Chapter 4, Motion in 2 Dimensions

1. Chapter 4, Motion in 2 Dimensions

2. Position, Velocity, Acceleration • Just as in 1d, in 2d, an object’s motion is completely known if it’s position, velocity, & acceleration are known. • Position Vector r • In terms of unit vectors discussed last time, for an object at position (x,y) in x-y plane: r  x i + y j For an object moving: r depends on time t: r = r(t) = x(t) i + y(t) j

3. Suppose that an object moves from point A(ri) to point B(rf) in the x-y plane: • The Displacement Vector is: Δr = rf - ri If this happens in in a time Δt = tf - ti • The Average Velocity is: vavg  (Δr/Δt) Obviously, this in the same direction as the displacement. It is independent of the path between A & B

4. As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define The Instantaneous Velocity  Velocity at Any Instant of Time.  average velocity over an infinitesimally short time • Mathematically, the instantaneous velocity is: v  lim∆t  0[(∆r)/(∆t)] ≡ (dr/dt) lim ∆t  0  ratio (∆r)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.  The instantaneous velocity v≡time derivative of displacementr

5. Instantaneous Velocityv ≡ (dr/dt). • The magnitude |v| of vector v ≡ speed. As motion progresses, the speed & direction of v can both change. For an object moving from A (vi) to B (vf) in the x-y plane: Velocity Change Δv = vf - vi This happens in time Δt = tf - ti • Average Acceleration aavg  (Δv/Δt) As both the speed & direction of v change, over an arbitrary path

6. As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define The Instantaneous Acceleration  Acceleration at Any Instant of Time.  average acceleration over infinitesimally short time • Mathematically, instantaneous acceleration is: a ≡ lim∆t  0[(∆v)/(∆t)] ≡ (dv/dt) lim ∆t  0  ratio (∆v)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.  The instantaneous acceleration a≡time derivative of velocityv

7. 2d Motion, Constant Acceleration • It can be shown that: Motion in the x-y plane can be treated as 2 independent motions in the x & y directions.  So, motion in the x direction doesn’t affect the y motion & motion in the y direction doesn’t affect the x motion.

8. Suppose an object moves from A (ri,vi), to B (rf,vf), in the x-y plane. So, it’s position clearly changes with time: The acceleration a is constant, so, as in 1d, we can write (vectors!):  rf = ri + vit + (½)at2 • If its velocity also changes with time, it is accelerating. If it’s acceleration a is constant, so, as in 1d, we can write (vectors!):  vf = vi + at

9. Summary • For 2d motion when the acceleration a is constant, we have:  rf = ri + vit + (½)at2, vf = vi + at • For the Horizontal Part of the Motionwe have: xf = xi + vxit + (½)axt2,vxf = vxi + axt • For the Vertical Part of the Motionwe have: yf = yyi + vyit + (½)ayt2,vyf = vyi + ayt

10. Projectile Motion

11. A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

12. Projectile Motion • Projectile Motion Motion of an object that is projected into the air at an angle. • Near the Earth’s surface, the accelerationaon the projectile is downward and equal to a =g = 9.8 m/s2 Goal: Describe the motion after it starts. • Galileo:Analyzed horizontal & vertical components of motion separately. • Today: Displacement D & velocity v are vectors  Components of the motion can be treated separately

13. Equations Needed • One dimensional, constant acceleration equations for x&y separately! x part: Acceleration ax = 0! y part: Acceleration ay =  g! (+ if down is positive- if down is negative). Take down positive for a while! • The initial x & y components of the velocity: vxi & vyi. x motion: vxf = vxi = constant. xf = xi + vxi t y motion:vyf = vyf + gt, yf = yi + vyi t + (½)g t2 (vyf) 2 = (vyi)2 + 2g (yf - yi)

14. Projectile Motion • Simplest Example:A ball rolls across the table, to the edge & falls off edge to floor. It leaves the table at time t = 0. Analyze the x & y parts of the motion separately. • y Part of the Motion:Down is positive. The origin is at the table top: yi = 0. Initially, there is noycomponent of velocity: vyi = 0  vyf = gt, yf = (½)g t2 • x Part of the Motion:The origin is at the table top: xi = 0. There is NO x component of acceleration(!) ax = 0.Initially, the xcomponent of velocity is: vxi  vxf = vxi , xf = vxit

15. The Ball Rolls Across the Table & Falls Off t = 0 here  Can be understood by analyzing the horizontal & vertical motions separately. Take down as positive. The initial velocity has an x component ONLY! That is vyi = 0. vxi At any point, v has bothx&y components. The kinematic equations tell us that, at time t, vx = vxi, vy = gt xf = vxit yf = (½)gt2

16. Summary • A ball rolling across the table & falling. • The vector velocityvhas 2 Components: vxf = vxi , vyf = gt • The vector displacementDalso has 2 components: xf = vxit , yf = (½)gt2

17. The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. The photo shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

18. The PHYSICS of Projectile Motion The vertical (y) part of the motion: vyf = gt , yf = (½)g t2 Is the SAME as free fall!! The horizontal (x) part of the motion: vxf = vxi, xf = vxit Is motion at constant velocity!!  An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point! (the x & y motions are independent)

19. General Case: An object is launched at initial angle θ0with the horizontal. The analysis is similar to before, except the initial velocity has a vertical component vy0 0. Let up be positive now! vx0 = v0cosθ0 vy0 = v0sinθ0 The acceleration = g downward for the entire motion! The Parabolic shape of the path is real (neglecting air resistance!)

20. Projectile Motion Diagram

21. General Case:Take y positive upward& origin at the point where it is shot: xi = yi = 0 vxi = vicosθi, vyi = visinθi • Horizontal motion: NO ACCELERATION IN THE x DIRECTION! vxf = vxi , xf = vxi t • Vertical motion: vyf = vyi - gt , yf = vyi t - (½)gt2 (vyf) 2 = (vyi)2 - 2gyf • If y is positive downward, the - signs become + signs. ax = 0, ay = -g = -9.8 m/s2

22. Summary: Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where The acceleration is g and is down.

23. Projectile Motion: More Discussion • As we already said, the y-component of the velocity is zero at the maximum height of the trajectory • The acceleration stays the same throughout the trajectory

24. Range & Maximum Height • When analyzing projectile motion, two characteristics are of special interest: • The Range, R =the horizontal distance of the projectile • The Maximum Height h = the highest point.

25. Maximum Height • The Maximum Height hof a projectile is the height at which the y component of velocity vy = 0. • It can be obtained from the kinematic equation: (vyf) 2 = (vyi)2 + 2g (yf - yi) • Assuming yi = 0, setting vyf = 0 & putting yf = h: 0 = (vyi)2 + 2gh • So: h = [(vyi)2/(2g)] • Using vyi = vi sinθigives: • Note that this is valid only for symmetric motion

26. Range • The Range of a Projectile, R = xmax, is the maximum horizontal distance from the starting point. • It can be obtained from the kinematic equation R = xmax = vxitmax (1) with tmax = the time for it to hit the ground. • The time to hit tmax can be obtained by setting yf = 0 in the kinematic equation yf = vyi t - (½)gt2. This gives tmax = (2vyi/g)(2) • Putting (2) into (1) gives R = (2vxivyi/g) (3) • In (3), using vxi = vicosθi& vyi = visinθialong with the trig identity 2 sinθicosθi = sin(2θi) gives the final result: • Note that this is valid only for a symmetric trajectory

27. Ranges for Fixed vi & Changing θi

28. Range • Using differential calculus, setting (dR/dqi) = 0 & solving for qigives the result that • The maximum range R occurs at qi = 45o, • independent of the initial velocity vi. • Complementary angles will produce the same range • The maximum height will be different for the two angles • The times of the flight will be different for the two angles The Range is:

29. Non-Symmetric Projectile Motion • Follow the general rules for projectile motion • One way is to break the y-direction into parts • up and down or • symmetrical back to initial height and then the rest of the height • Apply the problem solving process to determine & solve the necessary equations