Chapter 6 Motion in Two Dimensions Section 6

Chapter 6 Motion in Two Dimensions

Section 6. 1 Projectile Motion n 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?

Section 6. 1 Projectile Motion Projectile - motion of objects given initial velocity that move only under the force of gravity. n Trajectory - the path of the projectile n

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

Horizontally Launched Projectiles gravity will not affect horizontal motion n horizontal displacement (x) n horizontal velocity (vx) n time (t) ¨ vx = x t ¨ x = v xt n

Vertical Motion vertical displacement (y) n initial vertical velocity (vyi) n final vertical velocity (vyf) n constant acceleration (g) n time (t) ¨ y = vyit + ½ gt 2 ¨ vyf = vyi + gt n

Objects Launched Horizontally n The initial vertical velocity (vyi) is zero!

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/s 2 y = vyit + ½ g t 2 (-78. 4 m) = 0 + ½ (-9. 8 m/s 2) t 2 78. 4 m (- 78. 4 m) = (- 4. 9 m/s 2) t 2 16 s 2 = t 2 t = 4. 00 s x = v xt x = (5. 0 m/s)(4. 0 s) x = 20. m

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 0. 74 m/s y = -1. 05 m g = -9. 80 m/s 2 y = vyit + ½ g t 2 y = 1. 05 m x=? m (-1. 05 m) = 0 + ½ (-9. 8 m/s 2) t 2 (-1. 05 m) = (-4. 9 m/s 2) t 2 t = 0. 463 s x = vxt x = (0. 74 m/s)(0. 46 s) x = 0. 34 m

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

n Objects Launched at an Angle vel o city vy = 0. 0 m/s Init ial viy θ vix range y = 0. 0 m

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 t 2 0 = (13. 5 m/s) t + ½ (- 9. 80 m/s 2) t 2 0 = [13. 5 m/s + - 4. 90 m/s 2 t] t -13. 5 m/s = -4. 9 m/s 2 t vy = 13. 5 m/s t = 2. 76 s vx = 23. 4 m/s y = 0. 0 m

Ex #3. Continued vy = 13. 5 m/s vx = 23. 4 m/s range Range (x) x = vxt x = (23. 4 m/s)(2. 76 s) x = 64. 584 m x = 64. 6 m y = 0. 0 m

Ex #3. Continued Maximum Height Use ½ of the “hang time. ” y = vyit + ½ g t 2 vy y = (13. 5 m/s)(1. 38 s) + ½ (-9. 8 m/s 2)(1. 38 s)2 y = 18. 6 m - 9. 33 m = 13. 5 m/s y = 9. 3 m ymax vx = 23. 4 m/s y = 0. 0 m

Section 6. 1 Projectile Motion n 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?

Section 6. 2 Circular Motion n 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?

Describing Circular Motion n 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!

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

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

Centripetal Acceleration (ac) n n n Acceleration is always toward the center of a circle. Also known as Radial Acceleration. ac or a. R Centripetal acceleration is directly proportional to the square of the speed and inversely proportional to the radius of the circle. a c= v 2 r

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

Derive Time! ac = v 2 and v = 2πr r T substitute v ac = (2πr/T)2/r ac = 4π2 r 2 / (r. T 2) ac = 4π2 r T 2

Centripetal Force (Fc) Force that is center seeking. n Using Newton’s 2 nd Law of Motion n ¨F = ma ¨ Fc = mac ¨F c= mac = mv 2 = m 4π2 r r T 2

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 m = 0. 013 kg r = 0. 93 m v = 2πr = 2(3. 14)(0. 93 m) T (1. 18 s) v = 4. 9 m/s Acceleration ac = v 2 = (4. 9 m/s)2 r (0. 93 m) ac = 26 m/s 2

Ex #4. Continued. m = 0. 013 kg r = 0. 93 m Force Fc = mac Fc = (0. 013 kg)(26 m/s 2) Fc = 0. 34 N

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? m = 45 kg 4. 1 m/s 6. 3 m r = 6. 3 m v = 4. 1 m/s Ff = F c Fc = mv 2 r Ff = (45 kg)(4. 1 m/s)2 (6. 3 m) Ff = 120 N

Centrifugal “Force” n 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. ”

Section 6. 2 Circular Motion n 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?

Section 6. 3 Relative Velocity n Essential Questions: ¨ What is relative velocity? ¨ How do you find the velocities of an object in different reference frames?

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

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

Combining Velocity Vectors n 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.

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

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

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? vb = 4. 0 m/s E vm = 0. 75 m/s N 0. 75 m/s 4. 0 m/s R 2 = v b 2 + v m 2 R 2 = (4. 0)2 + (0. 75)2 R = 4. 0697 = 4. 1 m/s θ = tan 1 -(y/x) = 10. 6197 = 11°

Section 6. 3 Relative Velocity n Did We Answer Our Essential Questions? ¨ What is relative velocity? ¨ How do you find the velocities of an object in different reference frames?