9/5/2013PHY 113 A Fall Lecture 41 PHY 113 A General Physics I 11 AM-12:15 PM TR Olin 101 Plan for Lecture 4: Chapter 4 – Motion in two dimensions 1.Position, velocity, and acceleration in two dimensions 2.Two dimensional motion with constant acceleration; parabolic trajectories 3.Circular motion

9/5/2013 PHY 113 A Fall Lecture 42 Updated schedule 4.1,4.12,4.35,4.60

9/5/2013PHY 113 A Fall Lecture 43 Tentative exam dates: 1.September 26, 2013 – covering Chap October 31, 2013 – covering Chap. 9-13, November 26, 2013 – covering Chap. 14, Final exam: December 12, 2013 at 9 AM

9/5/2013PHY 113 A Fall Lecture 44 From questions on Webassign #3 Problem 4 (3.24) Note: In this case the angle  is actually measured as north of east.  d1d1  11  

9/5/2013PHY 113 A Fall Lecture 45 From questions on Webassign #3 -- continued Problem 1 (3.11) magnitude m direction ° counterclockwise from the +x axis graphically. (d) Find A − 2B A 2B 

9/5/2013PHY 113 A Fall Lecture 46 From questions on Webassign #3 -- continued Problem 1 (3.29) F1F1 F2F2

9/5/2013PHY 113 A Fall Lecture 47 In the previous lecture, we introduced the abstract notion of a vector. In this lecture, we will use that notion to describe position, velocity, and acceleration vectors in two dimensions. iclicker exercise: Why spend time studying two dimensions when the world as we know it is three dimensions? A.Because it is difficult to draw 3 dimensions. B.Because in physics class, 2 dimensions are hard enough to understand. C.Because if we understand 2 dimensions, extension of the ideas to 3 dimensions is trivial. D.On Thursdays, it is good to stick to a plane.

9/5/2013PHY 113 A Fall Lecture 48 i j vertical direction (up) horizontal direction

9/5/2013PHY 113 A Fall Lecture 49 Vectors relevant to motion in two dimenstions Displacement: r(t) = x(t) i + y(t) j Velocity: v(t) = v x (t) i + v y (t) j Acceleration: a(t) = a x (t) i + a y (t) j

9/5/2013PHY 113 A Fall Lecture 410 Visualization of the position vector r(t) of a particle r(t 1 ) r(t 2 )

9/5/2013PHY 113 A Fall Lecture 411 Visualization of the velocity vector v(t) of a particle r(t 1 ) r(t 2 ) v(t)

9/5/2013PHY 113 A Fall Lecture 412 Visualization of the acceleration vector a(t) of a particle r(t 1 ) r(t 2 ) v(t 1 ) v(t 2 ) a(t 1 )

9/5/2013PHY 113 A Fall Lecture 413 Figure from your text:

9/5/2013PHY 113 A Fall Lecture 414 Visualization of parabolic trajectory from textbook

9/5/2013PHY 113 A Fall Lecture 415 The trajectory graphs of the motion of an object y versus x as a function of time, look somewhat like a parabola: iclicker exercise A.This is surprising B.This is expected C.No opinion

9/5/2013PHY 113 A Fall Lecture 416 Projectile motion (near earth’s surface) i j vertical direction (up) horizontal direction g = 9.8 m/s 2

9/5/2013PHY 113 A Fall Lecture 417 Projectile motion (near earth’s surface)

9/5/2013PHY 113 A Fall Lecture 418 Projectile motion (near earth’s surface)

9/5/2013PHY 113 A Fall Lecture 419 Projectile motion (near earth’s surface)

9/5/2013PHY 113 A Fall Lecture 420 Projectile motion (near earth’s surface) Trajectory equation in vector form: Aside: The equations for position and velocity written in this way are call “parametric” equations. They are related to each other through the time parameter. Trajectory equation in component form:

9/5/2013PHY 113 A Fall Lecture 421 Projectile motion (near earth’s surface) Trajectory path y(x); eliminating t from the equations: Trajectory equation in component form:

9/5/2013PHY 113 A Fall Lecture 422 Projectile motion (near earth’s surface) Summary of results iclicker exercise: These equations are so beautiful that A.They should be framed and put on the wall. B.They should be used to perfect my tennis/golf/basketball/soccer technique. C.They are not that beautiful.

9/5/2013PHY 113 A Fall Lecture 423 Diagram of various trajectories reaching the same height h=1 m: y x iclicker exercise: Which trajectory takes the longest time? A. Brown B. Green C. Same time for all.

9/5/2013PHY 113 A Fall Lecture 424 h=7.1m  i =53 o d=24m=x(2.2s)

9/5/2013PHY 113 A Fall Lecture 425 Problem solving steps 1.Visualize problem – labeling variables 2.Determine which basic physical principle(s) apply 3.Write down the appropriate equations using the variables defined in step 1. 4.Check whether you have the correct amount of information to solve the problem (same number of known relationships and unknowns). 5.Solve the equations. 6.Check whether your answer makes sense (units, order of magnitude, etc.). h=7.1m  i =53 o d=24m=x(2.2s)

9/5/2013PHY 113 A Fall Lecture 426 h=7.1m  i =53 o d=24m=x(2.2s)

9/5/2013PHY 113 A Fall Lecture 427 Review Position x(t), Velocity y(t), Acceleration a(t) in one dimension: Special case of constant acceleration a(t)=a 0 :

9/5/2013PHY 113 A Fall Lecture 428 Review – continued: Special case of constant acceleration a(t)=a 0 : initial velocity initial position

9/5/2013PHY 113 A Fall Lecture 429 Summary of equations – one-dimensional motion with constant acceleration iclicker question: Why did I show you part of the derivation of the last equation? A.Because professors like to torture physics students B.Because you will need to be able to prove the equation yourself C.Because the “proof” helps you to understand the meaning of the equation D.All of the above E.None of the above

9/5/2013PHY 113 A Fall Lecture 430 i j vertical direction (up) horizontal direction Review: Motion in two dimensions:

9/5/2013PHY 113 A Fall Lecture 431 Vectors relevant to motion in two dimenstions Displacement: r(t) = x(t) i + y(t) j Velocity: v(t) = v x (t) i + v y (t) j Acceleration: a(t) = a x (t) i + a y (t) j

9/5/2013PHY 113 A Fall Lecture 432 Projectile motion (constant acceleration) ( reasonable approximation near Earth’s surface ) i j vertical direction (up) horizontal direction

9/5/2013PHY 113 A Fall Lecture 433 Projectile motion (near earth’s surface)

9/5/2013PHY 113 A Fall Lecture 434 Projectile motion (near earth’s surface) Summary of component functions

9/5/2013PHY 113 A Fall Lecture 435 Uniform circular motion – another example of motion in two-dimensions animation from

9/5/2013PHY 113 A Fall Lecture 436 iclicker question: Assuming that the blue particle is moving at constant speed around the circle what can you say about its acceleration? A.There is no acceleration B.There is acceleration tangent to the circle C.There is acceleration in the radial direction of the circle D.There is not enough information to conclude that there is acceleration or not

9/5/2013PHY 113 A Fall Lecture 437 Uniform circular motion – continued

9/5/2013PHY 113 A Fall Lecture 438 Uniform circular motion – continued r In terms of time period T for one cycle: In terms of the frequency f of complete cycles:

9/5/2013PHY 113 A Fall Lecture 439

9/5/2013PHY 113 A Fall Lecture 440 iclicker exercise: During circular motion, what happens when the speed of the object changes? A.There is only tangential acceleration B.There is only radial acceleration C.There are both radial and tangential acceleration D.I don’t need to know this because it never happens.

9/5/2013PHY 113 A Fall Lecture 441 Circular motion