Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 2. Using Mathematical Tools in Physics The universe we live in is one of change and motion. Although we all have intuition about motion, based on our experiences, some of the important aspects of motion turn out to be rather subtle. Chapter Goals: To introduce the fundamental concepts of motion To represent motion in one dimension To define velocity and acceleration in mathematically meaningful ways, using calculus To solve and understand example problems

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Translational Motion Circular Motion Projectile Motion Rotational Motion

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Topics: The Particle Model Position and Time Motion Diagrams 3 Ways of Representing 1-dimensional (1-d) Motion Average and Instantaneous Velocity Units and Significant Figures Linear Acceleration The Case of Constant Acceleration Solving Problems in Physics

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. (Later in the course, we will consider “systems” of particles, and extended bodies) For now, our “particle” or “object” or “body” is represented by a moving single point in space Our questions are simple: Where is the body? When is it there? Where: position! (in meters, m) When: time! (in seconds, s)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. This is a motion diagram: Also called ‘stop-action’ footage

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Curvilinear motion can take place in 2 or 3 Dimensions, but for now we assume the motion is strictly along a straight line! Where is the body? When is it there? Where it is: position x, where x is a displacement from the space origin x = 0. [x] = length units: m, ft, cm, mm… When it’s there: time t, where t is a time interval from the origin of time t = 0. [t] = time units: s, hr, ms, ns…

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What kind of ‘representations’ of motion? Motion diagram: qualitative and pretty, like a movie of the motion Graphical Representation of x as a function of t: x(t): quantitative, visually informative to the trained eye, if labeled carefully Tabular Representation of t and x: include sufficient number of pairs of values to be informative Explicit functional form for x(t): nice for The theorists, and very handy for calculating numbers. Not always obtainable! {show active figure 0201}

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Note: this is a piecewise- Continuous function: OK!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

A tabular representation of this motion Position x [km]Clock timeTime t [min] :00 am0.0 10:10 am10.0 –10.010:20 am20.0 –20.010:30 am30.0 –20.010:40 am40.0 –15.010:50 am50.0 –10.011:00 am60.0

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An explicit function for x(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

The velocity concepts answer the questions… How fast does object move during a time interval? How fast does the object move at any time t? Crucial notion of the CHANGE in a quantity: The Delta operator  Time interval  t := t 2 – t 1 or  t := t later – t earlier or  t := t final – t initial or  t := t B – t A …. Time interval has the same dimensions as t. Often, t initial = 0 s and t final = t (‘present’ time) so if that is the case then  t = t Change is position is called displacement

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the average velocity is the slope of the line connecting A and B!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the (instantaneous) velocity Suppose we want to know the velocity v AT A Allow the time interval to shrink by letting t B  t A The displacement also shrinks, as x B := x(t B ) approaches x A :=x(t A )

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the (instantaneous) velocity In the limit of infinitesimal time interval  t and displacement  x, we have the derivative of x(t): [Show active figure 0203]

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Position–time graph for a particle having an x coordinate that varies with t (quadratically) as x = –4.0 t + 2.0t 2 Note that v(t) is a linear function if x(t) is quadratic!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Properties of position and the velocity functions v(t) is the slope of the tangent line to the graph of x(t) at time t x(t) must be a (piecewise) continuous function (no gaps or holes, passes vertical line test). Why? x(t) may not have any sharp corners, either. If it had them, what would that imply about v(t)? v(t) may be graphed as well. Can it have sharp corners? Must it be continuous? If there is an explicit function x(t), then all of the tricks from calculus may be used to get v(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Speed as distinct from velocity Average speed does not include information about direction of motion Distance is a loose concept: if you walk to the store which is 1km away ‘as the crow flies’, and then return home, your distance may be anywhere from 2 km on up (depending how you wander), but your displacement is ZERO!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Number Notation When writing out numbers with many digits, spacing in groups of three will be used Commas in US, periods or spaces elsewhere Standard international notation Section 1.1 Examples: (3 significant figures, probably) (7 significant figures, definitely) x 10 4 (5 significant figures, definitely)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. US Customary System Still used in the US, but text will use SI QuantityUS UnitSI Metric Unit LengthFoot (ft or ‘)Meter (m) MassslugKilogram (kg) TimeSecond (s) Section 1.1

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Metric Prefixes Prefixes correspond (usually) to 3 powers of 10 Each prefix has a specific name Each prefix has a specific abbreviation The prefixes can be used with any basic units They are multipliers of the basic unit Examples: –1 mm = m = m = one-thousandth m –1 ng = g = g Section 1.1

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Prefixes, cont. Section 1.1

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Fundamental and Derived Units Derived quantities can be expressed as a mathematical combination of fundamental quantities. Examples: –Area A product of two lengths e.g cm 2 –Velocity A ratio of a length to a time e.g mi/hr –Density A ratio of mass to volume e.g. kg/m 3 Section 1.1

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Three motion diagrams are shown. Which is a dust particle settling to the floor at constant speed, which is a ball dropped from the roof of a building, and which is a descending rocket slowing to make a soft landing on Mars? A. (a) is ball, (b) is dust, (c) is rocket B. (a) is ball, (b) is rocket, (c) is dust C. (a) is rocket, (b) is dust, (c) is ball D. (a) is rocket, (b) is ball, (c) is dust E. (a) is dust, (b) is ball, (c) is rocket

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank in order, from the most to the least, the number of significant figures in the following numbers. For example, if b has more than c, c has the same number as a, and a has more than d, you could give your answer as b > c = a > d. a b c d × 10 −10 A. a = b = d > c B. b = d > c > a C. d > c > b = a D. d > c > a > b E. b > a = c = d

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem A bicycle moves on a straight path so that its position as a function of time is x(t) = – 200 t+ 50 t 2, where all quantities are in m and s. a)What is the average velocity between t = 2 s and t = 4 s? b) What are the instantaneous velocities at those two times? c) Make graphs of x(t) and v(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The next question: how is the velocity changing? Velocity is ‘the rate of change with time’ of position So, acceleration is the rate of change with time of velocity Average acceleration (during a time interval) is defined as The dimensions of acceleration: Length per time, per time is the slope of the line on a graph of v(t) between two points

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

The (instantaneous) acceleration At any instant of time t, the acceleration is the slope of the tangent to the graph of v(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The full set of x(t), v(t) and a(t) graphs, obtained by eyeballing the slopes Note how a(t) may BE discontinuous!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How to ‘go the other way’ to get v and x from a? The interval dt is so short that a is essentially a constant, and so we know the (tiny) change dv We now ask about the overall change  v during a non-infinitesimal (‘finite’)time interval  t It must be the SUM of all the infinitesimal changes:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. For x, in exactly the same way,we can say that One says that x(t) is the ‘anti-derivative’ or ‘integral’ of v(t), and v(t) is the ‘anti-derivative’ or ‘integral’ of a(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Graphical interpretation of these ideas dx = v dt = the area of that skinny rectangle (width dt, height v(t)) under a graph of v(t) So  x is the area under v(t) during the interval  t And so  v is the area under a(t) during the interval  t

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration There are two ways to work out the set of equations that I call ‘the big five’ for this case. We assume initial time zero and present time t Initial position x 0 and present position x(t) :=x Initial velocity v 0 and present velocity v(t) :=v Acceleration does not change: call it simply a(t):=A therefore we have (duh!!) a = A (constant) (#1) This came about because of the LINEAR relation between v and t

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration From the definition of the acceleration we have Now insert #3 for v into #2 for x:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration Finally, it is nice to get an equation that does not involve t at all. To get it, solve #3 for t: Expanding the squared term and combining gives

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Extra-special case: zero acceleration This is synonymous with constant v = v 0 := V the big 5 collapse down into simplicity itself:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A different book’s version of these equations Note that v(t) is a linear function Note that x(t) is a quadratic function Note that v(x) is a square-root function

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: Two railroad engines on parallel tracks are both moving at 25 m/s. Engine A applies its brakes and comes to a stop in 100 seconds (assume constant acceleration), while engine B continues at constant velocity. a) What is the acceleration of engine b) How far do both engines travel during that time interval? c) Make three graphs of the three kinematical quantities, using color to distinguish the two engines’ graphs. d) Convert 25 m/s to mi/hr. {show active figures

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: the velocity function of an object is given by v(t) = b + c/t, where b = 5 m/s and c = 15 m. a)Find an expression for a(t) by taking a derivative. b)Find an expression for x(t) by taking an integral. Assume that x(0.5) = – 10 m. c) Graph all three functions, for the time domain 0.5 s< t < 5.0 s.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. v(t) = /t

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Chapter 2. Using Mathematical Tools in Physics Remaining Chapter 2 Goals: To learn about the important case of free fall in gravity To describe rotational motion about a fixed axis kinematically, introducing concepts of radian, angular position, angular velocity and angular acceleration

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Free fall in gravity near the Earth It is known (thanks to Galileo and his contemporaries) That near the Earth, IN THE ABSENCE OF ANY OTHER FORCES BUT GRAVITY, ALL OBJECTS ACCELERATE DOWNWARD AT THE SAME RATE: A = – 9.81 m/s 2 Our convention is to say that A = – g, where g = 9.81 m/s 2 Thus, our 1d coordinate system, called y now, has the positive y direction pointing UP

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens to the big five? Our convention is to say that A = – g, where g = 9.81 m/s 2 Thus, our 1d coordinate system, called y now, has the positive y direction pointing UP

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: An object is thrown directly upward from the ground at 15 m/s. It climbs vertically, stops for an instant, and returns to the ground. a)How high does it climb? How much time Is required to reach maximum height? b)What are the times for it to be at half of the maximum height? How fast is it moving when it is at that height?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. To find how high it climbs, note that v y = 0 at the top. Use #5 (and for simplicity take g = 10 m/s 2 ): To find the time needed, one could use the above result and plug into #4… but that requires one to solve a quadratic equation for t. Instead, just use #3: Note: #2 also can be used with the above result!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. To find how the time to be at half of the maximum height, (which is y = 6.12 m when rounded off), one can either first find how fast it is moving at that height (using #5), or resort to #4 and bite the quadratic bullet:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Slightly different language here: t i is the same as t 0 and t f is the same as plain vanilla t.