Chap. 3: Kinematics (2D) Recap: Kinematics (1D) 1. Vector Kinematics 2. Projectile Motion 3. Uniform Circular Motion 4.

Chap. 3: Kinematics (2D) Recap: Kinematics (1D) 1. Vector Kinematics 2. Projectile Motion 3. Uniform Circular Motion 4. Relative Velocity 1

Last, This and Next Weeks [Last Week] Chap. 1 and Chap. 2 [This Week] Chap. 3 Chap. 4 [Next Week] Chap. 5 Question? 2

Critical Thinker One would just plug in the numbers and if it didn't come out to be a correct answer then he/she would just change the positive to negative and so on. What s wrong with this? This is a typical practice of memorizing the problems. In the exam, you wouldn t necessarily expect to see the same problems. What will be the safest bet? First, see concept. This should take you to a proper setting. ISEE Critical Thinker Problem Laws, Principles (so-called formulae) Solution A Solution B Solution C Answer 5

Common Steps for Thinker ISEE Step 1: Draw a diagram (or picture) of the situation, with coordinate axes. Step 2: Think about which principle(s) of physics apply in this problem. ISEE Step 3: Write down principle(s). ISEE Step 4: Solve them. 6

Where should the professor be when you release the egg? A set of 1-D kinematic eqs. for motion of each of TWO bodies with constant acceleration Chap.2: Problem II-2 Egg Drop [Chap.3] a set of 2-D kinematic eqs. for motion of ONE body y = y 0 + v 0y t + ½ a y t 2 x = x 0 + v 0x t + ½ a x t 2 7

Vector Kinematics x = x 0 + v 0x t + ½ a x t 2 (1) v x = v 0x + a x t (2) v x 2 = v 0x 2 + 2a x (x x 0 ) (3) y = y 0 + v 0y t + ½ a y t 2 (4) v y = v 0y + a y t (5) v y 2 = v 0y 2 + 2a y (y y 0 ) (6) One Set of 2-D Kinematic Eqs. for Motion of One Body with Constant Acceleration 2 0 0 2 0 0 2 0 0 2 1 2 1 2 1 t a v y y t a v x x t a t v r r y y x x x component y component 8

Visual Vector Kinematics G I II III 9

How to study Chap. 3 HWs Version 2 Key Categories Examples Exercises, Problems I 1 body, with q = 0 3.6 10, 71 II 1 body, with q 3.7, 3.8, 3.9 19, 21, 47, 52, 54, 60, 62, 63, 64, 65, 67, 71 III 2 bodies with 2 a s 3.10 11, 51, 56 C Circular Motion 3.11, 3.12 25, 28, 30 R Relative Velocity 3.12, 3.13, 3.14 32, 34 10

Direction of Velocity Direction of Position 3 3 2 2 3 2 0 0 ) (0.15 m/s 6 1 (1.0 m/s) ) ( ) 0.5 m/s ( 2 1 (2.0 m) ) ( 6 1 2 1 t t t y t t x t b t a t v r r 11

Kinematic Eqs. are related by derivatives and integrals. Example 1(a): Motion along with x axis x(t) = (2.0 m) (1/2) (0.50 m/s 2 ) t 2 [Eq. 1] v x (t) = (0.50 m/s 2 ) t Initial position = 2.0 m Initial velocity = zero a x (t) = 0.50 m/s 2 Motion with constant acceleration. 12

Kinematic Eqs. are related by derivatives and integrals. Example 1(b): Motion along with y axis y(t) = (1.0 m/s) t + (1/6) (0.150 m/s 3 ) t 3 [Eq. 2] Initial position = zero v y (t) = (1.0 m/s) + (1/2) (0.150 m/s 3 ) t 2 Initial velocity = 1 m/s a y (t) = (0.150 m/s 3 ) t Motion with varying acceleration. 13

ONE 2D Motion (Eqs. 1 & 2) Study each motion between 0 s and 10 s using spreadsheet: Calculate position every 0.5 s and plot them. Calculate displacement Calculate average velocity Calculate velocity every 0.5 s Calculate acceleration every 0.5 s 14

position displacement average velocity velocity acceleration time (s) x (m) y (m) distance dx dy dx/dt dy/dt vx vy ax ay 0.00 2.00 0.00 2.00-0.063 0.503-0.125 1.006 0.000 1.000-0.500 0.000 0.50 1.94 0.50 2.00-0.188 0.522-0.375 1.044-0.250 1.019-0.500 0.075 1.00 1.75 1.03 2.03-0.313 0.559-0.625 1.119-0.500 1.075-0.500 0.150 1.50 1.44 1.58 2.14-0.438 0.616-0.875 1.231-0.750 1.169-0.500 0.225 2.00 1.00 2.20 2.42-0.563 0.691-1.125 1.381-1.000 1.300-0.500 0.300 2.50 0.44 2.89 2.92-0.688 0.784-1.375 1.569-1.250 1.469-0.500 0.375 3.00-0.25 3.68 3.68-0.813 0.897-1.625 1.794-1.500 1.675-0.500 0.450 3.50-1.06 4.57 4.69-0.938 1.028-1.875 2.056-1.750 1.919-0.500 0.525 4.00-2.00 5.60 5.95-1.063 1.178-2.125 2.356-2.000 2.200-0.500 0.600 4.50-3.06 6.78 7.44-1.188 1.347-2.375 2.694-2.250 2.519-0.500 0.675 5.00-4.25 8.13 9.17-1.313 1.534-2.625 3.069-2.500 2.875-0.500 0.750 5.50-5.56 9.66 11.15-1.438 1.741-2.875 3.481-2.750 3.269-0.500 0.825 6.00-7.00 11.40 13.38-1.563 1.966-3.125 3.931-3.000 3.700-0.500 0.900 6.50-8.56 13.37 15.87-1.688 2.209-3.375 4.419-3.250 4.169-0.500 0.975 7.00-10.25 15.58 18.65-1.813 2.472-3.625 4.944-3.500 4.675-0.500 1.050 7.50-12.06 18.05 21.71-1.938 2.753-3.875 5.506-3.750 5.219-0.500 1.125 8.00-14.00 20.80 25.07-2.063 3.053-4.125 6.106-4.000 5.800-0.500 1.200 8.50-16.06 23.85 28.76-2.188 3.372-4.375 6.744-4.250 6.419-0.500 1.275 9.00-18.25 27.23 32.78-2.313 3.709-4.625 7.419-4.500 7.075-0.500 1.350 9.50-20.56 30.93 37.15-2.438 4.066-4.875 8.131-4.750 7.769-0.500 1.425 10.00-23.00 35.00 41.88-2.563 4.441-5.125 8.881-5.000 8.500-0.500 1.500 15

Direction of velocity Kinematics Direction (2D) of acceleration 18

Projectile Motions: I, II & III I II III 21

Projectile Motion 2.00 m/s a = (0) i + (g) j Horizontal and vertical motions analyzed separately. 22

[Quick Quiz 1] Is this a motion with a constant acceleration or with a varying acceleration? Question: How fast must the motorcycle leave the cliff-top? 24

This figure tells you a lot! 27

Further Look at Projectile Motion (3) v y = 0 (1) Choose an origin & an x-y coordinate system (4) y = 0 (2) a x = 0 a y = g = 9.80 m/s 2 v x = constant 28

Projectile Motion [Quick Quiz 2] Is this a motion with a constant acceleration or with a varying acceleration? v? f ina l 29

x = v 0x t (1) v x = v 0x (2) v 2 x = v 2 0x (3) y = v 0y t + ½ (g) t 2 (4) v y = v 0y + (g) t (5) v 2 y v 2 0y = 2 (g) y (6) 30

Example 1: A Hunter A hunter aims directly at a target (on the same level) 65.0 m away. Note that the magnitude of gravitational acceleration on the Earth is g = 9.80 m/s 2. (a) (10 pts) If the bullet leaves the gun at a speed of 145 m/s, by how much will it miss the target? (b) (15 pts) At what angle should the gun be aimed so the target will be hit? 33

ISEE ISEE ISEE Example 1(a) Step 1: Draw a diagram (or picture) of the situation, with coordinate axes. y R = 65.0 m x t = 0, v 0 = 145 m/s d t = T Step 2: Think about which principle(s) of physics apply in this problem. ( kinematic eqs.) Step 3: Write down kinematic equations. Step 4: Solve them. 35

Example 1(b) Step 1: Draw a diagram (or picture) of the situation, with coordinate axes. y R = 65.0 m q x t = 0, v 0 = 145 m/s t = T Step 2: Think about which principle(s) of physics apply in this problem. ( kinematic eqs.) Step 3: Write down kinematic equations. Step 4: Solve them. 39

Example 1(b) Cont d COMMON MISTAKE #1 d tan q = d / R, where d is from part (a). R Accidentally, you get the same answer! ( q = 0.868 o ) Your claim: I should get a full credit because the final answer is correct. My response: No What is wrong with this? 40

Example 1(b) Cont d COMMON MISTAKE #1 Wrong True (W-T)/(T) R v 0 q q [m] [m/s] [deg] [deg] [%] 65.0 145 0.867886 0.868086 0.023% 65.0 80.0 2.84901 2.85609 0.25% 65.0 40.0 11.2583 11.7305 4.0% 65.0 26.0 25.2277 35.2214 28% 65.0 25.3 26.4543 42.1839 37% You will know the correct answer in a formula form later. 41

y Example 1(b) Cont d COMMON MISTAKE #2 t = 0, v 0 = 145 m/s How to find T? R = 65.0 m x t = T (a) = R/v 0 = 0.448 s Question: What s wrong with T (a) = R/v 0? Hint #1: Remember vector kinematics Hint #2: Remember the velocity is a vector quantity Hint #3: v 0x =? [145 m/s is a wrong choice.] 44

Example 1 - Summary (a) d = 0.985 m (b) q = 0.5 x sin 1 (9.80 x 65.0/145 2 ) = 0.868 Note that the following is a wrong approach for part (b): q = tan 1 (0.985/65.0) = 0.868 even though the answer numerically agrees with the correct one. Below is an exercise of two approaches by changing the magnitude of the velocity (v 0 ), but using the same distance (R) of 65.0 m, where you see a larger discrepancy as v 0 decreases. 45

(3) Use kinematic eqs. in x and y separately. V y = 0 (2) a x = 0 a y = g = 9.80 m/s 2 V x = constant (1) Choose an origin & an x-y coordinate system y = 0 y = 1.00m 47

A boy on a small hill aims his water-balloon slingshot horizontally, straight at second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. 48

A boy on a small hill aims his water-balloon slingshot upward, directly at second boy hanging from a tree branch. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. 49

200 m, given x?? Same Concept H? d, given H? 50

In-Class Test #1 6 Answer: 51

Example 1 +1 A boy on a small hill aims his water-balloon slingshot horizontally, straight at second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. A boy on a small hill aims his water-balloon slingshot upward, directly at second boy hanging from a tree branch. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. Motion of the water-balloon is exactly the same problem as in Example 1 Plus: the 2 nd object (2 nd boy hanging from a tree branch) Principle: Kinematic eqs. for each boy See the textbook carefully. 53

Example 2: A projectile is launched from ground level to the top of a cliff which is R = 195 m away and H = 155 m high. The projectile lands on top of the cliff T = 7.60 s after it is fired. Use 2sinq cosq = sin2q if necessary. The acceleration due to gravity is g = 9.80 m/s 2 pointing down. Ignore air friction. a. Find the initial velocity of the projectile (magnitude v 0 and direction q). b. Find a formula of tanq in terms of g, R, H and T. t = 7.6 s 54

Chap. 3: Kinematics (2D) - Part II Recap: Kinematics (1D) 1. Vector Kinematics 2. Projectile Motion 3. Uniform Circular Motion 4. Relative Velocity 55

Uniform Circular Motion 1. Kinematics 2. Coordinates: r-f 3. Newton s 2 nd Law (Future) 57

Kinematics: Acceleration Rate Change in Velocity t 1 t 1 t 2 t 2 Dt Center-seeking acceleration as Dt0 58

Center-seeking Acceleration v v l v l v v l v v l v v r t r t r t r r t D D D D D D D D D D D 0 r r a 2 rad ˆ v rˆ 59

Example Problem 1 The radius of the semicircular drive in front of the Administration Building is 200 m. How fast could a sports car negotiate the turn, assuming that it could achieve the magnitude of a radial acceleration a rad = 0.8g? Ignore the stop sign at the intersection with East Gate Drive. 60

Example Problem 2 People advocating space colonies say that it will be simple to simulate gravity. One need only build a space module in the form of a doughnut (torus) and let it revolve at an appropriate rate. Consider a torus in circumference C = 10,000 m; it could easily accommodate 25 dwellings plus landscaping and gardens. How many times would the torus have to revolve each hour in order that the magnitude of the radial acceleration at the rim equal g? a rad v 2 r rˆ 62

Problem: Relative Velocity in 1D V P/A = V P/B + V B/A A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. A motor scooter is being ridden on the flatcar. Find the velocity (magnitude and direction) of the scooter relative to the flatcar if its velocity relative to the observer on the ground is: a) 20.0 m/s to the right; b) 4.00 m/s to the left; c) zero. 64

V P/B Person V P/A =? V B/S V W/S =? q 2.5 m/s Raft V B/A Boat V B/W = 2.40 m/s Bank Shore 67