1. FE Review Dynamics G. Mauer UNLV Mechanical Engineering

2. X-Y Coordinates Point Mass Dynamics

4. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters. The travel time t to Point B is (A) t = 4 s (B) t = 1 s (C) t = 0.5 s (D) t = 2 s Use g = 10 m/s 2

5. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters. The travel time t to Point B is (A) t = 4 s (B) t = 1 s (C) t = 0.5 s (D) t = 2 s Use g = 10 m/s 2

6. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s. The start velocity v0 is (A) v 0 = 40 m/s (B) v 0 = 20 m/s (C) v 0 = 10 m/s (D) v 0 = 5 m/s Use g = 10 m/s 2

7. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s. The start velocity v0 is (A) v 0 = 40 m/s (B) v 0 = 20 m/s (C) v 0 = 10 m/s (D) v 0 = 5 m/s Use g = 10 m/s 2

8. 12.7 Normal and Tangential Coordinates u t : unit tangent to the path u n : unit normal to the path

9. Normal and Tangential Coordinates Velocity Page 53

10. Normal and Tangential Coordinates

12. Fundamental Problem 12.27 (A) constant (B) 1 m/s 2 (C) 2 m/s 2 (D) not enough information (E) 4 m/s 2 The boat is traveling along the circular path with  = 40m and a speed of v = 0.5*t 2, where t is in seconds. At t = 4s, the normal acceleration is:

13. Fundamental Problem 12.27 (A) constant (B) 1 m/s 2 (C) 2 m/s 2 (D) not enough information (E) 4 m/s 2 The boat is traveling along the circular path with  = 40m and a speed of v = 0.5*t 2, where t is in seconds. At t = 4s, the normal acceleration is:

14. Polar coordinates

17. Polar Coordinates Point P moves on a counterclockwise circular path, with r =1m,  dot (t) = 2 rad/s. The radial and tangential accelerations are: (A) a r = 4m/s 2 a  = 2 m/s 2 (B) a r = -4m/s 2 a  = -2 m/s 2 (C) a r = -4m/ s 2 a  = 0 m/s 2 (D) a r = 0 m/s 2 a  = 0 m/s 2

18. Polar Coordinates Point P moves on a counterclockwise circular path, with r =1m,  dot (t) = 2 rad/s. The radial and tangential accelerations are: (A) a r = 4m/s 2 a  = 2 m/s 2 (B) a r = -4m/s 2 a  = -2 m/s 2 (C) a r = -4m/ s 2 a  = 0 m/s 2 (D) a r = 0 m/s 2 a  = 0 m/s 2

19. Point B moves radially outward from center C, with r-dot =1m/s,  dot (t) = 10 rad/s. At r=1m, the radial acceleration is: (A) a r = 20 m/s 2 (B) a r = -20 m/s 2 (C) a r = 100 m/s 2 (D) a r = -100 m/s 2

20. Point B moves radially outward from center C, with r-dot =1m/s,  dot (t) = 10 rad/s. At r=1m, the radial acceleration is: (A) a r = 20 m/s 2 (B) a r = -20 m/s 2 (C) a r = 100 m/s 2 (D) a r = -100 m/s 2

21. Example cont’d: Problem 2.198 Sailboat tacking against Northern Wind 2. Vector equation (1 scalar eqn. each in i- and j-direction) 50 0 15 0 i

22. Given: r(t) = 2+2*sin(  (t)),  dot = constant The radial velocity is (A) 2+2*cos(  (t ))*  -dot, (B) -2*cos(  (t))*  -dot (C) 2*cos(  (t))*  -dot (D) 2*cos(  (t)) (E) 2*  +2*cos(  (t ))*  -dot

23. Given: r(t) = 2+2*sin(  (t)),  dot = constant The radial velocity is (A) 2+2*cos(  (t ))*  -dot, (B) -2*cos(  (t))*  -dot (C) 2*cos(  (t))*  -dot (D) 2*cos(  (t)) (E) 2*  +2*cos(  (t ))*  -dot

24. 2.9 Constrained Motion v A is given as shown. Find v B Approach: Use rel. Velocity: v B = v A +v B/A (transl. + rot.)

25. The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is (A) 6 m/s (B) 40 rad/s (C) -40 rad/s (D) 4 rad/s (E) none of the above

26. The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is (A) 6 m/s (B) 40 rad/s (C) -40 rad/s (D) 4 rad/s (E) none of the above

27. The rope length between points A and B is: (A) x A – x B + x c (B) x B – x A + 4x c (C) x A – x B + 4x c (D) x A + x B + 4x c Omit all constants!

28. The rope length between points A and B is: (A) x A – x B + x c (B) x B – x A + 4x c (C) x A – x B + 4x c (D) x A + x B + 4x c Omit all constants!

29. NEWTON'S LAW OF INERTIA A body, not acted on by any force, remains in uniform motion. NEWTON'S LAW OF MOTION Moving an object with twice the mass will require twice the force. Force is proportional to the mass of an object and to the acceleration (the change in velocity). F=ma.

30. Dynamics M1: up as positive: F net = T - m 1 *g = m 1 a1 M2: down as positive. F net = F = m 2 *g - T = m 2 a2 3. Constraint equation: a1 = a2 = a

31. Equations From previous: T - m 1 *g = m 1 a  T = m 1 g + m 1 a Previous for Mass 2: m 2 *g - T = m 2 a Insert above expr. for T m 2 g - ( m 1 g + m 1 a ) = m 2 a ( m2 - m1 ) g = ( m1 + m2 ) a ( m1 + m2 ) a = ( m2 - m1 ) g a = ( m 2 - m 1 ) g / ( m 1 + m 2 )

32. Rules 1. Free-Body Analysis, one for each mass 3. Algebra: Solve system of equations for all unknowns 2. Constraint equation(s): Define connections. You should have as many equations as Unknowns. COUNT!

33. M*g M*g*sin  -M*g*cos  j Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown.

34. Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Step 2: Apply Newton’s Law in each Direction: M*g M*g*sin  -M*g*cos  j N

35. Friction F =  k *N: Another horizontal reaction is added in negative x-direction. M*g M*g*sin  -M*g*cos  j N  k *N

36. Problem 3.27 in Book: Find accel of Mass A Start with: (A)Newton’s Law for A. (B)Newton’s Law for A and B (C) Free-Body analysis of A and B (D) Free-Body analysis of A

37. Problem 3.27 in Book: Find accel of Mass A Start with: (A)Newton’s Law for A. (B)Newton’s Law for A and B (C) Free-Body analysis of A and B (D) Free-Body analysis of A

38. Problem 3.27 in Book cont’d Newton applied to mass B gives:  Fu = 2T = m B *a B (B)  Fu = -2T + mB*g = 0 (C)  Fu = m B *g-2T = m B *a B  D  Fu = 2T- m B *g-2T = 0

39. Problem 3.27 in Book cont’d Newton applied to mass B gives:  Fu = 2T = m B *a B (B)  Fu = -2T + mB*g = 0 (C)  Fu = m B *g-2T = m B *a B  D  Fu = 2T- m B *g-2T = 0

40. Problem 3.27 in Book cont’d Newton applied to mass A gives:   Fx = T +F= m A *a x ;  Fy = N - m A *g*cos(30 o ) = 0 (B)  Fx = T-F= m A *a x  Fy = N- m A *g*cos(30 o ) = m A *a y (C)  Fx = T = m A *a x ;  Fy = N - m A *g*cos(30 o ) =0  D   Fx = T-F = m A *a x ;  Fy = N-m A *g*cos(30 o ) =0

41. Problem 3.27 in Book cont’d Newton applied to mass A gives:   Fx = T +F= m A *a x ;  Fy = N - m A *g*cos(30 o ) = 0 (B)  Fx = T-F= m A *a x  Fy = N- m A *g*cos(30 o ) = m A *a y (C)  Fx = T = m A *a x ;  Fy = N - m A *g*cos(30 o ) =0  D   Fx = T-F = m A *a x ;  Fy = N-m A *g*cos(30 o ) =0

42. Energy Methods

43. Only Force components in direction of motion do WORK

44. Work of Gravity

45. Work of a Spring

46. The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.

47. A car is traveling at 20 m/s on a level road, when the brakes are suddenly applied and all four wheels lock.  k = 0.5. The total distance traveled to a full stop is (use Energy Method, g = 10 m/s 2 ) (A) 40 m  20 m (C) 80 m (D) 10 m (E) none of the above

48. Collar A is compressing the spring after dropping vertically from A. Using the y-reference as shown, the work done by gravity (Wg) and the work done by the compression spring (Wspr) are (A) Wg <0, Wspr <0 (B) Wg >0, Wspr <0 (C) Wg 0 (D) Wg >0, Wspr >0 y

49. Conservative Forces A conservative force is one for which the work done is independent of the path taken Another way to state it: The work depends only on the initial and final positions, not on the route taken.

50. Conservative Forces T 1 + V 1 = T 2 + V 2

51. Potential Energy Potential energy is energy which results from position or configuration. An object may have the capacity for doing work as a result of its position in a gravitational field. It may have elastic potential energy as a result of a stretched spring or other elastic deformation.

52. Potential Energy elastic potential energy as a result of a stretched spring or other elastic deformation.

53. Potential Energy

54. y

55. A child of mass 30 kg is sliding downhill while the opposing friction force is 50 N along the 5m long incline (3m vertical drop). The work done by friction is (A) -150 Nm (B) 150 Nm (C) 250 Nm (D) -250 Nm (E) 500 Nm

56. (Use Energy Conservation) A 1 kg block slides d=4 m down a frictionless plane inclined at  =30 degrees to the horizontal. The speed of the block at the bottom of the inclined plane is (A) 1.6 m/s (B) 2.2 m/s (C) 4.4 m/s (D) 6.3 m/s (E) none of the above d h 

57. Angular Momentum Linear Momentum

58. Rot. about Fixed Axis Memorize!

59. Page 336: a t =  x r a n =  x (  x r)

60. Mathcad EXAMPLE

61. Mathcad Example part 2: Solving the vector equations

62. 16.4 Motion Analysis http://gtrebaol.free.fr/d oc/flash/four_bar/doc/

63. Approach 1.Geometry: Definitions Constants Variables Make a sketch 2. Analysis: Derivatives (velocity and acceleration) 3. Equations of Motion 4. Solve the Set of Equations. Use Computer Tools.

64. Example Bar BC rotates at constant  BC. Find the angular Veloc. of arm BC. Step 1: Define the Geometry

65. Example Bar BC rotates at constant  BC. Find the ang. Veloc. of arm BC. Step 1: Define the Geometry

66. Geometry: Compute all lengths and angles as f(  (t)) All angles and distance AC(t) are time-variant Velocities:  =  -dot is given.

67. Analysis: Solve the rel. Veloc. Vector equation conceptually Seen from O: vA =  x OA

68. Analysis: Solve the rel. Veloc. Vector equation Seen from O: vA =  x OA

69. Analysis: Solve the rel. Veloc. Vector equation Seen from C: v Collar +  BC x AC(t)  BC x AC(t) v A,rel

70. Analysis: Solve the rel. Veloc. Vector equation numerically

71. Here:  BC is given as -2 rad/s (clockwise). Find  OA

72. Analysis: Solve the rel. Veloc. Vector equation numerically

73. Recap: The analysis is becoming more complex. To succeed: Try Clear Organization from the start Mathcad Vector Equation = 2 simultaneous equations, solve simultaneously!

74. fig_05_011 16.6 Relative Velocity v A = v B + v A/B Relative Velocity v A = v B + v A/B = V B (transl) + v Rot v Rot =  x r

75. Seen from O: v B =   x r Seen from A: v B = v A +   x r B/A

76. Seen from O: v B =   x r Seen from A: v B = v A +   x r B/A

77. Rigid Body Acceleration Stresses and Flow Patterns in a Steam Turbine FEA Visualization (U of Stuttgart)

84. The instantaneous center of Arm BD is located at Point: (A) F (B) G (C) B (D) D (E) H

85. fig_06_002 Plane Motion 3 equations:  Forces_x  Forces_y  Moments about G

86. fig_06_002 Plane Motion 3 equations:  Forces_x  Forces_y  Moments about G

87. fig_06_005 Parallel Axes Theorem Pure rotation about fixed point P

88. Describe the constraint(s) with an Equation Constrained Motion: The system no longer has all three Degrees of freedom