Chapter 4 Two-Dimensional Kinematics

Transcription

1 Chapter 4 Two-Dimensional Kinematics Units of Chapter 4 Motion in Two Dimensions Projectile Motion: Basic Equations Zero Launch Angle General Launch Angle Projectile Motion: Key Characteristics 1

2 4-1 Motion in Two Dimensions If velocity is constant, motion is along a straight line: If velocity is not constant, there is acceleration. When acceleration and velocity are along the same line, motion is still along a straight line. (Chapter 2) What if velocity and acceleration are not along the same line? Motion will no longer be in 1D. 4-1 Motion in Two Dimensions V 1 V 2 V 1 V 2 V 3 V 3 V 3 = V 1 + V 2 Consider one motion has two parts of velocity. If V 1 and V 2 are not perpendicular to each other, they are related. V 2 has component along V 1 s direction V 1 has component along V 2 s direction If V1 V2, one East, one North, they never contribute to each other! V 2 V 1 Motions in two perpendicular directions are INDEPENDENT from each other. 2

3 4-1 Motion in Two Dimensions Motion in the x-and y-directions should be solved separately: 1.) Define Coordinate System x direction: horizontal y direction: (positive can be up or down) 4-2 Projectile Motion: Basic Equations Assumptions: ignore air resistance g = 9.81 m/s 2, downward ignore Earth s rotation x-direction is a horizontal direction. acceleration in x-direction is zero If y-axis points upward, acceleration in y-direction is m/s 2 3

4 4-2 Projectile Motion: Basic Equations The acceleration is independent of the direction of the velocity: 4-2 Projectile Motion: Basic Equations These, then, are the basic equations of projectile motion: (if up is defined to be positive) 4

5 4-3 Zero Launch Angle Launch angle: direction of initial velocity with respect to horizontal When the lunch angle is zero, the initial velocity in the y-direction is zero. Initial velocity is in the x-direction v 0x =v 0 y Example 1: A x 30m 0 B x v x0 t v y Question: 1.)Which one touches the ground first? (what if Ball B was thrown upward or downward?) Answer: 1.) They touch ground at the Same time. Because the motion in y direction (vertical) is INDEPENDENT to motion in x direction (horizontal). Question: 2.)For B from cliff to water level, what are the known items? v x0 ~Ball A falls from rest ~Ball B was shot out horizontally at v0= 10 m/s Define right to be +x Define up to be +y 5

6 Question: 3.) How long it takes for them to hit ground? Hint: Which direction s motion is limiting the traveling time before it reach the ground? Answer: 3.) The y direction. Because from roof to the ground, y changes. The magnitude of y determines the total time to fall. To find time, we use the equation without v fy y = v 0y t + ½ ayt 2 v 0y = 0 ; t 2 = 2 y/a y Don t memorize these conclusions. The don t work for different problems. LEARN METHORDS NOT RESULTS. Question: 4.)Where is Ball B when it hits the ground? Hint: Which direction s motion should you use to decide Ball B s position when it hits the ground? Answer: 4.) Use x direction! We know y= 0 at the end. We need to find x at the end. x = v 0x t = 10(m/s) x 2.47s = 24.7 m Don t worry about x = v 0x t + 1/2a x t 2, because a x = 0!!! x = v 0x t 6

7 Question: 5.)What is the final velocity of Ball B when hits the ground? Find both direction and speed? v fx = v 0x = 10m/s v x didn t change! v fy = v 0y + a y t= x 2.47 = m/s v = (v x2 + v y2 ) = ( ) = m/s q= tan -1 (vy /v x ) = tan -1 (-24.2/10) = -tan -1 (24.2/10)=-67.5o q 4-4 General Launch Angle Treat x(horizontal) and y(vertical) directions separately. When angle is not zero, v 0x = v 0 cos θ v 0y = v 0 sin θ If up is + y direction: 7

8 Example 2: v 0 = 5.00 m/s, q= 30º, x 0 =0, y 0 =0 v 0x = v 0 cos 30º= m/s v 0y = v 0 sin 30º= m/s Question: 1) Find total time of flight before landing on the ground? Which final point? Which direction? Which equation? Y direction limits travel time! y = v 0y t - ½ gt 2 = 0 v 0y t= ½ gt 2 so t=2 v 0y /g = 5.0/9.81= 0.51(s) (Don t memorize) Example 2: v 0 = 5 m/s, q= 30º, x 0 =0, y 0 =0 v 0x = v 0 cos 30º= m/s v 0y = v 0 sin 30º= m/s Question: 2) Distance of landing? Which final point? Consider Which direction? Which equation? X direction motion determines landing distance x= v 0x t Landing total time t= 0.51 (s) x = = 2.2 (m) (Don t memorize) 8

9 Example 2: v 0 = 5 m/s, q= 30º, x 0 =0, y 0 =0 Question: 3) Find Final velocity before landing on the ground, both direction and speed. Consider both x and y directions. v x =v 0x = 4.33 (m/s) constant. t=0.51 s v y = v y0 -gt= = -2.5 (m/s) Same magnitude as v y0. Opposite direction. Final velocity: v = (v x2 + v y2 ) = 5 m/s q= tan -1 (v y / v x ) = tan -1 (-2.5/ 4.33) = -30 v = (4.33 m/s)x ^ v ^ + (-2.5 m/s)y ^ 0 =(4.33 m/s)x + (2.5 m/s)y ^ Example 2: v 0 = 5 m/s, q= 30º, x 0 =0, y 0 =0 Notice the Symmetry of Projectile Motion: If it lands at the same height t of first (raising) half = t of second (falling) half. Why?(read it yourself) y = v 0y t + ½ at 2 = (v0 + at) t - ½ at 2 y = v fy t - ½ at 2 Page 100, CE, 5 & 6 Raising v fy = 0, a = -g; v y =0-v 0y ; t raise = v 0y /g Falling v 0y = 0, a = -g; v y =v fy -0=-v 0y t fall = v 0y /g y fall = v 0y t fall + ½ at fall2 = - ½ gt fall 2 y fall = - y raise y total = 0 t raise =t fall = (total time)/2 9

10 x for the first half = x for the second half v x0 t raise =v x0 t fall V fy is opposite to v0y because V fy =v top + at fall = 0 gt fall v fy =-v 0y v top =v 0y + at raise = 0 v 0y = -at raise = gt raise Final speed, v f = (v fx2 + v fy2 ) = (v 0x2 + v 0y2 ) = v 0 Final velocity has same magnitude (speed) as launch velocity, angle q, becomes -q. Example 2: v 0 = 5 m/s, q= 30º, x 0 =0, y 0 =0 v0x = v 0 cos30º= m/s v0y = v 0 sin30º= m/s Question: 4) Find maximum height it can reach. Which final point? Which direction? Which equation? Y direction determines maximum height v yf2 = v 0y2 + 2a y v yf2 = v 0y2-2g y y =( v yf2 -v 0y2 )/2g= /( )= 0.32 m 5) Time to reach maximum height? v fy = v 0y - gt raise =0 t raise = v 0y /g 10

11 4-4 General Launch Angle Snapshots of a trajectory; red dots are at t = 1 s, t = 2 s, and t = 3 s The path followed by a projectile is a parabola. 4-5 Projectile Motion: Key Characteristics Range: the horizontal distance a projectile travels If the initial and final elevation are the same: x = v 0x t = v 0x 2(v 0y /g) = v 0 cosq 2v 0 sinq/g =v 0 2 (2cosq sinq) /g =v 02 (sin2q)/ g Do not memorize. 11

12 4-5 Projectile Motion: Key Characteristics The range is a maximum when θ = 45, 2θ = 90 Range= v 02 (sin2q)/ g Summary of Chapter 4 Components of motion in the x-direction and y- directions can be treated independently In projectile motion, the acceleration is downward (g), in y direction. There is no acceleration in x-direction, (ignore air resistance). If the launch angle is zero, the initial velocity has only an x-component The path followed by a projectile is a parabola. The range is the horizontal distance the projectile travels 12

13 Projectile Motion: Problem solving: Notice that y is not zero, if it lands at different height. 1, Define x and y directions 2, Decompose initial v in x and y directions 3, Identify final point and label known and unknown. 4, Choose and solve equations. 13