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Chapter 4

Chapter 4. Motion in 2-D and 3-D. Chapter 4: Review. Our kinematic variables are now vectors:. Similarly to 1-D, we have. Treat each dimension separately Like multiple 1-D problems. v. a. r. Review. Example: (Problem 4.2). The position vector for an electron is

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Chapter 4

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  1. Chapter 4 Motion in 2-D and 3-D

  2. Chapter 4: Review • Our kinematic variables are now vectors: • Similarly to 1-D, we have • Treat each dimension separately • Like multiple 1-D problems

  3. v a r Review

  4. Example: (Problem 4.2) • The position vector for an electron is r = (5.0 m)i – (3.0 m)j + (2.0 m)k • Find the magnitude of r • Sketch the vector on a right handed coord system ^ ^ ^

  5. Example: (Problem 4.9) • A particle moves so that its position (in meters) as a function of time (in seconds) is r = i + 4 t2 j + t k. Write expressions for • its velocity as a function of time and • its acceleration as a function of time.

  6. Gravity is the source of acceleration:a = -gĵ ay = -g ax = 0 Projectile Motion Free fall with horizontal motion Conventional coordinate system: x  horizontal y  vertical (take "up“ as positive direction) z is not relevant (motion is 2-D)

  7. Projectile Motion (Continued) In both directions the acceleration is constant ax = 0 ay = -g vx = v0x constant x = x0 + v0xt vy = v0y - gt y = y0 + v0yt - ½gt2

  8. Projectile Motion (Continued) x and y motions can be treated independently Connected by time: x = x0 + v0xt y = y0 + v0yt - ½gt2 Since y(t) is a parabola and x is linear in time: y(x) is a parabola too

  9. vx vx=0 Motion in x and y direction Independent ay = -g ax = 0 y = y0 + v0y t – ½ gt2 x = x0 + v0x t vy(t) = v0y – gt vx(t) = v0x

  10. v0=21 m/s 10 ft q=35o 40 yds Example: Kicking a Field Goal • Does the ball make it over the goal post? (assuming it starts from ground level) • How long does it take?

  11. v0=21 m/s 3.0 m q=35o 36.6 m Solution: 1) Convert to standard units 2) Get x-y components v0x = 21 m/s cos35° = 17.2 m/s v0y = 21 m/s sin35° = 12.0 m/s 3) Tabulate known and unknown quantities

  12. Solution: 4) Write down expressions x(t) = x0 + v0xt y(t) = y0 + v0yt – ½gt2 Note: x0 = y0 = 0 5) Solve for missing quantities The kick is good!

  13. Basic Equations of Projectile Motion Horizontal - constant velocity (neglect air resistance) Vertical - constant acceleration g downward

  14. v0 V0y θ0 V0x Breaking Velocity into Components Take components of v0 v(t) can also be written in components

  15. Writing y as a Function of x Substituting in t = x/v0x For x0 =0: Since y0 =0, we have: y(x) is a parabola!

  16. Range Equation To find the range, we find where y = 0 We have the trivial solution x=0 Or this term equals 0

  17. Range Equation (Continued) To obtain a non-trivial solution, we solve:

  18. Range Equation (Continued) For given v0, maximum range is at 45° angle. Caution: Only valid when ‘launch’ and ‘landing’ are on the same level!

  19. Evel Knievel: Famous Jumps • Fountains of Caesar’s Palace,Las Vegas (151 feet) • in coma for thirty days afterwards • Jumping cars • 13 in 1970; 19 in 1971 • Injured jumping 13 Pepsi trucks • Jumped 50 Cars (stacked) in 1973 at LA coliseum (35,000 spectators) • Retired after accident while jumping a tank of live sharks in 1976. • Listed in Who’s Who and Guinness Book of World Records for having broken 35 bones.

  20. Caesar’s Palace Jump y y = 0 x 151 feet 46 m 0 Choose top of start ramp as the origin. x axis is horizontal, y axis is vertical

  21. So how fast was Evel going? Assume he took Physics 211, and used a 45° ramp: What direction is he going?

  22. Another Problem Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 15o above the horizontal. Ball 2 has twice the initial speed of ball 1. The first ball travels a distance D1 before being caught. How far does ball 2 travel before it is caught? a) D2=2D1 b) D2=4D1 c) D2=8D1

  23. Demo: Shoot Kenny Magnet Magnet releases when ball is fired Where do we aim?

  24. y0K θ x0K x=0y=0 Shoot Kenny If there is no gravity, shoot right at him! Ball Kenny xB(t)=v0x t xK(t)=x0K yB(t)=v0y t yK(t)=y0K To hit Kenny,

  25. yK θ = ? xK x=0y=0 v0 v0y q v0x Shoot Kenny With gravity? Ball Kenny Breaking v0 into components:

  26. Shoot Kenny At what time does xB=xK? At what time does yB=yK? To hit Kenny, these times must be equal: Same result as before!

  27. Shoot Kenny Why? The ball and Kenny fall with exactly the same acceleration

  28. Example: v0 A ball rolls off a table of height h. The ball has horizontal velocity v0 when it leaves the table. How far away does it strike the ground? How long does it take to reach the ground? h

  29. Example: Gerbil on the Fly A daredevil gerbil, shot from ground level, is observed to have velocity: when he is 9.1 meters above a flat field. 9.1 m What is his maximum height? How far is the total distance (horizontal range)? What is his velocity on impact?

  30. Gerbil: Maximum Height Set x = 0 and y = 0 where he starts from the ground Set t = 0 when y = 9.1 m. Then: y0 = 9.1 m v0y = 6.1 m/s At maximum height: vy = 0 m/s ay = -g

  31. 11 m x Gerbil: Horizontal Range Total time in air = twice the time to fall 11 m At maximum height, The time in the air will then be: On landing: x = v0xtair = (7.6)(3) = 23 meters

  32. vimpact Gerbil: Impact Velocity vx = v0x =7.6 m/s (does not change) vy = 0 at the top Impact velocity is: Initial velocity was:

  33. v a R Uniform Circular Motion Velocity has magnitude and direction, so changing direction also causes an acceleration. Consider the special case of constant speed around a circular path. The magnitude of the “centripetal” acceleration is then, pointing towards the center of the circle of radius R Period (the time to make one revolution):

  34. Example: A runner takes 12 seconds round a 180º circular curve at one end of a race track. The distance covered on the curve is 100 meters. What is her centripetal acceleration?

  35. v = ? R=? Solution Her speed will be given by: The circumference of half a circle is: Finally we can calculate the centripetal acceleration:

  36. vobject/observer = vobject/in-frame +vref. frame vboy vescalator Relative Velocity Observed velocity depends on velocity of observer! If your reference frame (coordinate system) is moving relative to a stationary observer with velocity = vref. frame and an object in your reference frame has vobject/in-frame, then according to observer, the object has velocity Example: A kid running up the down escalator

  37. vtoy/observer vtoy/in-car vcar Example: Toy in Car You are standing on the side of the road, and a man drives by with a young passenger: vcar vtoy/in-car From what you observe, she has quite an arm!

  38. Passing a signpost u x is position in road frame x’ is position in car frame x’=x-ut

  39. y vrow vriver vboat x q Relative Velocity: Rowing a Boat Michael can row a boat at vrow= 3 m/s, and he wants to go straight across a river which flows with vriver = 2 m/s. At what angle should he row? vboat = vrow + vriver Must have vboat only in y-direction to go straight across

  40. vrow vboat = vrow + vriver vboat q vriver He needs vrow,x = - vriver,x Rowing a Boat (Continued) vrow cos q = vriver Note: If vriver > vrow then he cannot go straight across

  41. Example: There is a moving sidewalk in the airport. John does not use it, and takes 150 seconds to walk the length of the terminal. Paul stands on the moving sidewalk, and takes 70 seconds. Ringo walks on the sidewalk at the same pace that John walked beside it. How long will it take Ringo to walk the length of the terminal? a) 150 s – 70 s b) 150 s + 70 s c) Something else

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