Physics 2210 Fall smartphysics 02 Kinematics in 2- and 3-d 08/31/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics 02 Kinematics in 2- and 3-d 08/31/2015

2 Supplemental Instruction Schedule Tuesdays 8:30-9:20 am..jwb 308 Wednesdays 3:00-3:50 pm JFB B-1 Thursdays 11:30am - 12:20 pm...lcb 121 Reminder of Important Dates Events Dates Labor Day holiday Monday, September 7 Tuition payment due Friday, September 4 Classes begin Monday, August 24 Last day to add without a permission code Sunday, August 30 Last day to add, drop (delete), elect CR/NC, or audit classes Friday, September 4 Last day to withdraw from classes Friday, October 23

3 Unit 02

4 Unit 02

5 More than 1D Spherical coordinate System used in Physics x = r sin θ cos φ y = r sin θ sin φ z= r cos θ θ : Polar Angle φ: Azimuthal Angle NOTE PHYSICS TEXTBOOKS GENERALLY USE THIS CONVENTION for θ and φ backward from MATH Sometimes δ =90 θ : Elevation Angle x East up i z z k (x, y, z) j The x, y and z axes must point in mutually perpendicular directions Position Vector r = xi + yj + zk y North A simple 3d vector visualizer can be found at

6 2D motion in horizontal Plane We also assume unit 02 to be mostly review. For extra help, you might try Khan s Academy: Cartesian coordinates ( x, y ) Polar coordinates ( r, φ ) x = r cos φ y = r sin φ j i Position Vector r = 3i + 4j Common Convention Horizontal plane +x: east +y: north Projectile Motion +x: down range +y: up

7 Example 2.1 (1/3) You spot a plane that is 1.44 km north, 2.7 km east, and at an altitude 5.12 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? (give answer in E of N) (%i1) /* this is a 3d problem */ y: 1.44; x: 2.7; z: 5.12; (%o1) 1.44 (%o2) 2.7 (%o3) 5.12 (%i4) r : sqrt(x^2+y^2+z^2); (%o4) Answer (a) 5.96 km A simple 3d vector visualizer can be found at ary/curriculum/precalculusdiscrete/demos/vector-3d/ (%i6) phi : atan2(y,x)*180/%pi, numer; (%o6) (%i7) 90-phi; (%o7) Answer (b) 61.9 degrees East of North

8 Example 2.1 (2/3) You spot a plane that is 1.44 km north, 2.7 km east, and at an altitude 5.12 km above your position. (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting be toward the east direction, be toward the north direction, and be in vertically upward. In other words, write down the x, y, and z projections of the position vector along the coordinate directions. Answer: X-projection: 2.7 km i X-component: 2.7 km y-projection: 1.44 km j y-component: 1.44 km Z-projection: 5.12km k Z-component: 5.12 km r = 2.7 km i km j km k

9 Example 2.1 (3/3) You spot a plane that is 1.44 km north, 2.7 km east, and at an altitude 5.12 km above your position. (d) At what elevation angle (above the horizontal plane of Earth) is the airplane? (give the answer in degrees above the horizon) (%i8) theta : acos(z/r)*180/%pi, numer; (%o8) (%i9) 90-theta; (%o9) (%i10) Answer: (d) elevation angle is 59.1 degrees above horizon

10 Rule of Thumb If you use Cartesian coordinates i.e. x, y (z) are mutually orthogonal, Then motion in x, y (and z) are independent for the cases of (a) Constant velocity (b) Constant acceleration

11 Projectile Motion: x and y are independent 25 fps

12 Poll A physics demo launches a ball horizontally while dropping a second ball vertically at exactly the same time. Which ball hits the ground first? A. Dropped ball B. Both hit at the same time. C. Launched (horizontally) ball

13 Projectile Motion Objects flying without power at the surface of the Earth: (neglecting air resistance) X (horizontal down range): constant velocity v xi = v i cosθ (θ: angle the projectile s velocity at time ti makes to horizontal) Y (up/dow): constant acceleration with magnitude of g =9.81m/s downward v yi = v i sinθ a y = - g (up is +y)

14 Poll Destroyer Enemy 1 Enemy 2 A destroyer simultaneously fires two shells with different initial speeds at two different enemy ships. The shells follow the parabolic trajectories shown. Which ship gets hit first? A. Enemy 2 B. Both ships are hit at about the same time. C. Enemy 1

15 (politically incorrect) Example 2.2 (1/4) From ground level, a hunter shoots at a monkey hanging in a tree. The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey hears the shot instantaneously (assume sound travels very fast), lets go of the tree branch, and falls. (a) Treating h, D, and v 0 (initial velocity of the bullet) as given parameters, at what angle θ above the ground should the hunter aim in order to hit the monkey? (b) What is the minimum v 0 required for success? Monkey starts dropping vertically at t = 0 y Gun fired from origin at t = 0 v 0 h O θ x D

16 (politically incorrect) Example 2.2 (2/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. At what angle θ above the ground should the hunter aim in order to hit the monkey (%i1) /* The bullet travels according to projectile motion: constant velocity in the x-direction (horizontal, right+) and constant acceleration at aby=-g in the y (vertical, up+) direction. Staring at t=0 from the origin: xb0=0 and yb0=0, the bullet's position is given by */ xb: vbx0*t; (%o1) (%i2) yb: vby0*t - g/2*t^2; t vbx0 g t (%o2) t vby (%i3) /* The monkey starts at initial position (D, h) from rest, so its position is given by */ xm: D; (%o3) D (%i4) ym: h - g/2*t^2; 2 g t (%o4) h

17 (politically incorrect) Example 2.2 (3/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. (a) At what angle θ above the ground should the hunter aim in order to hit the monkey? (%i5) /* First find time t1 at which the bullets reaches the horizontal position of the monkey */ t1: rhs(solve(xb = xm, t)[1]); D (%o5) ---- vbx0 (%i6) /* at t=t1, the bullet and monkey are at height yb1 and ym1, respectively, given by */ yb1: yb, t=t1; 2 D vby0 g D (%o6) vbx0 2 2 vbx0 (%i9) sol2: solve(yb1 = ym1,vby0); h vbx0 (%o9) [vby0 = ] D Answer (a) vby0/vbx0 = h/d = tan(theta) it means you aim right at the monkey i.e. theta = arctan(h/d)

18 (politically incorrect) Example 2.2 (3/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. (a) At what angle θ above the ground should the hunter aim in order to hit the monkey? (b) What is the minimum v 0 required for success? (%i10) /* we want to make sure yb1 > 0 *, so we solve for vby0 such that yb1=0 as the limiting case */ sol3: solve(yb1=0, vby0); g D (%o10) [vby0 = ] So the requirement is that 2*vBy0*vBx0 > g*d But vbx0 = v0*cos(theta) And vby0 = v0*sin(theta) 2 vbx0 where theta is NOW the optimized angle given by tan(theta)=h/d And 2*sin(theta)*cos(theta) = sin(2*theta) So the minimum v0 is given by v0_min = sqrt( g*d/sin(2*theta) ) = sqrt( g*d/(2*sin(theta)*cos(theta)) )

19 30 fps

20 Extra Example 2.1 (1/4) At a given time you are in ship A heading directly north at 5.0m/s Your radar shows ship B at a distance of 1200 m 30 degrees north of west from you headed directly east at 7.0m/s (a) Taking your current position to be the origin, write down the position of ship B in unit vector form (b) How close do the two ships come to one another (and when)? (%i1) /* t=0 at given moment */ va: 5; (%o1) 5 (%i2) vb: 7; (%o2) 7 (%i3) phi: (180-30)*%pi/180, numer; (%o3) (%i4) R: 1200; (%o4) 1200 (%i5) xb0: R*cos(phi); (%o5) (%i6) yb0: R*sin(phi); (%o6)

21 Extra Example 2.1 (2/3) Answer (a): r B = 1039mi + 600mj (%i7) xa: 0; (%o7) 0 (%i8) ya: va*t; (%o8) 5 t (%i9) xb: xb0 + vb*t; (%o9) 7 t (%i10) yb: yb0; (%o10) (%i11) r2: (xb-xa)^2+(yb-ya)^2; 2 2 (%o11) (7 t ) + ( t) (%i12) dr2dt: diff(r2, t); (%o12) 14 (7 t ) - 10 ( t) (%i13) expand(dr2dt=0); (%o13) 148 t = 0 (%i14) soln1: solve(dr2dt=0, t); (%o14) [t = ] (%i15) t1: rhs(soln1[1]), numer; (%o15)

22 Extra Example 2.1 (3/3) (%i18) sqrt(r2), t=t1, numer; (%o18) Answer: They come closest together at a distance of 115.8m (at t=138.85s )