1. Uniform Circular Motion (UCM)

2. Motion in a circle at a constant speed

3. Uniform Circular Motion (UCM) Motion in a circle at a constant speed

4. Uniform Circular Motion (UCM) Motion in a circle at a constant speed A

5. Uniform Circular Motion (UCM) Motion in a circle at a constant speed What is the direction of the instantaneous velocity at point A? A

6. Uniform Circular Motion (UCM) Motion in a circle at a constant speed Just like projectile motion, the direction of v is along the tangent-line at a point on the particle’s path A vAvA

7. Uniform Circular Motion (UCM) What is the direction of the instantaneous velocity at point B? A vAvA B

8. Uniform Circular Motion (UCM) A vAvA B vBvB

9. How should the length or magnitude of v A compare with the length or magnitude of v B ? Why? A vAvA B vBvB

10. Uniform Circular Motion (UCM) The length or magnitude of v A is the same as the length or magnitude of v B because the speed stays constant. A vAvA B vBvB

11. Uniform Circular Motion (UCM) The length or magnitude of v A is the same as the length or magnitude of v B because the speed stays constant. Even if ǀv A ǀ = ǀv B ǀ and the speed stays the same, why is UCM considered to be acceleration? A vAvA B vBvB

12. Why is UCM considered to be acceleration? Even if ǀv A ǀ = ǀv B ǀ and the speed stays the same, the velocity changes direction. This means there is a non-zero change- in velocity, ∆v. Since a = ∆v/∆t, whenever there is a non-zero change-in velocity, there is a non-zero acceleration. A vAvA B vBvB

13. Deriving a formula to find the magnitude of UCM acceleration

14. Consider a particle undergoing UCM up right

15. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM What is the direction of the instantaneous velocity at point S? up right S r

16. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Right and a little up up right S r vSvS

17. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM What is the direction of the instantaneous velocity at point F? up right S r F r vSvS

18. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Right and a little down up right S r F r vSvS vFvF

19. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM We want to create a a formula for instantaneous a = ∆v/∆t. What must be true about ∆t? up right S r F r vSvS vFvF

20. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM To calculate instantaneous acceleration ∆t must be very small, so… up right S r F r vSvS vFvF

21. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM To calculate instantaneous acceleration ∆t must be very small, so… S and F are actually so close together that the arc distance SF is very nearly the straight line distance SF up right S r F r vSvS vFvF

22. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM To calculate instantaneous acceleration ∆t must be very small, so… S and F are actually so close together that the arc distance SF is very nearly the straight line distance SF We can make a triangle SFO, but keep in mind angle O is actually very small. up right S r F r vSvS vFvF O

23. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM To calculate instantaneous acceleration ∆t must be very small, so… S and F are actually so close together that the arc distance SF is very nearly the straight line distance SF We can make a triangle SFO, but keep in mind angle O is actually very small. up right S r F r vSvS vFvF O

24. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM To calculate instantaneous acceleration ∆t must be very small, so… S and F are actually so close together that the arc distance SF is very nearly the straight line distance SF We can make a triangle SFO, but keep in mind angle O is actually very small. For ∆t very small, straight line SF ≈ ∆d up right S r F r vSvS vFvF O ∆d

25. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so... up right S r F r vSvS vFvF O ∆d

26. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) up right S r F r vSvS vFvF O ∆d

27. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail up right S r F r vSvS vFvF O ∆d

28. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail up right S r F r vSvS vFvF O ∆d vFvF

29. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail up right S r F r vSvS vFvF O ∆d vFvF -v S

30. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

31. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail Note Δ v occurs between S and F, which is the top of the circle. up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

32. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail Note Δ v occurs between S and F, which is the top of the circle. Therefore Δ v has a direction towards the centre of the circle. up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

33. Deriving a formula to find the magnitude of UCM acceleration Consider a particle undergoing UCM Since a = Δ v/∆t, we want to find Δ v But Δ v = v F - v S, so.. Δ v = v F + (- v S ) Now let's arrange tip-to-tail Note Δ v occurs between S and F, which is the top of the circle. Therefore Δ v has a direction towards the centre of the circle. The instantaneous acceleration is also toward the centre. Why? up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

34. Deriving a formula to find the magnitude of UCM acceleration Let's change our Δ v vector diagram to scalars only How does I-v S | compare with |v F |? up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

35. Deriving a formula to find the magnitude of UCM acceleration Let's change our Δ v vector diagram to scalars only Remember I-v S | = |v F | because the speed is constant. up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

36. Deriving a formula to find the magnitude of UCM acceleration Let's change our Δ v vector diagram to scalars only Remember I-v S | = |v F | because the speed is constant. Let's call I-v S | = |v F | = v (speed) up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv

37. Deriving a formula to find the magnitude of UCM acceleration Let's change our Δ v vector diagram to scalars only Remember I-v S | = |v F | because the speed is constant. Let's call I-v S | = |v F | = v (speed) Now we can change our Δ v vector diagram to a triangle with magnitudes only like this... up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv vFvF

38. Deriving a formula to find the magnitude of UCM acceleration Let's change our Δ v vector diagram to scalars only Remember I-v S | = |v F | because the speed is constant. Let's call I-v S | = |v F | = v (speed) Now we can change our Δ v vector diagram to a triangle with magnitudes only like this... up right S r F r vSvS vFvF O ∆d vFvF -v S ΔvΔv v vFvF v |Δ v |

39. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. S r F r O ∆d v v |Δ v |

40. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. If these triangles are similar, then what is |Δ v | = ? v S r F r O ∆d v v |Δ v |

41. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r S r F r O ∆d v v |Δ v |

42. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ? S r F r O ∆d v v |Δ v |

43. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r S r F r O ∆d v v |Δ v |

44. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | Δ t S r F r O ∆d v v |Δ v |

45. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ? Δ t S r F r O ∆d v v |Δ v |

46. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r S r F r O ∆d v v |Δ v |

47. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r But what is S r F r O ∆d v v |Δ v |

48. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r But what is So |a| = ? S r F r O ∆d v v |Δ v |

49. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r But what is So |a| = v v r S r F r O ∆d v v |Δ v |

50. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r But what is So |a| = v v or |a|= ? r S r F r O ∆d v v |Δ v |

51. Deriving a formula to find the magnitude of UCM acceleration We now have two similar triangles. Later for homework, see if you can prove it. |Δ v | = ∆d v r |Δ v | = ∆d v r But |a| = |Δ v | = ∆d v Δ t Δ t r But what is So |a| = v v or |a|= v 2 r r S r F r O ∆d v v |Δ v |

52. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM

53. We have derived |a|= v 2 r where v is the speed and r is the radius

54. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius

55. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the __________ of the _________.

56. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle.

57. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle. But a and Δv are always in the ___________ direction.

58. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle. But a and Δv are always in the same direction. Why?

59. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle. But a and Δv are always in the same direction. This is because a = Δv/Δt, and division of a vector by a positive scalar always results in a new vector in the same direction as the original.

60. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle. But a and Δv are always in the same direction. This is because a = Δv/Δt, and division of a vector by a positive scalar always results in a new vector in the same direction as the original. Therefore the direction of the instantaneous acceleration for UCM is toward the ______ of the ______.

61. Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM We have derived |a|= v 2 Memorize this! r where v is the speed and r is the radius As we determined earlier, for UCM, Δv is always directed toward the center of the circle. But a and Δv are always in the same direction. This is because a = Δv/Δt, and division of a vector by a positive scalar always results in a new vector in the same direction as the original. Therefore the direction of the instantaneous acceleration for UCM is toward the center of the circle.

62. A New Name for the Instantaneous Acceleration of UCM

63. A Latin adjective is used by physicists to help them remember that the direction of the instantaneous acceleration for UCM is always directed “towards the centre of the circle”. Do you know the name of this Latin adjective?

64. A New Name for the Instantaneous Acceleration of UCM A Latin adjective is used by physicists to help them remember that the direction of the instantaneous acceleration for UCM is always directed “towards the centre of the circle”. Instantaneous UCM acceleration is called centripetal acceleration from the Latin petare = “to seek” and centri = “the centre”

65. A New Name for the Instantaneous Acceleration of UCM A Latin adjective is used by physicists to help them remember that the direction of the instantaneous acceleration for UCM is always directed “towards the centre of the circle”. Instantaneous UCM acceleration is called centripetal acceleration from the Latin petare = “to seek” and centri = “the centre” The instantaneous acceleration for UCM will henceforth be referred to as “centripetal” acceleration with new symbol a c. The formula for the magnitude of the centripetal acceleration is now modified as: a c = v 2 / r

66. A New Name for the Instantaneous Acceleration of UCM A Latin adjective is used by physicists to help them remember that the direction of the instantaneous acceleration for UCM is always directed “towards the centre of the circle”. Instantaneous UCM acceleration is called centripetal acceleration from the Latin petare = “to seek” and centri = “the centre” The instantaneous acceleration for UCM will henceforth be referred to as “centripetal” acceleration with new symbol a c. The formula for the magnitude of the centripetal acceleration is now modified as: a c = v 2 / r Memorize please!

67. Other Formulas for UCM centripetal acceleration

68. We can use a c = v 2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called _____________ or the time to complete one cycle and it has the symbol _____.

69. Other Formulas for UCM centripetal acceleration We can use a c = v 2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called period or the time to complete one cycle and it has the symbol T.

70. Other Formulas for UCM centripetal acceleration We can use a c = v 2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called period or the time to complete one cycle and it has the symbol T. For example, the earth going around the sun is approximately UCM with a period of T = _________days.

71. Other Formulas for UCM centripetal acceleration We can use a c = v 2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called period or the time to complete one cycle and it has the symbol T. For example, the earth going around the sun is approximately UCM with a period of T = 365.25 days.

72. Other Formulas for UCM centripetal acceleration We can use a c = v 2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called period or the time to complete one cycle and it has the symbol T. For example, the earth going around the sun is approximately UCM with a period of T = 365.25 days. We can quickly derive a formula for a c in terms of r and T.

73. Deriving a Centripetal Acceleration Formula in terms of T and r

74. The basic constant speed formula is v = Δ d/ Δ t

75. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t For an object undergoing UCM, what formula can we use to calculate the distance Δ d once around the circumference of a circle if we know the radius “r”?

76. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Δ d = 2 π r

77. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Δ d = 2 π r What do we call the time interval Δ t for an object to complete one cycle or revolution?

78. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Δ d = 2 π r Δ t = T

79. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3

80. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1:

81. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4

82. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5

83. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5 Without simplifying, sub Eq #4 into Eq#5:

84. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5 Without simplifying, sub Eq #4 into Eq#5: a c = (2 π r/T) 2 / r

85. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5 Without simplifying, sub Eq #4 into Eq#5: a c = (2 π r/T) 2 / r With pencil, simplify to a two-story expression:

86. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5 Without simplifying, sub Eq #4 into Eq#5: a c = (2 π r/T) 2 / r With pencil, simplify to a two-story expression: a c = 4 π 2 r / T 2

87. Deriving a Centripetal Acceleration Formula in terms of T and r The basic constant speed formula is v = Δ d/ Δ t Eq #1 Δ d = 2 π r Eq #2 Δ t = T Eq #3 Without simplifying, sub equations #2 and #3 into #1: v = 2 π r/T Eq #4 We know a c = v 2 / r Eq #5 Without simplifying, sub Eq #4 into Eq#5: a c = (2 π r/T) 2 / r With pencil, simplify to a two-story expression: a c = 4 π 2 r / T 2 Memorize please!

88. Deriving a Centripetal Acceleration Formula in terms of f and r

89. For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) What is “ f ” called in physics?

90. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency.

91. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the ________.

92. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz).

93. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). How are frequency and period related in an equation?

94. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T

95. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6

96. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2

97. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2 We can write a c = 4 π 2 r T 2

98. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2 We can write a c = 4 π 2 r or = 4 π 2 r (1/T 2 ) T 2

99. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2 We can write a c = 4 π 2 r or = 4 π 2 r (1/T 2 ) T 2 How can we express a c in terms of frequency?

100. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2 We can write a c = 4 π 2 r or = 4 π 2 r (1/T 2 ) T 2 a c = 4 π 2 r f 2

101. Deriving a Centripetal Acceleration Formula in terms of f and r For UCM. sometimes we don't know v or T, but we know f, the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) f is called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s -1 or the Hertz (Hz). Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know a c = 4 π 2 r / T 2 We can write a c = 4 π 2 r or = 4 π 2 r (1/T 2 ) T 2 a c = 4 π 2 r f 2 Memorize please!

102. Review of the Circle Three for UCM

103. For UCM, what is the formula for the magnitude of the centripetal acceleration in terms of speed and radius?

104. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1

105. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 For UCM, what is the formula for the magnitude of the centripetal acceleration in terms of period and radius?

106. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2

107. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 For UCM, what is the formula for the magnitude of the centripetal acceleration in terms of frequency and radius?

108. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 a c = 4 π 2 r f 2 UCM circle equation #3

109. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 a c = 4 π 2 r f 2 UCM circle equation #3 The direction of a c is always directed towards the __________ of the _________.

110. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 a c = 4 π 2 r f 2 UCM circle equation #3 The direction of a c is always directed towards the centre of the circle.

111. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 a c = 4 π 2 r f 2 UCM circle equation #3 The direction of a c is always directed towards the centre of the circle. If we want acceleration in units of m/s 2, we must substitute ________ units into the formula.

112. Review of the Circle Three for UCM a c = v 2 / r UCM circle equation #1 a c = 4 π 2 r / T 2 UCM circle equation #2 a c = 4 π 2 r f 2 UCM circle equation #3 The direction of a c is always directed towards the centre of the circle. If we want acceleration in units of m/s 2, we must substitute MKS units into the formula.

113. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s?

114. Given:

115. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days

116. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days r = 384,000 km

117. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = ? r = 384,000 km

118. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km

119. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = ?

120. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m

121. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Unknown: ?

122. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Unknown: a c

123. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c

124. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c a c = 4 π 2 r / T 2

125. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c a c = 4 π 2 r / T 2 Sub:

126. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c a c = 4 π 2 r / T 2 Sub: a c = 4 π 2 ( 3.84X10 8 ) / ( 2.36X10 6 ) 2

127. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c a c = 4 π 2 r / T 2 Sub: a c = 4 π 2 ( 3.84X10 8 ) / ( 2.36X10 6 ) 2 a c = 2.72 X 10 -3 m/s 2 [toward the centre of the earth]

128. Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to- centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s? Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X10 6 s r = 384,000 km X 1000 m/km = 3.84X10 8 m Formula:Unknown: a c a c = 4 π 2 r / T 2 Sub: a c = 4 π 2 ( 3.84X10 8 ) / ( 2.36X10 6 ) 2 a c = 2.72 X 10 -3 m/s 2 [toward the centre of the earth] Note the moon is accelerating with a very tiny acceleration compared to projectiles near the earth's surface.

129. Try Example #2: A car is moving on a circular track of radius 0.522 km. The magnitude of its centripetal acceleration is 4.00 m/s 2. Find the speed of the car in km/h.

130. Given: r = 0.522 km = 0.522 km X 1000 m/km = 522 m a c = 4.00 m/s 2 Unknown: v =? Formula: a c = v 2 / r or v 2 = r a c or v = (ra c ) 1/2 Sub: v = (522 X 4.00) 1/2 = 45.7 m/s X 3.6 = 165 km/h too fast!

131. Example #3:The planet Mercury moves in an approximately circular path around the sun at an average distance of 5.8 X 10 10 m, accelerating centripetally at 0.040 m/s 2. What is the period of revolution about the sun in days?

132. Given: r = 5.8 X 10 10 m a c = 0.040 m/s 2 Unknown: T =? Formula: a c = 4 π 2 r / T 2 or T = ( 4 π 2 r / a c ) 1/2 Sub: T = ( 4 π 2 ( 5.8 X 10 10 )/ 0.040 ) 1/2 = 7.6 X 10 6 s =7.6 X 10 6 s X 1 hour/3600 s X 1 day/24 hours = 88 days

133. Example #4: A stone is whirled in UCM on a smooth sheet of ice. The stone traces out a circle of diameter 3.0 m as it accelerates centripetally at 93.0 m/s 2. Find the frequency of rotation in Hertz and in rpm (revolutions per minute)

134. Given: r = 3.0 m / 2 = 1.5 m a c = 93.0 m/s 2 Unknown: f = ? Formula: a c = 4 π 2 r f 2 or f = ( a c / (4 π 2 r)) 1/2 Sub: f = ( 93.0 / (4 π 2 (1.5))) 1/2 Answer: f = 1.25 Hz Or 1.25 rev/s X 60 s/ min = 75 rpm