2. Rotational Kinematics – its just like linear kinematics except stuff spins instead of going in a straight line…

3. Rotational Displacement,  Consider a disk that rotates from A to B: A B  Angular displacement  : Measured in revolutions, degrees, or radians. 1 rev = 360 0 = 2  rad The best measure for rotation of rigid bodies is the radian.

4. Definition of the Radian One radian is the angle  subtended at the center of a circle by an arc length s equal to the radius R of the circle. s

5. Example 1 A rope is wrapped many times around a drum of radius 50 cm. How many revolutions of the drum are required to raise a bucket to a height of 20 m? h = 20 m R  = 40 rad Now, 1 rev = 2  rad  = 6.37 rev

6. Example 2 A bicycle tire has a radius of 25 cm. If the wheel makes 400 rev, how far will the bike have traveled?  = 2513 rad s =  R = 2513 rad (0.25 m) s = 628 m

7. Angular Velocity Angular velocity,  is the rate of change in angular displacement. (radians per second.)  fAngular frequency f (rev/s).  f  Angular frequency f (rev/s). Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm):  Angular velocity in rad/s.  Angular velocity in rad/s. tttt

8. Example 3 A rope is wrapped many times around a drum of radius 20 cm. What is the angular velocity of the drum if it lifts the bucket to 10 m in 5 s? h = 10 m R  = 10.0 rad/s  tt 50 rad 5 s  = 50 rad

9. Those Oversized Tires! If you drive a vehicle with oversize tires, how is the speedometer affected?

10. Those Oversized Tires! The use of oversized tires has an impact on the speedometer reading. Your vehicle is actually covering greater distance due to the larger circumference of the oversized tires. Your speedometer was calibrated for the smaller tires sold on the vehicle so your speedometer is reading less speed and you are moving faster.

11. If your tires are 10% larger than the original tires on your truck you travel 10% more than your truck recognizes. This has two effects: Your actual speed is 10% MORE than what your speedometer is reflecting. (When your speedometer reads 60, you are actually traveling 66) 10=11, 40=44, 100=110 etc. Your actual mileage is 10% MORE than what your odometer is reflecting. (When your odometer reads 100 miles, you have actually traveled 110 miles.)

12. Example 4 In the previous example, what is the frequency of revolution for the drum? Recall that  = 10.0 rad/s. h = 10 m R f  = 95.5 rpm Or, since 60 s = 1 min:

13. Angular Acceleration Angular acceleration is the rate of change in angular velocity. (Radians per sec per sec.) The angular acceleration can also be found from the change in frequency, as follows:

14. Example 5 The block is lifted from rest until the angular velocity of the drum is 16 rad/s after a time of 4 s. What is the average angular acceleration? h = 20 m R  = 4.00 rad/s 2 0

15. Angular and Linear Speed From the definition of angular displacement: s =  R Linear vs. angular displacement v =  R Linear speed = angular speed x radius

16. Angular and Linear Acceleration: From the velocity relationship we have: v =  R Linear vs. angular velocity a =  R Linear accel. = angular accel. x radius a Δω∙R Δω

17. Example 6: R 1 = 20 cm R 2 = 40 cm   = 0;  f = 20 rad/s t = 4 s What is final linear speed at points A and B? Consider flat rotating disk: v Af = 4 m/s v Af =  Af R 1 = (20 rad/s)(0.2 m); v Af = 4 m/s v Bf = 8 m/s v Bf =  Bf R 2 = (20 rad/s)(0.4 m); v Bf = 8 m/s R1R1 R2R2 A B

18. Acceleration Example 7 R 1 = 20 cm R 2 = 40 cm What is the average angular and linear acceleration at B? R1R1 R2R2 A B   = 0;  f = 20 rad/s t = 4 s Consider flat rotating disk:  = 5.00 rad/s 2 a =  R = (5 rad/s 2 )(0.4 m) a  = 2.00 m/s 2

19. Angular vs. Linear Parameters Angular acceleration is the time rate of change in angular velocity. Recall the definition of linear acceleration a from kinematics. But, a =  R and v =  R, so that we may write: becomes

20. Trains ride on a pair of tracks. For straight-line motion, both tracks are the same lengths. But which track is longer for a curve, the one on the outside or the one on the inside of the curve?

21. The outside track is longer, just as a circle with a greater radius has a longer circumference

22. A Comparison: Linear vs. Angular d d

23. Linear Example: A car traveling initially at 20 m/s comes to a stop in a distance of 100 m. What was the acceleration? 100 m v o = 20 m/s v f = 0 m/s Select Equation: a = = 0 - v o 2 2s -(20 m/s) 2 2(100 m) a = -2.00 m/s 2 V 0 = 20m/s d = 100 m V f = 0 m/s

24. Example 8: A disk (R = 50 cm), rotating at 600 rev/min comes to a stop after making 50 rev. What is the acceleration? Select Equation:  = = 0 -  o 2 2  -(62.8 rad/s) 2 2(314 rad)  = -6.29 rad/s 2  o = 600 rpm  f = 0 rpm  = 50 rev 50 rev = 314 rad

25. Problem Solving Strategy:  Draw and label sketch of problem.  Indicate + direction of rotation.  List givens and state what is to be found. Given: ____, _____, _____ ( ,  ,  f, ,t) Find: ____, _____  Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

26. Example 9 A drum is rotating clockwise initially at 100 rpm and undergoes a constant counterclockwise acceleration of 3 rad/s 2 for 2 s. What is the angular displacement?  = -14.9 rad Given: Given:  o = -100 rpm; t = 2 s  = +3 rad/s 2  = -20.9 rad + 6 rad Net displacement is clockwise (-) R 

27. Bet you can’t Center of gravity or Center of mass is the point at the center of an object’s weight distribution, where the force of gravity can be considered to act.

28. If the Center of Mass is NOT supported, an object will topple.

29. A Force that cause rotation about an axis is TORQUE

30. TORQUE The pivot is in the hinges and the force needed to cause it to rotate is dependent on … 1) the amount of force 2) the angle of the force 3) the distance from the pivot Torque = Force perpendicular x lever arm  = F r  Units = N.m Pushing on a door is an example of a force that causes rotation.

31. The same torque can be produced by a large force with a short lever arm, or a small force with a long lever arm. Picture of lever arms from page 151 Although the magnitudes of the applied forces are the same, the torques are different. Only the component of forces perpendicular to the lever arm contribute to torque.

32. Let’s go back to the door If the force is perpendicular then the lever arm (r) is the distance from the force to the pivot. If the force is along the arm then, no rotation occurs so no torque.

33. If you apply the force at an angle, then the lever arm will be smaller and the torque less. So how can we use this… Say the bolt is too tight and you can’t get it to turn, what can you do?

34. Example 10 Example 10: What torque is necessary to lift a 10 N dumbbell by your muscles 0.80 m away?   = F r   = 10 N x 0.80 m  = 8 Nm  F  = 10 N r = 0.80m

35. Let’s review Equilibrium The condition of equilibrium exists where the resultant of all external forces is zero. We have used this relationship to solve problem where the objects were stationary or at rest.

36. First Condition of Equilibrium: A body is in translational equilibrium if and only if the vector sum of the forces acting upon it is zero. Σ F x = 0 and Σ F y = 0

37. Second Condition of Equilibrium: An object is in rotational equilibrium when the sum of the forces and torques acting on it is zero. Σ τ = 0 By choosing the axis of rotation at the point of application of an unknown force, problems may be simplified.

38. We will solve torque problems by using the idea that the clockwise torque balances the counterclockwise torque when an object is in rotational equilibrium. Counterclockwise TorqueClockwise Torque  = 0  clockwise =  counterclockwise

39. Example 11: The 60 kg diver is standing 1.20 m from the pivot. What force must be exerted by the bolt 0.80 m away from the pivot? Assume the board is supported at its center of mass. 0.8 m1.20 m  clockwise  counterclockwise = Fr ccw = Fr cw F (0.80 m) = (600 N)(1.20 m) F = 900 N r cw = 1.2 m r ccw = 0.8 m F cw = 600 N F ccw = ??

40. Example 12: What is the mass required to balance this system?

41. Example 13: A uniform beam weighing 200 N is held up by two supports A and B. Given the distances and forces listed find the forces exerted by the supports.

44. The dumbells below are all the same size, and the forces applied all have the same magnitude.Rank in order from largest to smallest, the torques created ( τ 1, τ 2 or τ 3 )

45. τ 1 > τ 2 = τ 3 An equal force will lead to a greater torque if applied at a right angle. A torque applied at 90 degrees – θ creates the same torque as 90 degrees + θ.

47. Comparing rotational and linear motion

48. Angular Velocity Angular velocity,  is the rate of change in angular displacement. (radians per second.)  fAngular frequency f (rev/s).  f  Angular frequency f (rev/s). Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm):  Angular velocity in rad/s.  Angular velocity in rad/s. tttt

49. Angular Acceleration Angular acceleration is the rate of change in angular velocity. (Radians per sec per sec.) The angular acceleration can also be found from the change in frequency, as follows:

50. Torque and Angular Acceleration When an object is subject to a net force, it undergoes an acceleration. (Newton’s 2 nd ) When a rigid object is subject to a net torque, it undergoes an angular acceleration When a rigid object is subject to a net torque, it undergoes an angular acceleration. Force and Linear Acceleration

51. Inertia of Rotation Consider Newton’s second law for the inertia of rotation F = 20 N a = 4 m/s 2 Linear Inertia, m = F/a m = = 5 kg 20 N 4 m/s 2 F = 20 N R = 0.5 m  = 2 rad/s 2 Force Force does for translation what torque torque does for rotation : Rotational Inertia, I I = = = 2.5 kg m 2 (20 N)(0.5 m) 2 rad/s 2 

52. Moment of Inertia This mass analog is called the moment of inertia, I, of the object Is defined relative to rotation axis SI units are kg m 2

53. More About Moment of Inertia I depends on both the mass and its distribution. If an object’s mass is distributed further from the axis of rotation, the moment of inertia will be larger.

54. Some dragsters are built so that the front wheels are far ahead of the rear wheels. The main reason for this is a. to streamline the dragster b. to provide better traction. c. to keep the front of the car down.

55. Dragsters baby!! The front wheels are so far forward to help prevent the front from raising off the ground. It can produce a counter torque to the driving force of the wheels.

56. Concept Question: Cylinder Race Two cylinders of the same size and mass roll down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which reaches the bottom first? 1) A 2) B 3) Both at the same time.

57. Concept Q. Ans.: Cylinder Race Answer 2: Because the moment of inertia of cylinder B is smaller, more of the mechanical energy will go into the translational kinetic energy hence B will have a greater center of mass speed and hence reach the bottom first.

58. Common Moments of Inertia Common moments of inertia are on page 251.

59. Example 14 A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. R I = mR 2 Hoop R I = ½mR 2 Disk I = 0.120 kg m 2 I = 0.0600 kg m 2

60. Reminders and info Should you take the AP test?

61. Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. x f R 4 kg     50 rad/s  = 40 N m A resultant force F produces negative acceleration a for a mass m. I m A resultant torque  produces angular acceleration  of disk with rotational inertia I.

62. Example 15 Treat the spindle as a solid cylinder. a) What is the moment of Inertia of the spindle? b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? d) What is the mass of the bucket? M

63. Solution a) What is the moment of Inertia of the spindle? Given: M = 5 kg, R = 0.6 m M = 0.9 kgm 2

64. Solution b) If the tension in the rope is 10 N, what is  ? Given: I = 0.9 kg m 2, T = 10 N, r = 0.6 m M  (0.6m)(10 N)/(0.9 kg∙m 2 )  = 6.67 rad/s 2 c) What is the acceleration of the bucket? Given: r=0.6 m,  = 6.67 rad/s a=4 m/s 2 a = (6.67 rad/s 2 )(0.6 m)

65. Solution d) What is the mass of the bucket? Given: T = 10 N, a = 4 m/s 2 M M = 1.72 kg

66. Concept Question: Cylinder Race Different Masses Two cylinders of the same size but different masses roll down an incline, starting from rest. Cylinder A has a greater mass. Which reaches the bottom first? 1) A 2) B 3) Both at the same time.

67. Concept Q. Ans.: Cylinder Race Different Masses The initial mechanical energy is all potential energy and hence proportional to mass. When the cylinders reach the bottom of the incline, both the mechanical energy consists of translational and rotational kinetic energy and both are proportional to mass. So as long as mechanical energy is constant, the final velocity is independent of mass.So both arrive at the bottom at the same time.

68. Other applications of Torque https://mail.google.com/mail/u/0/#inbox/14 a0b51b82af3d78?projector=1https://mail.google.com/mail/u/0/#inbox/14 a0b51b82af3d78?projector=1

69. Combined Rotation and Translation v cm First consider a disk sliding without friction. The velocity of any part is equal to velocity v cm of the center of mass.  v R P Now consider a ball rolling without slipping. The angular velocity  about the point P is same as  for disk, so that we write: Or

70. Two Kinds of Kinetic Energy  v R P Kinetic Energy of Translation: K = ½mv 2 Kinetic Energy of Rotation: K = ½I  2 Total Kinetic Energy of a Rolling Object: KE of center-of-mass motion KE due to rotation

71. Example 16 What is the kinetic energy of the Earth due to the daily rotation? Given: M earth =5.98 x10 24 kg, R earth = 6.63 x10 6 m. First, find  = 7.27 x10 -5 rad/s = 2.78 x10 29 J

72. Summary – Rotational Analogies QuantityLinearRotational DisplacementDisplacement x Radians  InertiaMass (kg)I (kg  m 2 ) ForceNewtons NTorque N·m Velocity v “ m/s ”  Rad/s Acceleration a “ m/s 2 ”  Rad/s 2 Momentum mv (kg m/s) I  (kg  m 2  rad/s)

73. Analogous Formulas Linear MotionRotational Motion F = ma  = I  K = ½mv 2 K = ½I  2 Work = Fx Work =  Power = Fv Power = I  Fx = ½mv f 2 - ½mv o 2  = ½I  f 2 - ½I  o 2

74. Example 17 A solid sphere rolls down a hill of height 40 m. What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!) v = 23.7 m/s

75. Angular Momentum Analogy between L and p Angular MomentumLinear momentum L = I  p = mv  =  L/  tF =  p/  t Conserved if no net outside torques Conserved if no net outside forces Rigid body Point particle

76. Example 18 A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

77. Solution Known: M, R, m, v 0 Find:  F First, find L 0 Next, find I tot Now, given I tot and L 0, find  = 2.71 rad/s

78. An ice skater spins with her arms folded. When she extends her arms outward her angular velocity a. increases b. decreases a. remains the same

79. If the skater extends her arms her radius becomes greater and she has a greater momentum of inertia. A greater momentum of inertia causes her to have less ω due to conservation of momentum. If her new I is greater, her new ω must be smaller.

80. Summary of Formulas: I =  mR 2 mgh o ½    ½mv o 2 = mgh f ½  f  ½mv f 2 Height?Rotation?velocity?Height?Rotation?velocity? Conservation: