Physics 201 –College Physics
Akhdiyor Sattarov
Lectures MWF 1:50-2:40
E-mail a-sattarov@physics.tamu.edu
Phone: 458-7967 office / 845-6015 lab
Office Hours: ? or by appointment,
Office: MPHYS 303
Web-site: http://people.physics.tamu.edu/a-sattarov/
• Text: Physics 8th ed by Young & Geller with Mastering Physics; PHYS 202
Lab Manual
• Optional: Student Solutions Manual, Student Student Guide
•
• Grading: 4 exams 60%; Final (comprehensive) 20%; Lab10%; Recitation
5%; Homework (Mastering Phys) 5%
• You must achieve 70% or better in the laboratory in order to pass the
course.
• If your grade on the Final Exam is higher than your lowest grade on one of
the four exams during the semester, your grade on the Final will replace that
one lowest exam grade in computing the course grade.
• Sept. 3 is last day to drop with no record.
• Nov. 5 is the last day to Q-drop.
• Final Exam is December 14 2010 3:30pm-5:30pm
The Goal of Physics
• To provide understanding of the physical world by developing
theories based on experiments
• Physical theories model real world and are statement of what has
been observed and provide prediction of new observations that can
be confirmed or turned down by experiment
• Broad and solidly established by experiment physical theory –
physical law
Physical Law
• True, there is no contradiction to observation
• Universal – apply everywhere in the universe
• Simple
• Stable – unchanged since first discovered
How do we analyze physical system?
• Simplification of a complicated system – depending on our goals we
exclude/neglect unimportant/weak effects – isolate system of
interest.
Example: A small rock thrown upward – we can neglect the air
resistance
• Very often we use point particle – idealized object that lacks spatial
extensions
Example:
Earth orbiting around Sun, we can treat Earth as point particle
because the radius of Earth much smaller than the dimensions of
the system
Standards and Units
• Physical quantity - Quantity used to describe an observation of
physical phenomenon
• Each physical quantity is represented with respect to reference
standard – Unit of the quantity
• Some Physical quantities are defined only by describing a
procedure for measuring them – operational definition
• Some physical quantities can be derived from other quantities
• Most fundamental quantities:
International System
Length
meter=distance traveled in vacuum by light
during 1/299792458 of sec
Time
second-9,192,631700 times the period of oscillations of
radiation from the cesium atom.
Mass
kilogram=90% Platinum 10% iridium alloy cylinder
h=d=0.03917m
Since we will work with very small and very big systems we have
to have conversion multipliers
Power of 10
10-6
10-3
10-2
103
106
109
Prefix
micromillicentikilomegagiga-
Abbreviation
m
m
c
k
M
G
Using prefixes
• 1cm=1 centimeter=1x10-2m=0.01m – thickness of a notebook
• 1fm=1 femtometer=10-15m – radius of a nucleus
• 1ns=1 nanosecond=10-9s – time required for light to travel about 1ft.
• 1ms=1 millisecond=10-3s - time required for sound to travel about 1ft.
• 1kg=1 kilogram=103g
• 1Mg=1000kg=106g – mass of water that has volume of 1m3 at 4oC
Dimensional analysis
In Physics, the word dimension denotes the physical nature of a
quantity
Example:
distance between 2 points can be measured in meters, centimeters,
feet etc. – different ways of expressing the dimension of length
• it is often necessary to derive mathematical expression or equation
or check its correctness. A useful procedure for doing this is called
dimensional analysis.
• Dimensional analysis makes use of the fact that dimensions can be
treated as algebraic quantities
Example: Volume of a cube of water, L=2m
Example: Find mass of water, density of water 997 kg/m3
Example: Express speed of light 3x108 m/s in km/h
V  L  L  L  2m  2m  2m  2  2  2  m  m  m  8m3
kg
kg
3
m    V  997 3  8m  997  8  3  m 3  7976 kg
m
m
m
km
3600 s
8 m
9 km
c  3 10
 3 10


 1.08 10
s
s 1000 m
h
h
8
Always use units in calculations!
You can not add or subtract quantities that have
different units
3kg+15m means something is wrong
It will help you to check dimensional consistency of your
result!!
2
3 2
3= ******
2
3
Mass=r *V=(997kg/m ) 8m
kg /m - something is wrong
Precision and Significant figures
• No physical quantity can be determined with complete accuracy
• Knowing experimental uncertainties in any measurements is very
important
• Accuracy of measurements depends on the sensitivity of the
devices, the skill of the investigator…
• In many cases result from one measurement is used in derivation of
other physical quantities
We have to develop basic and reliable method of keeping track of those
uncertainties in subsequent calculations.
Example: Let we have to measure area of a rectangular plate with a
meter stick. Suppose that we can measure particular side with
0.1cm accuracy. Suppose that side a=16.3cm and side b=4.5cm.
• Side a=16.3+/-0.1 cm, side b=4.5+/-0.1cm
• Mid Area=16.3cm x 4.5cm = 73.35cm2
• High Area=16.4cm x 4.6cm = 75.44cm2
• Low Area=16.2cm x 4.4cm = 71.28cm2
so our area 73+/-2cm – note that the answer has two significant figures
In our example First term had 3 and the second 2 significant figures
In multiplying (dividing) two or more quantities, the number of
significant figures in the final result is no greater than the number of
significant figures in the term that had fewest significant figures.
In addition (subtraction) two or more quantities the final result can have
no more decimal places than the term with fewest decimal places
Example: 128.???+5.35?=133.???
Scalar and vector
• Scalar physical quantity is a quantity described by single
number, examples are time, mass, density, charge etc.
• Vector quantity is a quantity that is described by a
magnitude and a direction.
• Graphically vector is represented as an arrow pointing in
given direction and having length that is proportional to
the magnitude of the vector.
• Symbolically vector is represented by a label with small
arrow sign over it.
• Example: Position vector – shows a direction and how
far from the reference point the object resides.
• Example: Displacement vector– shows change in
position, from starting point to final
A
r
Magnitude of a vector

A A
Let have several vectors
We say that vectors are parallel
A
D
We say that A=D
B
We say that vectors are antiparallel
C
E
We say A=-E
Product of a scalar and a Vector
Resulting Vector is collinear
(parallel or anti parallel) to the
original vector.
Adding vectors: Tail to tip method
•Draw the vectors, with proper scaling
•Draw the second one putting its tail to the
tip of the first one
A+B=B+A
•Draw the resultant from the tail of the first
vector to the tip of the second
A
•One can change the order of the vectors
B
   
A B  B  A
Parallelogram method
2 vectors are along two
sides of a parallelogram
A
B
Resulting vector along the
diagonal of the parallelogram
that starts at the tails of the
vectors
Multiplying sum of two vectors by scalar
s.B
s.A
A+B
A
B
(A+B)*s
Using two similar triangles, we find
that the bigger triangle is just scaled
by s.
 


s  ( A  B)  s  A  s  B
Sum of 3 vectors
  
( A  B)  C
  
  
A  (B  C)  ( A  B)  C
A
C
 
BC
B
Subtraction of vectors
   
A B  B C
Direction vector

A
Aˆ  
A
 
A  A  Aˆ
Note:
It is in the direction of the vector A, but has a unit length and it is
dimensionless.
Components of vectors (2d)
•
•
•
•
Let we have some vector
We define some reference frame
A tail of the vector positioned at O.
  
Define two component vectors: A  A  A
x
y



Ax  Ax , 0 or A  cos , 0 A  cos 0



Ay  0, Ay or 0 , A  sin  A  sin  90
 



A  Ax  Ay  Ax , Ay or A  cos , A  sin 

A
A  Ax2  Ay2
tan 

A 
Let define angle between
vector and positive x-direction
y
Ax
Ay
Ay
Ay=IAI sin
Components are not vectors
They
cancan
be
and negative,
A
vector
bey-component
represented
in vector
2 waysA
Ay –positive
of
depending on an angle
a) by its components
b) By its magnitude and angle with
positive x-direction (in 3d case also
angle with positive z-direction)
Ax
Q
x
Ax=IAI cos
Ax – x-component of vector A
Example:
A person starts from point A and arrives at point B. Find
components, magnitude of the position vectors and angle
between the vector and x-axis. Find the displacement vector.
Ax=2km
Y (km)
 A  tan
1
Ay=3km
Ay
Ax
 tan1 1.5
A
Bx=-2km
By=-1km
 B  tan1
By
Bx
 tan1 0.5
A
QB
QA
B
B
X (km)
Pay attention!! You may get
this angle
Multiplication by a scalar and addition of
vectors becomes very simple

s A 
s  Ax , s  Ay
x-componet

s A 
 
A B 
y-componet

( s  Ax )  ( s  Ay )  s  ( Ax  Ay )  s  A
2
2
2
2
2
Ax  Bx , Ay  By
x-componet
y-componet
Example 1.45: Vector A has components Ax=1.3cm, Ay=2.25cm;
vector B has components Bx=4.1cm and By=-3.75cm. Find
the components of the vector sum A+B;
The magnitude and direction of A+B;
The components of the vector difference B-A;
The magnitude and direction of B-A;
Example1.50 A postal employee drives a delivery truck along the
route shown in figure below. Use components to determine the
magnitude and direction of the truck’s resultant displacement.
Then check the reasonableness of your answer by sketching a
graphical sum.
Kinematics – describes the motion of object without causes that
leaded to the motion
We are not interested in details of the object (it can be car, person,
box etc..). We treat it as dimensionless point
We want to describe position of the object with respect to time –
we want to know position at any given time
Path (trajectory) – imaginary line along
which the object moves
x
t
Motion along a straight line
• We will always try to set up our reference frame in a such
way that motion is along or “x” or “y” coordinate axis. The
direction of the axis is up to us.
• Position vector then can be represented by a single
component, the other components are equal to zero.
y
t=t1
t=t2
Very often I will write rn instead of
r(tn), the same for the components of
the vector, for example, yn instead of
t=t3y(t )
n
x
r(t1)
r(t2)
r(t3)
x1=19m,t1=1s
x2=277m,t1=4s

r (t1 )  ( x1 , y1 , z1 )

r (t2 )  ( x2 , y2 , z2 )
y1  z1  0
y2  z2  0
Motion along a straight line
 
• x component of a displacement vector d  r2  r1 – for time
interval t1 t2 is equal Dx=x2-x1
• Note that we can define another reference frame, position
vector will be different in each frame, not a displacement vector
  
d  r '2 r '1
y
x
r(t1)
r(t2)