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Introduction to 2-Dimensional Motion

Introduction to 2-Dimensional Motion. Vectors . Two vectors are equal if they have the same magnitude (how big) and same direction. How to add vectors: Line them up head to tail Draw a vector that connects the tail of the first arrow to the head of the last arrow

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Introduction to 2-Dimensional Motion

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  1. Introduction to 2-Dimensional Motion

  2. Vectors • Two vectors are equal if they have the same magnitude (how big) and same direction. • How to add vectors: • Line them up head to tail • Draw a vector that connects the tail of the first arrow to the head of the last arrow • Make sure your calculator is set to DEGREES • Always use UNITS

  3. 2-Dimensional Motion • Definition: motion that occurs with both x and y components. • Example: • Playing pool . • Throwing a ball to another person. • Each dimension of the motion can obey different equations of motion.

  4. Solving 2-D Problems • Resolve all vectors into components • x-component • Y-component • Work the problem as two one-dimensional problems. • Each dimension can obey different equations of motion. • Re-combine the results for the two components at the end of the problem.

  5. Sample Problem • You run in a straight line at a speed of 5.0 m/s in a direction that is 40o south of west. • How far west have you traveled in 2.5 minutes? • How far south have you traveled in 2.5 minutes?

  6. Sample Problem • You run in a straight line at a speed of 5.0 m/s in a direction that is 40o south of west. • How far west have you traveled in 2.5 minutes? • How far south have you traveled in 2.5 minutes? v = 40 m/s

  7. Sample Problem • You run in a straight line at a speed of 5.0 m/s in a direction that is 40o south of west. • How far west have you traveled in 2.5 minutes? • How far south have you traveled in 2.5 minutes? v = 5 m/s, Q = 40o, t = 2.5 min = 150 s vx= v cosQvy= v sin Q vx= 5 cos 40vy= 5 sin 40 vx= vy= x = vxt y = vyt x = ( )(150) y = ( )(150) x = y = vx vy v = 5 m/s

  8. Sample Problem A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2. • What are the x and y positions at 5.0 seconds?

  9. Sample Problem A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2. • What are the x and y positions at 5.0 seconds? Vo,y = 6.2 m/s, Vo,x = 0 m/s, t = 5 s, ax = -4.4 m/s2, ay = 0 m/s2 x = ? y = ? x = vo,x + at y = vo,y + at x = 0 + (-4.4)(5) y = 6.2 + (0)(5) x = y =

  10. Sample Problem A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2. • What are the x and y components of velocity at this time?

  11. Sample Problem A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2. • What are the x and y components of velocity at this time? vx = vo,x+ axtvy = voy + axt vx = 0+ (-4.4)(5)vy = 6.2+ 0(5) vx = vy =

  12. Mechanics Quiz • A mass experiences a force vector with components 30N to the right, and 40 N down . ExplainN how to determine the magnitude and direction (angle) of the force vectors. • The resultant vector can found graphically or mathematically. • 30N2 + 40N2 = 2500N2 = 50N

  13. Projectiles

  14. Projectile Motion • Something is fired, thrown, shot, or hurled near the earth’s surface. • Horizontal velocity is constant. • Vertical velocity is accelerated. • Air resistance is ignored.

  15. 1-Dimensional Projectile • Definition: A projectile that moves in a vertical direction only, subject to acceleration by gravity. • Examples: • Drop something off a cliff. • Throw something straight up and catch it. • You calculate vertical motion only. • The motion has no horizontal component.

  16. 2-Dimensional Projectile • Definition: A projectile that moves both horizontally and vertically, subject to acceleration by gravity in vertical direction. • Examples: • Throw a softball to someone else. • Fire a cannon horizontally off a cliff. • Shoot a monkey with a blowgun. • You calculate vertical and horizontal motion.

  17. Horizontal Component of Velocity • Is constant • Not accelerated • Not influence by gravity • Horizontal component velocity vx = v(cosQ) • Follows equation: • x = Vo,xt

  18. Horizontal Component of Velocity

  19. Vertical Component of Velocity • Undergoes accelerated motion • Accelerated by gravity (9.8 m/s2 down) • Vertical component of velocity vy = v(sinQ) • Vy = Vo,y - gt • y = yo + Vo,yt - 1/2gt2 • Vy2 = Vo,y2 - 2g(y – yo)

  20. Horizontal and Vertical

  21. Horizontal and Vertical

  22. Zero Launch Angle Projectiles

  23. Launch angle • Definition: The angle at which a projectile is launched. • The launch angle determines what the trajectory of the projectile will be. • Launch angles can range from -90o (throwing something straight down) to +90o (throwing something straight up) and everything in between.

  24. vo Zero Launch angle • A zero launch angle implies a perfectly horizontal launch.

  25. Sample Problem • The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

  26. Sample Problem • The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops. yo = 108 m, y = 0 m, g = -9.8 m/s2, vo,x = 3.6 m/s v = ?

  27. Sample Problem • The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops. yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s v = ? Gravity doesn’t change horizontal velocity. vo,x = vx =3.6 m/s Vy2 = Vo,y2 - 2g(y – yo) Vy2 = (0)2 – 2(9.8)(0 – 108) Vy =

  28. Sample Problem • The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops. yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s v = ? Gravity doesn’t change horizontal velocity. vo,x = vx =3.6 m/s Vy2 = Vo,y2 - 2g(y – yo) v = Vy2 = (0)2 – 2(9.8)(0 – 108) Vy =

  29. Sample Problem • An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?

  30. Sample Problem • An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon? vo,x = 6.75 m/s, x = 8.95 m, y = 0 m, yo = 1.2, Vo,y = 0 m/s g = ? y = yo + Vo,yt - 1/2gt2 g = -2(y - yo - Vo,yt)/t2 g = -2[0 – 1.2 – (0)( )]/( )2 g = x = vo,xt t = x/vo,x t = 8.95/6.75 t =

  31. Sample Problem • Playing shortstop, you throw a ball horizontally to the second baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later. • How far were you from the second baseman? • What is the distance of the vertical drop? Should be able to do this on your own!

  32. General Launch Angle Projectiles

  33. vo  General launch angle • Projectile motion is more complicated when the launch angle is not straight up or down (90o or –90o), or perfectly horizontal (0o).

  34. vo  General launch angle • You must begin problems like this by resolving the velocity vector into its components.

  35. Vo,y = Vo sin  Vo,x = Vo cos  Resolving the velocity • Use speed and the launch angle to find horizontal and vertical velocity components Vo 

  36. Vo,y = Vo sin  Vo,x = Vo cos  Resolving the velocity • Then proceed to work problems just like you did with the zero launch angle problems. Vo 

  37. Sample problem • A soccer ball is kicked with a speed of 9.50 m/s at an angle of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?

  38. Sample problem • A soccer ball is kicked with a speed of 9.50 m/s at an angle of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air? vo = 9.5 m/s, q = 25o, g = -9.8 m/s2, Remember: because it lands at the same height: Dy = y – yo = 0 m and vy =- vo,y Find: Vo,y = Vo sin  and Vo,x = Vo cos  Vo,y = 9.5 sin 25 Vo,x = Vo cos  Vo,y = Vo,x = t = ? Vy = Vo,y - gt t = (Vy - Vo,y )/g t = [( ) – ( )]/9.8 don’t forget vy =- vo,y t =

  39. Sample problem • Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25o above the horizontal. When the snowballs land, is the speed of A greater than, less than, or the same speed of B? Verify your answer by calculation of the landing speed of both snowballs. • We’ll do this in class.

  40. Projectiles launched over level ground • These projectiles have highly symmetric characteristics of motion. • It is handy to know these characteristics, since a knowledge of the symmetry can help in working problems and predicting the motion. • Lets take a look at projectiles launched over level ground.

  41. y x Trajectory of a 2-D Projectile • Definition: The trajectory is the path traveled by any projectile. It is plotted on an x-y graph.

  42. y x Trajectory of a 2-D Projectile • Mathematically, the path is defined by a parabola.

  43. y x Trajectory of a 2-D Projectile • For a projectile launched over level ground, the symmetry is apparent.

  44. y x Range of a 2-D Projectile • Definition: The RANGE of the projectile is how far it travels horizontally. Range

  45. y x Maximum height of a projectile • The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward. Maximum Height Range

  46. y x Maximum height of a projectile • The vertical velocity component is zero at maximum height. Maximum Height Range

  47. y x Maximum height of a projectile • For a projectile launched over level ground, the maximum height occurs halfway through the flight of the projectile. Maximum Height Range

  48. y g g g g g x Acceleration of a projectile • Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.

  49. y x Velocity of a projectile • Velocity is tangent to the path for the entire trajectory. v v v vo vf

  50. y x Velocity of a projectile • The velocity can be resolved into components all along its path. vx vx vy vy vx vy vx vy vx

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