Physics 111: Lecture 1, Pg 1
Physics 111: Lecture 1Physics 111: Lecture 1“Mechanics for Physicists and Engineers”“Mechanics for Physicists and Engineers”
Agenda for TodayAgenda for Today
AdviceAdvice Scope of this courseScope of this course Measurement and UnitsMeasurement and Units
Fundamental unitsSystems of unitsConverting between systems of unitsDimensional Analysis
1-D Kinematics (review)1-D Kinematics (review)Average & instantaneous velocity and accelerationMotion with constant acceleration
Physics 111: Lecture 1, Pg 2
Course Info & AdviceCourse Info & Advice
See info on the World Wide Web (heavily used in Physics 111)Go to http://www.physics.uiuc.edu and follow “courses” link to the
Physics 111 homepage
Course has several components:Lecture: (me talking, demos and Active learning)Discussion sections (group problem solving)Homework sets, Web basedLabs: (group exploration of physical phenomena)If you miss a lab or discussion you should always try to make it up as
soon as possible in another section!!
The first few weeks of the course should be review, hence the pace is fast. It is important for you to keep up!
Physics 111: Lecture 1, Pg 3
Lecture OrganizationLecture Organization Three main components:Three main components:
Lecturer discusses class material
» Follows lecture notes very closely
Lecturer does as many demos as possible
» If you see it, you gotta believe it!
» Look for the symbol
Students work in groups on conceptual“Active Learning” problems
» Usually three per lecture
Physics 111: Lecture 1, Pg 4
Scope of Physics 111Scope of Physics 111
Classical Mechanics:Classical Mechanics:
Mechanics: Mechanics: How and why things workClassical: Classical:
» Not too fast (v << c)
» Not too small (d >> atom)
Most everyday situations can be described in these terms.Most everyday situations can be described in these terms.Path of baseballOrbit of planetsetc...
Physics 111: Lecture 1, Pg 5
How we measure things! All things in classical mechanics can be expressed in terms
of the fundamental units:fundamental units:
Length LMass MTime T
For example:Speed has units of L / T (i.e. miles per hour).Force has units of ML / T2 etc... (as you will learn).
UnitsUnits
Physics 111: Lecture 1, Pg 6
Length:Length:
DistanceDistance Length (m)Length (m)
Radius of visible universe 1 x 1026
To Andromeda Galaxy 2 x 1022
To nearest star 4 x 1016
Earth to Sun 1.5 x 1011
Radius of Earth 6.4 x 106
Sears Tower 4.5 x 102
Football field 1.0 x 102
Tall person 2 x 100
Thickness of paper 1 x 10-4
Wavelength of blue light 4 x 10-7
Diameter of hydrogen atom 1 x 10-10
Diameter of proton 1 x 10-15
Physics 111: Lecture 1, Pg 7
Time:Time:
IntervalInterval Time (s)Time (s)
Age of universe 5 x 1017
Age of Grand Canyon 3 x 1014
32 years 1 x 109
One year 3.2 x 107
One hour 3.6 x 103
Light travel from Earth to Moon 1.3 x 100
One cycle of guitar A string 2 x 10-3
One cycle of FM radio wave 6 x 10-8
Lifetime of neutral pi meson 1 x 10-16
Lifetime of top quark 4 x 10-25
Physics 111: Lecture 1, Pg 8
Mass:Mass:
ObjectObject Mass (kg)Mass (kg)
Milky Way Galaxy 4 x 1041
Sun 2 x 1030
Earth 6 x 1024
Boeing 747 4 x 105
Car 1 x 103
Student 7 x 101
Dust particle 1 x 10-9
Top quark 3 x 10-25
Proton 2 x 10-27
Electron 9 x 10-31
Neutrino 1 x 10-38
Physics 111: Lecture 1, Pg 9
Units...Units...
SI (Système International) Units:SI (Système International) Units:mks: L = meters (m), M = kilograms (kg), T = seconds (s)cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)
British Units:British Units:Inches, feet, miles, pounds, slugs...
We will use mostly SI units, but you may run across some problems using British units. You should know how to convert back & forth.
Physics 111: Lecture 1, Pg 10
Converting between different systems of unitsConverting between different systems of units
Useful Conversion factors:1 inch = 2.54 cm1 m = 3.28 ft1 mile = 5280 ft 1 mile = 1.61 km
Example: convert miles per hour to meters per second:
s
m4470
s
hr
3600
1
ft
m
283
1
mi
ft5280
hr
mi1
hr
mi 1 .
.
Physics 111: Lecture 1, Pg 11
This is a very important tool to check your workIt’s also very easy!
Example:Example:
Doing a problem you get the answer distance
d = vt 2 (velocity x time2)
Units on left side = L
Units on right side = L / T x T2 = L x T
Left units and right units don’t match, so answer must be Left units and right units don’t match, so answer must be wrong!!wrong!!
Dimensional Analysis Dimensional Analysis
Physics 111: Lecture 1, Pg 12
Lecture 1, Lecture 1, Act 1Act 1Dimensional AnalysisDimensional Analysis
The periodThe period PP of a swinging pendulum depends only on of a swinging pendulum depends only on the length of the pendulumthe length of the pendulum dd and the acceleration of and the acceleration of gravitygravity gg..Which of the following formulas forWhich of the following formulas for PP couldcould be be
correct ?correct ?
Pdg
2Pdg
2(a)(a) (b)(b) (c)(c)
Given: d has units of length (L) and g has units of (L / T 2).
P = 2 (dg)2
Physics 111: Lecture 1, Pg 13
Lecture 1, Lecture 1, Act 1Act 1 SolutionSolution
Realize that the left hand side P has units of time (TT ) Try the first equation
P dg2 2(a)(a) (b)(b) (c)(c)
(a)(a) LL
T
L
TT
2
2 4
4 Not Right !!Not Right !!
Pdg
2Pdg
2
Physics 111: Lecture 1, Pg 14
LL
T
T T
2
2
P dg2 2(a)(a) (b)(b) (c)(c)
(b)(b) Not Right !!Not Right !!
Try the second equation
Lecture 1, Lecture 1, Act 1Act 1 SolutionSolution
Pdg
2Pdg
2
Physics 111: Lecture 1, Pg 15
TT
TLL 2
2
P dg2 2(a)(a) (b)(b) (c)(c)
(c)(c) This has the correct units!!This has the correct units!!
This must be the answer!!This must be the answer!!
Try the third equation
Lecture 1, Lecture 1, Act 1Act 1 SolutionSolution
Pdg
2Pdg
2
Physics 111: Lecture 1, Pg 16
Motion in 1 dimensionMotion in 1 dimension In 1-D, we usually write position as x(t1 ).
Since it’s in 1-D, all we need to indicate direction is + or .
Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1
t
x
t1 t2
x
t
x1
x2some particle’s trajectory
in 1-D
Physics 111: Lecture 1, Pg 17
1-D kinematics1-D kinematics
tx
tt)t(x)t(x
v12
12av
t
x
t1 t2
x
x1
x2trajectory
Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:
t
Vav = slope of line connecting x1 and x2.
Physics 111: Lecture 1, Pg 18
Consider limit t1 t2
Instantaneous velocity v is defined as:
1-D kinematics...1-D kinematics...
dt)t(dx
)t(v
t
x
t1 t2
x
x1
x2
t
so v(t2) = slope of line tangent to path at t2.
Physics 111: Lecture 1, Pg 19
1-D kinematics...1-D kinematics...
tv
tt)t(v)t(v
a12
12av
Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:
And instantaneous acceleration a is defined as:
2
2
dt)t(xd
dt)t(dv
)t(a
dt)t(dx
)t(v using
Physics 111: Lecture 1, Pg 20
RecapRecap
If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!
adv
dt
d x
dt
2
2
vdx
dt
x x t ( )
x
a
vt
t
t
Physics 111: Lecture 1, Pg 21
More 1-D kinematicsMore 1-D kinematics
We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
2
1
t
t12 dttvtxtx )()()(
Graphically, this is adding up lots of small rectangles:
v(t)
t
+ +...+
= displacement
Physics 111: Lecture 1, Pg 22
High-school calculus:
Also recall that
Since a is constant, we can integrate this using the above rule to find:
Similarly, since we can integrate again to get:
1-D Motion with constant acceleration1-D Motion with constant acceleration
constt1n
1dtt 1nn
adv
dt
vdx
dt
0vatdtadtav
002
0 xtvat21
dt)vat(dtvx
Physics 111: Lecture 1, Pg 23
RecapRecap So for constant acceleration we find:
atvv 0
200 at
2
1tvxx
a const
x
a
v t
t
t
Planew/ lights
Physics 111: Lecture 1, Pg 24
Lecture 1, Lecture 1, Act 2Act 2Motion in One DimensionMotion in One Dimension
When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the at the highest point in its path?highest point in its path?
(a)(a) BothBoth v = 0v = 0 andand a = 0a = 0..
(b)(b) v v 0 0, but , but a = 0a = 0..
(c) (c) v = 0v = 0, but , but a a 0 0..
y
Physics 111: Lecture 1, Pg 25
Lecture 1, Lecture 1, Act 2Act 2Solution Solution
x
a
vt
t
t
Going up the ball has positive velocity, while coming down Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is it has negative velocity. At the top the velocity is momentarily zero.momentarily zero.
Since the velocity is Since the velocity is
continually changing there mustcontinually changing there must
be some acceleration.be some acceleration. In fact the acceleration is caused In fact the acceleration is caused
by gravity ( by gravity (g = 9.81 g = 9.81 m/sm/s22).). (more on gravity in a few lectures)(more on gravity in a few lectures)
The answer is (c) The answer is (c) v = 0v = 0, but , but a a 0 0. .
Physics 111: Lecture 1, Pg 26
Derivation:Derivation:
Plugging in for t:
atvv 0 200 at
21
tvxx
Solving for t:
avv
t 0
200
00 avv
a21
avv
vxx
)xx(a2vv 02
02
Physics 111: Lecture 1, Pg 27
Average VelocityAverage Velocity
Remember that atvv 0
v
t
t
v
vav
v0
vv2
1v 0av
Physics 111: Lecture 1, Pg 28
Recap:Recap: For constant acceleration:
atvv 0
200 at
2
1tvxx
a const
From which we know:
v)(v21
v
)x2a(xvv
0av
02
02
Washers
Physics 111: Lecture 1, Pg 29
Problem 1Problem 1
A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab
x = 0, t = 0ab
vo
Physics 111: Lecture 1, Pg 30
Problem 1...Problem 1...
A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel?
x = xf , t = tf
v = 0
x = 0, t = 0ab
v0
Physics 111: Lecture 1, Pg 31
Problem 1...Problem 1...
Above, we derived: v = v0 + at
Realize that a = -ab
Also realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or
tf = v0 /ab
Physics 111: Lecture 1, Pg 32
Problem 1...Problem 1...
To find stopping distance we use:
In this case v = vf = 0, x0 = 0 and x = xf
fb2
0 x)a(2v
b
20
f a2v
x
)x2a(xvv 02
02
Physics 111: Lecture 1, Pg 33
Problem 1...Problem 1...
So we found that
Suppose that vo = 65 mi/hr = 29 m/s Suppose also that ab = g = 9.81 m/s2
Find that tf = 3 s and xf = 43 m
b
20
fb
0f a
v
2
1x ,
a
vt
Physics 111: Lecture 1, Pg 34
Tips:Tips:
Read !Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem.
Watch your units !Always check the units of your answer, and carry the units
along with your numbers during the calculation.
Understand the limits !Many equations we use are special cases of more general
laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).
Physics 111: Lecture 1, Pg 35
Recap of today’s lectureRecap of today’s lecture Scope of this courseScope of this course Measurement and Units Measurement and Units (Chapter 1)(Chapter 1)
Systems of units (Text: 1-1)Converting between systems of units (Text: 1-2)Dimensional Analysis (Text: 1-3)
1-D Kinematics 1-D Kinematics (Chapter 2)Average & instantaneous velocity
and acceleration (Text: 2-1, 2-2)Motion with constant acceleration (Text: 2-3)
Example car problemExample car problem (Ex. 2-7)(Ex. 2-7)
Look at Text problems Chapter 2: # 6, 12, 56, 119 Chapter 2: # 6, 12, 56, 119
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