Physics 111: Lecture 1 “Mechanics for Physicists and Engineers” Agenda for Today Advice Scope of this course Measurement and Units Fundamental units Systems of units Converting between systems of units Dimensional Analysis 1-D Kinematics (review) Average & instantaneous velocity and acceleration Motion with constant acceleration 1

Course 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 based Labs: (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! 2

Lecture Organization 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

Scope of Physics 111 Classical Mechanics: Mechanics: How and why things work Classical: Not too fast (v << c) Not too small (d >> atom) Most everyday situations can be described in these terms. Path of baseball Orbit of planets etc...

Units How we measure things! All things in classical mechanics can be expressed in terms of the fundamental units: Length L Mass M Time 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).

Length: Distance 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

Time: Interval 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

Mass: Object 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

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: 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. 9

Converting between different systems of units Useful Conversion factors: 1 inch = 2.54 cm 1 m = 3.28 ft 1 mile = 5280 ft 1 mile = 1.61 km Example: convert miles per hour to meters per second:

Dimensional Analysis This is a very important tool to check your work It’s also very easy! 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 wrong!!

Lecture 1, Act 1 Dimensional Analysis The period P of a swinging pendulum depends only on the length of the pendulum d and the acceleration of gravity g. Which of the following formulas for P could be correct ? (a) P = 2 (dg)2 (b) (c) Given: d has units of length (L) and g has units of (L / T 2).

Lecture 1, Act 1 Solution Realize that the left hand side P has units of time (T ) Try the first equation (a) Not Right !! (a) (b) (c)

Lecture 1, Act 1 Solution Try the second equation (b) Not Right !! (a)

Lecture 1, Act 1 Solution Try the third equation (c) This has the correct units!! This must be the answer!! (a) (b) (c)

Motion 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 x some particle’s trajectory in 1-D x2 x x1 t1 t2 t t

1-D kinematics Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is: x trajectory x2 x Vav = slope of line connecting x1 and x2. x1 t1 t2 t t

1-D kinematics... Consider limit t1 t2 Instantaneous velocity v is defined as: x so v(t2) = slope of line tangent to path at t2. x2 x x1 t1 t2 t t

1-D kinematics... 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: using

Recap 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! x t v t a t

More 1-D kinematics We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we can integrate to obtain: Graphically, this is adding up lots of small rectangles: v(t) + +...+ = displacement t

1-D Motion with constant acceleration 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:

Recap So for constant acceleration we find: x t v t a t Plane w/ lights x t v t a t

Lecture 1, Act 2 Motion in One Dimension When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v  0, but a = 0. (c) v = 0, but a  0. y

Lecture 1, Act 2 Solution Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. Since the velocity is continually changing there must be some acceleration. In fact the acceleration is caused by gravity (g = 9.81 m/s2). (more on gravity in a few lectures) The answer is (c) v = 0, but a  0. x t v t a t

Useful Formula Solving for t: Plugging in for t:

Alternate (Calculus-based) Derivation (chain rule) (a = constant)

Recap: For constant acceleration: Washers From which we know:

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 x = 0, t = 0 ab vo

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? v0 ab x = 0, t = 0 v = 0 v t x = xf , t = tf

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 v v0 -ab tf t

Problem 1... To find stopping distance we use: In this case v = vf = 0, x0 = 0 and x = xf v v0 -ab tf t

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 29 m/s v - 9.81 m/s2 43 m 3s t

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). 26

Recap of today’s lecture Scope of this course Measurement and Units (Chapter 1) Systems of units (Text: 1-1) Converting between systems of units (Text: 1-2) Dimensional Analysis (Text: 1-3) 1-D Kinematics (Chapter 2) Average & instantaneous velocity and acceleration (Text: 2-1, 2-2) Motion with constant acceleration (Text: 2-3) Example car problem (Ex. 2-7) Look at Text problems Chapter 2: # 49, 54, 71, 122 27