Chapter 6 Motion in Two Dimensions

1. Chapter 6Motion in Two Dimensions Merrill Physics: Principles and Problems

2. A. Motion in two dimensions 1. Will use the kinematic equations: d = vt d = vot + 1/2 at 2 vf2 = vo2 + 2ad vf = vo + at d = {(vf + vo ) / 2}(t)

3. 2. Will use Newton’s Laws 1st: An object in motion remains in motion unless acted on by a net force 2nd : F = m a where F is net force 3. And coefficients of friction:  ff / F N

4. Projectile Motion Objectives: Understand that the vertical and horizontal components of a projectile are independent Determine the range, maximum height and flight time of projectiles *As you throw a ball, or any type of object, what shape does it make in the air?

5. Definitions used for Projectile Motion (Ballistics) a. Projectile - object that only has gravity acting on it b. Trajectory - path a projectile follows c. Shape of the trajectory depends on your point of view and the object.

6. (1) Top looking down on object’s path starting point ending point What force accelerates the object in the x-direction? (None once you let go or the bullet is fired) *this assumes no air resistance for simplicity So velocity is constant in the x-direction

7. (2) Side view of object’s motion start end What force accelerates in the y-direction? Gravity! So VY is a changing value.

8. 4. Three assumptions used in analyzing motion Assumptions: (1) acceleration due to gravity is 9.8 m/s2 down (2) air resistance is negligible (3) rotation of the earth does not affect motion This is not rocket science! …. Otherwise we would include these!

9. Independence of Motion in Two Dimensions What forces act on an object after it’s initial thrust if you ignore air resistance?

10. Projectiles launched at an angle Objectives: • Determine the range, maximum height and flight time of projectiles launched at an angle

11. Projectiles Launched Horizontally

12. Example: A stone is thrown horizontally off a 44m cliff at velocity of 15 m/s. How long will it take to hit the bottom? How far will it go horizontally if the air resistance is negligible? Given: vx = 15 m/s dx = vxt = 15m(t) voy= 0m/s dy = voyt + ½at2 ay = -9.8 m/s2 44m = ½(9.8m/s2)t2 * Time is the same in the vertical and horizontal directions! t2 = 2(44m)/9.8m/s2 t = 3s and 15m/s(3s) = 45m

13. Projectiles Launched at an Angle

14. b. (1) velocity in x-directionnever changes (2) velocity in y-direction changes vy-f = vy-o + g t vy-f2 = vy-o2 + 2 gdy (3) vR2 = vx2 + vy2

15. Projectile Motion Problem Solving a. Projectile is an object that only has gravity acting on it b. Break given initial velocity into x and y components. v at an angle q (1)horizontal motion: vx = v cosq (2)vertical motion: vy = v sinq c. use subscripts to identify x & y components of velocity, displacement, acceleration

16. d. Use independence of vertical and horizontal motion to solve problems e. Draw diagrams to aid you in visualizing what is occurring f. Write down given information with signs g. Note: ax = 0 ay = - 9.8 m/s2 - time is the same for both directions - vertical motion is symmetrical!

17. Velocity of the Projectile • The velocity of the projectile at any point of its motion is the vector sum of its x and y components at that point • Remember to be careful about the angle’s quadrant

18. Example: Object is launched at an angle of 60o at a velocity of 110 m/s. How far will it go if the air resistance is negligible? Given: vx = 110 cos 60 = 55 m/s voy= 110 sin 60 = 95.3 m/s vfy = - 95.3 m/s ay = -9.8 m/s2 * Commonality is with TIME. How long it is in the air is same for horizontal and vertical motion. Knowing time, you can find distance in x

19. vyf = v0y + ayt t = 19.4 sec use travel in the vertical direction to determine how long object is in flight. Remember, object is accelerated in the y-direction. Knowing time then can find out how far vx = dx / t since uniform motion in x-direct. dx = (55 m/s) (19.4 sec) = 1070 meters

20. i. Example Problem: A punter kicks a football at an angle of 30o with the horizontal at an initial speed of 20 m/s. Where should the punt returner position himself to catch the ball just before it strikes the ground?

21. Givens: vy = 20 m/s sin 30 = 10 m/s vx = 20 cos 30 = 17.3 m/s vy 0 = 10 m/s vyf =-10m/s vyf = vy 0 + at where a=-9.8m/s2 t = 2.04 sec dx = vx t =17.3 m/s (2.04s) = 35.3 m v 30

22. Projectile Motion Simulation labs • http://phet.colorado.edu/en/simulation/projectile-motion • http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html

23. Periodic Motion • Explain centripetal acceleration of an object in circular motion • Apply Newton’s Laws to circular motion • Define simple harmonic motion • Apply Newton’s Laws to harmonic motion

24. B. Periodic Motion 1. Uniform Circular Motion a. Consider a model airplane on control strings - the hand pulls on cable and keeps an inward force on the plane so it flies in a circle. Let go of the string and it flies in a straight line

25. A net force, acting perpendicular to a mass moving at constant speed, changes direction of motion of the mass. (1) Force is always 90o to direction of motion of the mass (2) Since it is perpendicular, cannot change the magnitude of velocity only direction. (3) Net force is centripetal force, Fc.

26. Change in direction is a change in velocity (vso object is accelerated (1)Fc is towards the center, so acceleration is towards the center. (2) Called centripetal acceleration.

27. Centripetal Acceleration Formulas: (1) ac = v2/r (2) Fc = mac = mv2/r

28. Key: Centripetal force is always caused by some other action - friction, pulling, pushing, etc. We define period of circular motion (T) to be the time to go around the circumference one time For uniform circular motion time to go around circle is constant

29. v = d/t (d = the circumference) where d = 2r & t = T (the period) thereforev2r/T and ac = v2/r = 4r/T2 Fnet = Fc = mac = mv2/r = m(4r/T2)

30. Example: A person is flying an 80.0g model airplane in a horizontal path at the end of a string 10.0m long. If the string is horizontal and exerts a 2.65N force on the hand, what is the speed of the plane? Givens: Formulas m= 0.080kg v = 2r/T r = 10.0 m Fc = mac = mv2/r V = (Fcr/m)1/2 = 18.2 m/s

31. A 200g object is tied on a cord and whirled in a horizontal circle of radius 1.20m at 3 rev/sec. The cord is horizontal. Determine acceleration and tension in the cord.

32. Givens: Formulas m= 0.200kg v = 2r/T r = 1.20m rev/sec (freq.)vm/s 1/ =  sec a. ac = v2 / r = 426 m/s2 b. Fc = m ac = 85 N

33. What is the maximum speedat which a car can round a curve of 25m radius on a level road if static between the tires and road is 0.80 Fc = Ff = static FN =static m(-g) Fc = mac = static m(-g) = mv2/r Solving for v: v = (rstatic-g)1/2 Therefore v = (25mx0.80x9.8m/s2)1/2 v = (196m2/s2)1/2 = 14 m/s * the maximum velocity only depends on the radius of the curve and static , not the mass!

34. 11.1Torque To make an object turn or rotate, apply a torque.

35. Torque Every time you open a door, turn on a water faucet, or tighten a nut with a wrench, you exert a turning force. Torque is produced by this turning force and tends to produce rotational acceleration. Torque is different from force. • Forces tend to make things accelerate. • Torques produce rotation.

36. Torque A torque produces rotation.

37. Changing Circular Motion (Torque) • The door is free to rotate about an axis through the hinge at point O • There are three factors that determine the effectiveness of the force in opening the door: • (1) The magnitude of the force • (2) The distance of the application of the force • (3) The angle at which the force is applied

38. Torque Magnitude of force Distance from axis Direction

39. Torque When a perpendicular force is applied, the lever arm is the distance between the doorknob and the edge with the hinges.

40. Torque When the force is perpendicular, the distance from the turning axis to the point of contact is called the lever arm. If the force is not at right angle to the lever arm, then only the perpendicular component of the force will contribute to the torque.

41. Torque The same torque can be produced by a large force with a short lever arm, or a small force with a long lever arm. The same force can produce different amounts of torque. Greater torques are produced when both the force and lever arm are large.

42. Torque t = Frsinθ Units – Nm If θ = 90o Then sinθ = 1 t = Fr or Fd where d = the lever arm length

43. Torque, t, is the tendency of a force to rotate an object about some axis t= Frsin tis the torque F is the force t symbol is the Greek tau rsin is the length of the lever arm acting perpendicular to the force where  is the angle between F and r SI unit is N.m

44. Torque Although the magnitudes of the applied forces are the same in each case, the torques are different.

45. Torque think! If you cannot exert enough torque to turn a stubborn bolt, would more torque be produced if you fastened a length of rope to the wrench handle as shown? Answer: No, because the lever arm is the same. To increase the lever arm, a better idea would be to use a pipe that extends upward.

46. Balanced Torques do the math! What is the weight of the block hung at the 10-cm mark?

47. Balanced Torques do the math! The block of unknown weight tends to rotate the system of blocks and stick counterclockwise, and the 20-N block tends to rotate the system clockwise. The system is in balance (static equilibrium) when the two torques are equal: counterclockwise torque = clockwise torque

48. Balanced Torques Rearrange the equation to solve for the unknown weight: The lever arm for the unknown weight is 40 cm. The lever arm for the 20-N block is 30 cm. The unknown weight is thus 15 N.

49. Torque problems • Complete Gantry Crane Worksheet (Due Thurs 1/7/2014)

50. Simple Harmonic Motion Objectives Identify objects in simple harmonic motion. Determine variables that affect the period of a pendulum.