1. Final Exam Review

2. Please Return Loan Clickers to the MEG office after Class! Today!

3. Always work from first Principles! Review

4. Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints Review

5. 1. Free-Body Review

6. 1. Free-Body B_x B_y mg

7. 2. Newton B_x B_y mg Moments about B: -mg*L/2 = IB*a with IB = m*L 2 / 3

8. 3. Constraint B_x B_y mg aG =  *L/2 = -g*3/4

9. 1. Free-Body mg A_y A_x N

10. mg A_y A_x N 2. Newton Moments about Center of Cylinder:A_x From triangle at left: Ax*(R-h) –b*mg = 0 acart*(R-h) –b*g = 0

11. mg A_y A_x N 2. Newton N = 0 at impending rolling, thus Ay = mg Ax = m*acart

12. Kinematics (P. 16-126) CTR

13. Kinematics (P. 16-126) CTR 4r -2r*i + 2r*j

14. Feedback Overall, when comparing traditional Homework formats with Mastering, I prefer (A) Paper submission of Homework (B) Electronic Submission

15. Feedback For me, the most useful benefit of Mastering is (A) Hints while developing the solution to a problem (B) Instant grading of results (C) Practice Exams

16. X-Y Coordinates Point Mass Dynamics

18. 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

19. 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

20. 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

21. 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

22. 12.7 Normal and Tangential Coordinates u t : unit tangent to the path u n : unit normal to the path

23. Normal and Tangential Coordinates Velocity Page 53

24. Normal and Tangential Coordinates

26. 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:

27. 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:

28. Polar coordinates

31. 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

32. 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

33. 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

34. 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

35. We Solve Graphically (Vector Addition) 12.10 Relative (Constrained) Motion vBvB vAvA v B/A

36. Example : Sailboat tacking against Northern Wind 2. Vector equation (1 scalar eqn. each in i- and j-direction) 50 0 15 0 i

37. 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

38. 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

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

40. 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

41. 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

42. 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!

43. 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!

44. Given: v0 = const. The vertical velocity component of point A (in y-direction) is (A)v A,y = v 0 *tan(  ) (B) v A,y = v 0 *cot(  ) (C) v A,y = v 0 *cos(  (D)v A,y = 2*v 0  

45. Given: v0 = const. The velocity of point A in vertical y-direction is (A)v A,y = v 0 *tan(  ) (B) v A,y = v 0 *cot(  ) (C) v A,y = v 0 *cos(  (D)v A,y = 2*v 0 (E) vA,y = v 0 /cos(  )    V y /v x =cot 

46. 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.

47. 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

48. 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 )

49. 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!

50. 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.

51. 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

52. 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

53. Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j- direction is (A) -mg*sin30 o (B) +mg*sin30 o (C) -mg*cos30 o (D) +mg*cos30 o (E) None of the above

54. Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j- direction is (A) -mg*sin30 o (B) +mg*sin30 o (C) -mg*cos30 o (D) +mg*cos30 o (E) None of the above mg mg*cos(  )

55. Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i- direction is (A) -mg*sin30 o (B) +mg*sin30 o (C) -mg*cos30 o (D) +mg*cos30 o (E) None of the above

56. Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i- direction is (A) -mg*sin30 o (B) +mg*sin30 o (C) -mg*cos30 o (D) +mg*cos30 o (E) None of the above mg*sin(  )

57. 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

58.  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 Newton applied to mass B gives:

59.   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 Newton applied to mass A gives:

60.   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 Newton applied to mass A gives:

61. Energy Methods

62. Only Force components in direction of motion do WORK

63. Work of Gravity

64. Work of a Spring

65. 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.

66. 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

67. 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

68. 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

69. 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

70. Potential Energy For any conservative force F we can define a potential energy function U in the following way: – The work done by a conservative force is equal and opposite to the change in the potential energy function. This can be written as: r1r1 r2r2 U2U2 U1U1

71. Hooke’s Law Force exerted to compress a spring is proportional to the amount of compression.

72. Conservative Forces: Gravity is a conservative force: Gravity near the Earth’s surface: A spring produces a conservative force:

73. (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 

74. (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 

75. 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 change of potential energy is (A) -882 Nm (B) 882 Nm (C) 1470 Nm (D) -1470 Nm (E) None of the above

76. 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 change of potential energy is Change of PE = mg(h final -h 0 ) = 30*9.8*(0-3) = -882 Nm (A) -882 Nm (B) 882 Nm (C) 1470 Nm (D) -1470 Nm (E) None of the above

77. Rot. about Fixed Axis Memorize!

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

79. Arm BD is rotating with constant  dot=  , while point D moves at vD*i. Seen from D, the velocity vector at B is: (A) vB = vD*i - BD*  *cos  i  BD  *sin  j (B) vB = vD*i - BD*  *cos  i  BD  *sin  j (C) vB = vD*i + BD*  *cos  i  BD  *sin  j (D) vB = - BD*  *cos  i  BD  *sin  j (E) none of the above

80. Arm BD is rotating with constant  dot=  , while point D moves at vD*i. Seen from D, the velocity vector at B is: (A) vB = vD*i - BD*  *cos  i  BD  *sin  j (B) vB = vD*i - BD*  *cos  i  BD  *sin  j (C) vB = vD*i + BD*  *cos  i  BD  *sin  j (D) vB = - BD*  *cos  i  BD  *sin  j (E) none of the above

81. Mathcad EXAMPLE

82. Mathcad Example part 2: Solving the vector equations

83. Mathcad Examples part 3 Graphical Solution

86. v B = 3 ft/s down,  = 60 o and v A = v B /tan  The relative velocity v A/B is found from vector eq. (A)v A = v B + v A/B, v A/B points  v A = v B + v A/B, v A/B points (C)  v B = v A + v A/B, v A/B points  D) V B = v B + v A/B, v A/B points vBvB vAvA x y

87. v B = 3 ft/s down,  = 60 o and v A = v B /tan  The relative velocity v A/B is found from vector eq. (A)v A = v B + v A/B, v A/B points  v A = v B + v A/B, v A/B points (C)  v B = v A + v A/B, v A/B points  D) V B = v B + v A/B, v A/B points vBvB vAvA x y vBvB vAvA v A/B

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

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

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

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

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

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

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

101. 6.78 Given: I_ G=m*k 2 =300*1.5^2 = 675 kgm^2. The angular accel of the rocket is Thrust T = 4 kN F_y = m*a = 300*8.69N (A) 0.102 rad/s 2 (B) 0.31 rad/s 2 (C) 3.1 rad/s 2 (D) 5.9 rad/s 2

102. 6.78 Given: I_ G= 675 kgm^2, m = 300 kg. The angular accel of the rocket is Thrust T = 4 kN F_y = m*a = 300*8.69N (A) 0.102 rad/s 2 (B) 0.31 rad/s 2 (C) 3.1 rad/s 2 (D) 5.9 rad/s 2 Answer: (sum of moments about G = I_G*alpha) 4000N * sin(1deg)*3m = 675*alpha alpha = 0.31 rad/s^2