1. PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4)
CHAPTER 18
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections )
Objectives:
a) Define the various ways a force and couple do work.
b) Apply the principle of work and energy to a rigid body.

2. APPLICATIONS
The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum.
The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of
its roller and wheels

3. KINETIC ENERGY
The kinetic energy of a rigid body in general plane motion is given by
T = 1/2 m (vG)2 + 1/2 IG ω2
Several simplifications can occur.
Pure Translation: the rotational kinetic energy is zero (ω = 0):
T = 0.5 m (vG)2

4. KINETIC ENERGY (continued)
2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy:
T = 0.5m(vG) IGw2
Since
vG = rGw, T = 0.5(IG + m(rG)2)w2 = 0.5IOw2
If the rotation occurs about the mass center (pure rotation), G, then vG=0
T = 0.5 IG w2

5. WORK OF A FORCE The work done by a force is
UF =  F•dr =  (F cos θ) ds
Constant force : UFc = (Fc cos θ)s s
Work of a weight: Uw = -WΔy.
Work of a spring force:
Us = -0.5k[(s2)2 – (s1)2]

6. FORCES THAT DO NO WORK
1. Reactions at fixed supports because the displacement at their point of application is zero.
2. Normal and frictional forces acting on bodies as they roll without slipping over a rough surface since there is no instantaneous displacement of the point in contact with ground.
3. Internal forces because they always act in equal and opposite pairs.

7. THE WORK OF A COUPLE
A body subjected to a couple produces work only when it undergoes rotation.
If the body rotates through an angular displacement dq, the work of the couple moment, M, is ò = q2 q1
M dq UM
constant M: UM = M (q2 – q1)

8. PRINCIPLE OF WORK AND ENERGY
The principle states: SU1-2 = T12
i.e. the work done by all external forces and couple moments equals the change in body’s kinetic energy (translational and rotational).
This equation is a scalar equation. It can be applied to a system of rigid bodies by summing contributions from all bodies.

9. EXAMPLE
Given:The disk weighs 40 N, has a radius of gyration (kG) 0.6 m. A 15 N.m moment is applied. The spring constant is 10N/m.
Find: Angular velocity of the disk when point G moves 0.5 m. The disk starts from rest and rolls without slipping. The spring is initially unstretched.

10. EXAMPLE (solution) Free body diagram of the disk:
Only the spring force and couple moment M do work.
Spring will stretch twice the amount of displacement of G, or 1 m. Why?

11. EXAMPLE (continued)
Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(q2 – q1)
U1-2 = -0.5(10)(12 – 0) + 15(0.5/0.8) = 4.4N.m
Kinematic relation: vG = r w = 0.8w
Kinetic energy: T1 = 0
T2 = 0.5m (vG) IG w2
T2 = 0.5(40/9.8)(0.8w) (40/9.8)(0.6)2w2
T2 = 2.0 w2
Work and energy: U1-2 = T12
4.4 = 2.0 w2
w = 1.5 rad/s

12. GROUP PROBLEM SOLVING
Given: A sphere weighing 10N rolls along a semicircular hoop. Its w equals 0 when q = 0.
Find: The angular velocity of the sphere when q = 45° if the sphere rolls without slipping.

13. GROUP PROBLEM SOLVING (continued)
Solution: Draw a FBD and calculate the vertical distance the mass center moves.
Now calculate the _______________ :

14. GROUP PROBLEM SOLVING (continued)
A kinematic equation for finding the velocity of the mass center is needed. It is
_________ energy:
Now apply the principle of work and energy equation:
w = 13.1 rad/s

15. End of
Let Learning Continue

16. PLANAR KINETICS OF A RIGID BODY: CONSERVATION OF ENERGY (Section 18.5)
Objectives:
Determine the potential energy of conservative forces.
Apply the principle of conservation of energy.

17. APPLICATIONS
Torsional spring at the top of the door winds up as the door is lowered. When the door is raised, the spring potential energy is transferred into gravitational potential energy of the door’s weight, thus making it easy to open.

18. CONSERVATION OF ENERGY
The conservation of energy theorem is a “simpler” energy method for solving problems. Once again, the problem parameter of distance is a key indicator of when conservation of energy is a good method for solving the problem.
If it is appropriate, conservation of energy is easier to use than the principle of work and energy.

19. CONSERVATIVE FORCES
A force F is conservative if the work done by the force is independent of the path.
Typical conservative forces encountered in dynamics are gravitational forces (i.e., weight) and elastic forces (i.e., springs).
Friction forces!!!
What is a common force that is not conservative?

20. CONSERVATION OF ENERGY
When a rigid body is acted upon by a system of conservative forces, the work done by these forces is conserved. Thus, the sum of kinetic energy and potential energy remains constant:
T1 + V1 = T2 + V2 = Constant
i.e. as a rigid body moves from one position to another when acted upon by only conservative forces, kinetic energy is converted to potential energy and vice versa.

21. GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy is a function of the height of the body’s center above or below a datum.
The gravitational potential energy is found by:
Vg = W yG
Gravitational potential energy is +ve when yG is +ve, since the weight has the ability to do +ve work
when the body is moved back to the datum.

22. ELASTIC POTENTIAL ENERGY
Spring forces are also conservative forces.
The potential energy of a spring force (F = ks) is found by the equation
Ve = ½ ks2
Notice that the elastic potential energy is always +ve

23. EXAMPLE 1
Given: The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when  = 0°. Neglect the mass of the pistons.
Find: The angular velocity of rod AB at  = 0° if the rod is released from rest when  = 30°.

24. EXAMPLE 1 (solution) Initial Position Final Position
Potential Energy: put datum in line with rod when =0°. Gravitational and elastic energy=0 at 2. => V2 = 0
Gravitational energy at 1: - (10)( 9.81) ½ (0.4 sin 30)
Elastic energy at 1: ½ (800) (0.4 sin 30)2
So V1 = J J = J

25. EXAMPLE 1 (continued) Initial Position Final Position
Kinetic Energy:
The rod released from rest (vG1=0, 1=0). Thus, T1 = 0. At 2, the angular velocity is 2 and the velocity is vG2 .

26. EXAMPLE 1 (continued) Thus,
T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(2)2
At 2, vA=0. Hence, vG2 = r  = 0.2 2 .
Then, T2 = 0.2  22 = 22
conservation of energy:
T1 + V1 = T2 + V2
=  => 2 = 4.82 rad/s

27. EXAMPLE 2
Given: The weight of the disk is 30 N and its kG equals 0.6 m. The spring has a stiffness of 2 N/m and an unstretched length of 1m.
Find: The angular velocity at the instant G moves 3 m to the left. The disk is released from rest in the position shown and rolls without slipping.

28. EXAMPLE 2 (solution)
Potential Energy: No changes in the gravitational potential energy
The elastic potential energy at 1: V1 = 0.5 k (s1)2 where s1 = 4 m.
Thus, V1 = ½ 2 (4)2 = 16 N.m.
The elastic potential energy at 2 is
V2 = ½ 2 (3)2 = 9 N.m.

29. EXAMPLE 2 (continued) Kinetic Energy:
Released from rest: vG1=1=0, T1=0
At 2, 2 and vG2:
T2 = ½ m (vG2)2 + ½ IG (2) 2
= ½ (30/9.8) (vG2)2 + ½ (30/9.8) (2) 2
Disk is rolling without slipping: vG2 = (0.75 2)
T2 = ½(30/9.8)(0.75 2)2 + ½(30/9.8) 0.62 (2)2 = 1.41 (2)2

30. EXAMPLE 2 (continued) conservation of energy: T1 + V1 = T2 + V2
= 1.41 22 + 9
Solving , 2 = 2.23 rad/s

31. GROUP PROBLEM SOLVING
Given: A 50 N bar is rotating downward at 2 rad/s.
The spring has an unstretched length of 2m and a spring constant of 12 N/m
Find: The angle (measured down from the horizontal) to which the bar rotates before it stops its initial downward movement.

32. GROUP PROBLEM SOLVING (solution)
Potential Energy:
Put the datum in line with the rod when  = 0.
Gravitational potential energy at 2:
Elastic potential energy at 2:
So, V2 =

33. GROUP PROBLEM SOLVING (continued)
Kinetic Energy:
At 1 (when  = 0):
At 2:

34. GROUP PROBLEM SOLVING (continued)
conservation of energy
Thus,  = 49.9 deg.

35. End of 18.5
Let Learning Continue

36. PLANAR KINETICS: IMPULSE AND MOMENTUM (Sections 19.1-19.2)
CHAPTER 19
PLANAR KINETICS: IMPULSE AND MOMENTUM (Sections )
Today’s Objectives:
a) Develop formulations for the linear and angular momentum of a body.
b) Apply the principle of linear and angular impulse and momentum.

37. APPLICATIONS
As the pendulum swings downward, its angular momentum and linear momentum both increase. By calculating its momenta in the vertical position, we can calculate the impulse the pendulum exerts when it hits the test specimen.
The space shuttle has several engines that exert thrust on the shuttle when they are fired. By firing different engines, the pilot can control the motion and direction of
the shuttle.

38. LINEAR AND ANGULAR MOMENTUM
The linear momentum of a rigid body is: L = m vG
The angular momentum of a rigid body is: HG = IG w

39. LINEAR AND ANGULAR MOMENTUM
(continued)
Translation.
When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because ω = 0.
Therefore:
L = m vG
HG = 0

40. LINEAR AND ANGULAR MOMENTUM
(continued)
Rotation about a fixed axis.
The body’s linear momentum and angular momentum are:
L = m vG
HG = IGw
HO = ( rG x mvG) + IGw = IO w

41. LINEAR AND ANGULAR MOMENTUM
(continued)
General plane motion.
The linear and angular momentum computed about G are required.
L = m vG
HG = IGω
The angular momentum about point A is
HA = IGω + (d)mvG

42. ò ò PRINCIPLE OF IMPULSE AND MOMENTUM
The principle is developed by combining the equation of motion with kinematics. The resulting equations allow a direct solution to problems involving force, velocity, and time.
Linear impulse-linear momentum equation:
L F dt = L2 or (mvG) F dt = (mvG)2 t2 t1 å ò
Angular impulse-angular momentum equation:
(HG) MG dt = (HG)2 or IGw MG dt = IGw2 t2 t1 å ò

43. PRINCIPLE OF IMPULSE AND MOMENTUM
(continued)
The previous relations can be represented graphically by drawing the impulse-momentum diagram. + =

44. EXAMPLE
Given: A disk weighing 50 N has a rope wrapped around it. The rope is pulled with a force P equaling 2 N.
Find: The angular velocity of the disk after 4 seconds if it starts from rest and rolls without slipping.

45. (P t) 2 r = (mvG)2 r + IGw2 = m r2 w2 + 0.5 m r2 w2 = 1.5 m r2 w2
EXAMPLE (solution)
(m vG)1
(m vG)2 G
IGw1
IG w2
P t
F t
N t
W t r + = A y x
Impulse-momentum diagram:
Kinematics: (vG)2 = r w2 t2 t1 å ò
Impulse & Momentum: (HA) MA dt = (HA)2
(P t) 2 r = (mvG)2 r + IGw2 = m r2 w m r2 w2 = 1.5 m r2 w2
3.5 rad/s
3(50/9.81)(0.6)
4(2)(4)
3mr
4 Pt w2 =

46. GROUP PROBLEM SOLVING Given: A gear set with: WA = 15 N WB = 10 N
kA = 0.5 m
kB = 0.35 m
M = 2(1 – e-0.5t) N.m
Find: The angular velocity of gear A after 5 seconds if the gears start turning from rest.
Plan: Time is a parameter, thus

47. GROUP PROBLEM SOLVING (continued)
Solution:
Impulse-momentum diagrams:
Gear A: y x
Gear B:

48. GROUP PROBLEM SOLVING (continued)
Kinematics:
Angular impulse & momentum relation for gear A about point A yields:
For gear B:

49. GROUP PROBLEM SOLVING (continued)
and wA = rad/s

50. End of
Let Learning Continue

51. CONSERVATION OF MOMENTUM (Section 19.3)
Objectives:
a) Understand the conditions for conservation of linear and angular momentum
b) Use the condition of conservation of linear/ angular momentum

52. APPLICATIONS
A skater spends a lot of time either spinning on the ice or rotating through the air. To spin fast, or for a long time, the skater must develop a large amount of angular momentum.
If the skater’s angular momentum is constant, can the skater vary his rotational speed? How?
The skater spins faster when the arms are drawn in and slower when the arms are extended. Why?

53. ò CONSERVATION OF LINEAR MOMENTUM
Recall that the linear impulse and momentum relationship is
L F dt = L2 or (m vG) F dt = (m vG)2 t2 t1 å ò
If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L1 = L2.
This equation is referred to as the conservation of linear momentum. The conservation of linear momentum equation can be used if the linear impulses are small or non-impulsive.

54. ò CONSERVATION OF ANGULAR MOMENTUM
The angular impulse-angular momentum relationship is:
(HG) MG dt = (HG)2 or IGw MG dt = IGw2 t2 t1 å ò
Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved . The resulting equation is referred to as the conservation of angular momentum or (HG)1 = (HG)2 .
If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the final linear or angular velocity of a body just after an event occurs.

55. EXAMPLE Given: A 10 kg wheel
(IG = kg·m2) rolls without slipping and does not bounce at A.
Find: The minimum velocity, vG, of the wheel to just roll over the obstruction at A.
Note: Since no slipping or bouncing occurs, the wheel pivots about point A. The force at A is much greater than the weight, and since the time of impact is very short, the weight can be considered non-impulsive. The reaction force at A, since we don’t know either its direction or magnitude, can be eliminated by applying the conservation of angular momentum equation about A.

56. EXAMPLE (solution) Impulse-momentum diagram: + =y
Impulse-momentum diagram: x + =
Conservation of angular momentum: (HA)1 = (HA)2
r ' (mvG)1 + IG w1 = r (mvG)2 + IG w2
( )(10)(vG) w1 = 0.2(10)(vG) w2
Kinematics: no slip, w = vG / r = 5 vG .Substituting and solving: (vG)2 = (vG)1

57. EXAMPLE (continued)
To complete the solution, conservation of energy can be used. Since it cannot be used for the impact (why?), it is applied just after the impact. In order to roll over
the bump, the wheel must go to position 3 from 2. When (vG)2 is a minimum, (vG)3 is zero. Why?
T2 + V2 = T3 + V3
{½(10)(vG)22 + ½(0.156)w22 } + 0 = (0.03)
Substituting w2 = 5 (vG)2 and (vG)2 = (vG)1 and solving yields: (vG)1 = m/s, (vG)2 = 0.65 m/s

58. GROUP PROBLEM SOLVING
Given: A slender rod (Wr = 5 N) has a wood block (Ww = 10 N) attached. A bullet (Wb = 0.2 N) is fired into the center of the block at 1000 m/s. Assume the pendulum is initially at rest and the bullet embeds itself into the block.
Find: The angular velocity of the pendulum just after impact.

59. GROUP PROBLEM SOLVING (continued)
Solution:
To use conservation of angular momentum,
First draw a FBD.
IA = ?
IA = 7.2kg.m2

60. GROUP PROBLEM SOLVING (continued)
Apply the conservation of angular momentum equation:
Solving yields:
w2 = 7.0 rad/s

61. End of 19.3
Let Learning Continue

62. CHAPTER 22 VIBRATIONS (section 22.1)
Objectives:
Discuss Undamped single degree of freedom (SDOF) vibration of a rigid body

63. Vibration definition
Vibration is the periodic motion of a body or system of connected bodies displaced from a position of equilibrium
Types of vibrations
Free vibrations: motion is maintained by conservative forces (like gravitational or elastic restoring forces).
Forced vibrations: motion is caused by external periodic or intermittent forces.

64. Undamped free response for SDOF systems
If you displace mass m a distance u from equilibrium position and release it, vibration occurs

65. Undamped free response for SDOF systems (cont):
Equilibrium eq:
A linear, constant coefficients, homogeneous, second order ordinary differential equation
A and B are found from initial conditions

66. Tn =natural period of vibration
Undamped free response for SDOF systems:
ωn =natural circular frequency of vibration
Tn =natural period of vibration
fn =1/Tn= natural cyclic frequency of vibration in hertz (cycles per second)

67. Solving initial conditions

68. Example 1

69. Example 1 (continued)

70. Example 1 (continued)
By analogy
By analogy

71. Example 2

72. Example 2 (continued)
From equilibrium

73. Example 2 (continued)
Thus:

74. Group Solving problem

75. Draw FBD
Write equation of motion

76. Use analogy

77. End of 22.1
Let Learning Continue

78. Energy method Example 1

81. Energy methods: Example 2

84. Undamped forced vibrations

85. Free Body Diagram

86. Equilibrium Equation

87. Solution for constant loading

88. Solving initial conditions

89. Sine curve

90. General solution is a linear combination
Viscous Damped free vibration :
c : viscous damping constant (in N s/m)
Define: viscous damping factor (no unit)
Use a trial solution x = A et :
Obtain two possible ’s :
and
are two solutions.
General solution is a linear combination 90

91. > Three cases: overdamped z 1, Critical damped = 1 <
underdamping
Case 1 :  > 1 (overdamping) :
1 and 2 < 0
x decays to zero without oscillation 91

92. Case 2:  = 1 (critical damping) : 1 = 2 = -n
A1 exp(-nt) is a solution
A2 t exp(-nt) is also a solution (prove !)
The general solution is x = (A1 + A2 t )exp(-nt)
x approaches zero quickly without oscillation. 92

93. Case 3: < 1 (underdamping) :
Define damped natural frequency
Period d = 2/d 93

94. It is because  = c/(2mn), measurement of  determines
the viscous damping constant c. 94

95. B. Viscous Damped Forced vibration
The equation of motion : 95

96. 96

97. The resonance frequency is
Maximum M occurs at:
The resonance frequency is 97

98. tan  = (3)   n-, tan +,   /2(-)
Consider the following regions:
(1)  is small, tan > 0,   0+, xp in phase with the driving force
(2)  is large, tan < 0,   0-,  = , xp lags the driving force by 90o
(3)   n-, tan +,   /2(-)
  n+, tan -,   /2(+) 98

99. If b2 is replaced by Fo/m:
If the driving force is not applied to the mass, but is applied to the base of the system:
If b2 is replaced by Fo/m:
This can be used as a device to detect earthquake. 99

100. Example m =45 kg, k = 35 kN/m, c = 1250 N.s/m, p = 4000 sin (30 t) Pa,
A= 50 x 10-3 m2.
Determine : (a) steady-state displacement
(b) max. force transmitted to the base.
100

101. The amplitude of the steady-state vibration is:
101

102. The force transmitted to the base is :
For max Ftr :
102

103. End of 22.2
Let Learning Continue

104. Newton Laws, Energy Theorems and Momentum Principles
Integration Principles

105. Group Working Problems Problem 1: Chapters 17, 18 and 19
Given: The 30-kg disk is pin connected at its center. If it starts from rest,
Find:
the number of revolutions it must make and the time needed to attain an angular velocity of 20rad/s?

106. Problem 1: Newton Laws Find angular acceleration: (11.7rad/s2)
Find angle (17.1 radians)
Find number of revolutions: (2.73)
Find time (1.71 sec)

107. Problem 1: Energy Theorems
Find angle (17.1 radians)
Find number of revolutions: (2.73)
Find angular acceleration: (11.7rad/s2)
Find time (1.71 sec)

108. Problem 1: Momentum Principle
Find time (1.71 sec)
Find angular acceleration: (11.7rad/s2)
Find angle (17.1 radians)
Find number of revolutions: (2.73)

109. Problem 2: Chapter 22 in view of Chapters 17, 18 and 19
Given: a 10-kg block suspended from a cord wrapped around a 5kg disk.
Find: natural period of vibration

110. Problem 2: Newton Laws
Find differential equation
Find period (1.57s)

111. Problem 2: Energy Theorems: use conservation of energy
Write conservation of energy equation
Differentiate equation
Find period (1.57s)

112. Problem 2: Momentum Principle: use M=d(HG)/dt
Write M=d(HG)/dt for a displacement θ from equilibrium position
Differentiate equation
Find period (1.57s)

113. End of Integration Principles
Let Learning Continue