CHAPTER 3 MOTION IN A PLANE

1. CHAPTER 3 MOTION IN A PLANE

2. Goals for Chapter 3 • To study position, velocity, and acceleration vectors. • To frame two-dimensional motion as it occurs in the motion of projectiles. • To restrain two-dimensional motion to a circular path and understand uniform circular motion. • To study the new concept of one motion frame relative to another.

3. Position vectors can specify the location and displacement of a point in an x-y coordinate system

4. Displacement Vector Δr = rf – ri SI unit: meter, m rf = ri + Δr

5. Average Velocity Vector Δr/Δt vav = SI unit; meter/second, m/s

6. Velocity In A Plane vav = (r2-r1) / (t2–t1) = Δr / Δt Components of the average velocity vectors are: vav,x= Δx / Δt and vav,y= Δy / Δt

7. A MODEL CAR

8. At every point along the path, the instantaneous velocity vector is tangent to the path.

9. ACCELERATION IN A PLANE

10. ACCELERATION IN A PLANE aav = v2 – v1 / (t2 – t1) = Δv / Δt

12. When a particle moves in a curved path, it always has nonzero acceleration, even when it moves with constant speed. More generally, acceleration is associated with change of speed, change of direction of velocity, or both.

14. Constant-Acceleration Equations of Motion

15. Constant-Acceleration Equations of Motion in Two-Dimensions

16. Independence Of Horizontal And Vertical Motions

18. PROJECTILE MOTION A projectile is any object that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance. In studying projectile motion we make the following assumptions: Air resistance is ignored. The acceleration of gravity is constant, downward, and has a magnitude equal to g = 9.81 m/s2. The Earth’s rotation is ignored. The Earth’s curvature is ignored.

21. Equations for Projectile Motion (assuming that ax = 0, ay = -g) Constant-Acceleration Equations of Motion in Two-Dimensions vx = v0x + axt vx = v0x (ALL THE TIME(1) x = x0 + v0xt + (½ )axt2 x = x0 + v0xt (2) vy = v0y + ayt vy = v0y – gt (3) y = y0 + v0yt + (½ )ayt2 y = y0 + v0yt - (½ )gt2 (4) vy2 = voy2 -2g(y-y0) (5)

22. Two types of projectile problems • 1) Zero angle: the y-component of the initial velocity is zero. The x-component of the initial velocity has the same magnitude as the initial speed. • 2) general angle: the initial velocity v0 makes angle θ0 with the positive x axis and the initial velocity componants are: vox = v0 cosθ0 voy = v0sinθ0

23. Determination of key itemsfor projectiles • x = (vocos o)t •  = tan-1(vy/vx) • y = (vosin o)t - ½gt2 • vx = vocoso • vy = vosino- gt

24. The contestant in the figure shoots a paintball horizontally at a speed of 75m/s from a point 1.50 m above the ground. The ball misses its target and hits the ground.a) For how many seconds is the ball in the air?b) Find the maximum horizontal displacement (range)

25. A home-run baseball is a hit with an initial speed v0 = 37.0 m/s at an initial angle θ0 = 53.1o. Assume the ball is struck at the ground level.a) Find the ball’s position, and the magnitude and direction of its velocity when t = 2s.b) Find the time when the ball reaches the highest point of its flight and find its height and this point.c) Find the horizontal range R

26. For a field goal attempt, a football is kicked from a point on the ground that is a horizontal distance d from the goalpost. For the attempt to be successful, the ball must clear the crosbar, 10 ft (3.06m) above the ground. The ball leaves the kicker foot with an initial speed of 20.0 m/s at an angle of 30o above the horizontal. Will the ball clear the crossbar?What is the distance d between kicker and goal post if the ball barely clears the crossbar

27. A grasshopper leaps into the air from the edge of a vertical cliff (see figure). Use information from the figure to find:a) the initial speed of the grasshopperb) the height of the cliff

28. UNIFORM CIRCULAR MOTION

30. Uniform Circular Motion

31. Centripetal Acceleration arad = v2/R