Chapter 3: Kinematics (2D) Part I

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1 Chapter 3: Part I Recap: Kinematics (1D) 1. Vector Kinematics 2. Projectile Motion 3. Uniform Circular Motion 4. Relative Velocity Vector Kinematics One Set of 2-D 2 Kinematic Eqs.. for Motion of One Body with Constant Acceleration 1 2 x = x0 + v0 x + a x t r r r 1 r 2 2 = 0 + v0 t + a t y = y0 + v0 y + a y t 2 x component y component x = x 0 0x t x t 2 (1) 0 + v 0x t + ½ a x t 2 (1) y = y 0 0y t y t 2 (4) 0 + v 0y t + ½ a y t 2 (4) v x 0x x t (2) x = v 0x + a x t (2) v y 0y y t (5) y = v 0y + a y t (5) v x2 0x2 2a x (x 0 ) (3) x2 = v 0x2 + 2a x (x x 0 ) (3) v y2 0y2 2a y (y 0 ) (6) y2 = v 0y2 + 2a y (y y 0 ) (6) G Vector Kinematics I II III 1

2 How to study Chap. 3 HWs Key Concepts Examples Problems I 1 body, with θ = *, 10*, 11, 67 II 1 body, with θ *, 21, 47*, 52*, 60, 61, 63, 64, 71 III 2 bodies with 2 a s *, 83* * is HW problem. r r r 1 r 2 1 r 3 = 0 + v0 t + a t + b t x( t) = (2.0 m) + ( 0.5 m/s ) t y( t) = (1.0 m/s) t + (0.15 m/s ) t 6 3 Direction of Velocity Kinematics Direction (2D) of Position 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 2

3 ay ax vy vx dy/dt dx/dt dy dx distance y (m) x (m) time (s) acceleration velocity average velocity displacement position Direction of velocity Direction of acceleration Projectile Motion Projectile Motion 2.00 m/s a = (0) i +( g) j Horizontal and vertical motions analyzed separately.

4 How fast must he motorcycle leave the cliff-top? H R? This figure tells you a lot! 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 4

5 Chapter 3: Part II 1. Vectors (and Scalars) 2. Projectile Motion 3. Uniform Circular Motion 4. Relative Velocity 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? 5

6 ISEE Example 1(a) Step 1: Draw a diagram (or picture) of the situation, with coordinate axes ISEE ISEE 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. 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? ISEE Example 1(b) Step 1: Draw a diagram (or picture) of the situation, with coordinate axes ISEE ISEE 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. 6

7 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. 200 m, given x?? Same Concept H? d, given H? 7

8 Example 2: A projectile is launched from ground level to the top of a cliff which is R =195 maway and H =155 mhigh. The projectile lands on top of the cliff T = 7.60 s after it is fired. Use 2sinθ cosθ = sin2θ 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 θ). b. Find a formula of tanθ in terms of g, R, H and T. t = 7.6 s Uniform Circular Motion 1. Kinematics 2. Coordinates: r-φ 3. Newton s 2 nd Law (Future) Kinematics: Acceleration Rate Change in Velocity t 1 t 1 t 2 t 2 t Center-seeking acceleration as t 0 8

9 Center-seeking Acceleration rˆ v l l = v = v v r r v l v l v v t = = v t r t r t r 0 a r r rad v = r 2 ˆ 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. 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? 9

10 Problem: Relative Velocity in 1D 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. V P/B Person V P/A =? V B/S V W/S =? θ Bank 2.5 m/s Raft V B/A Shore Boat V B/W = 2.40 m/s 10