTRANSLATIO
for forces) of balance (i.e. 0
dt
Ld
dt
net
net
REGAN PHY34210
134The Centre of Gravity
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.
REGAN PHY34210
135ElasticityA solid is formed when the atoms which make up the solid take upregular spacings known as a LATTICE. In a lattice, the atoms take upa repetitive arrangement whereby they are separated by a fixed, welldefined 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 extremelyrigid (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 ->L/L(stretching) ; (2) shearing stress -> x/L (shearing) ; and (3) hydraulic stress -> V/V (3-D compression).
REGAN PHY34210
136
.extension) theis length, orginal theis (object theof
the toingcorrespond by definedquantity
essdimensionl a is thisstress, eFor tensil n.deformatiounit theis The
point).at that pressure thedefines also (This
area thelarly toperpendicu applied force theof magnitude theis where
,by given isn compressioor tension simplefor object an on The
ΔLL
h the lengt change infractionalL
ΔL
STRAIN
AFA
FSTRESS
STRESS and STRAIN are PROPORTIONAL TO EACH OTHER.The constant of proportionality which links these two quantities isknow as the MODULUS OF ELASTICITY, where STRESS = MODULUS x STRAIN
REGAN PHY34210
137
V
VBPB
ΔVVV
ΔV
P STRESSHYDRAULICl
xG
A
F(G) MODULUS, SHEAR
l
Δx
FA
FSHEARING
L
ΔLE
A
F
EDULUSYOUNG'S MO
by defined is )( MODULUS BULK The
change. volume theis and volumeinitial theis where
, as defined isstrain The area).unit per force (i.e.
, pressure fluid theas defined is
where theto
leading, now isstrain The area. theof plane theto
parallel is but , still is stress the,For
by defined is stress ecompressiv
or efor tensil )( The
LL+L
F
F
L
x F
F
V
V
V-V
REGAN PHY34210
138
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 BEYONDTHE YIELD STRENGTH, Sy,of the specimen,it will become PERMANENTLY DEFORMED.
If the stress is increased further, it will ultimatelyreach its ULTIMATE STRENGTH, Su, where the specimen breaks/ruptures.
Strain (l/l)
Stre
ss (
F/A
)
Sy (perm. deformed)
Su (rupture)
REGAN PHY34210
139Example 1: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 ?
28
23
4
2102.2
105.9
102.6
Nm
m
N
r
F
A
Fstress
(b) If the Young’s modulus for steel is 2.2 x 1011 Nm-2, what
are the elongation and strain on the cylinder ?
%11.0101.181.0
109.8
109.8102.2
102.281.0
modulus, sYoung' of definition theFrom
44
4211
28
m
m
l
lstrain
mNm
NmmΔl
stressE
lΔl
A
F
l
lE
l =81cm
F=62kN
F=62kN
A
REGAN PHY34210
14014: GravitationIsaac Newton (1665) proposed a FORCE LAW which described the mutual attraction of all bodies with mass to each other. He proposedthat each particle attracts any other particle via the GRAVITATIONAL FORCE with magnitude given byG=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).
221
r
mmGF
m1
m2
rF
F
The two particles m1 and m2 mutually attract with a force of magnitude, F. m1 attracts m2 with equal magnitudebut opposite sign to the attraction of m2 to m1. Thus, F and -F form a third force pair, which only depends on the separation of the particles, r, not their specific positions. 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’.
REGAN PHY34210
141Gravitation Near the Earth’s Surface
from. dropped isobject an at which height'' on the depends
gravity todueon accelerati theThus . by
given is magnitude whose, ON,ACCELERATI NALGRAVITATIO awith
gravityunder centre searth' the toaccelerate it will released, is particle theIf
:bygiven is centre searth' thefrom
, distance aat , mass of particle aon earth thefrom force nalgravitatio the
of magnitude the, mass of SPHERE UNIFORMa isearth theAssuming
22
2
r
GMa
r
MmGmaF
a
r
MmGF
rm
M
gggrav
g
grav
The earth can be thought of a nest of shells, and thus all its masscan be thought of as being positioned at it centre as far as bodieswhich lie outside the earth’s surface are concerned.
average ag at earth’s surface = 9.83 ms-2 altitude = 0 kmag at top of Mt. Everest = 9.80ms-2 altitude = 8.8 kmag 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 joiningthe north and south poles. Objects on the earth surface anywhere apartthese 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.
REGAN PHY34210
143Centripetal Acceleration at Earth’s Surface
justified. well)least!at earth (on usually is
onaccelerati nalgravitatio theequals weight the
Assuming . tocompared small very i.e.
)89 (cf. 034.0
3600s24
rads 2from estimated becan
m.106.37 equator, at the max. is
velocity.rotational theis and
around rotatesobject which theradius theis
weight, the toequal is force, normal The
from
isobject surface aon force normal The
222
6
22
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g
-gcentr
earth
gg
grnet
ma
ms.mamsRa
Δt
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RRR
R
RagRωmma mg
mgN
RmmaNmaF
N
‘above’ view,looking from pole,
R
m
N
S
mR
mag
REGAN PHY34210
144Gravitation Inside the Earth‘A uniform shell of matter exerts no NET force on a particle locatedinside 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.
In the example on the left, for = M/V = constanta planet of radius, R and total mass M. An object of mass m, which burrows downwards such that it is now ata 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 ofthe planet between radii r and the outer radius R.
R
r
m
krFeimrG
r
rG
r
mGMF
rVM
netins
ins
.. 3
4 is shell theinside mass
the todue particle by the dexperience force thesince and
2
334
2
334
NonetForce
Netforce
REGAN PHY34210
145Gravitational Potential Energy
. separation infiniteat zero be todefined and
,expression by the defined isenergy potential nalgravitatio The
)(rr
MmGU
R
GMmW UUUWU
RmW
R
GMm
R
GMm
r
GMmdr
rGMmdr
r
MmGW
rdrFWdrr
MmGrdrF -
drrFrdrFrdrFW
RR
RRR
R
R
,
,expression general by the related are done work andenergy potential Since
. out to distance a from mass, a move toREQUIRED WORK
01
. . 1180coscos
,cos. , . is done work general,In
22
20
PROOF
REGAN PHY34210
146Potential Energy and Force
2 true.also is converse theforce, the
for expression thefromenergy potential nalgravitatio thederivecan weSince
TAKEN. PATH THE NOT positions, final and initial on the dependonly
energy potential grav. in the changes and force veconservati a isGravity
r
MmG
r
MmG
dr
d
dr
dUF
Escape Speed (Velocity)
12
2
2.11; 2
02
1
energy ofon conservati of principle thefrom Thus ).for ion configurat
energy potential zero at the and velocity zero (i.e. 0 distance, infiniteAt
. energy, potential nalgravitatio The; 2
1,energy kinetic The
distance. infiniteat 0 down to slowsit until gravity,against speed,constant
with up move object to thecauses This . SPEED, ESCAPE
an has radius ofplanet mass a leaving projectile massA
kms vR
GMv
R
GMmmvUK
r
UKR
GMmUmvK
vv
v
RMm
earthescescesc
REGAN PHY34210
147Johannes 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.
a ab
b
REGAN PHY34210
148The Law of Orbits
f f’
r
RaRp
a
ea ea
M
m
If M >> m,the centre of mass of the planet-sun system is approximately at the centre of the sun. The orbit is described by the length of the semi-major axis, a and the eccentricityparameter, e. The eccentricity is definedby the fact that the eachfocus f and f’ are distanceea 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).
REGAN PHY34210
149
a
f f’
r
RaRp
ea ea
20
20
0minmax
20
20
22
22
0min
0max
0
1
2 AXIS MAJOR-SEMI theoflength The
1
2
1
1
1
1 2 axismajor ofLength
21 equation, by the defined is Ellipse
sincos ordinates-coCartesian In
1r ,
1
cos1
, focus, theas ordinates-co theoforigin the take weIf
,by defined isty eccentrici theellipse,an for generalIn minmax
minmax
rA
rrRRrraA
ryxrx
θrθ, yr, xyxr
rrr
rr
f
aP
RR
RR
rrrr
pa
pa
f f’
r
y
x
REGAN PHY34210
150
constant. conserved, is if Thus, . 22
1
is,star thearoundplanet theof mom. ang. The
sun). thearoundplanet of velocity rotational (i.e.,
planet andsun theconnecting line rotating theof speedangular theis where2
1
2
1
0, and 0 As
. of luessmaller vafor
exact more becomes expression This
.2
1heightbase
2
1
elyapproximat isout swept area The
. base and , height, of TRIANGLE a isout swept area of wedgethe
assuming ESTIMATED becan , in timeout swept area theis theIf
2
2
22
dt
dAL
m
Lr
dt
dA
mrmrrmvrrpL
rdt
dr
dt
dA
Δt
ΔA
ΔA
ΔA
θ.r.rA
rsr
ΔAtA
The Law of Areas
pp
M
mr
REGAN PHY34210
151The Law of Periods
r
M
m
system).solar for myr103.0( ,for constant .,.
4 predicts law ofion Exact vers
axis.major -semi ellipse,For .4
4
get in we ngsubstituti ,2
,revolution of period The
law,2 sNewton' using orbit,circular aFor
3-234-3
2
2
3
2
32
22
23
2322
222
2
nd
3
2
aT
grav
mMa
Tei
mMG
π
a
T
arrGM
TT
πrGM
ω
πT
ωrGMmrωr
MmG
mrωr
rωm
r
vmmg
r
MmGmaF
REGAN PHY34210
152Satellites, Orbits and Energies
a
GMmEra
Kr
GMm
r
GMm
r
GMmE
KU E
UK
r
GMmmvK
r
vm
r
GMmF
Ur
GMmU
2 ie. ,for length) axismajor -(semi substituteorbit ellipticalan For
ENERGY KINETIC THE OF NEGATIVE the toequal isenergy total thei.e.22
by given isenergy mechanical totalThe
orbit.circular ain satellite afor 2
Therefore,
22
1
is law2 sNewton' viaSATELLITE,ORBITING
CICRULARLYA OF ENERGY KINETIC The
separation infinitefor 0 ,
bygiven is system ofenergy potential The
22
2
nd
rEn
ergy
K(r)
Etot(r)
=-K(r)
REGAN PHY34210
153Example 1:A satellite in a circular orbit at an altitude of 230km above the earth’ssurface as a period of 89 minutes. From this information, calculatethe mass of the earth ?
kgM
s
m
skgmT
hR
G
πM
skgmG
RhRr
mMrGM
T
earth
earth
24
2
36
21311
2
2
32
21311-
6
32
2rd
106
6089
106.6
1067.6
44
106.67 and
radius searth' them106.37 where
assuming 4
:law 3 sKepler' From
REGAN PHY34210
15415: Fluids
.density uniform aFor . i.e., ,element
volume theof size theand mass theof ratio theis :)( DENSITY
VM
VM ρΔV
Δm
Fluids (liquids and gases), by contrast with solids, have the ability to FLOW. Fluids push to the boundary of the object which holds them.
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 ofthe expression, p = F /A as A is made as small as possible.If the force is UNIFORM over a FLAT AREA, A, we can write p=F/AThe pressure in a fluid has the same value no matter what directionthe pressure WITHIN the fluid is measured. Pressure is a SCALARquantity (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
155
BAROMETER) MERCURY thebehind (Principle
pressure catmospheri depth, level, surface If
. and where gives forces theBalancing
cylinder. of volume cylinderin water of mass The
. area, sectional-cross andly respective and depthsat bottom and
h top water witofcylinder a of bottom and topat the forces theare and
balance)it on forces theandy (stationar M,EQUILIBRIU STATICin For water
0
0121
2112211212
221112
21
21
21
ghpp
pphyy
gyyppgyyAApApmgFF
ApFApFmgFF
, VyyρAVm
Ay y
FF
h
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.
Fluids at Rest
y=0AIR
WATER
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.
y1
y2
A
mgF2
F1
REGAN PHY34210
156Pascal’s Principle‘ 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.
This is the basis behind the concept of the HYDRAULIC LEVER. A downward force on one platform (the ‘input piston’) causes a change in pressure of the INCOMPRESSIBLE LIQUID, resulting in the movement of a second platform (the ‘output piston’).For equilibrium, there must be a downward force due to a load on the output piston which balances the upwardforce, Fo.
di
do
Fi Fo
Ai
Ao
We can write Pascal’s principle as p=pext, i.e. the change in pressure in the liquid equals the change in the applied external pressure.
Fo
load
REGAN PHY34210
157The Hydraulic Lever
iio
ii
i
oioo
o iioo
iioooii
o
i
i
oio
o
o
i
i
o
i
dFA
Ad
A
AFdFW
ddAAA
AdddAdAV
Vd
d
A
AFF
A
F
A
Fp
F
F
by given is WORK OUTPUT The
piston.input than thedistancesmaller a movespiston output thei.e.,
, if
then,pistons,both at displaced is fluid ibleincompress theof
, volumesame thesuch that distance a upwards movespiston output the
, distance adown movespiston input theIf
liquid theof pressure in the change a produces
piston handright on the load thefrom , force
downward theand , force applied The
With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force over a smaller distance.
di
do
Fi Fo
Ai
Ao
Fo
load
REGAN PHY34210
158Archimedes’ Principle ‘when a body is fully or partially submerged in a fluid, a BUOYANTFORCE, Fb from the surrounding fluid acts on the body. The force isdirected upwards and has a magnitude equal to the weight, mfg of the fluid that has been displaced by the body.
If a body submerged in a fluid has a greater density that then fluid, thereis a net force downwards (Fg>Fb), while if the density is less than the fluid, there will be net force upwards (since Fb<Fg).
This net upward bouyant force exists because the water pressure aroundthe submerged body increases with depth below the surface (p=gh). Thus the pressure at the bottom of the object is larger than at the top.
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!
REGAN PHY34210
159Flow of Ideal Fluids In MotionThe flow of real fluids is very complicated mathematically. Oftenmatters 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 themoving fluid does not change in magnitude or direction. B) INCOMPRESSIBLE FLOW: This assumes the fluid has constantand 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 thatthis does not mean that it can not move in a circular path).
REGAN PHY34210
160The Equation of Continuity
constant RATE,FLOW MASS the
constant, is fluid theofdensity theIf
RATE.FLOW VOLUME theis hereconstant w
asrewritten becan This .CONTINUITY OF EQUATION
segment, tube theof endsboth toApplying
2211
2211
AvRR
RAvR
vAvA
tvAtvAV
Vm
VV
Everyday experience tells us that the velocity of a fluid emerging froma 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).
v1 v2
A1
A2
If we have a tube of cross-sectional area, A1, which narrows to area A2. In a time interval,t, a volume V of fluid enters the tube, with velocity v1. Since the fluid is ideal, and thusincompressible, 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 (V) is given by the product of the cross-sectional area (A) and the length it flows (x). Also, by definition, v=x/t thus,
REGAN PHY34210
161Bernoulli’s Equation
211221
222
12
212
1121
221
2222
121
212
11
becomesequn sBernoulli' ,0rest at fluidsFor
. toreduces equn. sBernoulli' ,0ythen
flow, itsin elevation changenot does fluid theIf constant.
aswritten -re becan which
yygppvv
vpvpyy
gyvp
gyvpgyvp
If the speed of a fluid element increases as it travels along a horizontalstreamline, the pressure of the fluid must decrease and vice versa.
y1
y2
p1 ,v1
p2 ,v2If an ideal (incompressible) fluid flows through a tube at a steady rate. If in time t , a volume of fluid, V enters the tube and an identical volume emerges from the other end. If y1, v1 and p1 are elevation, speed and pressure of the 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,
REGAN PHY34210
162Proof of Bernoulli’s Equation
equationsBernoullivvppyyg
vvVppΔVyyVgΔKWWW
ΔVppΔVpΔVpW
VpΔxpAxFW
-yyVgyymgW
m
ΔtρΔVΔm
vvρΔVΔmvΔmvΔK
ΔK
ei
KW
pg
p
g
' gives,which
theorem
kinetic- workThe . is done work NET The
by given is done work thegeneral,in Since,
end. emerging at the fluid forwardpush tosystem theBY and entrance the
at tubehe through tfluid thepushingby system theON done also isWork
.directions oppositein are force nalgravitatio andmotion thesince arises
sign The . i.e. level.output the
input to thefromlift verticalits during , mass ofelement fluid on the
force grav. by the done work the todue is system on the done work The
. interval, in timeoutput theleaving and entering mass fluid theis
tube. theof ends ebetween th speed
fluidin change thefrom resulting ,energy kineticin
change theequals system on the donenet work the,..
. states oremenergy the kinetic- workThe
21
222
11212
21
222
11212
1212
1212
21
222
1212
1222
1
y1
y2
p1 ,v1
p2 ,v2
REGAN PHY34210
16316: Oscillations
motion. theof PHASE thecalled is t
.0 at time particle theof velocity andnt displaceme
by the determined is which motion, theof ANGLE PHASE theis
.2hat relation t the toleads this, 2coscos Since
. coscosby defined FREQUENCY ANGULAR
.) of value(maximum AMPLITUDE theis
wherecosby given is , particle a ofnt displaceme
theof dependence time theMOTION, HARMONIC SIMPLEFor
. wheren,oscillatio one complete to time theis , period, The
2
1
t
f
Ttxtx
xx
txtxx
TT
Tπ
mm
m
m
f
Oscillations describe motions which are repetitive. An importantproperty of oscillatory motion is its FREQUENCY, f, which describesthe 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.
REGAN PHY34210
164
AMPLITUDE. VELOCITY thecalled is where
sinsin cos
mm
mmm
vx
tvtxdt
txd
dt
tdxtv
The Acceleration of Simple Harmonic Motion
SHMfor equation signature theis which ,
amplitude.on accelerati theasknown is
cossin
2
2
2
txta
ax
txdt
txd
dt
tdvta
mm
mm
‘ 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 velocity of a particle undergoing simple harmonic motion can be found by differentiating the displacement, x(t) with respect to time.
Note, in SHM, the magnitude of the velocity is greatest when thedisplacement is smallest and vice versa, since cos()=sin(+/2)
The Velocity of Simple Harmonic Motion
The acceleration can be found by differentiating the velocity in SHM,
REGAN PHY34210
165The Force Law for Simple Harmonic Motion
k
m
ω
πT
m
k
mk
kxx-ωmmaFmaF
22
is oscillatorlinear afor n oscillatio of PERIOD theThus
by constant spring theofstrength the torelated is oscillator
harmonic simple afor frequency angular that thegives thisarranging-Re
Motion, Harmonic Simple afor then
CONSTANT, SPRING for the ngSubstituti FORCE.RESTORING a of
idea with theCONSISTENT is This !LAW SHOOKE' iswhich
law, 2 sNewton' From
2
2nd
‘ 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’
). ofpower other somean (rather th toalproportion is , force
the wheresystem a describes OSCILLATOR HARMONIC LINEARA
xxxF
REGAN PHY34210
166Energy in Simple Harmonic Motion
2222
222
22222
2222
222
2
1sincos
2
1
1,sincos since then , SHM,for that recalling
sin2
1cos
2
1
as defined is E, ENERGY, MECHANICAL The
sin2
1
2
1
bygiven is system theof ENERGY KINETIC The
cos2
1
2
1
by given is oscillatorlinear a of ENERGY POTENTIAL The
mm
mm
m
m
kxttkxKUE
mωk
txmtxkKUE
txmmvtK
txkkxtU
Therefore, the mechanical energy of a linear oscillator is constant and time independent.
REGAN PHY34210
167Angular Simple Harmonic MotionAn ANGULAR SIMPLE HARMONIC PENDULUM (also known as a TORSION PENDULUM) is an angularversion of the linear simple harmonic oscillator. The disk oscillates in the horizontal plane, with the reference lineoscillating 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 torsionwire also gives rise to the RESTORING TORQUE, .
-m
+m
referenceline, =0
torsion wire
fixed end
κ
IπT
θ-τ
2by given is oscillator harmonic simpleangular an of
PERIOD thecase, oscillator harmonic simple th theanalogy wiBy
CONSTANT. TORSION thecalled is where, by given
TORQUE RESTORING a causes 0)(at position rest its from
anglean gh disk throu therotating motion, harmonic simpleangular For
REGAN PHY34210
168Simple Pendulums
l
m
l
Fg
T
s=r
Fgsin Fgcos
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.
)dependence (no 2 , 2
sin
radians,in nt,displacemeangular smallFor
. nt,displacemeangular at pendulum theof
on acceleratiangular inertia, of mom.
sin law,2 sNewton' From
. nt,displacemeangular theREDUCE to
acts , torque that theindicatessign The
sinsin
22
nd
mg
lTmlI
lmg
IT
I
lmg
I
lmglmgI
I
lmgI
lmgFlrFFrτ g
simple pendulums have SHM ONLY for small values of .
REGAN PHY34210
169Real (‘Physical’) PendulumsReal 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
Fg
s=r
Fgsin Fgcos
cog
O
3
8
3
22
32 2 since
3
1 rod, theof end oneat pivot afor that gives
THEOREM AXIS PARALLEL The . is mass) (of centre ethrough th
,length of rod uniform aFor case.each for differs pendulums realFor
). pendulum, simple (For the . 2by given is period the
,amplitudes smallfor pendulum, physical aFor
2
2
2
2
22
22
121
0
2121
2
T
lπg
g
l
mg
ml
mgh
IT
mlmmlI
mlI
lI
mlImgh
IT
l
l
i.e. can measure g directly usingFocault’s pendulum
REGAN PHY34210
170Damped Simple Harmonic MotionIf the motion of an oscillator is reduced as a result of an external force, the oscillator and its motion are described as DAMPED.
.const oscillator damped afor amplitude The
. then ,i.e., small, isconstant damping theIf
SHM.in as damping) (no 0For .oscillator damped for thefrequency
angular theis 2
where'cosby given is
equation aldifferentiorder 2 for thissolution The . 0
law,2 sNewton' From
motion. theopposes force that thisindicatessign
minus The CONSTANT. DAMPING a is where,
, then system, goscillatin theof theof
velocity the toalproportion is force damping theIf
22212
22
nd2
2
nd
mbtmmec
mbtm
mbtm
net
d
exktEex
ω'km b
m
kω'b
m
b
m
kω'textx
dt
dxbkx
dt
xdm
bvkxmaF
b
bvF
m
watertank
spring,constant, kFk=-kx
Fb=-bv
REGAN PHY34210
171Forced Oscillations and ResonancesIf a body oscillates without an external force on the body, the bodyis 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 forcedoscillations, 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).
REGAN PHY34210
17217: Waves - Part 1Waves describe situations where the energy of the system is spreadout over the space through which it passes. This is in contrast to particles which imply a tiny concentration of matter which is capableof 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). Examplesof 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.
REGAN PHY34210
173Transverse and Longitudinal Waves
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 thewave 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 anoscillating string is perpendicular to the direction of travel of the wave,the wave said to be TRANSVERSE (i.e. a transverse wave, such aswaves on a string.)
By contrast, if the displacement is parallel to the direction of motionof the wave (as in sound waves), the motion is described at LONGITUDINAL (i.e., transmitted via a longitudinal wave such as sound).
REGAN PHY34210
174Wavelength and FrequencyTo 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.
tkxytxy
xt
y
m sin),( is string thealong
position at element an for , timeoffunction
a as ,nt displaceme the wave,sinusoidal aFor
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. HRW p374
REGAN PHY34210
175Wavelength and Angular Wave Number The wavelength , of a wave is the distance
(parallel to the waves direction of travel) between repetitions of the shape of the wave.
2 R, WAVENUMBEANGULAR The
2 if thusradians,
2every itself repeats sinfunction theSince
sinsin0,0,
thush, wavelengtsingle a of endsboth at
same theisy nt,displaceme the,definitionBy
sinsin0,
as 0for tseen becan The
k
kkkxkx
xkykxyxyxy
kxytkxytxy
he waveshape of t
mm
mm
HRW p374
REGAN PHY34210
176
Period, Angular Frequency and FrequencyWe can monitor the time dependence of the displacement of a fixed
position on a vibrating string. This can be done by taking x=0.
tyty
tytyty
x
m
mm
sin,0 sinsin
sin0sin,0
then is timeoffunction a as 0
at string theofnt displaceme The
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,
π
ω
Tf
TπωT
TtyTtyty mmm
2
1 as defined isfrequency The
2 Frequency,Angular 2 if truebeonly can This
sinsinsin 111
HRW p375
REGAN PHY34210
177Speed of a Travelling Wave
fT
vT
πωk
kv
dt
dx ω-
dt
dxk
constωtkx
tkxyy
y
dt
dx
Δt
Δxv
x
T
m
2
2
2
and 2
callingRe
0 gives timeto
respect with atingdifferenti . phase, The
.sin sinceconstant remain must waveform
theof phase then the,nt displaceme same thehaving
by defined is which waveformmoving on thepoint fixed
a take weIf . of ratio theas SPEED WAVE
thedefinecan wedirection,- in the travels wave theIf
REGAN PHY34210
178Wave Speed on a Stretched StringFor 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.
vR
vlμma
R
lmaF
a
lμΔm
ττθθτF
F
θR
Δl
Rv
RΔl
2
toleads
. thuscircle, ain moveselement string The
string. theofdensity linear theis where,
element theof mass The .2sin2 is, force thisof magnitude
of force restoring theproviding sum, components verticalThe cancel.
components horizontal theends, two thely totangential
pulls string in the tension the toequal magnitude
a with force a If .2 anglean subtending radius
of arccircular a forms ,length ofelement string smallA
2
O
R
l
v
The speed of a wave on an ideal stretched string only dependson the string’s tension and linear density.
F
REGAN PHY34210
179Energy and Power of a Travelling Wave
2221
21222
4122
21
22
2122
21
22
21
221
2 i.e., waveby the nsmittedenergy tra mechanical
total which therate average theis POWER AVERAGE theequal, are
energies potential and kinetic average thesystem, goscillatinan for Recalling
cos of average cos
bygiven is nsportedenergy tra kineticat which rate AVERAGE the
coscos
thusiselement string a ofenergy kinetic heat which t rate The
. cosby given isenergy
kinetic the, density,linear the Using.cos
element. string goscillatin theof SPEED TRANSVERSE theis
where is mass ofelement string a ofenergy kinetic The
mave
mavem
mm
m
m
dtdy
yvdt
dKP
Xyvtkxyμvdt
dK
tkxyμvtkxydt
dxμ.
dt
dK
tkxyμ.dxdK
μ.dx dmtkxyu
u
uudmdKdm
REGAN PHY34210
180The Principle of SuperpositionThe principle of superposition states that when several effects occur simultaneously, their net effect is the sum of the individual effects. Mathematically, this means, txytxytxy ,,,' 11 Overlapping waves add algebraically to produce a RESULTANT or NET WAVE.Note however, that overlapping 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 thedisplacement observed for individual waves. By contrast, if they are completely out of phase (peaks of one wave matched by troughs of theother), they will completely cancel out resulting in a ‘flat’ string.
HRW p383
REGAN PHY34210
181 tkxytxytkxytxy mm sin, and sin, If 21
These waves have the same frequency determined by , wavelength from and amplitude ym . They differ only by the phase constant . From the principle of superposition,
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.
HRW p384
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REGAN PHY34210
182PhasorsWaves 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 originof 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, its projection, y1 onto the vertical axisvaries sinusoidally between +ym and -ym.2 waves which travel along the same string in the same direction can be added using a PHASOR DIAGRAM.
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REGAN PHY34210
183Standing WavesIf 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 patternsdo not move in the x-direction (i.e. they are stationary left to right).
If two sinusoidal waves of the same amplitude and wavelength travel inopposite directions along a stretched string, their intereference with each other produces a standing wave.
HRW
REGAN PHY34210
184Analysis of Standing Waves
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REGAN PHY34210
185Standing Waves and ResonanceA 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) andreflected 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
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At certain frequencies, such behaviour results in STANDING WAVEPATTERNS 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.)
REGAN PHY34210
187Example 1: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 ?
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REGAN PHY34210
188Example 2: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 constantsare 0 and /3 respectively. What are the amplitude, y’ and phase constant of the resulting wave ?
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REGAN PHY34210
18918: Waves - Part 2
Sound Waves: can be generally definedas longitudinal waves whose oscillationsare parallel to the direction of travel through some medium (such as air).
If a point source, P, emits soundwaves, wavefronts and rays describe thedirection of travel of the waves. Wavefronts correspondto 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, wavefrontsspread 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.
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REGAN PHY34210
190Speed of SoundThe speed of any mechanical wave depends on the physical propertiesof the medium through which it travels. As a sound wave passes through air, we can associate a potential energywith 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).
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REGAN PHY34210
191The Doppler EffectThe 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.
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REGAN PHY34210
192Example 1A rocket moves at a speed of 242 m/s through stationary air directly towards a stationary pole while emitting sound waves at a sourcefrequency of f =1250Hz.
(a) What frequency is measured by a detector attached to the pole ?
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