Physics 121 for Majors 121M Tutors Tutorial Lab N-304 ESC Ethan Fletcher: M 1pm 3pm, T 3-6 pm, Th 3-10 pm, W 7-9pm, F 3pm 6-10 pm Spencer Vogel: M 1-4pm, W 1-5pm, F1-3 pm Schedule Do Post-Class Check #4 Do Pre-Class Quiz #5 HW #3 is due Friday Quiz #1 is due Saturday Lab #1 is due Monday Last Class Graphing position, velocity, acceleration Blinking light plots - relating the real world to graphs Derivative rules Integrals as anti-derivatives Differential Equations in Mathematica Key Idea! If you know any one of these x(t) v(t) a(t) you know them all because you can differentiate and integrate. Derivatives Understand the derivative as a slope or rate. dat n dt = Ant 1
Integrals Understand the integral as the antiderivative. At dt = A t n + 1 + C Not Crucial Differential equations Solving differential equations in Mathematica Spreadsheet problems Class 4 Integration Working with Acceleration Today s Class Integrals as cumulative sums and areas under curves Specific example of constant acceleration Free fall in one dimension How to solve physics problems: SCEVUR Two Velocities What is each expression? v = v = average velocity instantaneous velocity Section 1 More about Integration When is the average velocity the same as the instantaneous velocity? 2
Spreadsheet vs Integrals We found that given the acceleration, we could solve for v(t) and x(t) using either integrals, differential equations, or a spreadsheet. a t = dv v t = dx dt dt v = a t x = v t v t = a t dt x t = v t dt Convince yourself that these three forms are all equivalent! Spreadsheet When we use v = a t, we assume that a is constant in time. Otherwise, what value of a would we use? If a is not constant, we can add many little changes of velocity to get the final result: v v = v a t We call this a cumulative sum. Spreadsheet Let s see how cumulative sums works in a spreadsheet. Make spreadsheet. How are cumulative sums involved in the spreadsheet? How are cumulative sums related to integrals? Two Symbols Where do the symbols and come from? Integrals and Sums v v = v a t v v = dv = a(t) dt The last integral is an example of a definite integral. It means that t ranges from t initial to t final. We ll soon see what that means. 3
Area under a Curve Area under a ¼ Circle The equation of a ¼ circle in the first quadrant is y x = R x where R is the circle s radius. What is its area? Area under a ¼ Circle The equation of a ¼ circle in the first quadrant is y x = R x where R is the circle s radius. What is its area? We can approximate its area by the area of a number of rectangles. Area under a ¼ Circle A y x The expression is more accurate if we have more, thinner rectangles. Area under a ¼ Circle A = y x dx Area under a ¼ Circle A = y x dx It becomes perfect if we have an infinite number of infinitely small rectangles. The numbers on the integral sign mean that x goes from 0 to R. This kind of integral is called a definite integral. 4
Area under a ¼ Circle A = y x dx Indefinite Integrals Let s start with the usual integral, an indefinite integral : Important conclusion: Integrals can be graphically interpreted as areas under curves! A(x) = y x dx = That s not an integral we memorized, so let s put it in Mathematica: R x dx Definite Integrals A x = y x dx A = y x dx = A R A(0) The definite integral can be found by first finding the indefinite integral. Mathematica did not add a constant of integration, so we add that to the Mathematica result. Note that the definite integral is NOT a function of x! The indefinite integral is a function of x. Definite Integrals A x = 1 2 x R x + 1 2 R tan x A = y x dx R x = A R A 0 + C = 0 + 1 2 R tan R 0 + C 0 1 2 R tan 0 C A x = 1 2 x R x + 1 2 R tan x R x + C = 1 2 R π 2 0 = 1 4 πr Definite Integrals Important conclusion: To evaluate a definite integral, evaluate the indefinite integral at the upper limit and then subtract off the indefinite integral evaluated at the lower limit. Note that the constant of integration may be ignored. Section 2 Constant Acceleration 5
Equations of Motion 1 The equations of motion can be obtained by integration: v t = adt = at + v x t = v t dt = at + v dt = 1 2 at + v t + x Equations of Motion 2 A useful equation is obtained by eliminating t: v t = at + v t = v v a x t = 1 2 at + v t + x Here v 0 is the initial velocity and x 0 is the initial position. x t = 1 2 a v v a v v + v + x a A Little Algebra x = 1 2 a v v a x x = 1 2 v v a v v + v + x a + 2v v v 2a Know These!!! v t = at + v x t = 1 2 at + v t + x 2a x x = v 2vv + v + 2vv 2v = v v v v = 2a x x v v = 2a x x Sometimes the last one gets forgotten, but it will save you doing a lot of algebra if you memorize it! Section 3 Free Fall (1-D) Gravitational Acceleration Galileo s result for free-falling objects can be simply expressed as follows: In the absence of air all objects fall with the acceleration of g = 9.80 m/s 2 on the surface of the earth. (Up is positive and g is positive!) In practice g varies a little with location, and air can significantly affect free fall, especially of lighter objects. 6
Free Fall Positive y Up v t = gt + v y t = 1 2 gt + v t + y Free Fall Positive y Down Sometimes it s easier to take positive y to be downward. It s your choice! v t = gt + v v v = 2g y y y t = 1 2 gt + v t + y v v = 2g y y Section 4 Solving Physics Problems (SCEVUR) Math Is Only a Tool Never say, What equation do I use? as you begin to solve a problem! The most important things to learn are ideas and concepts. Equations are a good shorthand way to represent ideas. If you understand an idea, the equation comes easily. An equation without a concept is just frustration! Physics is Just Story Problems! Physics homework is expected to be centered in working story problems. They usually are oversimplified. Physics used to be called natural philosophy. The ideas and the logic are more important than solving problems. Still, most of our homework and tests are based on solving problems. Even educators usually think of the main purpose of physics as learning problem-solving skills. Problem-solving is hard to teach. The best we can do is give you some guidelines to make it easier. SCEVUR A proven approach Read the problem carefully. Think about what the problem is asking. Draw a Sketch. Most problems deal with interrelationships of Concepts. What are the fundamental ideas that are involved? Do you understand the definitions? Now you can ask what Equations relate the ideas! What are the Variables in the equation? Which are known and which are unknown? If you have N unknowns, you need N equations. Do you need more equations or a different equation? Keep track of Units! Does the answer seem Reasonable? 7
SCEVUR A proven approach Sketch Concepts Equations Variables Units Reasonable Up and Down Balls You throw two balls one up and one down with velocities of 3.60 m/s. You release each ball 2.00 m above the floor. What is the velocity of each ball as it hits the floor? How long does it take each ball to hit the floor? Up and Down Balls: S Sketch: Where is zero? Which way is positive? Up and Down Balls: C Up and Down Balls: E Concepts: Free fall with an initial velocity Signs of velocities How does each ball move? Equations: Velocity: v v = 2a x x Time: v t = at + v 8
Up and Down Balls: V Known: g, y, v Unkown: v, v, t, t Write equations in terms of these variables: Up and Down Balls: U Units: Use SI units m, m/s, m/s 2 Up: v v = 2g 0 y v = gt + v Down: v v = 2g 0 y v = gt v Up: Now Do It! v v = 2g 0 y v = gt + v Down: v v = 2g 0 y v = gt v v = v = v + 2gy = 7.22 m/s t = v v = 0.370 s g t = v v = 1.10 s g Up and Down Balls: R Reasonable: Do negative times make sense? What went wrong? Up: Now Do It Right! v v = 2g 0 y v = gt + v Down: v v = 2g 0 y v = gt v Section 5 Examples v = v = v + 2gy = 7.22 m/s t = v v = 1.10 s g t = v v = 0.370 s g 9
Colliding Colliding Two bumper cars are at rest 24.0 m away from each other. (That s the distance between front bumpers.) On a signal, the drivers depress their accelerators and move directly toward each other. One has an acceleration of 0.825 m/s 2 and the other has an acceleration of 0.667 m/s 2. Where do they hit? When do they hit? What are their velocities when they collide? Sketch rest 24.0 m accelerations= 0.825 m/s 2 and --0.667 m/s 2. (Why minus?) v 1 (0)=0 x 1 (0)=0 x 2 (0)=24.0 m v 2 (0)=0 a 1 = +0.825 m/s 2 a 2 = 0.667 m/s 2 Concepts Constant acceleration position, velocity, acceleration v 1 (0)=0 x 1 (0)=0 x 2 (0)=24.0 m v 2 (0)=0 a 1 = +0.825 m/s 2 a 2 = 0.667 m/s 2 x 1 = v 1 x 2 v2 Equations v 1 (0)=0 x 1 (0)=0 x 2 (0)=24.0 m v 2 (0)=0 a 1 = +0.825 m/s 2 a 2 = 0.667 m/s 2 Variables v 1 (0)=0 x 1 (0)=0 x 2 (0)=24.0 m v 2 (0)=0 a 1 = +0.825 m/s 2 a 2 = 0.667 m/s 2 For each car: x = x + v t + 1 2 at v = v + at v v = 2a x x x = 1 2 a t v = a t v = 2a x Unknowns: x, t, v, v x = d + 1 2 a t v = a t v = 2a x d 10
Units v 1 (0)=0 x 1 (0)=0 x 2 (0)=24.0 m v 2 (0)=0 a 1 = +0.825 m/s 2 a 2 = 0.667 m/s 2 We ll use SI units everywhere that s usually easiest. (kg, m, s) Solution x = 1 2 a t v = a t Unknowns: x, t, v, v x = 1 2 a t = d + 1 2 a t (a a )t = 2d x = d + 1 2 a t v = a t -- Need 4 equations t = 2d a a = 48.0 m = 5.67 s 1.49 m/s Mathematica Solution x = 1 2 a t v = a t x = d + 1 2 a t v = a t Falling (HW3) Falling When a real ball falls in the atmosphere, the acceleration of the ball isn t just g. The acceleration of a ping-pong ball can be written as: a = g + cv where c is a constant and v is the velocity. Falling Or we could also combine the two equations: d y = g + c dy dt dt We can rewrite this as: dv = g + cv v dt = dy dt 11
Falling Recall that y is the same as dy/dt. You can put this into Mathematica in either of the following ways: y t = g + c y t y 0 = 0 y 0 = 0 v t = g + cv v = y (t) y 0 = 0 v 0 = 0 Spreadsheet Another approach is to put this into a spreadsheet. We ll avoid the form with second derivatives. dv = g + cv v = dy dt dt Then we ll write these in terms of Δ s. v = g + cv t y = v t This is only accurate when t is small. Spreadsheet y y + v t v v + g + v t t t + t This means that if we know the old values of t, y, and v, we can find the new values a time Δt later. So if we know y and v at time 0, we can find the position and velocity of the object at all later times (because what was new is old after each step)! Section 6 Recap Big Ideas Integrals can be used to sum the contributions of many small things Integrals can be used to find areas under curves Free fall is the most important example of constant acceleration Know the constant acceleration equations thoroughly! 12