1. Chapter 7 Rotational Motion and the Law of Gravity

2. 7–1 Measuring Rotational Quantities

3. Rotational Motion Motion of a body that spins about an axis.

4. Axis of rotation Is the line about which the rotation occurs.

5. Angles can be measured in radians (rads). Radians – an angle whose arc length is equal to its radius, which approximately equal to 57.3 degrees.

6. 1 radian = 57.3 o s  = ---- r

7. 1 radian = 57.3 o s  = ---- r s = arc length r = length of radius  = angle of rotation

8. The angle of 360 o is one revolution. (1 rev = 360 o ) One revolution is equal to the circumference of the circle of rotation.

9. Circumference is 2  r so, 1 rev = 360 o = 2  rads Radius is half the diameter (d).

11. Angular Displacement (  ) The angle through which a point, line, or body is rotated in a specified direction and about a specified axis.

12. Describes how far an object has rotated. Change Angular in arc length Displacement = -------------- radius

14. Counterclockwise (CCW) rotation is considered (+) Clockwise (CW) rotation is considered (-)

15. Ex 1: John Glenn, in 1962, circled the earth 3 times in less than 5 hours. If his distance from the center of the earth was 6560 km, what arc length did he travel through?

16. G:  = 3 rev = 6  rads, r = 6560km U:  s = ? E:  s =  r S:  s = (6  )(6560) S:  s = 123653 km

17. Angular Speed  The rate at which a body rotates about an axis, usually expressed in radians per second. (rads/sec)

18. Angular Average displacement Angular = ------------------- Speed time interval

20. Ex 2: In 1975, an ultra-fast centrifuge attained an average angular speed of 2.65 x 10 4 rads/sec. What was the angular displacement after 1.5 sec?

21. G:  avg = 2.65 x 10 4 rads/sec,  t = 1.5 sec U:  = ? E:  =  avg  t S:  = (2.65 x 10 4 )(1.5) S:  = 39750 radians

22. How fast is the earth rotating? The Earth rotates at a moderate angular velocity of 7.2921159 ×10 -5 rads/s

23. Ex 3: A water supply tunnel in NYC is 1.70 x 10 5 m long and a radius of 2 m. If an electronic eye sensor moves around the perimeter of the tunnel at an average angular speed of 5.9 rads/sec. How long will it take the beetle to travel the equivalent of the length of the tunnel?

24. G:  avg = 5.9 rads/sec, r = 2 m,  s = 1.70 x 10 5 m U:  t = ? E:  t =  /  avg We need to find .

25.  =  s / r  = 1.70 x 10 5 m / 2 m  = 85000 rads

26. S:  t = 85000 rads 5.9 rads/sec

27. S:  t = 85000 rads 5.9 rads/sec S:  t = 14406.8 sec

28. Angular Acceleration  The time rate of change of angular speed, expressed in rads/sec 2.

29.  f  i   t

30.  f  final angular speed  i  initial angular speed  angular acceleration  t = elapsed time

31. Equation is rearranged just like the linear acceleration equation: ALWAYS multiply by ‘  t’ first! Then isolate what you are looking for.

32. Ex 4: A brake is applied to a flywheel for 5 seconds, reducing the angular velocity of the wheel uniformly from 2.2 rads/sec to 1.8 rads/sec. What is the angular acceleration?

33. G:  t = 5 sec,  2  rads/sec, and  1  rads/sec U:  E:  2  1   t

34. 1.8 rads/sec – 2.2 rads/sec  = ------------------------ 5 sec  = - 0.08 rads/sec 2

35. In comparing angular and linear quantities, they are similar. LinearAngular Rotational Displacement  x (m)  rads  Velocity v (m/s)  rads/s  Acceleration a  m/s 2  rads/s 2 

36. Looking back to Ch 2 equations:  x = v i  t + ½ a  t 2  x = ½ (v i + v f )  t v f = v i + a  t v f 2 = v i 2 + 2a  x

37. We apply angular notation to these equations and get:  =  i  t + ½   t 2  = ½ (  i +  f )  t  f =  i +  t  f 2 =  i 2 + 2 

38. Ex 5: A wheel on a bike moves through 11.0 rads in 2 seconds. What is the angular acceleration of the initial angular speed is 2.0 rads/sec?

39. G:  = 11 rads,  t = 2 sec,  i = 2 rads/sec U:  = ? E:  =  i  t + ½  t 2  = 2(  i  t) /  t 2 S:  = 2[11 – (2)(2)] /2 2 S:  = 3.5 rads/sec 2

40. 7-2 Tangential and Centripetal Acceleration

41. Objects in motion have a tangential speed. Tangent – a line that lies in a plane of circle that intersects at one point.

42. Tangent line

44. Tangential Speed (v t ) The instantaneous linear speed of an object directed along the tangent to the object’s path.

45. r = distance from axis  = angular speed v t = tangential speed

46. Ex 6: The radius of a CD is 0.06 m. If a microbe riding on the disc’s rim has a tangential speed of 1.88 m/s, what is the microbe’s angular speed?

47. G: r =0.06 m, v t =1.88 m/s U:  = ? E:  = v t / r S:  = 1.88 m/s / 0.06 m S:  = 31.3 rads/sec

48. Tangential Acceleration (a t ) The instantaneous linear acceleration of an object directed along the tangent to the objects circular path

49. a t = tangential acceleration r = distance from axis  = angular acceleration

50. Ex 7: To ‘crack’ a whip requires its tip to move at supersonic speed. This was achieved with a whip 56.24 m. If the tip moved in a circle and it accelerated from 6.0 rads/sec to 6.3 rads/sec in 0.6 sec. What would the tangential acceleration be?

51. G: r =56.24 m,  t =0.6 sec  i = 6 rads/sec,  f = 6.3 rads/sec, U: a t = ? E: a t = r 

52. We need to find angular acceleration (a) E:  f  i  t

53. S:   0.6 S:  rads/sec 2

54. S: a t = (56.24)(0.5) S: a t = 28.12 m/s 2

55. Centripetal Acceleration (a c ) Acceleration directed toward the center of a circular path.

56. r = distance from axis (m) a c = centripetal accel. (m/s 2 ) v t = tangential speed (m/s)

57. r = distance from axis (m) a c = centripetal accel. (m/s 2 )  = angular speed (rads/sec)

58. Ex 8:Calculate the orbital radius of the Earth, if its tangential speed is 29.7 km/s and the centripetal acceleration acting on the Earth is 5 x 10 -3 m/s 2 ?

59. G: v t =29.7 km/s = 29700 m/s a c = 5 x 10 -3 m/s 2 U: r = ? E: v t 2 r = ---- acac

60. S:29700 2 r = ------------ 5 x 10 -3 S: r =1.764 x 10 11 m

61. Tangential and centripetal accelerations are perpendicular. acac atat a total

62. The total acceleration can be found by using the Pythagorean Theorem.

64. If you need to find the angle  use the inverse tangent function (tan -1 )

65. 7-3 Causes of Circular Motion

66. If a mass undergoes an acceleration, there must be a force causing it. (Newton’s 2 nd Law)

67. A force directed toward the center is necessary for circular motion.

68. This force causing circular motion is called Centripetal Force (F c ) r m v FcFc

69. With a centripetal force, an object in motion will be accelerated and change its direction.

70. Work Done by the Centripetal Force Since the centripetal force on an object is always perpendicular to the object’s velocity, the centripetal force never does work on the object - no energy is transformed.

71. Without a centripetal force, an object in motion continues along a straight- line path.

75. “Centrifugal Force  “centrifugal force” is a fictitious force - it is not an interaction between 2 objects, and therefore not a real force. Nothing pulls an object away from the center of the circle.

76. “Centrifugal Force” What is erroneously attributed to “centrifugal force” is actually the action of the object’s inertia - whatever velocity it has (speed + direction) it wants to keep.

77. From Newton’s 2 nd Law F = ma F c = ma c

78. From Newton’s 2 nd Law F = ma F c = ma c

79. From Newton’s 2 nd Law F = ma F c = ma c

80. Ex 9: Joe attaches a 0.5 kg mass to a 1m rope. Joe swings the rope in a horizontal circle above his head with a speed of 4 m/s. What force is exerted to keep the mass moving in a circle?

81. G:m = 0.5 kg v t = 4 m/s r = 1 m U: F c = ? E:F c = m(v t 2 / r)

82. S: F c =.5 kg x(4 m/s) 2 /(1 m) S: F c = 8 N

83. Gravitational Force The mutual force of attraction between particles of matter.

84. This is a field force that always exists between any two objects, regardless of their sizes.

85. The gravitational force (F g ) depends upon the distance between the two masses.

86. Newton’s Law of Universal Gravitation

87. m 1 = mass 1 (kg) m 2 = mass 2 (kg) r = distance between center of masses. (m) G = Constant of Universal Gravitation.

88. Units will cancel out with others leaving just force units of Newtons (N).

89. The universal gravitational constant (G) can be used to find the gravitational force between any two particles.

90. This is what is known as the inverse-square law. The force between the two masses decreases as the masses move farther apart.

91. The acceleration due to gravity, ‘g’, decreases with distance from the earth’s center as 1/d 2.

92. As an object moves away from the center of the Earth, by a distance equivalent to the earth’s radius.

93. Distance Force of Gravity RERE FgFg 2 R E 1/4 F g 3 R E 1/9 F g 4 R E 1/16 F g 5 R E 1/25 F g

94. Ex 10: What is the force of attraction between Adam and Jason, gravitational of course. If they are 12 m apart and Jason mass is 82 kg and Adam is 74 kg.

95. G: m 1 = 82 kg, m 2 = 74kg, r = 12 m, G = 6.673x 10 -11 n-m 2 /kg 2 U: F g = ?

96. G: m 1 = 82 kg, m 2 = 74kg, r = 12 m, G = 6.673x 10 -11 n-m 2 /kg 2 U: F g = ? E:

98. S: F g = 2.81 x 10 -9 N

99. Ex 11: The sun has a mass 2 x 10 30 kg and a radius of 7 x 10 5 km. What mass must be located on the sun’s surface for gravitational force of 470 N to exist between the mass and the sun?

100. G: m 1 = 2 x 10 30 kg, r =7 x 10 5 km=7 x 10 8 m G = 6.673x 10 -11 n-m 2 /kg 2 F g = 470 N U: m 2 = ?

101. E:

103. S: m 2 = 1.7 kg

105. PlanetRadius (m)Mass (kg)g (m/s 2 ) Mercury2.43 x 10 6 3.2 x 10 23 3.61 Venus6.073 x 10 6 4.88 x10 24 8.83 Mars3.38 x 10 6 6.42 x 10 23 3.75 Jupiter6.98 x 10 7 1.901 x 10 27 26.0 Saturn5.82 x 10 7 5.68 x 10 26 11.2 Uranus2.35 x 10 7 8.68 x 10 25 10.5 Neptune2.27 x 10 7 1.03 x 10 26 13.3 Pluto1.15 x 10 6 1.2 x 10 22 0.61