2. Physics 1501: Lecture 3, Pg 1 Physics 1501: Lecture 3 Today’s Agenda l Announcements : çLectures posted on: www.phys.uconn.edu/~rcote/www.phys.uconn.edu/~rcote/ çHomeworks 01: on MasteringPhysics Due next Monday at11:00AM Due next Monday at11:00AM l You need to register at MasteringPhysics.com l Labs START next week (Sept. 12)

3. Physics 1501: Lecture 3, Pg 2 Today’s Topic : l Finish Chapter 2 çReview of 1-D motion çFree Fall l Strategy for Problem Solving l Begin Chapter 3 çReview of vectors »Coordinate systems »Math with vectors »Unit vectors

4. Physics 1501: Lecture 3, Pg 3 Recap of 1-D motion l For constant acceleration: l From which we know:

5. Physics 1501: Lecture 3, Pg 4 Lecture 3, Act 1 Motion in One Dimension l 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

6. Physics 1501: Lecture 3, Pg 5 Lecture 3, Act 1 Solution x t v t l Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. l Since the velocity is continually changing there must be some acceleration. be some acceleration. çIn fact the acceleration is caused by gravity (g = 9.81 m/s 2 ). ç(more on gravity in a few lectures) The answer is (c) v = 0, but a  0. a t

7. Physics 1501: Lecture 3, Pg 6 Free Fall l When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity. l The acceleration caused by gravity is typically written as g l Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.

8. Physics 1501: Lecture 3, Pg 7 Gravity facts: l g does not depend on the nature of the material! çGalileo (1564-1642) figured this out without fancy clocks & rulers! l demo - feather & penny in vacuum l Nominally, g = 9.81 m/s 2 çAt the equatorg = 9.78 m/s 2 çAt the North poleg = 9.83 m/s 2 l More on gravity in a few lectures!

9. Physics 1501: Lecture 3, Pg 8 Problem : l On a bright sunny day you are walking around the campus watching one of the many construction sites. To lift a bunch of bricks from a central area, they have brought in a helicopter. As the pilot is leaving, she accidentally releases the bricks when they are 1000 m above the ground. The worker below is getting ready to walk away in 10 seconds. Does he live?

10. Physics 1501: Lecture 3, Pg 9 Problem Solution Method: Five Steps: (Similar to IDEA) 1) Focus the Problem - Draw a picture – what are we asking for? 2) Describe the physics -What physics ideas are applicable -What are the relevant variables known and unknown 3) Plan the solution -What are the relevant physics equations 4) Execute the plan -Solve in terms of variables -Solve in terms of numbers 5) Evaluate the answer -Are the dimensions and units correct? -Do the numbers make sense?

11. Physics 1501: Lecture 3, Pg 10 Problem: 1. We need to find the time it takes for the brick to hit the ground. 1000 m

12. Physics 1501: Lecture 3, Pg 11 Problem: 2. Free Fall Constant acceleration at value g l Choose coordinate system l Variables: h – height of helicopter = 1000 m g – acceleration due to gravity = 9.81 m/s 2 t – time to drop – goal m – mass of brick – unknown v o – initial velocity = 0 why? 1000 m y = 0 y

13. Physics 1501: Lecture 3, Pg 12 Problem: 3. Next write position for constant acceleration: 1000 m y = 0 y Realize that v 0y = 0 Solve for when y=0 Why ??

14. Physics 1501: Lecture 3, Pg 13 Problem: y 0 = 1000 m y y = 0 Solve for t in symbols: 4. Substitute y=0 and v o =0 Solve for t in numbers:

15. Physics 1501: Lecture 3, Pg 14 Problem: y 0 = 1000 m y y = 0 5. Does it make sense? Units worked out to be ‘s’ That’s correct for time It takes 14.3 s. That means that the man escapes !!!

16. Physics 1501: Lecture 3, Pg 15 Tips: l Read ! çBefore 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. l Watch your units ! çAlways check the units of your answer, and carry the units along with your numbers during the calculation.

17. Physics 1501: Lecture 3, Pg 16 Coordinate Systems / Chapter 3 l In 1 dimension, only 1 kind of system, çLinear Coordinates (x) +/- l In 2 dimensions there are two commonly used systems, çCartesian Coordinates (x,y)  Circular / polar Coordinates (r,  ) l In 3 dimensions there are three commonly used systems, çCartesian Coordinates(x,y,z)  Cylindrical Coordinates (r, ,z)  Spherical Coordinates(r,  )

18. Physics 1501: Lecture 3, Pg 17 Vectors (review): l In 1 dimension, we can specify direction with a + or - sign. l In 2 or 3 dimensions, we need more than a sign to specify the direction of something: r l To illustrate this, consider the position vector r in 2 dimensions. Example Example: Where is Boston? New York ç Choose origin at New York ç Choose coordinate system ç Boston is 212 miles northeast of New York or Boston is 150 miles north and 150 miles east of New York Boston New York r

19. Physics 1501: Lecture 3, Pg 18 Vectors... l There are two common ways of indicating that something is a vector quantity: A çBoldface notation: A ç“Arrow” notation: A A =A A

20. Physics 1501: Lecture 3, Pg 19 Vectors: definition l A vector is composed of a magnitude and a direction çexamples: displacement, velocity, acceleration çmagnitude of A is designated |A| çusually carries units l A vector has no particular position l Two vectors are equal if their directions and magnitudes match. A B C A = C A = B, B = C

21. Physics 1501: Lecture 3, Pg 20 Vectors and scalars: l A scalar is an ordinary number. ça magnitude without a direction çmay have units (kg) or be just a number çusually indicated by a regular letter, no bold face and no arrow on top. Note: the lack of specific designation of a scalar can lead to confusion l The product of a vector and a scalar is another vector in the same direction but with modified magnitude. A B A = -0.75 B

22. Physics 1501: Lecture 3, Pg 21 Lecture 3, ACT 2 Vectors and Scalars A) my velocity (3 m/s) C) my destination (the pub - 100,000 m) B) my acceleration downhill (30 m/s 2 ) D) my mass (150 kg) Which of the following is not a vector ? (For bonus points, which answer has a reasonable magnitude listed ?) While I conduct my daily run, several quantities describe my condition

23. Physics 1501: Lecture 3, Pg 22 Lecture 3, Answer Lecture 3, ACT 1 Answer Which of the following is not a vector ? (For bonus points, which answer has a reasonable magnitude listed ?) While I conduct my daily run, several quantities describe my condition D) my mass (150 kg) Answer: There is no way to assign a direction to a mass Can only be a scalar A)Velocity has direction (3 m/s east) B)Acceleration has direction C)Destination = position and has a direction ( pub is 100,000 m east of my house).

24. Physics 1501: Lecture 3, Pg 23 Lecture 3, ACT 2 Bonus Answer A) my velocity (3 m/s) C) my destination (the pub - 100,000 m) B) my acceleration downhill (30 m/s 2 ) D) my mass (150 kg) ~ 6 mi/hr or 10 min miles 0 to 60 mi/hr in 1 seconds (3 times g) 100 km is about 60 mi Pounds are a measure of force On the earth’s surface, 1 kg weighs 2.2 lbs 150 kg weighs about 330 lbs Answer = (A)

25. Physics 1501: Lecture 3, Pg 24 Converting Coordinate Systems In circular coordinates the vector R = (r,  ) l In Cartesian the vector R = (r x,r y ) = (x,y) We can convert between the two as follows: In cylindrical coordinates, r is the same as the magnitude of the vector r x = x = r cos  r y = y = r sin   arctan( y / x ) y x (x,y)  r ryry rxrx

26. Physics 1501: Lecture 3, Pg 25 Vector addition: l The sum of two vectors is another vector. A = B + C B C A B C

27. Physics 1501: Lecture 3, Pg 26 Vector subtraction: l Vector subtraction can be defined in terms of addition. B - C B C B -C-C = B + (-1)C

28. Physics 1501: Lecture 3, Pg 27 Unit Vectors: Unit Vector l A Unit Vector is a vector having length 1 and no units. l It is used to specify a direction. uU l Unit vector u points in the direction of U. u çOften denoted with a “hat”: u = ûUû x y z i j k i, j, k l Useful examples are the cartesian unit vectors [ i, j, k ] ç point in the direction of the x, y and z axes. R = r x i + r y j + r z k

29. Physics 1501: Lecture 3, Pg 28 Vector addition using components: CAB l Consider C = A + B. Ciji jij (a) C = (A x i + A y j ) + (B x i + B y j ) = (A x + B x )i + (A y + B y )j Cij (b) C = (C x i + C y j ) l Comparing components of (a) and (b): ç C x = A x + B x ç C y = A y + B y C BxBx A ByBy B AxAx AyAy

30. Physics 1501: Lecture 3, Pg 29 Lecture 3, ACT 3 Vector Addition l Vector A = {0,2,1} l Vector B = {3,0,2} l Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C? (a) {3,-4,2} (b) {4,-2,5}(c) {5,-2,4} (a) {3,-4,2} (b) {4,-2,5} (c) {5,-2,4}

31. Physics 1501: Lecture 3, Pg 30 Lecture 3, ACT 3 Solution D = (A X i + A Y j + A Z k) + (B X i + B Y j + B Z k) + (C X i + C Y j + C Z k) = (A X + B X + C X )i + (A Y + B Y + C Y )j + (A Z + B Z + C Z )k = (0 + 3 + 1)i + (2 + 0 - 4)j + (1 + 2 + 2)k = {4,-2,5} Answer is (b) l Vector A = {0,2,1} l Vector B = {3,0,2} l Vector C = {1,-4,2}