Physics Chapter 8- Rotational of a Rigid Body

1. 1 CHAPTER 8: Rotation of a rigid body (8 Hours)

2. 2 At the end of this chapter, students should be able to: a) Define and use : angular displacement (  ) average angular velocity (  av ) instantaneous angular velocity (  ) average angular acceleration (  av ) instantaneous angular acceleration (  ). Learning Outcome: 8.1 Rotational kinematics (2 hours)

3. 3 b) Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use; c) Use equations for rotational motion with constant angular acceleration; , , .

4. 4 8.1 Rotational kinematics a)( i ) Angular displacement,  is defined as an angle through which a point or line has been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian ( rad ) . Figure 7.1 shows a point P on a rotating compact disc (CD) moves through an arc length s on a circular path of radius r about a fixed axis through point O . Figure 7.1

5. 5 From Figure 7.1, thus Others unit for angular displacement is degree ( ) and revolution (rev) . Conversion factor : Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise . Negative – if the rotational motion is clockwise . OR where

6. 6 Average angular velocity,  av is defined as the rate of change of angular displacement . Equation : Instantaneous angular velocity,  is defined as the instantaneous rate of change of angular displacement . Equation : a)(ii)(iii) Angular velocity where

7. 7 It is a vector quantity . The unit of angular velocity is radian per second (rad s -1 ) Others unit is revolution per minute (rev min 1 or rpm) Conversion factor: Note : Every part of a rotating rigid body has the same angular velocity . Direction of the angular velocity Its direction can be determine by using right hand grip rule where Thumb : direction of angular velocity Curl fingers : direction of rotation

8. 8 Figures 7.2 and 7.3 show the right hand grip rule for determining the direction of the angular velocity. Figure 7.2 Figure 7.3

9. 9 The angular displacement,  of the wheel is given by where  in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine a. the angle,  in degree, at time 2.2 s and 4.8 s, b. the distance that a particle on the rim moves during that time interval, c. the average angular velocity, in rad s 1 and in rev min 1 (rpm), between 2.2 s and 4.8 s, d. the instantaneous angular velocity at time 3.0 s. Example 1 :

10. 10 Solution : a. At time, t 1 =2.2 s : At time, t 2 =4.8 s :

11. 11 Solution : b. By applying the equation of arc length, Therefore c. The average angular velocity in rad s 1 is given by

12. 12 Solution : c. and the average angular velocity in rev min 1 is d. The instantaneous angular velocity as a function of time is At time, t =3.0 s : OR

13. 13 A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the diver’s average angular (rotational) velocity during a dive. (Given g = 9.81 m s 2 ) Solution : Example 2 : water

14. 14 Solution : From the diagram, Thus Therefore the diver’s average angular velocity is

15. 15 Average angular acceleration,  av is defined as the rate of change of angular velocity . Equation : Instantaneous angular acceleration,  is defined as the instantaneous rate of change of angular velocity . Equation : a)(iv)(v) Angular acceleration where

16. 16 Figure 7.4 It is a vector quantity . The unit of angular acceleration is rad s 2 . Note: If the angular acceleration,  is positive , then the angular velocity,  is increasing . If the angular acceleration,  is negative , then the angular velocity,  is decreasing . Direction of the angular acceleration If the rotation is speeding up ,  and  in the same direction as shown in Figure 7.4.

17. 17 Figure 7.5 If the rotation is slowing down ,  and  have the opposite direction as shown in Figure 7.5. Example 3 : The instantaneous angular velocity,  of the flywheel is given by where  in radian per second and t in seconds. Determine a. the average angular acceleration between 2.2 s and 4.8 s, b. the instantaneous angular acceleration at time, 3.0 s.

18. 18 Solution : a. At time, t 1 =2.2 s : At time, t 2 =4.8 s : Therefore the average angular acceleration is

19. 19 Solution : b. The instantaneous angular acceleration as a function of time is At time, t =3.0 s :

20. 20 Exercise 8.1a : 1. If a disc 30 cm in diameter rolls 65 m along a straight line without slipping, calculate a. the number of revolutions would it makes in the process, b. the angular displacement would be through by a speck of gum on its rim. ANS. : 69 rev; 138 rad 2. During a certain period of time, the angular displacement of a swinging door is described by where  is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration a. at time, t =0, b. at time, t =3.00 s. ANS. : 5.00 rad, 10.0 rad s 1 , 4.00 rad s 2 ; 53.0 rad, 22.0 rad s 1 , 4.00 rad s 2

21. 21 At the end of this chapter, students should be able to: Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use ; Learning Outcome: 8.1.b) Relationship between linear and rotational motion (½ hour)

22. 22 8.1.b) Relationship between linear and rotational motion Relationship between linear velocity, v and angular velocity,  When a rigid body is rotates about rotation axis O , every particle in the body moves in a circle as shown in the Figure 7.6. P O Figure 8.6

23. 23 Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by The direction of the linear (tangential) velocity always tangent to the circular path . Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle , r is changing depend on the position of the particle . and Simulation 8.1

24. 24 P O If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 7.7. Relationship between tangential acceleration, a t and angular acceleration,  Figure 8.7

25. 25 The components are tangential acceleration, a t and centripetal acceleration, a c given by but The vector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, a given by and its magnitude, and Vector form

26. 26 At the end of this chapter, students should be able to: Write and use equations for rotational motion with constant angular acceleration; Learning Outcome: 8.1.c) Rotational motion with uniform angular acceleration (1/2 hour)

27. 27 8.1.c) Rotational motion with uniform angular acceleration Table 8.1 shows the symbols used in linear and rotational kinematics. Table 8.1 Linear motion Quantity Rotational motion Displacement Initial velocity Final velocity Acceleration Time

28. 28 Table 8.2 shows the comparison of linear and rotational motion with constant acceleration. Linear motion Rotational motion where  in radian. Table 8.2

29. 29 A car is travelling with a velocity of 17.0 m s 1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s 2 for 5.00 s, calculate a. the number of revolutions of the wheels during this period, b. the angular speed of the wheels after 5.00 s. Solution : a. The initial angular velocity is and the angular acceleration of the wheels is given by Example 4 :

30. 30 Solution : a. By applying the equation of rotational motion with constant angular acceleration, thus therefore b. T he angular speed of the wheels after 5.00 s is

31. 31 The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h -1 to 35.0 km h -1 . The wheels have a diameter of 70 cm. a. Calculate the angular acceleration. b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. Solution : Example 5 :

32. 32 Solution : a. The initial angular speed of the wheels is and the final angular speed of the wheels is therefore b. The car stops thus Hence and

33. 33 A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s -1 . The angular acceleration of the blade is 0.750 rev s -2 . Determine a. the angular velocity after 4.00 s, b. the number of revolutions for the blade turns in this time interval, c. the tangential speed of a point on the tip of the blade at time, t =4.00 s, d. the magnitude of the resultant acceleration of a point on the tip of the blade at t =4.00 s. Solution : a. Given t =4.00 s, thus Example 6 :

34. 34 Solution : b. The number of revolutions of the blade is c. The tangential speed of a point is given by

35. 35 Solution : d. The magnitude of the resultant acceleration is

36. 36 Calculate the angular velocity of a. the second-hand, b. the minute-hand and c. the hour-hand, of a clock. State in rad s -1 . d. What is the angular acceleration in each case? Solution : a. The period of second-hand of the clock is T = 60 s, hence Example 7 :

37. 37 Solution : b . The period of minute-hand of the clock is T = 60 min = 3600 s, hence c . The period of hour-hand of the clock is T = 12 h = 4.32 10 4 s, hence d. T he angular acceleration in each cases is zero .

38. 38 A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s 1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s 2 , calculate the distance travelled by the coin before coming to rest. Solution : The radius of the coin is Example 8 :

39. 39 Solution : The initial speed of the point at the edge the coin is and the final speed is The linear acceleration of the point at the edge the coin is given by Therefore the distance travelled by the coin is

40. 40 Exercise 8.1b&c : 1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev min -1 about its central axis. Determine a. its angular speed, b. the tangential speed at a point 3.00 cm from its centre, c. the radial acceleration of a point on the rim, d. the total distance a point on the rim moves in 2.00 s. ANS. : 126 rad s 1 ; 3.77 m s 1 ; 1.26  10 3 m s 2 ; 20.1 m 2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculate a. its angular velocity in rad s  1 , b. the linear speed and the radial acceleration of a point on the edge of the grinding wheel. ANS. : 262 rad s 1 ; 46 m s 1 , 1.2  10 4 m s 2

41. 41 Exercise 8.1b&c : 3. A rotating wheel required 3.00 s to rotate through 37.0 revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s -1 . Calculate the constant angular acceleration of the wheel. ANS. : 13.6 rad s 2 4. A wheel rotates with a constant angular acceleration of 3.50 rad s  2 . a. If the angular speed of the wheel is 2.00 rad s  1 at t =0, through what angular displacement does the wheel rotate in 2.00 s. b. Through how many revolutions has the wheel turned during this time interval? c. What is the angular speed of the wheel at t = 2.00 s? ANS. : 11.0 rad; 1.75 rev; 9.00 rad s 1

42. 42 Exercise 8.1b&c : 5. A bicycle wheel is being tested at a repair shop. The angular velocity of the wheel is 4.00 rad s -1 at time t = 0 , and its angular acceleration is constant and equal  1.20 rad s -2 . A spoke OP on the wheel coincides with the +x-axis at time t = 0 as shown in Figure 7.8. a. What is the wheel’s angular velocity at t = 3.00 s? b. What angle in degree does the spoke OP make with the positive x-axis at this time? ANS. : 0.40 rad s 1 ; 18 Figure 8.8

43. 43 At the end of this chapter, students should be able to: a) Define and use torque, b) State and use conditions for equilibrium of rigid body: Examples of problems : Fireman ladder leaning on a wall, see-saw, pivoted / suspended horizontal bar. Sign convention for moment or torque : + ve : anticlockwise  ve : clockwise Learning Outcome: 8.2 Equilibrium of a uniform rigid body (2 hours)

44. 44 8.2 Equilibrium of a rigid body Non-concurrent forces is defined as the forces whose lines of action do not pass through a single common point. The forces cause the rotational motion on the body. The combination of concurrent and non-concurrent forces cause rolling motion on the body. ( translational and rotational motion) Figure 5.11 shows an example of non-concurrent forces. Figure 8.2

45. 45 Torque (moment of a force), The magnitude of the torque is defined as the product of a force and its perpendicular distance from the line of action of the force to the point (rotation axis) . OR Because of where r : distance between the pivot point (rotation axis) and the point of application of force. Thus where where OR

46. 46 It is a vector quantity . The dimension of torque is The unit of torque is N m ( newton metre ), a vector product unlike the joule (unit of work) , also equal to a newton metre , which is scalar product . Torque is occurred because of turning (twisting) effects of the forces on a body. Sign convention of torque: Positive - turning tendency of the force is anticlockwise . Negative - turning tendency of the force is clockwise . The value of torque depends on the rotation axis and the magnitude of applied force .

47. 47 Case 1 : Consider a force is applied to a metre rule which is pivoted at one end as shown in Figures 5.12a and 5.12b. Figure 8.12a Figure 8.12b Pivot point (rotation axis) (anticlockwise) (anticlockwise) Point of action of a force Line of action of a force

48. 48 O Figure 8.13 Case 2 : Consider three forces are applied to the metre rule which is pivoted at one end (point O) as shown in Figures 5.13. Caution : If the line of action of a force is through the rotation axis then Therefore the resultant (nett) torque is and Simulation 8.2

49. 49 Determine a resultant torque of all the forces about rotation axis, O in the following problems. a. Example 4 : O

50. 50 b. Example 4 : O

51. 51 O Solution : a. Force Torque (N m),  o =Fd=Fr sin  The resultant torque: (clockwise)

52. 52 Solution : b. Force Torque (N m),  o =Fd=Fr sin  The resultant torque: (clockwise) O

53. 53 8.2 Equilibrium of a rigid body Rigid body is defined as a body with definite shape that doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it . If the rigid body is in equilibrium , means the body is translational and rotational equilibrium . There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must be zero. OR

54. 54 The vector sum of all external torques acting on a rigid body must be zero about any rotation axis . This ensures rotational equilibrium . This is equivalent to the three independent scalar equations along the direction of the coordinate axes, Centre of gravity, CG is defined as the point at which the whole weight of a body may be considered to act . A force that exerts on the centre of gravity of an object will cause a translational motion .

55. 55 Figures 5.14 and 5.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere rod – refer to the midway point between its end . sphere – refer to geometric centre . CG CG Figure 5.14 Figure 5.15

56. 56 5.3.4 Problem solving strategies for equilibrium of a rigid body The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components and to determine the torque by each force. Apply the condition for equilibrium of a rigid body : Solve the equations for the unknowns. ; and

57. 57 A hanging flower basket having weight, W 2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W 1 as shown in Figure 5.16. If the mass of the beam is 3.0 kg, calculate a. the weight, W 1 needed, b. the force exerted on the beam at point O. (Given g =9.81 m s 2 ) Example 5 : 35 cm 75 cm Figure 5.16

58. 58 Solution : The free body diagram of the beam : Let point O as the rotation axis. 0.75 m CG 0.35 m 0.55 m 0.55 m 0.20 m Force y-comp. (N) Torque (N m),  o =Fd=Fr sin 

59. 59 Solution : Since the beam remains at rest thus the system in equilibrium. a. Hence b. and

60. 60 A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 5.17. The height of the end A of the ladder is 8.0 m from the rough floor. a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B. b. If the ladder is just on the verge of slipping when the firefighter is 7.0 m up the ladder , Calculate the coefficient of static friction between ladder and floor. (Given g =9.81 m s 2 ) Example 6 : A B smooth wall rough floor Figure 5.17

61. 61 Solution : a. The free body diagram of the ladder : Let point B as the rotation axis. A B CG Force x-comp. (N) y-comp. (N) Torque (N m),  B =Fd=Fr sin 

62. 62 Solution : Since the ladder in equilibrium thus Horizontal force: Vertical force:

63. 63 A B Solution : b. The free body diagram of the ladder : Let point B as the rotation axis. Force x-comp. (N) y-comp. (N) Torque (N m),  B =Fd=Fr sin 

64. 64 Solution : Consider the ladder stills in equilibrium thus

65. 65 Figure 5.18 A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 5.18. A cable at an angle 30 with the beam helps to support the light. a. Sketch a free body diagram of the beam. b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole. (Given g =9.81 m s 2 ) Example 7 :

66. 66 Solution : a. The free body diagram of the beam : b. Let point O as the rotation axis. Force x-comp. (N) y-comp. (N) Torque (N m),  o =Fd=Fr sin  O CG

67. 67 Solution : b. The floodlight and beam remain at rest thus i. ii.

68. 68 Solution : b. ii. Therefore the magnitude of the force is and its direction is given by from the +x-axis anticlockwise

69. 69 Exercise 8.2 : Use gravitational acceleration, g = 9.81 m s  2 1. Figure 5.19 shows the forces, F 1 =10 N, F 2 = 50 N and F 3 = 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle  is 30. Calculate the resultant torque about point D. ANS. : -3.7 N m D A B C Figure 8.19

70. 70 Figure 5.20 Exercise 8.2 : 2. A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 5.20. The fulcrum is under the centre of gravity of the board. Determine a. the magnitude of the force exerted by the fulcrum on the board, b. where the father should sit from the fulcrum to balance the system. ANS. : 1128 N; 1.31 m

71. 71 3. A traffic light hangs from a structure as show in Figure 5.21. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. ANS. : 248 N; 197 N, 248 N Figure 5.21 Exercise 8.2 :

72. 72 4. A uniform 10.0 N picture frame is supported by two light string as shown in Figure 5.22. The horizontal force, F is applied for holding the frame in the position shown. a. Sketch the free body diagram of the picture frame. b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F . ANS. : 1.42 N, 11.2 N; 7.20 N Exercise 8.2 : Figure 5.22

73. 73 8.3 Rotational Dynamics (1 hour) Centre of mass ( CM ) is defined as the point at which the whole mass of a body may be considered to be concentrated . Its coordinate ( x CM , y CM ) is given the expression below: where

74. 74 Two masses, 2 kg and 4 kg are located on the x -axis at x =2 m and x =5 m respectively. Determine the centre of mass of this system. Solution : Example 9 : 2 m from m 1 OR

75. 75 A system consists of three particles have the following masses and coordinates : (1) 2 kg, (1,1) ; (2) 4 kg, (2,0) and (3) 6 kg, (2,2). Determine the coordinate of the centre of mass of the system. Solution : The x coordinate of the CM is Example 10 :

76. 76 Solution : The y coordinate of the CM is Therefore the coordinate of the CM is

77. 77 Figure 7.9 shows a rigid body about a fixed axis O with angular velocity   . is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis . Moment of inertia, I O Figure 7.9

78. 78 OR It is a scalar quantity . Moment of inertia, I in the rotational kinematics is analogous to the mass, m in linear kinematics. The dimension of the moment of inertia is M L 2 . The S.I. unit of moment of inertia is kg m 2 . The factors which affect the moment of inertia, I of a rigid body: a. the mass of the body, b. the shape of the body, c. the position of the rotation axis . where

79. 79 Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Diagram Equation Hoop or ring or thin cylindrical shell Solid cylinder or disk CM CM

80. 80 CM Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Diagram Equation Uniform rod or long thin rod with rotation axis through the centre of mass. CM Solid Sphere

81. 81 Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Diagram Equation Hollow Sphere or thin spherical shell CM Table 7.3

82. 82 Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 7.10. The spheres are connected by light rods . Determine the moment of inertia of the system about an axis a. through point O, b. along the line AB. Example 11 : O A B Figure 7.10

83. 83 Solution : a. rotation axis about point O, Since r 1 = r 2 = r 3 = r 4 = r thus and the connecting rods are light therefore O

84. 84 Solution : b. rotation axis along the line AB, r 1 = r 2 = r 3 = r 4 = r= 0.6 m therefore A B

85. 85 Parallel-Axis Theorem (Steiner’s Theorem) States that the moment of inertia, I about any axis parallel to and a distance, d away from the axis through the centre of mass, I CM is given by where

86. 86 Determine the moment of inertia of a solid cylinder of radius R and mass M about an axis tangent to its edge and parallel to its symmetry axis as shown in Figure 7.11. Given the moment of inertia of the solid cylinder about axis through the centre of mass is Example 12 : Figure 7.11 CM

87. 87 CM Solution : From the diagram, d = R By using the parallel axis theorem, CM Initial Final

88. 88 Relationship between torque,  and angular acceleration,  Consider a force, F acts on a rigid body freely pivoted on an axis through point O as shown in Figure 7.12. The body rotates in the anticlockwise direction and a nett torque is produced. Torque,  O Figure 7.12

89. 89 A particle of mass, m 1 of distance r 1 from the rotation axis O will experience a nett force F 1 . The nett force on this particle is The torque on the mass m 1 is The total (nett) torque on the rigid body is given by and and

90. 90 From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration . Note : is analogous to the

91. 91 Forces, F 1 = 5.60 N and F 2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 7.13. Calculate, a. the nett torque on the disc. b. the magnitude of angular acceleration influence by the disc. ( Use the moment of inertia, ) Example 13 : Figure 7.13

92. 92 Solution : a. The nett torque on the disc is b. By applying the relationship between torque and angular acceleration,

93. 93 A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is 0.050 kg m 2 . A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure 7.14. Assume the string does not slip on the wheel. a. Sketch a free body diagram of the wheel and the block. b. Calculate the magnitude of the angular acceleration of the wheel. Example 14 : Figure 7.14

94. 94 Solution : a. Free body diagram : for wheel, for block,

95. 95 Solution : b. F or wheel, For block, By substituting eq. (1) into eq. (2), thus (1) (2) and

96. 96 An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 7.15. The pulley has a moment of inertia 0.020 kg m 2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determine a. the linear acceleration of the object, b. the angular acceleration of the pulley, c. the tension in the string, d. the liner velocity of the object, e. the distance travelled by the object. (Given g = 9.81 m s -2 ) Example 15 : Figure 7.15

97. 97 Solution : a. Free body diagram : for pulley, for block, and (1) (2)

98. 98 Solution : a. By substituting eq. (1) into eq. (2), thus b. By using the relationship between a and  , hence

99. 99 Solution : c. From eq. (1), thus d. By applying the equation of liner motion, thus e. The distance travelled by the object in 0.3 s is (downwards)

100. 100 Exercise 8.3 : Use gravitational acceleration, g = 9.81 m s  2 1. Three odd-shaped blocks of chocolate have following masses and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m); 0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m). Determine the coordinates of the centre of mass of the system of three chocolate blocks. ANS. : (0.044 m, 0.056 m) 2. Figure 7.16 shows four masses that are held at the corners of a square by a very light frame. Calculate the moment of inertia of the system about an axis perpendicular to the plane a. through point A, and b. through point B. ANS. : 0.141 kg m 2 ; 0.211 kg m 2 Figure 7.16

101. 101 Exercise 8.3 : 3. A 5.00 kg object placed on a frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 7.17. The pulley has a radius of 0.250 m and moment of inertia I . The block on the table is moving with a constant acceleration of 2.00 m s  2 . a. Sketch free body diagrams of both objects and pulley. b. Calculate T 1 and T 2 the tensions in the string . c. Determine I . ANS. : 10.0 N, 70.3 N; 1.88 kg m 2 Figure 7.17

102. 102 At the end of this chapter, students should be able to: a) Solve problems related to : Rotational kinetic energy, work, power, Learning Outcome: 8.4 Work and Energy of Rotational Motion(2 hours)

103. 103 Rotational kinetic energy and power Rotational kinetic energy, K r Consider a rigid body rotating about the axis OZ as shown in Figure 7.18. Every particle in the body is in the circular motion about point O. O Z Figure 7.18

104. 104 The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by and

105. 105 From the formula for translational kinetic energy, K tr After comparing both equations thus For rolling body without slipping , the total kinetic energy of the body, K is given by  is analogous to v I is analogous to m where

106. 106 A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25  to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate a. the total kinetic energy of the sphere, b. the linear speed of the sphere, c. the angular speed about the centre of mass. (Given the moment of inertia of solid sphere is and the gravitational acceleration, g = 9.81 m s 2 ) Example 16 :

107. 107 Solution : a. From the principle of conservation of energy,

108. 108 Solution : b. The linear speed of the sphere is given by c. By using the relationship between v and  , thus and

109. 109 The pulley in the Figure 7.19 has a radius of 0.120 m and a moment of inertia 0.055 g cm 2 . The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s 2 ) Example 17 : Figure 7.19

110. 110 Solution : The moment of inertia of the pulley, Initial Final

111. 111 Solution : By using the principle of conservation of energy, thus

112. 112 Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 7.20. The work done by the tangential force is given by Work, W Figure 7.20 and

113. 113 If the torque is constant thus Work-rotational kinetic energy theorem states where is analogous to the

114. 114 From the definition of instantaneous power, Caution : The unit of kinetic energy, work and power in the rotational kinematics is same as their unit in translational kinematics. Power, P and and is analogous to the

115. 115 A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m 2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determine a. the rotational kinetic energy of the merry-go-round, b. the work done by the force on the merry-go-round, c. the average power supplied by the force. (Given g = 9.81 m s 2 ) Solution : Example 18 :

116. 116 Solution : a. By applying the relationship between nett torque and angular acceleration, thus Use the equation of rotational motion with uniform angular acceleration, Therefore the rotational kinetic energy for 15.0 s is

117. 117 Solution : b. The angular displacement,  for 15.0 s is given by By applying the formulae of work done in rotational motion, thus c. The average power supplied by the force is

118. 118 At the end of this chapter, students should be able to: Define and use the formulae of angular momentum, State and use the principle of conservation of angular momentum Learning Outcome: 8.5 Conservation of angular momentum (1 hour)

119. 119 Conservation of angular momentum Angular momentum, is defined as the product of the angular velocity of a body and its moment of inertia about the rotation axis . OR It is a vector quantity. Its dimension is M L 2 T  1 The S.I. unit of the angular momentum is kg m 2 s  1 . where is analogous to the

120. 120 The relationship between angular momentum, L with linear momentum , p is given by vector notation : magnitude form : Newton’s second law of motion in term of linear momentum is hence we can write the Newton’s second law in angular form as and states that the vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum . where

121. 121 states that the total angular momentum of a system about an rotation axis is constant if no external torque acts on the system . OR Therefore Principle of conservation of angular momentum If the and

122. 122 A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s -1 about the rotation axis as shown in Figure 7.21. A 50 kg bag of sand falls onto the disc at the edge of the wooden disc. Calculate, a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle) b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same? (Use the moment of inertia of disc is ) Example 19 : Before After Figure 7.21

123. 123 Solution : a. The moment of inertia of the disc, The moment of inertia of the bag of sand, By applying the principle of conservation of angular momentum,

124. 124 Solution : b. The initial rotational kinetic energy, The final rotational kinetic energy, thus It is because the energy is lost in the form of heat from the friction between the surface of the disc with the bag of sand.

125. 125 A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer. Solution : The answer is hard-boiled egg . Example 20 :

126. 126 Solution : Reason Raw egg : When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases because of From the principle of conservation of angular momentum, If the I is increases hence its angular velocity,  will decreases. Hard-boiled egg : The position of the yolk of a hard-boiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.

127. 127 Before After A student sits on a freely rotating stool holding two weights, each of mass 3.00 kg as shown in Figure 7.22. When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad s  1 . The moment of inertia of the student and stool is 3.00 kg m 2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. Determine a. the new angular speed of the student. b. the kinetic energy of the rotating system before and after he pulls the weights inward. Example 21 : Figure 7.22

128. 128 Before After Solution :

129. 129 Solution : a. The moment of inertia of the system initially is The moment of inertia of the system finally is By using the principle of conservation of angular momentum, thus

130. 130 Solution : b. The initial rotational kinetic energy is given by and the final rotational kinetic energy is

131. 131 Exercise 8.5 : Use gravitational acceleration, g = 9.81 m s  2 1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m 2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s  1 relative to the Earth. a. In the what direction and with what value of angular speed does the turntable rotate? b. How much work does the woman do to set herself and the turntable into motion? ANS. : 0.360 rad s 1 ,U think; 99.9 J

132. 132 Exercise 8.5 : 2. Determine the angular momentum of the Earth a. about its rotation axis (assume the Earth is a uniform solid sphere), and b. about its orbit around the Sun (treat the Earth as a particle orbiting the Sun). Given the Earth’s mass = 6.0 x 10 24 kg, radius = 6.4 x 10 6 m and is 1.5 x 10 8 km from the Sun. ANS. : 7.1 x 10 33 kg m 2 s  1 ; 2.7 x 10 40 kg m 2 s  1 3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is ) ANS. : 4.71 x 10  6 kg m 2 s  1

133. 133 Linear Motion Relationship Rotational Motion Summary:

134. 134 THE END… Next Chapter… CHAPTER 9 : Simple Harmonic Motion