Projectile Motion: Relationships and Independence

1. Chapter 6 Motion in Two Dimensions

2. Section 6.1 Projectile Motion • Essential Questions: • How are the vertical and horizontal motions of a projectile related? • What are the relationships between a projectile’s height, time in the air, initial velocity, and horizontal distance traveled?

3. Section 6.1 Projectile Motion • Projectile - motion of objects given initial velocity that move only under the force of gravity. • Trajectory - the path of the projectile

4. Independence of Motion in Two Dimensions • The horizontal and vertical velocities of a projectile are independent. • The shape of the trajectory depends on the viewpoint of the observer.

5. Horizontally Launched Projectiles • gravity will not affect horizontal motion • horizontal displacement (x) • horizontal velocity (vx) • time (t) • vx = x t • x = vxt

6. Vertical Motion • vertical displacement (y) • initial vertical velocity (vyi) • final vertical velocity (vyf) • constant acceleration (g) • time (t) • y = vyit + ½ gt2 • vyf = vyi + gt

7. Objects Launched Horizontally • The initial vertical velocity (vyi) is zero!

8. Ex #1. A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. Find time and how far the stone is from the base of the cliff. 5.0 m/s vxi = 5.0 m/s vyi = 0 y = -78.4 m g = -9.80 m/s2 y = vyit + ½ g t2 (-78.4 m) = 0 + ½ (-9.8 m/s2) t2 (- 78.4 m) = (- 4.9 m/s2) t2 16 s2 = t2 t = 4.00 s x = vxt x = (5.0 m/s)(4.0 s) x = 20. m 78.4 m

9. Ex #2. You are preparing breakfast and slide a plate on the countertop. Unfortunately, you slide it too fast, and it flies off the end of the countertop. If the countertop is 1.05 m above the floor and it leaves the top at 0.74 m/s, how long does it take to fall, and how far from the end of the counter does it land? vxi = 0.74 m/s vyi = 0 y = -1.05 m g = -9.80 m/s2 y = vyit + ½ g t2 (-1.05 m) = 0 + ½ (-9.8 m/s2) t2 (-1.05 m) = (-4.9 m/s2) t2 t = 0.463 s x = vxt x = (0.74 m/s)(0.46 s) x = 0.34 m 0.74 m/s y = 1.05 m x = ? m

10. Objects Launched at an Angle • We will use vertical and horizontal components. • At the highest point of the trajectory, the vertical velocity is zero (the object must stop to change direction). • The horizontal distance from the beginning to the end is called range (R). • The total air time is called “hang time.”

11. Objects Launched at an Angle vy = 0.0 m/s viy Initial velocity θ vix range y = 0.0 m y = 0.0 m

12. Ex #3. A player kicks a football from ground level with a velocity of 27.0 m/s and at an angle of 30.0 above the horizontal. Find “hang time”, range, and maximum height. x - component y - component vx = (27.0) cos 30˚ vy = (27.0) sin 30˚ vx = 23.4 m/s vy = 13.5 m/s “Hang Time” y = vyit + ½ g t2 0 = (13.5 m/s) t + ½ (- 9.80 m/s2) t2 0 = [13.5 m/s + - 4.90 m/s2 t] t -13.5 m/s = -4.9 m/s2 t t = 2.76 s vy = 13.5 m/s vx = 23.4 m/s y = 0.0 m

13. Ex #3. Continued Range (x) x = vxt x = (23.4 m/s)(2.76 s) x = 64.584 m x = 64.6 m vy = 13.5 m/s range vx = 23.4 m/s y = 0.0 m

14. Ex #3. Continued Maximum Height Use ½ of the “hang time.” y = vyit + ½ g t2 y = (13.5 m/s)(1.38 s) + ½ (-9.8 m/s2)(1.38 s)2 y = 18.6 m - 9.33 m y = 9.3 m vy = 13.5 m/s ymax vx = 23.4 m/s y = 0.0 m

15. Section 6.1 Projectile Motion • Did We Answer Our Essential Questions? • How are the vertical and horizontal motions of a projectile related? • What are the relationships between a projectile’s height, time in the air, initial velocity, and horizontal distance traveled?

16. Section 6.2 Circular Motion • Essential Questions: • Why is an object moving in a circle at a constant speed accelerating? • How does centripetal acceleration depend upon the object’s speed and the radius of the circle? • What causes centripetal acceleration?

17. Describing Circular Motion • One might think that twirling an object in a circle does not accelerate due to the speed not changing. But it IS! Change in direction is a change in velocity!

18. Uniform Circular Motion The movement of an object at a constant speed around a circle with a fixed radius.

19. Circular Motion • The radius and speed are constant, so the velocity is tangent to the curve.

20. Centripetal Acceleration (ac) • Acceleration is always toward the center of a circle. • Also known as Radial Acceleration. • ac or aR • Centripetal acceleration is directly proportional to the square of the speed and inversely proportional to the radius of the circle. • ac= v2 r

21. Period of a Revolution (T) • The time needed to make a complete revolution. • Circumference of the circle is 2 π r which represents the distance around a circle. • v = 2πr T

22. Derive Time! ac = v2 and v = 2πr r T substitute v ac = (2πr/T)2/r ac = 4π2r2 / (rT2) ac = 4π2r T2

23. Centripetal Force (Fc) • Force that is center seeking. • Using Newton’s 2nd Law of Motion • F = ma • Fc = mac • Fc= mac = mv2 = m4π2r r T2

24. Ex #4. A 0.013 kg rubber stopper is attached to a 0.93 m length of string. The stopper is swung in a horizontal circle, making one revolution in 1.18 s. Find the speed, acceleration, and force. m = 0.013 kg r = 0.93 m T = 1.18 s Speed v = 2πr = 2(3.14)(0.93 m) T (1.18 s) v = 4.9 m/s Acceleration ac = v2 = (4.9 m/s)2 r (0.93 m) ac = 26 m/s2 m = 0.013 kg r = 0.93 m

25. Ex #4. Continued. Force Fc = mac Fc = (0.013 kg)(26 m/s2) Fc = 0.34 N m = 0.013 kg r = 0.93 m

26. Ex #5. A 45 kg merry-go-round worker stands on the ride’s platform 6.3 m from the center. If his speed as he goes around the circle is 4.1 m/s, what is the force of friction necessary to keep him from falling off the platform? 4.1 m/s 6.3 m m = 45 kg r = 6.3 m v = 4.1 m/s Ff = Fc Fc = mv2 r Ff= (45 kg)(4.1 m/s)2 (6.3 m) Ff= 120 N

27. Centrifugal “Force” Recall that centripetal acceleration is always directed toward the center of the circle, but we still feel the outward “force.” This outward “force,” called centrifugal force, is a fictitious force. What we are really feeling is the object that we are in also being directed inward, which makes us feel this outward “force.”

28. Section 6.2 Circular Motion • Did We Answer Our Essential Questions? • Why is an object moving in a circle at a constant speed accelerating? • How does centripetal acceleration depend upon the object’s speed and the radius of the circle? • What causes centripetal acceleration?

29. Section 6.3 Relative Velocity • Essential Questions: • What is relative velocity? • How do you find the velocities of an object in different reference frames?

30. Relative Motion in One Dimension A coordinate system from which motion is viewed is a reference frame.

31. Different Reference Frames Depending on the reference frame chosen, displacement, velocity, and acceleration may change.

32. Combining Velocity Vectors When an object moves in a moving reference frame, you add the velocities if they are in the same direction. You subtract one velocity from the other if they are in opposite directions.

33. Ex #6: You are riding in a box car moving slowly at 15.0 m/s. You walk to the front of the boxcar at 1.2 m/s relative to the car. What is your speed relative to the ground? Both are going in the same direction, so we add the velocities. 15.0 m/s + 1.2 m/s 16.2 m/s

34. Ex #7: Rafi is pulling a toy wagon through a neighborhood at a speed of 0.75 m/s. A caterpillar in the wagon is crawling toward the rear of the wagon at a rate of 2.0 cm/s. What is the caterpillar’s velocity relative to the ground? The caterpillar is moving opposite of the wagon, so we will subtract the two velocities. 0.75 m/s – 0.020 m/s 0.73 m/s or 73 cm/s

35. Ex #8: Ana and Sandra are riding on a ferry boat traveling east at 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water? R2 = vb2 + vm2 R2 = (4.0)2 + (0.75)2 R = 4.0697 = 4.1 m/s θ = tan1-(y/x) = 10.6197 = 11° 0.75 m/s 4.0 m/s vb = 4.0 m/s E vm = 0.75 m/s N

36. Section 6.3 Relative Velocity • Did We Answer Our Essential Questions? • What is relative velocity? • How do you find the velocities of an object in different reference frames?