1. Biology – Premed Windsor University School of Medicine and Health Sciences J.C. Rowe Course Instructor.

2. Pre Med – Physics Chapter 4 Rotation and Circular Motion “ The view of your life & choices may change when you look at it from different angles “ Claude Emanuel

4. Our Solar System

7. Chapter 4 Rotational & Circular Motion

8. Circular Motion Terms  The point or line that is the center of the circle is the axis of rotation (pivot point)  If the axis of rotation is inside the object, the object is rotating (spinning).  If the axis of rotation is outside the object, the object is revolving.

10. Linear/Tangential Velocity  Objects moving in a circle still have a linear velocity = distance/time.  This is often called tangential velocity, since the direction of the linear velocity is tangent to the circle. v

11. Rotational/Angular Velocity  Objects moving in a circle also have a rotational or angular velocity, which is the rate angular position changes.  Rotational velocity is measured in degrees/second, rotations/minute (rpm), etc.  Common symbol,  (Greek letter omega)

12. Rotational/Angular Velocity

13. Rotational velocity = Change in angle Time (s)

14. Rotational & Linear Velocity  If an object is rotating:  All points on the object have the same rotational (angular) velocity.  All points on the object do not have the same linear (tangential) velocity.

15. Rotational & Linear Velocity  Linear velocity of a point depends on:  The rotational velocity of the point.  More rotational velocity means more linear velocity.  The distance from the point to the axis of rotation.  More distance from the axis means more linear velocity.

16. Rotational & Linear Velocity  In symbols: v = r  v r 

17. Acceleration  As an object moves around a circle, its direction of motion is constantly changing.  Therefore its velocity is changing.  Therefore an object moving in a circle is constantly accelerating.

18. Centripetal Acceleration  The acceleration of an object moving in a circle points toward the center of the circle.  This is called a centripetal (center pointing) acceleration. a

20. Centripetal Acceleration  The centripetal acceleration depends on:  The speed of the object.  The radius of the circle. A cent = v2v2 r

21. Centripetal Force  Newton’s Second Law says that if an object is accelerating, there must be a net force on it.  For an object moving in a circle, this is called the centripetal force.  The centripetal force points toward the center of the circle.

22. Centripetal Force  In order to make an object revolve about an axis, the net force on the object must pull it toward the center of the circle.  This force is called a centripetal (center seeking) force. F net

23. Centripetal Force  Centripetal force on an object depends on:  The object’s mass - more mass means more force.  The object’s speed - more speed means more force.  And…

24. Centripetal Force  The centripetal force on an object also depends on:  The object’s distance from the axis (radius).  If linear velocity is held constant, more distance requires less force.  If rotational velocity is held constant, more distance requires more force.

25. Centripetal Force  In symbols: F cent = mv 2 r = mr  2

26. 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. v F cent

27. “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.

28. “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.

29. 29 Circular Motion

30. Conceptual Physics Chapter 1030 Rotation and Revolution  Any object that is turning does so about an imaginary straight line called the axis.  If the axis is located within the turning body (internal), the motion is called a rotation.  If the axis is located outside of the turning body (external), the motion is called a revolution.

31. Conceptual Physics Chapter 1031 Rotation and Revolution The disc ROTATES about its internal axis While the bug REVOLVES about an external axis ROTATION REVOLUTION

32. Conceptual Physics Chapter 1032 Tangential Speed  Linear speed is the distance traveled per unit time. s = d t If an object travels once around a circular path, it travels a distance equal to the circumference of the circle. r 2πr2πr T =

33. The time it takes to complete precisely one circular path is called the period and is represented by T. The tangential speed of the body is the linear speed along the circular path. The circumference can be found by 2πr, where r is the radius of the circle.

34. Conceptual Physics Chapter 1034 Tangential Speed The linear speed of the ball at any given instant is always directed tangent to the circular path.

35. Conceptual Physics Chapter 1035 Rotational Speed  The rotational speed or angular speed is the number of rotations per unit time.  Rotational speed is commonly measured in RPM (rotations/revolutions per minute).  We use the Greek letter omega (ω) to represent rotational speed.  Example: ω = 30 RPM

36. Conceptual Physics Chapter 1036 Rotational Speed Two objects placed on the same rotating platform will have the same rotational speed. The object furthest from the center of rotation will have the greatest tangential speed. v = r·ω

37.  A boy has a rock and a pebble rotating on a CD. He notices that the pebble which is sitting on the outer rim of the CD is 5 times the distance from the center compared to the rock which has a tangential velocity of 6m/s and a rotational velocity of 4 rpms. Find the rotational velocity and tangential velocity of the rock.

38. The tangential speed and the rotational speed are related by …and the greater the radial distance, the greater the tangential speed will be. The faster the platform spins, the greater the tangential speed will be at any point on the platform.

39. Conceptual Physics Chapter 1039 Question On a merry-go-round, the horses along the outer rail are located three times farther from the axis of rotation than the horses along the inner rail. If a boy sitting on one of the inner horses has a rotational speed of 4 RPM and a tangential speed of 2 m/s, what will be the tangential speed and rotational speed of his sister sitting on one of the outer horses?

40. Conceptual Physics Chapter 1040 In order for an object to move along a circular path, there must be a force acting on the object to change its direction of motion. Centripetal Force This force must be directed toward the center of the circle and it is called the centripetal force (centripetal means “center-seeking”).

41. Conceptual Physics Chapter 1041 Centripetal Force  The sideways acting friction between the tires of a car and the road keeps the car moving safely along a circular curve. ¤The car door exerts an inward normal force on the passenger in a vehicle that is rounding a left-hand turn.

42. Conceptual Physics Chapter 1042 Centripetal Force If the road is slick or friction is not great enough, the car will have a tendency to skid off tangent to the curve.

43. Conceptual Physics Chapter 1043 Centripetal Force The earth exerts an inward gravitational force on the moon as it travels along its circular orbit about the earth.

44. Conceptual Physics Chapter 1044 Centripetal Force  The spinning drum in a washing machine exerts an inward force on the clothes inside of it. ¤ The holes in the spinning drum prevent it from exerting an inward force on the water and the water will consequently fly off tangent to the drum wall.

45. Conceptual Physics Chapter 1045 Centripetal Force A conical pendulum is a bob held in a circular path by a string attached above. The string of a conical pendulum sweeps out a cone.

46.  T x is the net force on the bob–the centripetal force!  The vector T can be resolved into two perpendicular components, T x (horizontal), and T y (vertical). Only two forces act on the bob: mg, the force due to gravity, and T, tension in the string. The vertical component of the normal force n y is equal and opposite to mg, and the horizontal component of the normal force n x is the centripetal force that keeps the vehicle in a circular path.

47. Conceptual Physics Chapter 1047 Centripetal Force Skidding is reduced on high-speed roads by banking the turns. This is called superelevation. The inward component of the normal force adds to friction to create a greater centripetal force. Suppose the speed of the vehicle is such that the vehicle has no tendency to slide down the curve or up the curve. At that speed, friction plays no role in keeping the vehicle on the track. Only two forces act on the vehicle, the weight, mg, and the normal force n (the support force of the road surface).

48. Super Elevation  Defined: the amount by which the outer edge of a curve on a road or railroad is banked above the inner edge Banked Turns - thus allowing vehicles to maneuver through the curve at higher speeds than would otherwise be possible if the surface was flat or level

49. Conceptual Physics Chapter 1049 ¤The centripetal force prevents an object from continuing along a straight line. When the centripetal force vanishes or is reduced, the object will fly off tangent to the circular path. Centripetal Force  There is no centrifugal (outward) force!

50. Conceptual Physics Chapter 1050 Centripetal Force  Centripetal force is not a new type of force. It is any force that happens to cause an object to move along a circular path. It can be provided by gravity, friction, tension, normal force, electrical force or any combination of these.

51. Conceptual Physics Chapter 1051 Centripetal Acceleration  Since a body undergoing uniform circular motion maintains a constant speed, we must find the acceleration of this body using a c = v2v2 r ¤This is called the centripetal acceleration.

52. Conceptual Physics Chapter 1052 Centripetal Acceleration  The centripetal acceleration and the centripetal force are related by Newton’s second law: F c = ma c ¤Both the force that causes circular motion and the acceleration that results will always be directed inward.

53. Conceptual Physics Chapter 1053 Centripetal Acceleration Although the speed of an object undergoing uniform circular motion remains constant, the body accelerates. The velocity and acceleration vectors are always perpendicular to each other.

54. Conceptual Physics Chapter 1054 Centripetal Force Suppose a ladybug is placed in the bottom of a can being whirled in a circle. Centripetal Force The string pulls inward on the can and the bottom of the can pulls inward on the feet of the ladybug. This same spinning motion can be used to generate a simulated gravity in space.

55. Conceptual Physics Chapter 1055 Simulated Gravity Even though a space station may be in free fall, the occupants of the space station feel a simulated gravity from the spinning motion. At the correct rotational speed this microgravity will feel identical to the gravitational pull on earth.

56. Conceptual Physics Chapter 1056 Simulated Gravity Space stations can either be of a modest radius with a rather large rotational speed or could be larger to allow for a reduced rotational speed.

57. Conceptual Physics Chapter 1057 Simulated Gravity To produce a rotational speed that could be acclimated to by most humans, the space station would have to be nearly 2 km in diameter. The effect of the simulated gravity varies directly with the distance from the axis and with the rotational speed of the space station.

58. Conceptual Physics Chapter 1058 Question How would one’s weight be affected if the earth were to begin spinning faster on its axis?

59. Torque  Consider force required to open door. Is it easier to open the door by pushing/pulling away from hinge or close to hinge? close to hinge away from hinge Farther from from hinge, larger rotational effect! Physics concept: torque

60. Torque TTorque,, is the tendency of a force to rotate an object about some axis  is the torque dd is the lever arm (or moment arm) FF is the force Door example:

61. Lever Arm  The lever arm, d, is the shortest (perpendicular) distance from the axis of rotation to a line drawn along the the direction of the force  d = L sin Φ  It is not necessarily the distance between the axis of rotation and point where the force is applied

62. Direction of Torque TTorque is a vector quantity TThe direction is perpendicular to the plane determined by the lever arm and the force DDirection and sign: IIf the turning tendency of the force is counterclockwise, the torque will be positive IIf the turning tendency is clockwise, the torque will be negative Direction of torque: out of pageUnitsSI Newton meter (Nm) US Customary Foot pound (ft lb)

63. An Alternative Look at Torque TThe force could also be resolved into its x- and y- components TThe x-component, F cos Φ, produces 0 torque TThe y-component, F sin Φ, produces a non-zero torque F is the force L is the distance along the object Φ is the angle between force and object L

64. Let’s watch the movie!

65. ConcepTest 1 You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door? 1. No 2. Yes Please fill your answer as question 29 of General Purpose Answer Sheet

66. ConcepTest 1 You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door? 1. No 2. Yes Please fill your answer as question 30 of General Purpose Answer Sheet

67. ConcepTest 1 You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door? 1. No 2. Yes

68. ConcepTest 2 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements: Please fill your answer as question 31 of General Purpose Answer Sheet

69. ConcepTest 2 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements: Please fill your answer as question 32 of General Purpose Answer Sheet

70. ConcepTest 2 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements: 2, 1, 4, 3 or 2, 4, 1, 3

71. What if two or more different forces act on lever arm?

72. Net Torque  The net torque is the sum of all the torques produced by all the forces  Remember to account for the direction of the tendency for rotation  Counterclockwise torques are positive  Clockwise torques are negative

73. Example 1: Given: weights: w 1 = 500 N w 2 = 800 N lever arms: d 1 =4 m d 2 =2 m Find:  = ? 1. Draw all applicable forces 2. Consider CCW rotation to be positive 500 N 800 N 4 m2 m Rotation would be CCW N Determine the net torque:

74. Where would the 500 N person have to be relative to fulcrum for zero torque?

75. Example 2: Given: weights: w 1 = 500 N w 2 = 800 N lever arms: d 1 =4 m  = 0 Find: d 2 = ? 1. Draw all applicable forces and moment arms 500 N 800 N 2 md 2 m According to our understanding of torque there would be no rotation and no motion! N’ y What does it say about acceleration and force? Thus, according to 2 nd Newton’s law  F=0 and a=0!

76. Torque and Equilibrium FFirst Condition of Equilibrium TThe net external force must be zero TThis is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium TThis is a statement of translational equilibrium SSecond Condition of Equilibrium TThe net external torque must be zero TThis is a statement of rotational equilibrium

77. Axis of Rotation SSo far we have chosen obvious axis of rotation IIf the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque TThe location of the axis of rotation is completely arbitrary OOften the nature of the problem will suggest a convenient location for the axis WWhen solving a problem, you must specify an axis of rotation OOnce you have chosen an axis, you must maintain that choice consistently throughout the problem

78. Center of Gravity  The force of gravity acting on an object must be considered  In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point

79. Calculating the Center of Gravity 1. The object is divided up into a large number of very small particles of weight (mg) 2. Each particle will have a set of coordinates indicating its location (x,y) 3. The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm 4. We wish to locate the point of application of the single force, whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles. TThis point is called the center of gravity of the object

80. Coordinates of the Center of Gravity TThe coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object TThe center of gravity of a homogenous, symmetric body must lie on the axis of symmetry. OOften, the center of gravity of such an object is the geometric center of the object.

81. Example: Given: masses:m 1 = 5.00 kg m 2 = 2.00 kg m 3 = 4.00 kg lever arms: d 1 =0.500 m d 2 =1.00 m Find: Center of gravity Find center of gravity of the following system:

83.  Circus Physics: Centre of mass  Video Time

84. Experimentally Determining the Center of Gravity  The wrench is hung freely from two different pivots  The intersection of the lines indicates the center of gravity  A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object’s center of gravity

85. Equilibrium, once again  A zero net torque does not mean the absence of rotational motion  An object that rotates at uniform angular velocity can be under the influence of a zero net torque  This is analogous to the translational situation where a zero net force does not mean the object is not in motion

86. Example of a Free Body Diagram  Isolate the object to be analyzed  Draw the free body diagram for that object  Include all the external forces acting on the object

87. Suppose that you placed a 10 m ladder (which weights 100 N) against the wall at the angle of 30°. What are the forces acting on it and when would it be in equilibrium? Example

88. Example: Given: weights: w 1 = 100 N length: l=10 m angle:  =30°  = 0 Find: f = ? n=? P=? 1. Draw all applicable forces Torques:Forces:  2. Choose axis of rotation at bottom corner (  of f and n are 0!) Note: f =  s n, so mg

89. Friction  Where F f is the magnitude of friction, μ is the coefficient of friction and  F n is the magnitude of normal force. The Normal force = Weight of the given body. Hence Normal force F n is given by

90. So far: net torque was zero. What if it is not?

91. Torque and Angular Acceleration WWhen a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration TThe angular acceleration is directly proportional to the net torque TThe relationship is analogous to ∑F = ma NNewton’s Second Law

92. torque  dependent upon object and axis of rotation. Called moment of inertia I. Units: k kk kg m2 The angular acceleration is inversely proportional to the analogy of the mass in a rotating system

93. Moment of Inertia the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. The Moment of inertia depends on the shape of the body and may be different around different axes of rotation.

96. Example: Moment of Inertia of a Uniform Ring  Image the hoop is divided into a number of small segments, m 1 …  These segments are equidistant from the axis

97. Other Moments of Inertia

98. Newton’s Second Law for a Rotating Object TThe angular acceleration is directly proportional to the net torque TThe angular acceleration is inversely proportional to the moment of inertia of the object TThere is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object. TThe moment of inertia also depends upon the location of the axis of rotation

99. Let’s watch the movie!

100. Example: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m and weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg

101. Example: Given: masses: M = 5 kg weight:w = 9.8 N radius: R=0.2 m Find: Forces=? 1. Draw all applicable forces Forces:Torques: Tangential acceleration at the edge of flywheel (a=a t ): N T T Mg mg

102. ConcepTest 3 A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed? Please fill your answer as question 33 of General Purpose Answer Sheet 1. (a) 2. (b) 3. no difference 4. The answer depends on the rotational inertia of the dumbbell.

103. ConcepTest 3 A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed? Please fill your answer as question 34 of General Purpose Answer Sheet 1. (a) 2. (b) 3. no difference 4. The answer depends on the rotational inertia of the dumbbell.

104. ConcepTest 3 A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed? Force acts the same time: change of momentum is the same. Thus CM speed is the same as well. 1. (a) 2. (b) 3. no difference 4. The answer depends on the rotational inertia of the dumbbell.

105. Return to our example: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg If flywheel initially at rest and then begins to rotate, a torque must be present: Define physical quantity:

106. Angular Momentum SSimilarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum AAngular momentum is defined as L = I ω IIf the net torque is zero, the angular momentum remains constant CConservation of Linear Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero. TThat is, when (compare to )

107. Return to our example once again: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg Each small part of the flywheel is moving with some velocity. Therefore, each part and the flywheel as a whole have kinetic energy! Thus, total KE of the system: