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PHY 113 A General Physics I 11 AM-12:15 PM TR Olin 101 Plan for Lecture 4: Chapter 4 – Motion in two dimensions Position, velocity, and acceleration in two dimensions Two dimensional motion with constant acceleration; parabolic trajectories Circular motion. Updated schedule.
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PHY 113 A General Physics I • 11 AM-12:15 PM TR Olin 101 • Plan for Lecture 4: • Chapter 4 – Motion in two dimensions • Position, velocity, and acceleration in two dimensions • Two dimensional motion with constant acceleration; parabolic trajectories • Circular motion PHY 113 A Fall 2013 -- Lecture 4
Updated schedule 4.1,4.12,4.35,4.60 PHY 113 A Fall 2013 -- Lecture 4
Tentative exam dates: • September 26, 2013 – covering Chap. 1-8 • October 31, 2013 – covering Chap. 9-13, 15-17 • November 26, 2013 – covering Chap. 14, 19-22 • Final exam: December 12, 2013 at 9 AM PHY 113 A Fall 2013 -- Lecture 4
From email questions on Webassign #3 Problem 4 (3.24) Note: In this case the angle f is actually measured as north of east. f d1 b q1 f=b+q1 PHY 113 A Fall 2013 -- Lecture 4
From email questions on Webassign #3 -- continued Problem 1 (3.11) graphically. A q 2B • (d) Find A− 2B PHY 113 A Fall 2013 -- Lecture 4
From email questions on Webassign #3 -- continued Problem 1 (3.29) F1 F2 PHY 113 A Fall 2013 -- Lecture 4
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? • Because it is difficult to draw 3 dimensions. • Because in physics class, 2 dimensions are hard enough to understand. • Because if we understand 2 dimensions, extension of the ideas to 3 dimensions is trivial. • On Thursdays, it is good to stick to a plane. PHY 113 A Fall 2013 -- Lecture 4
j vertical direction (up) horizontal direction i PHY 113 A Fall 2013 -- Lecture 4
Vectors relevant to motion in two dimenstions Displacement: r(t) = x(t) i + y(t) j Velocity: v(t) = vx(t) i + vy(t) j Acceleration: a(t) = ax(t) i + ay(t) j PHY 113 A Fall 2013 -- Lecture 4
Visualization of the position vector r(t) of a particle r(t2) r(t1) PHY 113 A Fall 2013 -- Lecture 4
Visualization of the velocity vector v(t) of a particle v(t) r(t2) r(t1) PHY 113 A Fall 2013 -- Lecture 4
Visualization of the acceleration vector a(t) of a particle v(t2) v(t1) r(t2) r(t1) a(t1) PHY 113 A Fall 2013 -- Lecture 4
Figure from your text: PHY 113 A Fall 2013 -- Lecture 4
Visualization of parabolic trajectory from textbook PHY 113 A Fall 2013 -- Lecture 4
The trajectory graphs of the motion of an object y versus x as a function of time, look somewhat like a parabola: • iclicker exercise • This is surprising • This is expected • No opinion PHY 113 A Fall 2013 -- Lecture 4
j vertical direction (up) Projectile motion (near earth’s surface) g = 9.8 m/s2 horizontal direction i PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) Trajectory equation in vector form: Trajectory equation in component 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. PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) Trajectory equation in component form: Trajectory path y(x); eliminating tfrom the equations: PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) Summary of results • iclicker exercise: • These equations are so beautiful that • They should be framed and put on the wall. • They should be used to perfect my tennis/golf/basketball/soccer technique. • They are not that beautiful. PHY 113 A Fall 2013 -- Lecture 4
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. PHY 113 A Fall 2013 -- Lecture 4
h=7.1m qi=53o d=24m=x(2.2s) PHY 113 A Fall 2013 -- Lecture 4
Problem solving steps • Visualize problem – labeling variables • Determine which basic physical principle(s) apply • Write down the appropriate equations using the variables defined in step 1. • Check whether you have the correct amount of information to solve the problem (same number of known relationships and unknowns). • Solve the equations. • Check whether your answer makes sense (units, order of magnitude, etc.). h=7.1m qi=53o d=24m=x(2.2s) PHY 113 A Fall 2013 -- Lecture 4
h=7.1m qi=53o d=24m=x(2.2s) PHY 113 A Fall 2013 -- Lecture 4
Review Position x(t), Velocity y(t), Acceleration a(t) in one dimension: Special case of constant acceleration a(t)=a0: PHY 113 A Fall 2013 -- Lecture 4
Review – continued: Special case of constant acceleration a(t)=a0: initial position initial velocity PHY 113 A Fall 2013 -- Lecture 4
Summary of equations – one-dimensional motion with constant acceleration • iclicker question: • Why did I show you part of the derivation of the last equation? • Because professors like to torture physics students • Because you will need to be able to prove the equation yourself • Because the “proof” helps you to understand the meaning of the equation • All of the above • None of the above PHY 113 A Fall 2013 -- Lecture 4
Review: Motion in two dimensions: j vertical direction (up) horizontal direction i PHY 113 A Fall 2013 -- Lecture 4
Vectors relevant to motion in two dimenstions Displacement: r(t) = x(t) i + y(t) j Velocity: v(t) = vx(t) i + vy(t) j Acceleration: a(t) = ax(t) i + ay(t) j PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (constant acceleration) (reasonable approximation near Earth’s surface) j vertical direction (up) horizontal direction i PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) PHY 113 A Fall 2013 -- Lecture 4
Projectile motion (near earth’s surface) Summary of component functions PHY 113 A Fall 2013 -- Lecture 4
Uniform circular motion – another example of motion in two-dimensions animation from http://mathworld.wolfram.com/UniformCircularMotion.html PHY 113 A Fall 2013 -- Lecture 4
iclicker question: • Assuming that the blue particle is moving at constant speed around the circle what can you say about its acceleration? • There is no acceleration • There is acceleration tangent to the circle • There is acceleration in the radial direction of the circle • There is not enough information to conclude that there is acceleration or not PHY 113 A Fall 2013 -- Lecture 4
Uniform circular motion – continued PHY 113 A Fall 2013 -- Lecture 4
Uniform circular motion – continued r In terms of time period T for one cycle: In terms of the frequency f of complete cycles: PHY 113 A Fall 2013 -- Lecture 4
iclicker exercise: During circular motion, what happens when the speed of the object changes? There is only tangential acceleration There is only radial acceleration There are both radial and tangential acceleration I don’t need to know this because it never happens. PHY 113 A Fall 2013 -- Lecture 4
Circular motion PHY 113 A Fall 2013 -- Lecture 4