Tutorial Solutions

1. Tutorial 0
Basic Vector Calculations
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

2. Concepts
Dot product
Cross product
𝑭𝒊𝒏𝒅 𝑨 × 𝑩
|
Ƹ𝑖 Ƹ𝑗 ෠𝑘
1 2 3
−2 0 1
| = |
2 3
0 1
| Ƹ𝑖 − |
1 3
−2 1
| Ƹ𝑗 + |
1 2
−2 0
|෠𝑘
= 2 Ƹ𝑖 − 7 Ƹ𝑗 + 4෠𝑘
Ԧ𝐴 = 1 Ƹ𝑖 + 2 Ƹ𝑗 + 3෠𝑘
𝐵 = −2 Ƹ𝑖 + 0 Ƹ𝑗 + 1෠𝑘
𝑭𝒊𝒏𝒅 𝑨 ⋅ 𝑩
Ԧ𝐴 ⋅ 𝐵 = (1)(−2) + (2)(0) + (3)(1) = 1
Ԧ𝐴 = 1 Ƹ𝑖 + 2 Ƹ𝑗
𝐵 = −2 Ƹ𝑖 + 0 Ƹ𝑗
𝑭𝒊𝒏𝒅 𝑨 ⋅ 𝑩
Ԧ𝐴 = 2.24∡63.4
𝐵 = 2∡180
Ԧ𝐴 ⋅ 𝐵 = (2.24)(2)cos(180 − 63.4) = −2.01
𝑭𝒊𝒏𝒅 𝑨 × 𝑩
Ԧ𝐴 × 𝐵 = (2.24)(2)sin(180 − 63.4)෠𝑘 = 4෠𝑘
2D vectors 3D vectors

3. Question 1
| Ԧ𝐴| = 𝑥2 + 𝑦2 = 22 + −1 2 = 2.236
𝜃 = tan−1
𝑦
𝑥
= tan−1
−1
2
= −26.57
3

4. Question 1
Ԧ𝐴 ⋅ 𝐵 = (2.236) ⋅ (3.606) ⋅ cos(123.69 + 26.56) = −7
4

5. Question 1
Ԧ𝐴 × 𝐵 = 2.236 ⋅ 3.606 ⋅ sin(123.69∘
+ 26.56∘
)𝑘 = 4𝑘
𝐵 × Ԧ𝐴 = −4𝑘
5

6. Question 1
6
Ԧ𝐶 × 𝐵 = |
Ƹ𝑖 Ƹ𝑗 ෠𝑘
0 0 0.3
−2 3 0
| = |
0 0.3
3 0
| Ƹ𝑖 − |
0 0.3
−2 0
| Ƹ𝑗 + |
0 0
−2 3
|෠𝑘
= − 0.9 Ƹ𝑖 − 0.6 Ƹ𝑗
ˆˆ ˆ ˆ ˆ( 0.9 0.6 ) 0.18 0.27 0C i j i j k     

7. Question 1

8. Question 2
8

9. Question 3
)𝑥 = 𝑟ℎ𝑜 ⋅ cos(𝜃
)𝑦 = 𝑟ℎ𝑜 ⋅ sin(𝜃
Recall:
9

10. Question 4
F
F
F
T
F
10

11. Question 5
11

12. Tutorial 1
Position vector, velocity and
acceleration
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

13. Question 1

14. 14
Question 1
Point attached – non rotating
Reference (absolute)
Body attached – rotating

15. Question 1 (a)
Reference (absolute)
Inertial reference frame
Both points appear to rotate when viewed in
a reference frame that is placed outside the disk

16. Question 1 (b)
ra
rb
rb/aA
B

17. Question 1 (b)
ra
rb
rb/a
The distance between the points A and B does not change

18. Question 1 (b)
ra
rb
rb/a
The reference frame can translate, but not rotate (point attached)
The distance between the points A and B does not change

19. Question 1 (b)
ra
rb
rb/a
The reference frame can translate, but not rotate (point attached)
The distance between the points A and B does not change
𝐫𝐛 = 𝐫𝐚 + 𝐫𝐛/𝐚

20. Question 1 (b)
ra
rb
rb/a
The reference frame can translate, but not rotate (point attached)
The distance between the points A and B does not change
𝐫𝐛 = 𝐫𝐚 + 𝐫𝐛/𝐚

21. Question 1 (b)
ra
rb
rb/a
The reference frame can translate, but not rotate (point attached)
The distance between the points A and B does not change
𝐫𝐛 = 𝐫𝐚 + 𝐫𝐛/𝐚
Object is taken to be rotating about the reference system

22. Question 1 (b)
ra
rb
rb/a
The reference frame can translate, but not rotate
The distance between the points A and B does not change
𝐫𝐛 = 𝐫𝐚 + 𝐫𝐛/𝐚
Object is taken to be rotating about the reference system

23. Question 1(c)
rb
The reference frame can translate, but not rotate (point attached)
The distance between the points B and O does not change
Object is taken to be rotating about the reference system

24. Question 1(d)
The reference frame can translate AND rotate (body attached)
A
B
The distance to points A and B does not change

25. Question 1(d)
A
B
Point fixed
If your reference frame does not rotate, but just gets centred on
your point of interest, the other points appears to rotate around
it
Body fixed
If consider a rotating reference frame, then all points (that are
fixed to the disc, or to the frame) are obviously and by definition
stationary inside it, and there is no "relative motion" between
them in such a reference frame.

26. 26
Question 1 Recall
Point attached – non rotating
Reference (absolute)
Body attached – rotating

27. Question 2

28. Question 2

29. Question 2

30. Question 2

31. Question 2

32. Question 2

33. Question 3

34. Question 3

35. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡

36. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡

37. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

38. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

39. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

40. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

41. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

42. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

43. Question 3
)𝑟cos(𝜔𝑡
)𝑟sin(𝜔𝑡
)𝑙2 − 𝑟2sin2(𝜔𝑡

44. Question 3 – Steps for velocity derivative

45. Question 3 – Steps for acceleration derivative
Velocity is found to be….

46. Question 4
Ԧ𝑟 = 8𝑡2 Ƹ𝑖 + (𝑡3 + 5) Ƹ𝑗

47. Question 4
Ԧ𝑟 = 8𝑡2 Ƹ𝑖 + (𝑡3 + 5) Ƹ𝑗
𝑑Ԧ𝑟
𝑑𝑡
= 16𝑡 Ƹ𝑖 + 3𝑡2 Ƹ𝑗
𝑑Ԧ𝑟
𝑑𝑡
| 𝑡=3𝑠 = 16(3) Ƹ𝑖 + 3(3)2 Ƹ𝑗
𝑑2
Ԧ𝑟
𝑑𝑡2
= 16 Ƹ𝑖 + 6𝑡 Ƹ𝑗
𝑑2
Ԧ𝑟
𝑑𝑡2
| 𝑡=3𝑠 = 16 Ƹ𝑖 + 6(3) Ƹ𝑗

48. Question 4
Ԧ𝑟 = 8𝑡2 Ƹ𝑖 + (𝑡3 + 5) Ƹ𝑗
𝑑Ԧ𝑟
𝑑𝑡
= 16𝑡 Ƹ𝑖 + 3𝑡2 Ƹ𝑗
𝑑Ԧ𝑟
𝑑𝑡
| 𝑡=3𝑠 = 16(3) Ƹ𝑖 + 3(3)2 Ƹ𝑗
𝑑2
Ԧ𝑟
𝑑𝑡2
= 16 Ƹ𝑖 + 6𝑡 Ƹ𝑗
𝑑2
Ԧ𝑟
𝑑𝑡2
| 𝑡=3𝑠 = 16 Ƹ𝑖 + 6(3) Ƹ𝑗
Ԧ𝑟 = 8𝑡2 Ƹ𝑖 + (𝑡3 + 5) Ƹ𝑗
)∴ 𝑥 = 8𝑡2
𝑦 = (𝑡3
+ 5
Since 𝑥 = 8𝑡2 ⇒ 𝑡 =
𝑥
8
∴ 𝑦 =
𝑥
8
3
+ 5 = 0.125𝑥 Τ3 2
+ 5

49. 49
Question 4
𝑦 =
𝑥
8
3
+ 5 = 0.125𝑥 Τ3 2
+5

50. Tutorial 2
Simple motion of a particles
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

51. Question 1
15m
5m/s
9.81m/s2
𝑣2
= 𝑢2
+ 2𝑎𝑠
𝑣 = )52 + 2(9.81)(15 = 17.9 Τ𝑚 𝑠
𝑎 =
𝑣 − 𝑢
𝑡
⇒ 𝑡 =
𝑣 − 𝑢
𝑎
𝑡 =
17.9 − 5
9.81
= 1.31𝑠
CONSTANT ACCELERATION PROBLEM
+ve

52. Question 2

53. 53
Question 2
Position
Velocity
Acceleration
Position
Velocity
Acceleration

54. Question 3
VARIABLE ACCELERATION PROBLEM

55. 55
Question 3
Position
VelandAcc
Acceleration
Velocity

56. 56
Question 4
Recall…
VARIABLE ACCELERATION PROBLEM

57. 57
Question 4

58. N-t coordinate system
• Rectangular coordinate
• Polar coordinate
• Normal and tangential coordinate
Normal and tangential coordinate
• t-axis points tangential to the path in the direction of Velocity
• n-axis points perpendicular to t-axis, towards centre of curvature
• Body fixed frame, hence moves and rotates with the particle
Velocity
• ALWAYS tangential to path
ሿԦ𝑣 = 𝑣 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛[𝐴𝐿𝑊𝐴𝑌𝑆!
Acceleration
• Has a tangential and normal component to path
Ԧ𝑎 = 𝑎 𝑡 Ƹ𝑒𝑡 + 𝑎 𝑛 Ƹ𝑒 𝑛
𝐶𝑢𝑟𝑣𝑖𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = ሶ𝑣 𝑡 Ƹ𝑒𝑡 +
𝑣 𝑡
2
𝜌
Ƹ𝑒 𝑛
𝑃𝑢𝑟𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = 𝑟 ሷ𝜃 Ƹ𝑒𝑡 + 𝑟 ሶ𝜃2 Ƹ𝑒 𝑛
Ԧ𝑣 = 𝜔𝑟 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛

59. 59
Equations of motion in N-t coordinate system
Acceleration
• Has a tangential and normal component to path
Ԧ𝑎 = 𝑎 𝑡 Ƹ𝑒𝑡 + 𝑎 𝑛 Ƹ𝑒 𝑛
𝐶𝑢𝑟𝑣𝑖𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = ሶ𝑣 𝑡 Ƹ𝑒𝑡 +
𝑣 𝑡
2
𝜌
Ƹ𝑒 𝑛
𝑃𝑢𝑟𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = 𝑟 ሷ𝜃 Ƹ𝑒𝑡 + 𝑟 ሶ𝜃2 Ƹ𝑒 𝑛

60. Question 5

61. Question 5

62. Question 5
Cross product form is
useful when there is
rotation around
multiple axis

63. Question 5
Total acceleration of point A and B

64. N-t coordinate system
• Rectangular coordinate
• Polar coordinate
• Normal and tangential coordinate
Normal and tangential coordinate
• t-axis points tangential to the path in the direction of Velocity
• n-axis points perpendicular to t-axis, towards centre of curvature
• Body fixed frame, hence moves and rotates with the particle
Velocity
• ALWAYS tangential to path
ሿԦ𝑣 = 𝑣 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛[𝐴𝐿𝑊𝐴𝑌𝑆!
Acceleration
• Has a tangential and normal component to path
Ԧ𝑎 = 𝑎 𝑡 Ƹ𝑒𝑡 + 𝑎 𝑛 Ƹ𝑒 𝑛
𝐶𝑢𝑟𝑣𝑖𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = ሶ𝑣 𝑡 Ƹ𝑒𝑡 +
𝑣 𝑡
2
𝜌
Ƹ𝑒 𝑛
𝑃𝑢𝑟𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = 𝑟 ሷ𝜃 Ƹ𝑒𝑡 + 𝑟 ሶ𝜃2 Ƹ𝑒 𝑛
Ԧ𝑣 = 𝜔𝑟 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛

65. 65
Equations of motion in N-t coordinate system
Acceleration
• Has a tangential and normal component to path
Ԧ𝑎 = 𝑎 𝑡 Ƹ𝑒𝑡 + 𝑎 𝑛 Ƹ𝑒 𝑛
𝐶𝑢𝑟𝑣𝑖𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = ሶ𝑣 𝑡 Ƹ𝑒𝑡 +
𝑣 𝑡
2
𝜌
Ƹ𝑒 𝑛
𝑃𝑢𝑟𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = 𝑟 ሷ𝜃 Ƹ𝑒𝑡 + 𝑟 ሶ𝜃2 Ƹ𝑒 𝑛
Ԧ𝑣 = 𝜔𝑟 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛

66. Question 6
ሷ𝜃(𝑡) = 5

67. Question 6
ሷ𝜃(𝑡) = 5

68. Question 6
ሷ𝜃(𝑡) = 5

69. Tutorial 3
Planar curvilinear motion
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

70. 70
Question structure
•Fixed reference frame – Q1 and Q5
•N-t reference frame – Q2 and Q3
•R-θ reference frame – Q4 and Q6

71. 71
Be careful when calculating vector direction!
Ԧ𝐴 = 5 Ƹ𝑖 + 10 Ƹ𝑗
| Ԧ𝐴| = 52 + 102 = 11.2
∠ Ԧ𝐴 = tan−1
10
5
= 63.40
𝐵 = −5 Ƹ𝑖 − 10 Ƹ𝑗
Ԧ𝐴 = 11.2∠63.40
Ԧ𝐴 = 5 Ƹ𝑖 + 10 Ƹ𝑗
|𝐵| = 52 + 102 = 11.2
∠𝐵 = tan−1
−10
−5
= 63.40
𝑩 = 𝟏𝟏. 𝟐∠𝟔𝟑. 𝟒 𝟎
𝐵 = −5 Ƹ𝑖 − 10 Ƹ𝑗
5
10
-10
Recall….
|𝑉𝑒𝑐𝑡𝑜𝑟| = 𝑥2 + 𝑦2
∠𝑉𝑒𝑐𝑡𝑜𝑟 = tan−1
𝑦
𝑥
𝑩 = 𝟏𝟏. 𝟐∠𝟏𝟖𝟎 𝟎 + 𝟔𝟑. 𝟒 𝟎
𝜃
𝜃
−5
𝑩 = 𝟏𝟏. 𝟐∠𝜋 + 𝜃

72. 72
෠𝑘 ×. . .
Ƹ𝑖
Ƹ𝑗
෠𝑘 × Ƹ𝑗 = − Ƹ𝑖

73. 73
෠𝑘 ×. . .
Ƹ𝑖
෠𝑘 × Ƹ𝑗 = − Ƹ𝑖
Rotates the axis CCW by 90o

74. 74
෠𝑘 ×. . .
Ƹ𝑖
෠𝑘 × − Ƹ𝑖= - Ƹ𝑗

75. 75
CONSTANT ACCELERATION PROBLEM
Projectile motion

76. Question 1
𝑥 𝑝 = 0
𝑦𝑝 = 0
Initial projectile position Initial target position
𝑥 𝑡 = 𝐿
)𝑦𝑡 = 𝐿tan(𝜃
Initial projectile velocity
൯ሶ𝑥 𝑝 = Ԧ𝑣0cos(𝜃
൯ሶ𝑦𝑝 = Ԧ𝑣0sin(𝜃
Initial target velocity
ሶ𝑥 𝑡 = 0
ሶ𝑦𝑡 = 0
Show that when 𝑥 𝑝 = 𝐿, 𝑦𝑝 = 𝑦𝑡
൯𝑥 𝑝(𝑡) = Ԧ𝑣0cos(𝜃 𝑡
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃 t −
1
2
𝑔𝑡2
Position of projectile at time, t
Position of target at time, t
𝑥 𝑡(𝑡) = 𝐿
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔𝑡2
Time taken by projectile to travel the horizontal distance, L
𝑥 𝑝(𝑡) = 𝐿 = Ԧ𝑣0cos(𝜃)𝑡 ⇒ 𝑡| 𝑥 𝑝=𝐿 =
𝐿
)Ԧ𝑣0cos(𝜃
Vertical position of projectile and target at time, txp=L
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃
𝐿
)Ԧ𝑣0cos(𝜃
−
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2

77. Question 1
𝑥 𝑝 = 0
𝑦𝑝 = 0
Initial projectile position Initial target position
𝑥 𝑡 = 𝐿
)𝑦𝑡 = 𝐿tan(𝜃
Initial projectile velocity
൯ሶ𝑥 𝑝 = Ԧ𝑣0cos(𝜃
൯ሶ𝑦𝑝 = Ԧ𝑣0sin(𝜃
Initial target velocity
ሶ𝑥 𝑡 = 0
ሶ𝑦𝑡 = 0
Show that when 𝑥 𝑝 = 𝐿, 𝑦𝑝 = 𝑦𝑡
൯𝑥 𝑝(𝑡) = Ԧ𝑣0cos(𝜃 𝑡
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃 t −
1
2
𝑔𝑡2
Position of projectile at time, t
Position of target at time, t
𝑥 𝑡(𝑡) = 𝐿
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔𝑡2
Time taken by projectile to travel the horizontal distance, L
𝑥 𝑝(𝑡) = 𝐿 = Ԧ𝑣0cos(𝜃)𝑡 ⇒ 𝑡| 𝑥 𝑝=𝐿 =
𝐿
)Ԧ𝑣0cos(𝜃
Vertical position of projectile and target at time, txp=L
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃
𝐿
)Ԧ𝑣0cos(𝜃
−
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2

78. Question 1
𝑥 𝑝 = 0
𝑦𝑝 = 0
Initial projectile position Initial target position
𝑥 𝑡 = 𝐿
)𝑦𝑡 = 𝐿tan(𝜃
Initial projectile velocity
൯ሶ𝑥 𝑝 = Ԧ𝑣0cos(𝜃
൯ሶ𝑦𝑝 = Ԧ𝑣0sin(𝜃
Initial target velocity
ሶ𝑥 𝑡 = 0
ሶ𝑦𝑡 = 0
Show that when 𝑥 𝑝 = 𝐿, 𝑦𝑝 = 𝑦𝑡
൯𝑥 𝑝(𝑡) = Ԧ𝑣0cos(𝜃 𝑡
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃 t −
1
2
𝑔𝑡2
Position of projectile at time, t
Position of target at time, t
𝑥 𝑡(𝑡) = 𝐿
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔𝑡2
Time taken by projectile to travel the horizontal distance, L
𝑥 𝑝(𝑡) = 𝐿 = Ԧ𝑣0cos(𝜃)𝑡 ⇒ 𝑡| 𝑥 𝑝=𝐿 =
𝐿
)Ԧ𝑣0cos(𝜃
Vertical position of projectile and target at time, txp=L
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃
𝐿
)Ԧ𝑣0cos(𝜃
−
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2

79. Question 1
𝑥 𝑝 = 0
𝑦𝑝 = 0
Initial projectile position Initial target position
𝑥 𝑡 = 𝐿
)𝑦𝑡 = 𝐿tan(𝜃
Initial projectile velocity
൯ሶ𝑥 𝑝 = Ԧ𝑣0cos(𝜃
൯ሶ𝑦𝑝 = Ԧ𝑣0sin(𝜃
Initial target velocity
ሶ𝑥 𝑡 = 0
ሶ𝑦𝑡 = 0
Show that when 𝑥 𝑝 = 𝐿, 𝑦𝑝 = 𝑦𝑡
൯𝑥 𝑝(𝑡) = Ԧ𝑣0cos(𝜃 𝑡
𝑦𝑝(𝑡) = Ԧ𝑣0sin 𝜃 t −
1
2
𝑔𝑡2
Position of projectile at time, t
Position of target at time, t
𝑥 𝑡(𝑡) = 𝐿
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔𝑡2
Time taken by projectile to travel the horizontal distance, L
𝑥 𝑝(𝑡) = 𝐿 = Ԧ𝑣0cos(𝜃)𝑡 ⇒ 𝑡| 𝑥 𝑝=𝐿 =
𝐿
)Ԧ𝑣0cos(𝜃
Vertical position of projectile and target at time, txp=L
𝑦𝑝(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2
𝑦𝑡(𝑡) = 𝐿tan(𝜃) −
1
2
𝑔
𝐿
)Ԧ𝑣0cos(𝜃
2
𝑦𝑡(𝑡) = 𝑦𝑝(𝑡)

80. 80
[21:50]

81. Question 5
Rocket position components w.r.t to A-xy
)𝑦𝑟 = btan(𝜃
Rocket velocity components w.r.t to A-xy
Rocket acceleration components w.r.t to A-xy
ሶ𝑥 𝑟 = 0 (Since 𝑏 is a constant)
ሶ𝑦𝑟 =
)𝑑(𝑏tan𝜃
𝑑𝑡
=
)𝑑(𝑏tan𝜃
𝑑𝜃
𝑑𝜃
𝑑𝑡
= 𝑏 sec2
𝜃 ሶ𝜃
𝑥 𝑟 = 𝑏
ሷ𝑥 𝑟 = 0
ሷ𝑦𝑟 =
𝑑 ሶ𝑦𝑟
𝑑𝑡
=
൯𝑑(𝑏(sec2
𝜃) ሶ𝜃
𝑑𝑡 = 𝑏(sec2
𝜃) ሷ𝜃 + 2𝑏(sec2
𝜃tan𝜃) ሶ𝜃2
x
y
Fixed
reference frame

82. N-t coordinate system
• Rectangular coordinate
• Polar coordinate
• Normal and tangential coordinate
Normal and tangential coordinate
• t-axis points tangential to the path in the direction of Velocity
• n-axis points perpendicular to t-axis, towards centre of curvature
• Body fixed frame, hence moves and rotates with the particle
Velocity
• ALWAYS tangential to path
ሿԦ𝑣 = 𝑣 Ƹ𝑒𝑡 + 0 Ƹ𝑒 𝑛[𝐴𝐿𝑊𝐴𝑌𝑆!
Acceleration
• Has a tangential and normal component to path
Ԧ𝑎 = 𝑎 𝑡 Ƹ𝑒𝑡 + 𝑎 𝑛 Ƹ𝑒 𝑛
𝐶𝑢𝑟𝑣𝑖𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = ሶ𝑣 𝑡 Ƹ𝑒𝑡 +
𝑣 𝑡
2
𝜌
Ƹ𝑒 𝑛
𝑃𝑢𝑟𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
Ԧ𝑎 = 𝑟 ሷ𝜃 Ƹ𝑒𝑡 + 𝑟 ሶ𝜃2 Ƹ𝑒 𝑛

83. 83
Question 2
Setup n-t reference frame Express the given vectors in n-t reference frame
Ԧ𝑣𝑡 = 120 Ƹ𝑒𝑡 Ԧ𝑣 𝑛 = 0 Ƹ𝑒 𝑛
Ԧ𝑎 𝑡 = 21cos(60) Ƹ𝑒𝑡 Ԧ𝑎 𝑛 = 21sin(60) Ƹ𝑒 𝑛
Ԧ𝑎 𝑛 =
𝑣 𝑡
2
𝜌
⇒ 𝜌 =
1202
)21sin(60
= 792𝑚
Since in a n-t reference frame…

84. 84
Question 3
Setup n-t reference frame
Ԧ𝑎 𝑡
Ԧ𝑎 𝑛
Ԧ𝑣 𝑡
Ƹ𝑒 𝑛
Ƹ𝑒𝑡
Express the given vectors in n-t reference frame
𝑣 𝑡 = 72𝑘 Τ𝑚 ℎ = 20 ΤƸ𝑒𝑡(𝑚 𝑠)
𝑎 𝑡 = −1.25 ΤƸ𝑒𝑡(𝑚 𝑠2)
Given
𝜌 = 350𝑚
Since in an n-t reference frame…
Ԧ𝑎 𝑛 =
𝑣 𝑡
2
𝜌
=
202
350
= 1.1429𝑚/𝑠2
Magnitude of acceleration vector at t=0
|𝑎| = 𝑎 𝑛
2
+ 𝑎 𝑡
2
= 1.14292 + 1.252 = 1.694 Τ𝑚 𝑠2
Magnitude of acceleration vector at t=4
𝑣 𝑡(𝑡) = 𝑣0 + 𝑎 𝑡 𝑡
𝑣 𝑡(4𝑠) = 20 − 1.25 ⋅ 4 = 15 Τ𝑚 𝑠
Ԧ𝑎 𝑛(4𝑠) =
152
350
= 0.6429 Τ𝑚 𝑠2
|𝑎| = 𝑎 𝑡
2
+ 𝑎 𝑛
2
= 1.406 Τ𝑚 𝑠2

85. r-θ coordinate system
• Rectangular coordinate
• Polar coordinate
• Normal and tangential coordinate
r-θcoordinate
• r-axis points in the direction of increasing radius
• θ-axis points perpendicular to r-axis, increasing angular displacement
• Body fixed frame, hence moves and rotates with the particle
• er is the unit vector in the radial direction
• eθ is the unit vector in the theta direction
Change of basis
Ƹ𝑒 𝑟 = cos𝜃 Ƹ𝑖 + sin𝜃 Ƹ𝑗
Ƹ𝑒 𝜃 = −sin𝜃 Ƹ𝑖 + cos𝜃 Ƹ𝑗
Velocity
𝑣 = ሶ𝑟 Ƹ𝑒 𝑟 + 𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Acceleration
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2
Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Relative velocity to inertial frame
Relative acceleration to inertial frame
Coriolis acceleration
Entrained velocity
Entrained acceleration
If the distance to the rotation center is not
consistent, you must include Coriolis item in
calculating the absolute particle acceleration.

86. 86
Question 4
• Draw inertial reference frame (Rectangular or Polar)
• Draw body-fixed frame (N-t, r- θ)
• Specify unit vector direction
Ƹ𝑒 𝑟
Ƹ𝑒 𝜃
• Specify velocity and acceleration components along unit vector directions
𝑉𝐵/𝑓
𝑉𝜃
𝑣 = ሶ𝑟 Ƹ𝑒 𝑟 + 𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Recall…
ሶ𝑟 = 60(2)𝑡 − 20(3)𝑡2
ሶ𝑟| 𝑡=1 = 60(2) − 20(3) = 60
𝑟 = 60𝑡2 − 20𝑡3 𝜃 = 2𝑡2
ሶ𝜃 = 4𝑡
ሶ𝜃| 𝑡=1 = 4
𝑣 = 60 Ƹ𝑒 𝑟 + (40)(4) Ƹ𝑒 𝜃= 60 Ƹ𝑒 𝑟 + 160 Ƹ𝑒 𝜃
• Find velocity and acceleration components along unit vector directions
𝑗
𝑖

87. 87
Question 4
• Draw inertial reference frame (Rectangular or Polar)
• Draw body-fixed frame (N-t, r- θ)
• Specify unit vector direction
Ƹ𝑒 𝑟
Ƹ𝑒 𝜃
• Specify velocity and acceleration components along unit vector directions
𝑉𝐵/𝑓
𝑉𝜃
𝑣 = ሶ𝑟 Ƹ𝑒 𝑟 + 𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Recall…
ሶ𝑟 = 60(2)𝑡 − 20(3)𝑡2
ሶ𝑟| 𝑡=1 = 60(2) − 20(3) = 60
𝑟 = 60𝑡2 − 20𝑡3 𝜃 = 2𝑡2
ሶ𝜃 = 4𝑡
ሶ𝜃| 𝑡=1 = 4
𝑣 = 60 Ƹ𝑒 𝑟 + (40)(4) Ƹ𝑒 𝜃= 60 Ƹ𝑒 𝑟 + 160 Ƹ𝑒 𝜃
• Find velocity and acceleration components along unit vector directions
|𝑣| = 602 + 1602 = 170.9𝑚 Τ𝑚 𝑠
∠𝑣 = tan−1
60
160
= 20.6 𝑜 𝑀𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑓𝑟𝑜𝑚ෝ𝑒θ
Ƹ𝑒 𝑟
Ƹ𝑒 𝜃
𝑉𝐵/𝑓
𝑉𝜃
𝑉
𝜃| 𝑡=1 = 2 1 2
= 2𝑟𝑎𝑑 = 114.60
[ෝ𝑒r 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑓𝑟𝑜𝑚 𝑖ሿ
∠𝑣 = 114.60
+ (900
−20.60
) = 1840
measured from 𝑖
𝑖
𝑗
𝑗
𝑖
• Find velocity and acceleration components w.r.t to inertial reference frame
170.9mm/s∠1840

88. 88
Question 4
• Draw inertial reference frame (Rectangular or Polar)
• Draw body-fixed frame (N-t, r- θ)
• Specify unit vector direction
Ƹ𝑒 𝑟
Ƹ𝑒 𝜃
• Specify velocity and acceleration components along unit vector directions
Recall…
ሶ𝑟 = 60(2)𝑡 − 20(3)𝑡2
ሶ𝑟| 𝑡=1 = 60(2) − 20(3) = 60
𝑟 = 60𝑡2 − 20𝑡3 𝜃 = 2𝑡2
ሶ𝜃 = 4𝑡
ሶ𝜃| 𝑡=1 = 4
• Find velocity and acceleration components along unit vector directions
𝑖
𝑗
• Find velocity and acceleration components w.r.t to inertial reference frame
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2
Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃
ሷ𝑟 = 120 − 60(2)𝑡 ሷ𝜃 = 4
ሷ𝑟| 𝑡=1 = 0(zero relative to rod) ሷ𝜃| 𝑡=1 = 4
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2 Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃
𝑎 = (0 − 40 ⋅ 42
) Ƹ𝑒 𝑟 + (40 ⋅ 4 + 2 ⋅ 60 ⋅ 4) Ƹ𝑒 𝜃
𝑎 = −640 Ƹ𝑒 𝑟 + 640 Ƹ𝑒 𝜃
|𝑎| = 6402 + 6402 = 905.1𝑚 Τ𝑚 𝑠2
∠𝑎 = tan−1 −640
640
= −450 measured from ෝ𝑒θ
𝜃| 𝑡=1 = 2 1 2 = 2𝑟𝑎𝑑 = 114.60
∠𝑎 = 450 + 114.60 + 900 = 249.60
𝑎 = 905.1𝑚 Τ𝑚 𝑠2
∠249.60
Ƹ𝑒 𝑟
Ƹ𝑒 𝜃
𝑗
𝑖
Ԧ𝑎

89. 89
Question 6
𝑟 = 2𝑏cos𝜃 𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 ሶ𝜃 𝑖𝑠 𝑐𝑜𝑛𝑠tan𝑡
ሶ𝑟 = −2𝑏 ሶ𝜃 sin𝜃
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2
Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Acceleration of a point in r-theta coordinate is given by….
ሷ𝑟 = −2𝑏 ሶ𝜃2cos𝜃 [since ሷ𝜃 = 0]
ሷ𝜃 = 0
Ԧ𝑎 𝐵 = (−2𝑏 ሶ𝜃2cos𝜃 − 2𝑏cos𝜃 ⋅ ሶ𝜃2) Ƹ𝑒 𝑟 + (2𝑏cos𝜃 ⋅ 0 + 2(−2𝑏sin𝜃) ሶ𝜃) Ƹ𝑒 𝜃
൯∴ Ԧ𝑎 𝐵 = −4𝑏 ሶ𝜃2
(cos𝜃 Ƹ𝑒 𝑟 + sin𝜃 Ƹ𝑒 𝜃
Thus the magnitude of the acceleration is a constant 4𝑏 ሶ𝜃2
∠from er = tan−1
sin𝜃
cos𝜃
= tan−1
sin𝜃
cos𝜃
= 𝜃 + 𝝅
Magnitude of the acceleration vector direction is
Ԧ𝑎
Angle of the acceleration vector measured from i is
𝑖
𝑗
𝜃
𝜃
er
e 𝜃
∠from 𝑖 = 𝜃 + (𝜋 + 𝜃) = 2𝜃 + 𝜋
൯∴ Ԧ𝑎 𝐵 = 4𝑏 ሶ𝜃2
∠(2𝜃 + 𝜋
ሷ𝑟 = −2𝑏 ሶ𝜃2
cos𝜃 − 2𝑏 ሷ𝜃sin𝜃
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2 Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃

90. 90
Question 6
Ԧ𝑎
𝑖
𝑗
𝜃
𝜃
er
e 𝜃
Velocity
𝑣 = ሶ𝑟 Ƹ𝑒 𝑟 + 𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Acceleration
𝑎 = ሷ𝑟 − 𝑟 ሶ𝜃2 Ƹ𝑒 𝑟 + 𝑟 ሷ𝜃 + 2 ሶ𝑟 ሶ𝜃 Ƹ𝑒 𝜃
Relative velocity to inertial frame
Relative acceleration to inertial frame
Coriolis acceleration
Entrained velocity
Entrained acceleration
𝑟 = 2𝑏cos𝜃 𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 ሶ𝜃 𝑖𝑠 𝑐𝑜𝑛𝑠tan𝑡
ሶ𝑟 = −2𝑏 ሶ𝜃 sin𝜃
ሷ𝑟 = −2𝑏 ሶ𝜃2
cos𝜃 [since ሷ𝜃 = 0]
ሷ𝜃 = 0
ሷ𝑟 = −2𝑏 ሶ𝜃2
cos𝜃 − 2𝑏 ሷ𝜃sin𝜃
൯𝑉 Τ𝐵 𝑓 = ሶ𝑟 Ƹ𝑒 𝑟 = −2𝑏 ሶ𝜃sin(𝜃 Ƹ𝑒 𝑟
Ԧ𝑎 Τ𝐵 𝑓 = ሷ𝑟 Ƹ𝑒 𝑟 = −2𝑏 ሶ𝜃2
cos(𝜃) Ƹ𝑒 𝑟

91. Tutorial 5
Planar motion of rigid slab
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

92. 92
Question 1
൯𝑉𝐴 = 900 Ƹ𝑖(𝑚 Τ𝑚 𝑠
Ԧ𝑟𝐵𝐴 = 300∠ − 30 𝑜
)Ԧ𝑟𝐵𝐴 = 300(cos(−30) Ƹ𝑖 + sin(−30) Ƹ𝑗
)Ԧ𝑟𝐵𝐴 = 300(0.87 Ƹ𝑖 − 0.5 Ƹ𝑗
൯𝑉𝐵 = 𝑉𝐵(cos(−70) Ƹ𝑖 + sin(−70) Ƹ𝑗
൯𝑉𝐵 = 𝑉𝐵(0.34 Ƹ𝑖 − 0.94 Ƹ𝑗
𝑉𝐵 = 𝑉𝐴 + 𝜔 × Ԧ𝑟𝐵𝐴
൯𝑉𝐵(0.34 Ƹ𝑖 − 0.94 Ƹ𝑗) = 900 Ƹ𝑖 + 𝜔෠𝑘 × 300(0.87 Ƹ𝑖 − 0.5 Ƹ𝑗
Equating i components
0.34𝑉𝐵 − 150𝜔 = 900
Equating j components
0.94𝑉𝐵 + 259.8𝜔 = 0
Solving these equations simultaneously
𝜔 = −3.68𝑟𝑎 Τ𝑑 𝑠
𝑉𝐵 = 1017𝑚 Τ𝑚 𝑠
Therefore
𝑉𝐵 = 1017∠ − 700(mm/s)
൯𝜔 = −3.68෠𝑘(𝑟𝑎 Τ𝑑 𝑠

93. 93
Question 2
𝑉𝐵
𝑉𝐷
i
j
Ԧ𝑟𝐵𝐴 = −300 Ƹ𝑖 − 100 Ƹ𝑗
൯𝜔 𝐵𝐴 = 3෠𝑘(𝑟𝑎 Τ𝑑 𝑠
𝑉𝐵 = 𝜔 𝐵𝐴 × Ԧ𝑟𝐵𝐴
𝑉𝐵 = 300 Ƹ𝑖 − 900 Ƹ𝑗 = 948.7∠ − 71.60

94. 94
Question 2
𝑉𝐵𝑉𝐷
i
j
Ԧ𝑟𝐵𝐴 = −300 Ƹ𝑖 − 100 Ƹ𝑗
൯𝜔 𝐵𝐴 = 3෠𝑘(𝑟𝑎 Τ𝑑 𝑠
𝑉𝐵 = 𝜔 𝐵𝐴 × Ԧ𝑟𝐵𝐴
𝑉𝐵 = 300 Ƹ𝑖 − 900 Ƹ𝑗 = 948.7∠ − 71.60
𝑉𝐵𝐷
𝛽
𝛽
𝛽 = tan−1
100
300
= 18.430
∴ |𝑉𝐷| = 𝑉𝐵cos(𝛽) = 900𝑚 Τ𝑚 𝑠
∴ |𝑉𝐵𝐷| = 𝑉𝐵sin(𝛽) = 300𝑚 Τ𝑚 𝑠
Based on vector diagram
൯𝑉𝐷 = −900 Ƹ𝑗(𝑚 Τ𝑚 𝑠
൯𝑉𝐵𝐷 = 300 Ƹ𝑖(𝑚 Τ𝑚 𝑠
Since
Ԧ𝑟 𝐷𝐸 = −225 Ƹ𝑖 𝑎𝑛𝑑 Ԧ𝑟𝐵𝐷 = 200 Ƹ𝑗
𝑉𝐷 = 𝜔 𝐷𝐸
෠𝑘 × Ԧ𝑟 𝐷𝐸
−900 Ƹ𝑗 = 𝜔 𝐷𝐸
෠𝑘 × −225 Ƹ𝑖
−900 Ƹ𝑗 = 𝜔 𝐷𝐸
෠𝑘 × −225 Ƹ𝑖
−900 Ƹ𝑗 = −225 ⋅ 𝜔 𝐷𝐸 Ƹ𝑗
𝜔 𝐷𝐸 =
−900
−225
= 4
𝜔 𝐷𝐸 = 4෠𝑘
300 Ƹ𝑖 = 𝜔 𝐵𝐷
෠𝑘 × 200 Ƹ𝑗
𝑉𝐵𝐷 = 𝜔 𝐵𝐷 × Ԧ𝑟𝐵𝐷
300 Ƹ𝑖 = −𝜔 𝐵𝐷 ⋅ 200 Ƹ𝑖
𝜔 𝐵𝐷 =
−300
200
= −1.5
𝜔 𝐵𝐷 = −1.5෠𝑘

95. 95
Question 2
i
j
𝛽
The angular accelerations are determined by simultaneously
solving the component equations the relative acceleration
equation
From part (a), we know that…
𝜔 𝐵𝐷 = −1.5෠𝑘
𝜔 𝐷𝐸 = 4෠𝑘
Since link AB has a constant angular velocity, we can apply
relative acceleration equation with 𝛼 𝐴𝐵 = 0 𝑟𝑎 Τ𝑑 𝑠2
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐴 + Ԧ𝛼 𝐴𝐵 × Ԧ𝑟𝐵𝐴 − 𝜔 𝐴𝐵
2
Ԧ𝑟𝐵𝐴
)Ԧ𝑎 𝐵 = −𝜔2
Ԧ𝑟𝐵𝐴 = − 3 2
(−300 Ƹ𝑖 − 100 Ƹ𝑗) = 2700 Ƹ𝑖 + 900 Ƹ𝑗(𝑚 Τ𝑚 𝑠2
For bar BD…
Ԧ𝑎 𝐷 = Ԧ𝑎 𝐵 + Ԧ𝛼 𝐵𝐷 × Ԧ𝑟 𝐷𝐵 − 𝜔 𝐵𝐷
2
Ԧ𝑟𝐷𝐵
൯Ԧ𝑎 𝐷 = (2700 Ƹ𝑖 + 900 Ƹ𝑗) + 𝛼 𝐵𝐷
෠𝑘 × (−200 Ƹ𝑗) − 1.5 2
(−200 Ƹ𝑗
Ԧ𝑎 𝐷 = (2700 Ƹ𝑖 + 900 Ƹ𝑗) + 200 Ƹ𝑖𝛼 𝐵𝐷 + 450 Ƹ𝑗
For bar DE…
Ԧ𝑎 𝐷 = Ԧ𝑎 𝐸 + Ԧ𝛼 𝐷𝐸 × Ԧ𝑟 𝐷𝐸 − 𝜔 𝐷𝐸
2
Ԧ𝑟 𝐷𝐸
൯Ԧ𝑎 𝐷 = 𝛼 𝐷𝐸
෠𝑘 × (−225 Ƹ𝑖) − 4 2
(−225 Ƹ𝑖

96. 96
Question 2
i
j
𝛽
Since link AB has a constant angular velocity, we can apply
relative acceleration equation with 𝛼 𝐴𝐵 = 0 𝑟𝑎 Τ𝑑 𝑠2
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐴 + Ԧ𝛼 𝐴𝐵 × Ԧ𝑟𝐵𝐴 − 𝜔 𝐴𝐵
2
Ԧ𝑟𝐵𝐴
)Ԧ𝑎 𝐵 = −𝜔2
Ԧ𝑟𝐵𝐴 = − 3 2
(−300 Ƹ𝑖 − 100 Ƹ𝑗) = 2700 Ƹ𝑖 + 900 Ƹ𝑗(𝑚 Τ𝑚 𝑠2
For bar BD…
Ԧ𝑎 𝐷 = Ԧ𝑎 𝐵 + Ԧ𝛼 𝐵𝐷 × Ԧ𝑟 𝐷𝐵 − 𝜔 𝐵𝐷
2
Ԧ𝑟𝐷𝐵
൯Ԧ𝑎 𝐷 = (2700 Ƹ𝑖 + 900 Ƹ𝑗) + 𝛼 𝐵𝐷
෠𝑘 × (−200 Ƹ𝑗) − 1.5 2(−200 Ƹ𝑗
Ԧ𝑎 𝐷 = (2700 Ƹ𝑖 + 900 Ƹ𝑗) + 200 Ƹ𝑖𝛼 𝐵𝐷 + 450 Ƹ𝑗
For bar DE…
Ԧ𝑎 𝐷 = Ԧ𝑎 𝐸 + Ԧ𝛼 𝐷𝐸 × Ԧ𝑟 𝐷𝐸 − 𝜔 𝐷𝐸
2
Ԧ𝑟 𝐷𝐸
൯Ԧ𝑎 𝐷 = 𝛼 𝐷𝐸
෠𝑘 × (−225 Ƹ𝑖) − 4 2(−225 Ƹ𝑖
Ԧ𝑎 𝐷 = −225𝛼 𝐷𝐸 Ƹ𝑗 + 3600 Ƹ𝑖
Equating bar BD and DE…
2700 + 200𝛼 𝐵𝐷 Ƹ𝑖 + 900 + 450 Ƹ𝑗 = −225𝛼 𝐷𝐸 Ƹ𝑗 + 3600 Ƹ𝑖
2700 + 200𝛼 𝐵𝐷 = 3600 ⇒ 𝛼 𝐵𝐷 = 4.5
Equate i and j vector components…
900 + 450 = −225𝛼 𝐷𝐸 ⇒ 𝛼 𝐷𝐸 = −6

97. 97
Question 3
Angular velocity of crank AB
i
j
𝑉𝐷
𝑉𝐵
𝐼𝐶
Using the known velocity directions of point D and B
𝑉𝐷 = 𝜔 𝐵𝐸 × 𝑟 Τ𝐷 𝐼𝐶
𝜔 𝐵𝐸 =
−120 Ƹ𝑖
−100 Ƹ𝑖
= 1.2𝑟𝑎 Τ𝑑 𝑠
൯𝜔 𝐵𝐸 = 1.2෠𝑘(𝑟𝑎 Τ𝑑 𝑠
𝑉𝐵 = 1.2෠𝑘 × −225 Ƹ𝑖 = −270 Ƹ𝑗
𝜔 𝐴𝐵 =
𝑉𝐵
𝑟𝐵𝐴
=
−270 Ƹ𝑗
−100 Ƹ𝑖
= 2.7෡𝑘 𝑟𝑎𝑑/𝑠
Velocity of point E
𝛽 = tan−1
100
225
= 23.960
𝜙
=
350
225
⋅ 100 = 155.6
𝜙 = tan−1
155.6
125
= 51.20
𝛾
𝐿 𝐵𝐸 = 383.03𝑚𝑚
)sin(23.960
𝑟 Τ𝐸 𝐼𝐶
=
)sin(180 − 51.20
383.03
|𝑟 Τ𝐸 𝐼𝐶| = 199.56𝑚𝑚
𝑟 Τ𝐸 𝐼𝐶 = 199.56𝑚𝑚∠51.20
𝑟 Τ𝐸 𝐼𝐶 = 125.0 Ƹ𝑖 + 155.5 Ƹ𝑗
𝑉𝐸 = 𝜔 𝐵𝐸 × 𝑟 Τ𝐸 𝐼𝐶
𝑉𝐸 = 239.5∠141.20

98. 98
Question 4
𝑉𝐸
0.7m/s
i
j
𝑉𝐴
𝐼𝐶
Ԧ𝑟𝐸/𝐼𝐶
8m
5m
𝜓 = 1800
− 36.90
− 22.60
− 67.40
= 53.10
)sin(53.1
13
=
)sin(59.50
Ԧ𝑟 Τ𝐸 𝐼𝐶
Ԧ𝑟 Τ𝐸 𝐼𝐶 = 13 ⋅
)sin(59.50
)sin(53.10
൯Ԧ𝑟 Τ𝐸 𝐼𝐶 = −14 Ƹ𝑗(𝑚
3m
4m
𝑉𝐸 = 𝜔 × Ԧ𝑟 Τ𝐸 𝐼𝐶
0.7 Ƹ𝑖 = 𝜔෠𝑘 × −14 Ƹ𝑗
0.7 Ƹ𝑖 = 14𝜔 Ƹ𝑖
𝜔 =
0.7
14
= 0.05𝑟𝑎 Τ𝑑 𝑠 𝑐𝑐𝑤

99. 99
Question 4
𝑉𝐸
0.7m/s
i
j
𝑉𝐴
𝐼𝐶
Ԧ𝑟𝐸/𝐼𝐶
8m
5m
𝜓 = 1800
− 36.90
− 22.60
− 67.40
= 53.10
)sin(53.1
13
=
)sin(59.50
Ԧ𝑟 Τ𝐸 𝐼𝐶
Ԧ𝑟 Τ𝐸 𝐼𝐶 = 13 ⋅
)sin(59.50
)sin(53.10
൯Ԧ𝑟 Τ𝐸 𝐼𝐶 = −14 Ƹ𝑗(𝑚
3m
4m
𝑉𝐸 = 𝜔 × Ԧ𝑟 Τ𝐸 𝐼𝐶
0.7 Ƹ𝑖 = 𝜔෠𝑘 × −14 Ƹ𝑗
0.7 Ƹ𝑖 = 14𝜔 Ƹ𝑖
𝜔 =
0.7
14
= 0.05𝑟𝑎 Τ𝑑 𝑠 𝑐𝑐𝑤
൯Ԧ𝑟 Τ𝐴 𝐼𝐶 = −8 Ƹ𝑖 − 7 Ƹ𝑗(𝑚
𝑉𝐴 = 𝜔 × Ԧ𝑟 Τ𝐴 𝐼𝐶
𝑉𝐴 = 0.05෠𝑘 × (−8 Ƹ𝑖 − 6 Ƹ𝑗)
𝑉𝐴 = 0.35 Ƹ𝑖 − 0.4 Ƹ𝑗
|𝑉𝐴| = 0.5 Τ𝑚 𝑠
∠𝑉𝐴 = tan−1
−0.4
0.3
= −53.130
𝑉𝐵 = 0.806 Τ𝑚 𝑠 ∠ − 29.740
𝑉𝐷 = 0.3 Τ𝑚 𝑠 ∠00

100. 100
Question 5
൯𝑉𝐵 = 𝑉𝑐 = −1.5 Ƹ𝑗( Τ𝑚 𝑠
൯𝑉𝐴 = 𝑉𝐷 = −1.8 Ƹ𝑗( Τ𝑚 𝑠
)Ԧ𝑎 𝐵
𝑡
= Ԧ𝑎 𝑐 = −0.45 Ƹ𝑗( Τ𝑚 𝑠2
)Ԧ𝑎 𝐴
𝑡
= Ԧ𝑎 𝐷 = 0.6 Ƹ𝑗( Τ𝑚 𝑠2
𝐴
𝐺
𝑉𝐵 = 𝑉𝐴 + 𝑉 Τ𝐵 𝐴 = 𝑉𝐴 + 𝜔 × Ԧ𝑟𝐵𝐴
−1.5 Ƹ𝑗 = −1.8 Ƹ𝑗 + 𝜔෠𝑘 × −1.2 Ƹ𝑖
−1.5 Ƹ𝑗 = −1.8 Ƹ𝑗 − 1.2𝜔 Ƹ𝑗
Equating j components
𝜔 =
−1.5 + 1.8
−1.2
= −0.25
𝜔 = −0.25෠𝑘
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐴 + Ԧ𝛼 × Ԧ𝑟𝐵𝐴 − 𝜔2
Ԧ𝑟𝐵𝐴
൯−0.45 Ƹ𝑗 + x 𝑏 Ƹ𝑖 = 0.6 Ƹ𝑗 + x 𝑎 Ƹ𝑖 + 𝛼෠𝑘 × −1.2 Ƹ𝑖 − −0.25 2
(−1.2 Ƹ𝑖
−0.45 Ƹ𝑗 + x 𝑏 Ƹ𝑖 = 0.6 Ƹ𝑗 + x 𝑎 Ƹ𝑖 − 1.2𝛼 Ƹ𝑗 − 0.075 Ƹ𝑖
Equating j components and solving for 𝜶
𝛼 =
−0.45 − 0.6
−1.2
= 0.875𝑟𝑎 Τ𝑑 𝑠2
൯Ԧ𝛼 = 0.875෠𝑘(𝑟𝑎 Τ𝑑 𝑠2
Ԧ𝑎 𝐵 = 𝑎 𝐺 + Ԧ𝛼 𝐵𝐺 × Ԧ𝑟𝐵𝐺 − 𝜔 𝐵𝐺
2
Ԧ𝑟𝐵𝐺
൯−0.45 Ƹ𝑗 + 𝐱 Ƹ𝑖 = 𝑎 𝐺 Ƹ𝑗 + 0.875෠𝑘 × −0.6 Ƹ𝑖 − −0.25 2(−0.6 Ƹ𝑖
−0.45 Ƹ𝑗 + 𝐱 Ƹ𝑖 = (−0.875 ⋅ 0.6 + 𝑎 𝐺) Ƹ𝑗 + 0.0375 Ƹ𝑖
Equating i components
𝐱 = 0.0375m/s2
)∴ Ԧ𝑎 𝐵 = 0.0375 Ƹ𝑖 − 0.45 Ƹ𝑗 = 0.452∠ − 85.29( Τ𝑚 𝑠2

101. Tutorial 6
Motion relative to a body attached rotating
reference
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

102. 102
Translating, but not rotating reference frame
Key points
Translating and rotating reference frame
𝑉𝐴 = 𝑉𝐵 + 𝑉 Τ𝐴 𝐵
Ԧ𝑎 𝐴 = Ԧ𝑎 𝐵 + Ԧ𝑎 Τ𝐴 𝐵
𝑉𝐴 = 𝑉𝐵 + 𝜔 × Ԧ𝑟𝐴𝐵 + 𝑉𝑟𝑒𝑙
Normal to r
Along increasing s direction, it’s the relative
velocity measured from x-y frame.
Velocity of point B measured from X-Y frame
𝑽 𝒓𝒆𝒍 is always tangent to path and points
in the direction of increasing s
𝝎 × 𝒓 𝑨𝑩 is normal position vector

103. 103
Key points
Translating and rotating reference frame
Ԧ𝑎 𝐴 = Ԧ𝑎 𝐵 + Ԧ𝛼 × Ԧ𝑟𝐴𝐵 + 𝜔 × (𝜔 × Ԧ𝑟𝐴𝐵) + 2𝜔 × Ԧ𝑣 𝑟𝑒𝑙 + Ԧ𝑎 𝑟𝑒𝑙
acceleration of point B measured from X-Y frame
Coriolis acceleration
Acceleration of point A measured from x-y frame
Normal to r vector
Tangent to
r vector
Ԧ𝑎 𝑟𝑒𝑙 𝑡 = ሷ𝑠 = ሶ𝑉𝑟𝑒𝑙
Ԧ𝑎 𝑟𝑒𝑙 𝑛 =
𝑉𝑟𝑒𝑙
2
𝜌
(appears when particle moves in curvilinear or circular motion
with respect to point P)
Radius r is measured from the centre of rotation of the rotating reference frame
to particle A
𝜔 × (𝜔 × Ԧ𝑟𝐴𝐵) represents the normal component of acceleration of point P with
respect B. This vector is directed along r and points towards centre of rotation.
Ԧ𝛼 × Ԧ𝑟𝐴𝐵 represents the tangential component of acceleration of point p with
respect to b. This vector is directed perpendicular to r.
𝜔 is the angular velocity of the rotating reference frame

104. 104
Key points
Ԧ𝑎 𝐴 = Ԧ𝑎 𝐵 + Ԧ𝛼 × Ԧ𝑟𝐴𝐵 + 𝜔 × (𝜔 × Ԧ𝑟𝐴𝐵) + 2𝜔 × Ԧ𝑣 𝑟𝑒𝑙 + Ԧ𝑎 𝑟𝑒𝑙
acceleration of point B measured from X-Y frame
Coriolis acceleration
Acceleration of point A measured from x-y frame
Normal to r vector
Tangent to
r vector
𝑇ℎ𝑒𝑟𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
even if an object travels at a constant
velocity in circular or curvilinear
motion (be it inertial or rotating
reference frame).
This acceleration vector will point
towards the centre of rotation.

105. 105
Question 1
𝑉𝑃 = 𝑉𝐴 + 𝜔 𝐴𝑃
2
× Ԧ𝑟𝐴𝑃 + 𝑉𝑟𝑒𝑙 = 𝑉𝐴 + 𝜔 𝐴𝑃
2
× Ԧ𝑟𝐴𝑃+ 𝑉𝑟𝑒𝑙
200
)sin(400
=
𝑟𝑃𝐵
)sin(1200
=
𝑟𝑃𝐴
)sin(200
Ԧ𝑟𝑃𝐴 = 106.42𝑚𝑚∠1200
Ԧ𝑟𝑃𝐵 = 269.46𝑚𝑚∠1600
൯𝑉𝑃 = 𝜔 𝐴𝑃 × Ԧ𝑟𝑃𝐴 = −6෠𝑘 × 106.42∠1200
= −6෠𝑘 × (−53.21 Ƹ𝑖 + 92.16 Ƹ𝑗
൯𝑉𝑃 = 552.96 Ƹ𝑖 + 319.26 Ƹ𝑗 = 638.5∠300
(𝑚 Τ𝑚 𝑠
From the diagram…
𝑉𝑃
′
= 𝑉𝑃 sin(500)∠700 = 638.5sin(50)∠700 = 489.12∠700
(angle measured from horizontal axis)
𝑉 Τ𝑃 𝑓 = 𝑉𝑃 cos(500)∠ − 200 = 638.5cos(50)∠ − 200 = 410.42∠ − 200
For frame f2…
𝑉𝑃
′
= 𝜔 𝑃𝐵 × Ԧ𝑟𝑃𝐵 = 𝜔 𝑃𝐵
෠𝑘 × 269.46∠1600
൯𝑉𝑃
′
= 𝜔 𝑃𝐵
෠𝑘 × (−253.21 Ƹ𝑖 + 92.16 Ƹ𝑗
𝑉𝑃
′
= −253.21𝜔 𝑃𝐵 Ƹ𝑗 − 92.16𝜔 𝑃𝐵 Ƹ𝑖
from f1, since
𝑉𝑃
′
= 489.12∠700
𝑉𝑃
′
= 167.29 Ƹ𝑖 + 459.62 Ƹ𝑗
𝜔 𝑃𝐵 =
169.29
−92.16
= −1.82෠𝑘
Alternatively, equating j component…
𝜔 𝑃𝐵 =
459.62
−253.21
= −1.82෠𝑘
For frame f1…
𝑉𝑃 = 𝑉𝐵 + 𝜔 𝐵𝑃
2
× Ԧ𝑟𝐵𝑃+ 𝑉𝑟𝑒𝑙
𝑉𝑟𝑒𝑙= 410.42∠ − 200
Equating i component…

106. 106
Question 2
𝐿𝑒𝑡 𝑉𝑟𝑒𝑙 = 𝑉 Τ𝐵 𝑓
𝑉 Τ𝐵 𝑓 = −2 Ƹ𝑖
Ԧ𝑟𝐵𝑂 = 2 Ƹ𝑗
𝜔 = 5෠𝑘
𝑉𝐵 = 𝑉𝑂 + 𝜔 × Ԧ𝑟𝐵𝑂 + 𝑉 Τ𝐵 𝑓
𝑉𝐵 = 5෠𝑘 × 2 Ƹ𝑗 − 2 Ƹ𝑖 = −12 Ƹ𝑖
𝐿𝑒𝑡 𝑉𝑟𝑒𝑙 = 𝑉 Τ𝐵 𝑓
𝑉 Τ𝐵 𝑓 = −2 Ƹ𝑖
Ԧ𝑟𝐵𝑂 = (1.1547 Ƹ𝑖 + 2 Ƹ𝑗)
𝜔 = 5෠𝑘
𝑉𝐵 = 𝑉𝑂 + 𝜔 × Ԧ𝑟𝐵𝑂 + 𝑉 Τ𝐵 𝑓
𝑉𝐵 = 5෠𝑘 ×= 1.1547 Ƹ𝑖 + 2 Ƹ𝑗 − 2 Ƹ𝑖 = −12 Ƹ𝑖 + 5.77 Ƹ𝑗

107. 107
Question 3
Ԧ𝑟𝐵𝐴 = 6∠300
= 5.2 Ƹ𝑖 + 3 Ƹ𝑗
𝑉𝑟𝑒𝑙 = 0.15∠ − 1500
𝜔 = −0.075෠𝑘
𝑉𝐵 = 𝑉𝐴 + 𝜔 × Ԧ𝑟𝐵𝐴 + 𝑉𝑟𝑒𝑙
𝑉𝐵 = −0.075෠𝑘 × 6∠300
+ 0.15∠ − 1500
= 0.475∠ − 78.430
Point B has:
• Constant rate of retraction
• Constant angular velocity
• Linear motion with respect to frame-f
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐴 + Ԧ𝛼 × Ԧ𝑟𝐵𝐴 + 𝜔 × (𝜔 × Ԧ𝑟𝐵𝐴) + 2𝜔 × Ԧ𝑣 𝑟𝑒𝑙 + Ԧ𝑎 𝑟𝑒𝑙
൯𝜔 × (𝜔 × Ԧ𝑟𝐵𝐴) = −0.075෠𝑘 × (−0.075෠𝑘 × (5.2 Ƹ𝑖 + 3 Ƹ𝑗)
𝜔 × (𝜔 × Ԧ𝑟𝐵𝐴) = −0.0292 Ƹ𝑖 − 0.0169 Ƹ𝑗
𝑉𝑟𝑒𝑙 = 0.15∠ − 1500
= −0.121 Ƹ𝑖 − 0.070 Ƹ𝑗
2𝜔 × 𝑉𝑟𝑒𝑙 = −0.0112 Ƹ𝑖 + 0.0195 Ƹ𝑗
∴ Ԧ𝑎 𝐵 = (−0.0292 Ƹ𝑖 − 0.0169෡𝑗) + (−0.0112 Ƹ𝑖 + 0.0195 Ƹ𝑗)
∴ Ԧ𝑎 𝐵 = −0.0404 Ƹ𝑖 + 0.0026 Ƹ𝑗

108. 108
Question 4
Point B has
• Constant rate of retraction
• Constant angular velocity
• Linear motion with respect to frame-f
𝑉𝐵 = 𝑉𝑂 + 𝜔 × Ԧ𝑟𝐵𝑂 + 𝑉𝑟𝑒𝑙
𝑉𝐵 = −2.4෠𝑘 × (−250 Ƹ𝑖 − 188 Ƹ𝑗) + 375 Ƹ𝑖
൯𝑉𝐵 = −76.2 Ƹ𝑖 + 600 Ƹ𝑗(𝑚 Τ𝑚 𝑠
X
Y
x
yf
Radius r is measured from the centre of
rotation of the rotating reference frame
to particle A

109. 109
Question 4
)Ԧ𝑎 𝐵 = 𝜔 × (𝜔 × Ԧ𝑟𝐵𝑂) + (2𝜔 × 𝑉𝑟𝑒𝑙
)Ԧ𝑎 𝐵 = 𝜔 × (𝜔 × Ԧ𝑟𝐵𝑂) + (2𝜔 × 𝑉𝑟𝑒𝑙
൯Ԧ𝑎 𝐵 = −2.4෠𝑘 × (−2.4෠𝑘 × (−250 Ƹ𝑖 − 188 Ƹ𝑗)) + (2 ⋅ −2.4෠𝑘 × 375 Ƹ𝑖
Ԧ𝑎 𝐵 = 1440.06 Ƹ𝑖 − 717.2 Ƹ𝑗
• Constant rate of retraction
• Constant angular velocity
• Linear motion with respect to frame-f
x
y f

110. 110
Question 5
17.32
10.00
10.00
𝑉𝑃 = 𝜔 × Ԧ𝑟𝑃𝑂 + 𝑉𝑟𝑒𝑙
𝑉𝑃 = 0.3෠𝑘 × (17.32 Ƹ𝑖 − 30 Ƹ𝑗) + 10∠ − 600
൯𝑉𝑃 = 0.3෠𝑘 × (17.32 Ƹ𝑖 − 30 Ƹ𝑗) + (5.00 Ƹ𝑖 − 8.66 Ƹ𝑗
൯𝑉𝑃 = (9𝑖 + 5.2 Ƹ𝑗) + (5.00 Ƹ𝑖 − 8.66 Ƹ𝑗
𝑉𝑃 = 14𝑖 − 3.46 Ƹ𝑗
• Constant angular velocity
• Circular motion with respect to rotating reference frame x”-y”
Y
X
Radius r is measured from the centre of rotation of the rotating reference frame
to particle A

111. 111
Question 5
17.32
10.00
10.00
൯2𝜔 × 𝑉𝑟𝑒𝑙 = 2 ⋅ 0.3෠𝑘 × (5.00 Ƹ𝑖 − 8.66 Ƹ𝑗) = (5.20 Ƹ𝑖 + 3.00 Ƹ𝑗
𝜔 × 𝜔 × Ԧ𝑟𝑃𝑂 = 0.3෠𝑘 × (0.3෠𝑘 × (17.32 Ƹ𝑖 − 30 Ƹ𝑗)) = −1.56 Ƹ𝑖 + 2.70 Ƹ𝑗
Ԧ𝑎 𝑟𝑒𝑙 =
𝑉𝑟𝑒𝑙
2
𝑟
∠2100 =
102
20
∠2100 = 5∠2100 = −4.33 Ƹ𝑖 − 2.50 Ƹ𝑗
Ԧ𝑎 𝑝 = 3.28∠102.20
𝑚 Τ𝑚 𝑠
Y
X
Summing these three acceleration components gives…

112. Tutorial 7
Kinetics of particles (Newton’s Second Law)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

113. 113
Key point
෍ 𝐹𝑜𝑟𝑐𝑒 = 𝑚 ⋅ Ԧ𝑎

114. 114
Key point
෍ 𝐹𝑜𝑟𝑐𝑒 = 𝑚 ⋅ Ԧ𝑎
In a given direction Absolute acceleration along that direction
Ԧ𝑎 𝐴 = Ԧ𝑎 𝐵 + Ԧ𝑎 Τ𝐴 𝐵
Ԧ𝑎 𝐴 = Ԧ𝑎 𝐵 + Ԧ𝛼 × Ԧ𝑟𝐴𝐵 + 𝜔 × (𝜔 × Ԧ𝑟𝐴𝐵) + 2𝜔 × Ԧ𝑣 𝑟𝑒𝑙 + Ԧ𝑎 𝑟𝑒𝑙
Translating reference frame:
Translating and rotating reference frame:

115. 115
Question 1
𝑉 = 3𝑚/𝑠x
y
෍ 𝐹𝑦 = 𝑚 ⋅ 𝑎 𝑦
𝑁 − 𝑚𝑔 = −𝑚
𝑉2
𝜌
𝑁 = −𝑚
𝑉2
𝜌
+ 𝑚𝑔
𝑁 = −(2)
3 2
1.8
+ (2)(9.81) = 9.62𝑁
N will be zero at point of losing contact
0= −𝑚
𝑉2
𝜌
+ 𝑚𝑔
Solving for V
𝑉max = 𝑔𝜌 = 9.81 ⋅ 1.8 = 4.2 Τ𝑚 𝑠

116. 116
Question 2
For block A
For block B
Belt direction
Slip direction
Equations of Motion along i-axis
0.2𝑚 𝐴 𝑔 − 𝐹 = 𝑚 𝐴 𝑎 𝑥
0.1𝑚 𝐵 𝑔 + 𝐹 = 𝑚 𝐵 𝑎 𝑥
0.1(30)𝑔 + 𝐹 = (30)𝑎 𝑥
0.2(24)𝑔 − 𝐹 = (24)𝑎 𝑥
Solve for F and ax
𝐹 = 13.07𝑁
𝑎 = 1.417 Τ𝑚 𝑠2

117. 117
Question 3
Ԧ𝑎 𝐴
Ԧ𝑎 𝐵/𝐴
The acceleration of B is composed of the acceleration
of A plus the acceleration of B relative to A.
Apply equation of motion along i direction for block A
𝑇sin(250) = 𝑚 𝐴 𝑎 𝐴
Apply equation of motion along i and j direction for block B
൧𝑇(sin(250
) Ƹ𝑖 + cos(250
) Ƹ𝑗) − 𝑚 𝐵 𝑔 Ƹ𝑗 = 𝑚 𝐵[𝑎 𝐴(−𝑖) + 𝑎 Τ𝐵 𝐴(cos(250
) Ƹ𝑖 − sin(250
) Ƹ𝑗)
Equating i and j components for block B
൯𝑇sin(250
) = −𝑚 𝐵 𝑎 𝐴 + 𝑎 Τ𝐵 𝐴 𝑚 𝐵cos(250
൯𝑇cos(250
) − 𝑚 𝐵 𝑔 = −𝑎 Τ𝐵 𝐴 𝑚 𝐵sin(250
250
250

118. 118
Question 3
𝑇sin(250) = 𝑚 𝐴 𝑎 𝐴
Three unknows and three equations
൯𝑇sin(250
) = −𝑚 𝐵 𝑎 𝐴 + 𝑎 Τ𝐵 𝐴 𝑚 𝐵cos(250
൯𝑇cos(250
) − 𝑚 𝐵 𝑔 = −𝑎 Τ𝐵 𝐴 𝑚 𝐵sin(250
𝑇 = 117.6𝑁
𝑎 𝐴 = 2.49 Τ𝑚 𝑠2
൯𝑚 𝐴 𝑎 𝐴 = −𝑚 𝐵 𝑎 𝐴 + 𝑎 Τ𝐵 𝐴 𝑚 𝐵cos(250
Therefore….
𝑚 𝐴 𝑎 𝐴
)sin(250 cos(250) − 𝑚 𝐵 𝑔 = −𝑎 Τ𝐵 𝐴 𝑚 𝐵sin(250)
Rearranging and substituting…
20𝑎 𝐴 + 15𝑎 𝐴 = 13.6𝑎 Τ𝐵 𝐴
2.14(20)𝑎 𝐴 − 15𝑔 = −6.3𝑎 Τ𝐵 𝐴
Simplifying…
35𝑎 𝐴 = 13.6𝑎 Τ𝐵 𝐴
42.8𝑎 𝐴 − 147.15 = −6.3𝑎 Τ𝐵 𝐴
Substituting…
42.8
13.6
35
𝑎 Τ𝐵 𝐴 − 147.15 = −6.3𝑎 Τ𝐵 𝐴
16.63𝑎 Τ𝐵 𝐴 − 147.15 = −6.3𝑎 Τ𝐵 𝐴
Solve for 𝑎 Τ𝐵 𝐴
𝑎 Τ𝐵 𝐴 = 6.4 Τ𝑚 𝑠2
Therefore…

119. 119
Question 4

120. 120
Question 4

121. 121
Question 4
Only sum the non-constant lengths in next time instant!

122. 122
Question 4

123. 123
Question 4

124. 124
Question 5
mg N
Ƹ𝑒𝑡
The acceleration of C is
composed of the
acceleration of A plus the
acceleration of C relative
to A.
aA
Writing equation of motion along shaft AB
෍
Ƹ𝑒 𝑡
Ԧ𝐹 = 𝑚 ⋅ 𝑎 𝐶 Ƹ𝑒 𝑡
−𝑚 𝐶 𝑔cos(450
) = 𝑚 𝐶 𝑎 Τ𝐶 𝐴 + 𝑚 𝐶 𝑎 𝐴cos(450
)
൯−𝑔cos(450
) = 𝑎 Τ𝐶 𝐴 + 𝑎 𝐴cos(450
∴ 𝑎 Τ𝐶 𝐴 = −9.81cos(450
) − 4cos(450
) = −9.765
∴ Ԧ𝑎 Τ𝐶 𝐴 = −9.765 Ƹ𝑒𝑡
∴ Ԧ𝑎 Τ𝐶 𝐴 = 9.765∠ − 1350
Ԧ𝑎 𝐶 = Ԧ𝑎 𝐴 + Ԧ𝑎 Τ𝐶 𝐴
∴ Ԧ𝑎 𝐶 = 4∠900
+ 9.765∠ − 1350
Ԧ𝑎 𝐶 = −6.91 Ƹ𝑖 − 2.90 Ƹ𝑗 = 7.49∠ − 157.20
Negative sign indicates assumed direction of aC/A
was wrong

125. 125
Question 5
mg N
Ƹ𝑒 𝑛
The acceleration of C is
composed of the
acceleration of A plus the
acceleration of C relative
to A.
aA
Writing equation of motion perpendicular to shaft AB
෍
Ƹ𝑒 𝑛
Ԧ𝐹 = 𝑚 ⋅ 𝑎 𝐶 Ƹ𝑒 𝑛
)𝑁 − 𝑚 𝐶 𝑔cos(450) = 𝑚 𝐶 Ԧ𝑎 𝑎cos(450
)𝑁 = 𝑚 𝐶( Ԧ𝑎 𝑎cos(450) + 𝑔cos(450)
𝑁 = (2)(4cos(450) + 9.81cos(450)) = 19.530
൯𝑁 = 19.53∠1350
(𝑁

126. Tutorial 8
Kinetics of particles (Auxiliary Principles)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

127. 127
Conservation of energy in conservative field
(applies only to conservative system, i.e no external force)
Key points
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
where
T : Kinetic energy of a particle at a given position
V: Potential energy of a particle at a given position
𝑉 = 𝑚𝑔ℎ
𝑉 =
1
2
𝑘 𝑥 𝑠𝑡𝑟𝑒𝑐ℎ𝑒𝑑 − 𝑥 𝑢𝑛𝑠𝑡𝑟𝑒𝑐ℎ𝑒𝑑
2
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
where
𝛥𝐾. 𝐸 = 𝐾. 𝐸2 − 𝐾. 𝐸1
𝛥𝑃. 𝐸 = 𝑃. 𝐸2 − 𝑃. 𝐸1
Principle of linear Impulse and Momentum
Impulse in a given direction
Momentum in that direction
Principle of angular Impulse and Momentum
Impulse in a given direction
Momentum in that direction
Datum can be placed at any convenient location, but points above datum has
to be taken positive gravitation potential energy (vice versa for points below
datum)
Conservation of energy (General form)
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
඲෍ 𝑀𝑑𝑡 = 𝛥 ෍
𝑖
𝐼𝑖 𝜔 𝑎𝑏𝑠,𝑖
Work done is zero if the
particle or the system is not
subjected to any external
force or moment
Impulse is zero if the
particle or the system is not
subjected to any external
force or moment
𝑇 =
1
2
𝑚𝑉𝑎𝑏𝑠
2

128. 128
Key points
Ask yourself:
Am I applying conservation of energy /momentum to a particle or a system?
Particle
(ignore external forces/moment)
System
(ignore internal forces/moment)

129. 129
Question 1
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Datum
Based on defined datum, K.E and P.E of block A is zero
1
2
Block A, position 1
Block A, position 2
𝑃. 𝐸: 𝑚 𝐴 𝑔ℎ = 12(𝑔)(1.5𝑚) = 176.58𝐽
𝐾. 𝐸:
1
2
𝑚 𝐴 𝑉𝐴
2
=
1
2
(12) 1.4 2
= 11.76𝐽
∴ 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 11.76 + 176.58 = 188.34𝐽
∴ 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 188.34𝐽
Since
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
𝑇𝐴 =
𝑊𝐴
Ԧ𝑠 𝐴
=
188.34
1.5
= 125.56𝑁
+s12kg
15kg
Work done is non-zero since particle “A” is subjected to
tension force which acts along displacement direction

130. 130
Question 1
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Datum
K.E is zero as Block B is at rest
2
1
Block B, position 1
Block B, position 2
𝑃. 𝐸: 𝑚 𝐵 𝑔ℎ = 15(𝑔)(1.5𝑚) = 220.73𝐽
𝐾. 𝐸:
1
2
𝑚 𝐵 𝑉𝐵
2
=
1
2
(15) 1.4 2
= 14.7𝐽
∴ 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 14.7 + −220.73 = −206.03𝐽
Since
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
𝑇𝐵 =
𝑊𝐵
Ԧ𝑠 𝐵
=
−206.03
−1.5
= 137.35𝑁
Based on defined datum: P.E is zero
+s12kg
15kg
Work done is non-zero since particle “B” is subjected to
tension force which acts along displacement direction

131. 131
Question 1
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Datum
K.E is zero as Block B is at rest
2
1
Block B, position 1
Block B, position 2
P. E is zero based on defined datum
𝐾. 𝐸:
1
2
𝑚 𝐵 𝑉𝐵
2
=
1
2
(15) 1.4 2
= 14.7𝐽
∴ 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 14.7 + −220.73 = −206.03𝐽
Since
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
𝑇𝐵 =
𝑊𝐵
Ԧ𝑠 𝐵
=
−206.03
−1.5
= 137.35𝑁
Based on defined datum: P.E is zero
+s
𝑃. 𝐸 = 𝑚 𝐵 𝑔ℎ = (15)𝑔(−1.5𝑚) = −220.73
12kg
15kg

132. 132
Question 1
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Datum
System at state 1
+s
𝐾. 𝐸1 = 0 (𝑏𝑜𝑡ℎ 𝑏𝑙𝑜𝑐𝑘𝑠 𝑎𝑟𝑒 𝑎𝑡 𝑟𝑒𝑠𝑡)
)𝑃. 𝐸1 = 220.73𝐽 (𝑑𝑢𝑒 𝑡𝑜 𝑏𝑙𝑜𝑐𝑘 𝐵
System at state 2
𝐾. 𝐸2 = 11.76+ 14.7=26.46J
𝑃. 𝐸2 = 176.58𝐽(due to block A)
∴ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = (26.46 − 0) + (176.58 − 220.73) = −17.69𝐽
Based on the conservation equation, work done between states
State 1 State 2
∴ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = (𝐾. 𝐸2 − 𝐾. 𝐸1) + (𝑃. 𝐸2 - 𝑃. 𝐸1)
12kg
15kg

133. 133
Question 1
𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑁𝑒𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑆𝑡𝑎𝑡𝑒 2 − 𝑁𝑒𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑆𝑡𝑎𝑡𝑒 1
Datum
System at state 1
+s
𝐾. 𝐸1 = 0 (𝑏𝑜𝑡ℎ 𝑏𝑙𝑜𝑐𝑘𝑠 𝑎𝑟𝑒 𝑎𝑡 𝑟𝑒𝑠𝑡)
)𝑃. 𝐸1 = 220.73𝐽 (𝑑𝑢𝑒 𝑡𝑜 𝑏𝑙𝑜𝑐𝑘 𝐵
System at state 2
𝐾. 𝐸2 = 11.76+ 14.7=26.46J
𝑃. 𝐸2 = 176.58𝐽(due to block A)
∴ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 26.46 + 176.58 − (0 + 220.73) = −17.69𝐽
Based on the conservation equation, energy difference between states
State 1 State 2
∴ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = (𝐾. 𝐸2 + P. 𝐸2) - (𝐾. 𝐸1 + 𝑃. 𝐸1)
12kg
15kg

134. Key points
134
Conservation applied to particle A Conservation applied to particle B
Datum and positions 1 and 2 can be defined separately for each particle
Conservation applied to a system
Datum has to be common between states

135. 135
Question 2
Knowing that the trailer and the load A are moving together:
൯𝑉𝑡𝑟𝑎𝑖𝑙𝑒𝑟 = 25 Ƹ𝑖( Τ𝑚 𝑠
൯𝑉𝑙𝑜𝑎𝑑 = 25 Ƹ𝑖( Τ𝑚 𝑠
Static frictional force between the load and the trailer is
Ԧ𝐹𝑓 = −𝜇 𝑠 ⋅ 𝑁 Ƹ𝑖 = −𝜇 𝑠 𝑚 𝐴 𝑔 Ƹ𝑖
Using Principle of Impulse and Linear Momentum
൯−𝜇 𝑠 𝑚 𝐴 𝑔𝛥𝑡 Ƹ𝑖 = 𝑚 𝐴(𝑉𝑙𝑜𝑎𝑑,2 − 𝑉𝑙𝑜𝑎𝑑,1 Ƹ𝑖
−𝜇 𝑠 𝑚 𝐴 𝑔𝛥𝑡 = −𝑚 𝐴 𝑉𝑙𝑜𝑎𝑑,1 Ƹ𝑖
൯𝑉𝑙𝑜𝑎𝑑/𝑡𝑟𝑎𝑖𝑙𝑒𝑟 = 0 Ƹ𝑖( Τ𝑚 𝑠
𝛥𝑡 =
𝑚 𝐴 𝑉𝑙𝑜𝑎𝑑,1
𝜇 𝑠 𝑚 𝐴 𝑔
=
𝑉𝑙𝑜𝑎𝑑,1
𝜇 𝑠 𝑔
=
25
0.4)(9.81
= 6.371𝑠
Integral sign can be dropped off since we are dealing with a constant force w.r.t time
Summation sign in L.H.S can be dropped off since we are dealing with single force
i
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
Summation sign in R.H.S can be dropped off since the principal is applied to a single particle
Ԧ𝐹𝛥𝑡 = 𝛥𝑚𝑉

136. 136
Question 3
Principle of angular Impulse and Momentum
Impulse Momentum
35cos(300
) ⋅ 1.2)𝛥𝑡 = 𝐼(𝜔2 − 𝜔1
𝛥𝑡 =
𝐼 ⋅ 𝜔2
35cos(300) ⋅ 1.2
30
10 N
35 N150
N 2
1
O
Considering all the forces acting on B at the moment when the cord is broken
(at position 2), and using Newton’s second law in normal direction, we have
෍ 𝐹 𝑛 = 𝑚𝑎 𝑛 = 𝑚𝜔2
2
𝑟
ቇ150 − 10 − 35 ⋅ sin(300
) =
50
𝑔
𝜔2
2
(1.2)
𝜔2 = 4.4753𝑟𝑎 Τ𝑑 𝑠
𝑉2 = 1.2 ⋅ 4.4753 = 5.37 Τ𝑚 𝑠
𝑛
)𝑇 ⋅ 𝑟𝛥𝑡 = 𝐼(𝜔2 − 𝜔1
඲෍ 𝑀𝑑𝑡 = 𝛥 ෍
𝑖
𝐼𝑖 𝜔 𝑎𝑏𝑠,𝑖

137. 137
Question 3
Principle of angular Impulse and Momentum
Impulse Momentum
35cos(300) ⋅ 1.2)𝛥𝑡 = 𝐼(𝜔2 − 𝜔1
𝛥𝑡 =
𝐼 ⋅ 𝜔2
35cos(300) ⋅ 1.2
𝛥𝑡 =
𝐼 ⋅ 𝜔2
35cos(300) ⋅ 1.2
𝛥𝑡 =
𝑚𝑟2 𝜔2
35 ⋅ cos(300) ⋅ 1.2
=
50
𝑔
⋅ 1.22 ⋅ 4.4753
35 ⋅ cos(300) ⋅ 1.2
= 0.903𝑠
඲෍ 𝑀𝑑𝑡 = 𝛥 ෍
𝑖
𝐼𝑖 𝜔 𝑎𝑏𝑠,𝑖
)𝑇 ⋅ 𝑟𝛥𝑡 = 𝐼(𝜔2 − 𝜔1

138. Tutorial 9
Kinetics of particles (Conservation of Energy and
Momentum)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

139. 139
Conservation of energy in conservative field
(applies only to conservative system, i.e no external force)
Key points
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
where
T : Kinetic energy of a particle at a given position
V: Potential energy of a particle at a given position
𝑉 = 𝑚𝑔ℎ
𝑉 =
1
2
𝑘 𝑥 𝑠𝑡𝑟𝑒𝑐ℎ𝑒𝑑 − 𝑥 𝑢𝑛𝑠𝑡𝑟𝑒𝑐ℎ𝑒𝑑
2
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
where
𝛥𝐾. 𝐸 = 𝐾. 𝐸2 − 𝐾. 𝐸1
𝛥𝑃. 𝐸 = 𝑃. 𝐸2 − 𝑃. 𝐸1
Principle of linear Impulse and Momentum
Impulse in a given direction
Momentum in that direction
Principle of angular Impulse and Momentum
Impulse in a given direction
Momentum in that direction
Datum can be placed at any convenient location, but points above datum has
to be taken positive gravitation potential energy (vice versa for points below
datum)
Conservation of energy (General form)
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
඲෍ 𝑀𝑑𝑡 = 𝛥 ෍
𝑖
𝐼𝑖 𝜔 𝑎𝑏𝑠,𝑖
Work done is zero if the
particle or the system is not
subjected to any external
force or moment
Impulse is zero if the
particle or the system is not
subjected to any external
force or moment
𝑇 =
1
2
𝑚𝑉𝑎𝑏𝑠
2

140. 140
Key points
Ask yourself:
Am I applying conservation of energy /momentum to a particle or a system?
Particle
(ignore external forces/moment)
System
(ignore internal forces/moment)

141. 141
Question 1
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
1
2
Datum
+s
At position 1
)𝐾. 𝐸1 = 0 (𝑠𝑡𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑡
𝑃. 𝐸 1,𝑠𝑝𝑟𝑖𝑛𝑔 =
1
2
𝑘 𝑅2 + 𝑅2 − 𝑅 2 =
1
2
𝑘( 2𝑅 − 𝑅)2
At position 2
𝐾. 𝐸2 =
1
2
𝑚𝑉2
2
𝑃. 𝐸 2,𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = 0
𝑃. 𝐸 2,𝑠𝑝𝑟𝑖𝑛𝑔 =
1
2
𝑘 𝑅 − 𝑅 2 = 0
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒1 → 2 =
1
2
𝑚𝑉2
2
− 𝑚𝑔𝑅 −
1
2
𝑘 2𝑅 − 𝑅 2 = 0
At position 3
𝐾. 𝐸3 =
1
2
𝑚𝑉3
2
𝑃. 𝐸 3,𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = −𝑚𝑔R
𝑃. 𝐸 3,𝑠𝑝𝑟𝑖𝑛𝑔 =
1
2
𝑘 𝑅 − 𝑅 2 = 0
𝑃. 𝐸 1,𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 = 𝑚𝑔𝑅
𝑉2 =
𝑘
𝑚
2𝑅 − 𝑅
2
+ 2𝑔𝑅 𝑉2 = 0.1716
𝑘𝑅2
𝑚
+ 2𝑔𝑅
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒1 → 3 =
1
2
𝑚𝑉3
2
−
1
2
𝑘( 2𝑅 − 𝑅)+( − 𝑚𝑔R − mgR) = 0
𝑉3 = 0.1716
𝑘𝑅2
𝑚
+ 4𝑔𝑅

142. 142
Question 1
1
2
Datum
+s
𝑉C = 0.1716
𝑘𝑅2
𝑚
+ 4𝑔𝑅
+↑ ෍ 𝐹 = 𝑚 ⋅ 𝑎↑
𝑁 − 𝑚𝑔 = 𝑚 ⋅ 𝑎↑
𝑁 − 𝑚𝑔 = 𝑚 ⋅
𝑉3
2
𝑅
𝑁 = 𝑚
𝑉3
2
𝑅
+ 𝑔
𝑁 = 𝑚
0.1716
𝑘𝑅2
𝑚
+ 4𝑔𝑅
𝑅
+ 𝑔
)𝑁 = 𝑚(0.1716𝑘𝑅 + 5𝑔

143. 143
The cord must be in tension in order for the bob to make
a circular path about peg B.
T
mg
Ԧ𝑎 𝑇 + 𝑚𝑔 = 𝑚 ⋅ Ԧ𝑎↓
+↓ ෍ 𝐹 = 𝑚 ⋅ Ԧ𝑎↓
𝑇 + 𝑚𝑔 = 𝑚
𝑉2
𝑅
𝑉
Therefore, the minimum velocity to describe a circle
about the peg:
R
From the diagram above, the radius of the described
circle as function of L and a is:
)𝐿 = 2𝑅 + (𝑎 − 𝑅
𝐿 = 𝑅 + 𝑎 ⇒ 𝑅 = 𝐿 − 𝑎
𝑉min = 𝑔𝑅
Therefore, the minimum velocity to describe a circle
about the peg as a function of L and a is:
𝑉min = )𝑔(𝐿 − 𝑎
Question 2

144. 144
R
Datum
1
+s
2
Apply conservation of energy equation for points 1 and 2
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
0 + 0 =
1
2
𝑚𝑉min
2
− 𝑚𝑔(2𝑎 − 𝐿)
𝑉min = )2𝑔(2a − 𝐿
)𝑔(𝐿 − 𝑎 = )2𝑔(2a − 𝐿
)𝑔(𝐿 − 𝑎) = 2𝑔(2a − 𝐿
𝐿 − 𝑎) = 2(2a − 𝐿
𝑎 = 0.6𝐿
Question 2

145. 145
෍ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Datum
System at state 1
+s
𝐾. 𝐸1 = 0 (𝑏𝑜𝑡ℎ 𝑏𝑙𝑜𝑐𝑘𝑠 𝑎𝑟𝑒 𝑎𝑡 𝑟𝑒𝑠𝑡)
)𝑃. 𝐸1 = 220.73𝐽 (𝑑𝑢𝑒 𝑡𝑜 𝑏𝑙𝑜𝑐𝑘 𝐵
System at state 2
𝐾. 𝐸2 = 11.76+ 14.7=26.46J
𝑃. 𝐸2 = 176.58𝐽(due to block A)
∴ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = (26.46 − 0) + (176.58 − 220.73) = −17.69𝐽
Based on the conservation equation, work done between states
State 1 State 2
∴ 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = (𝐾. 𝐸2 − 𝐾. 𝐸1) + (𝑃. 𝐸2 - 𝑃. 𝐸1)
12kg
15kg
Recall…tutorial 8: question 1

146. 146
Question 3
𝑉𝐶
𝑉𝐵
State 1 State 2
System at state 1
𝐾. 𝐸1 = 0 (𝑏𝑜𝑡ℎ 𝑐𝑎𝑟𝑡 𝑎𝑛𝑑 𝑏𝑙𝑜𝑐𝑘 𝑎𝑟𝑒 𝑎𝑡 𝑟𝑒𝑠𝑡)
𝑃. 𝐸1 =
1
2
𝑘𝑥 𝑑𝑒𝑓𝑜𝑟𝑚𝑒𝑑
2
=
1
2
(300) 0.2 𝑑𝑒𝑓𝑜𝑟𝑚𝑒𝑑
2
= 6𝐽
System at state 2
𝐾. 𝐸2 =
1
2
𝑚 𝐶 𝑉𝐶
2
+
1
2
𝑚 𝐵 𝑉𝐵
2
= 37.5𝑉𝐶
2
+ 25𝑉𝐵
2
)𝑃. 𝐸2 = 0 (𝑠𝑖𝑛𝑐𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑢𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = (37.5𝑉𝐶
2
+ 25𝑉𝐵
2
) + (−6) = 0

147. 147
Question 3
𝑉𝐶
𝑉𝐵
State 1 State 2
Treating the cart, block and the spring as system, the force exerted by the spring to the block is internal, thus can be ignored.
Applying conservation of linear momentum between state 1 and 2
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
0 = 𝛥𝑚 𝐶 𝑉𝐶 + 𝛥𝑚 𝐵 𝑉𝐵
Since mass of Cart and Block B does not change between state 1 and 2
0 = −75𝑉𝐶,2 + 50𝑉𝐵,2
−75𝑉𝐶,2 + 50𝑉𝐵,2=0
37.5𝑉𝐶
2
+ 25𝑉𝐵
2
= 6
Solving these equations yields
𝑉𝐵 = 0.375𝑖( Τ𝑚 𝑠) 𝑉𝐶 = −0.253𝑖( Τ𝑚 𝑠)
Ƹ𝑖
൯0 = 𝑚 𝐶(𝑉𝐶,2 − 𝑉𝐶,1) + 𝑚 𝐵(𝑉𝐵,2 − 𝑉𝐵,1
൯𝑉 Τ𝐵 𝐶 = 𝑉𝐵 − 𝑉𝐶 = 0.632 Ƹ𝑖( Τ𝑚 𝑠

148. 148
Question 4
Treating the particle P and triangular block B as a system
Apply conservation of Liner momentum along i direction
Impulse Momentum
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
Impulse term can dropped off since there is no external force that acts on the system
along the i direction
𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖 = 0
Since mass of particle P and Block B are constant
𝑚 𝐵(𝑉𝐵,2 − 𝑉𝐵,1) + 𝑚 𝑃(𝑉𝑃,2 − 𝑉𝑃,1) = 0
Since these masses start from rest, momentum equation simplifies to
𝑚 𝐵 𝑉𝐵,2 + 𝑚 𝑃 𝑉𝑃,2 = 0
Absolute velocity of Particle, P is
𝑉𝑃,2 = 𝑉𝐵,2 + 𝑉 Τ𝑃 𝐵,2
Substituting the absolute velocity into momentum equation along i direction
−20𝑉𝐵,2 + 4(−𝑉𝐵,2 + 𝑉 Τ𝑃 𝐵,2cos(300)) = 0
൯24𝑉𝐵,2 = 4𝑉 Τ𝑃 𝐵,2cos(300
𝑉 Τ𝑃 𝐵,2 =
6𝑉𝐵,2
)cos(300
Ƹ𝑖
Ƹ𝑗
𝑚 𝐵 𝑉𝐵,2 + 𝑚 𝑃 𝑉𝑃,2 = 0
→
+

149. 149
Question 4
Treating the particle P and triangular block B as a system
Conservation of Energy principal
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 0 (Work done is zero since there is
no external force or moment acts
on the system)
𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 =
1
2
𝑚 𝑃 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝑃
2
𝑠𝑡𝑎𝑡𝑒 2
− 0 + 𝑚 𝑃 𝑔ℎ 𝑠𝑡𝑎𝑡𝑒 1= 0
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 =
1
2
𝑚 𝐵 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝑃
2
𝑠𝑡𝑎𝑡𝑒 2
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 =
1
2
𝑚 𝐵 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝐵 + 𝑉 Τ𝑃 𝐵
2
𝑠𝑡𝑎𝑡𝑒 2
Since 𝑉𝑃 = 𝑉𝐵 + 𝑉 Τ𝑃 𝐵
𝑉𝐵 = −𝑉𝐵 Ƹ𝑖
𝑉𝑃/𝐵 = 𝑉 Τ𝑃 𝐵∠ − 300
𝑉𝐵 + 𝑉 Τ𝑃 𝐵
2
= 𝑉𝐵
2
+ 2𝑉𝐵 𝑉 Τ𝑃 𝐵 )cos(300 + 𝑉 Τ𝑃 𝐵
2
Ƹ𝑖
Ƹ𝑗
𝑉𝐵 + 𝑉 Τ𝑃 𝐵
2
= 𝑉𝐵,2
2
− 2𝑉𝐵,26𝑉𝐵,2 +
6𝑉𝐵,2
)cos(300
2
= 𝑉𝐵,2
2
− 12𝑉𝐵,2
2
+ 48𝑉𝐵,2
2
= 37𝑉𝐵,2
2
൯𝑉 Τ𝑃 𝐵
2
= 𝑉 Τ𝑃 𝐵
2
cos2
(300
) + 𝑉 Τ𝑃 𝐵
2
sin2
(300
𝑉 Τ𝑃 𝐵
2
= 𝑉 Τ𝑃 𝐵
2
(cos2
(300
) + sin2
(300
)) = 𝑉 Τ𝑃 𝐵
2

150. 150
Question 4
Treating the particle P and triangular block B as a system
Conservation of Energy principal
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 0 (Work done is zero since there is
no external force or moment acts
on the system)
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 =
1
2
𝑚 𝐵 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝑃
2
𝑠𝑡𝑎𝑡𝑒 2
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 =
1
2
𝑚 𝐵 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝐵 + 𝑉 Τ𝑃 𝐵
2
𝑠𝑡𝑎𝑡𝑒 2
Since 𝑉𝑃 = 𝑉𝐵 + 𝑉 Τ𝑃 𝐵
Ƹ𝑖
Ƹ𝑗
𝑉𝐵 + 𝑉 Τ𝑃 𝐵
2
= 𝑉𝐵,2
2
− 2𝑉𝐵,26𝑉𝐵,2 +
6𝑉𝐵,2
)cos(300
2
= 𝑉𝐵,2
2
− 12𝑉𝐵,2
2
+ 48𝑉𝐵,2
2
= 37𝑉𝐵,2
2
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 =
1
2
(20)𝑉𝐵
2
+
1
2
(4)37𝑉𝐵,2
2
𝑠𝑡𝑎𝑡𝑒 2
8𝑔 𝑠𝑡𝑎𝑡𝑒 1 = 10𝑉𝐵
2
+ 74𝑉𝐵,2
2
𝑠𝑡𝑎𝑡𝑒 2
𝑉𝐵 = −0.967 Ƹ𝑖( Τ𝑚 𝑠) 𝑉 Τ𝑃 𝐵 = 6.697( Τ𝑚 𝑠)∠ − 300
൯𝑉𝑃 = −0.967 Ƹ𝑖 + 6.697∠ − 300
( Τ𝑚 𝑠
𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 =
1
2
𝑚 𝑃 𝑉𝐵
2
+
1
2
𝑚 𝑃 𝑉𝑃
2
𝑠𝑡𝑎𝑡𝑒 2
− 0 + 𝑚 𝑃 𝑔ℎ 𝑠𝑡𝑎𝑡𝑒 1= 0

151. Tutorial 10
Kinetics of particles (Conservation of Energy and
Momentum)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

152. 152
Key points
Principle of linear Impulse and Momentum
Impulse in a given direction
Momentum in that direction
Principle of angular Impulse and Momentum
Impulse in a given direction
Momentum in that direction
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
඲෍ 𝑀𝑑𝑡 = 𝛥 ෍
𝑖
𝐼𝑖 𝜔 𝑎𝑏𝑠,𝑖
Impulse is zero if the
particle or the system is not
subjected to any external
force or moment
Collision
• Elastic: Objects collide and move away from each other
• Inelastic: Objects collide and stick together
• Momentum is conserved in both elastic and inelastic collision, but
K.E is conserved only in perfectly elastic collision
• Coefficient of restitution is a ratio of relative velocity of the particles
after collision to before collision
Elastic
(e=1)
Inelastic
(e = 0)
Max K.E is lost K.E conserved
Momentum conserved

153. 153
Key points
Ask yourself:
Am I applying conservation of energy /momentum to a particle or a system?
Particle
(ignore external forces/moment)
System
(ignore internal forces/moment)

154. 154
Bonus question
Q: Assuming that all of the kinetic energy of the surging train was imparted to the stationary train, calculate
the average retarding force experienced by the stationary train given that it moved 10.7 meters forward before
stopping. Assume the weight of the surging train to be 140 tonne and it was moving at 20m/s before collision.
Joo Koon train collision
Treating stationary train as a particle, assuming an elastic collision
𝐾. 𝐸 𝑚𝑜𝑣𝑖𝑛𝑔 𝑡𝑟𝑎𝑖𝑛 = 𝐾. 𝐸𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑡𝑟𝑎𝑖𝑛
∴ 𝐾. 𝐸𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑡𝑟𝑎𝑖𝑛 =
1
2
𝑚𝑉2
=
1
2
(140 ⋅ 103
) 20 2
= 28𝑀𝐽
Since 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Ԧ𝐹 ⋅ Ԧ𝑠
Ԧ𝐹 =
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
Ԧ𝑠
Ԧ𝐹 =
28𝑀𝐽
10.7
= 2.62𝑀𝑁

155. 155
Question 1
Consider the mass A and B as a system
State 1
(Before contact)
State 2
(Mass B stops sliding)
Ƹ𝑖
඲෍ Ԧ𝐹𝑑𝑡 = 𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖
Apply principal of impulse and linear momentum to system
Since system is not subjected to any external forces
𝛥 ෍
𝑖
𝑚𝑖 𝑉𝑎𝑏𝑠,𝑖 = 0
𝛥𝑚 𝐴 𝑉𝐴 + 𝛥𝑚 𝐵 𝑉𝐵 = 0
൫𝑚 𝐴 + 𝑚 𝐵)𝑉𝑓𝑖𝑛𝑎𝑙 − (𝑚 𝐴 𝑉𝐴 + 𝑚 𝐵 𝑉𝐵) = 0
Perfectly inelastic collision (Final velocity is common)
𝑉𝑓𝑖𝑛𝑎𝑙 =
𝑚 𝐴 𝑉𝐴 + 𝑚 𝐵 𝑉𝐵
𝑚 𝐴 + 𝑚 𝐵
=
30 ⋅ 0 Ƹ𝑖 + 12 ⋅ 2.5 Ƹ𝑖
30 + 12
= 0.71 Ƹ𝑖

156. 156
Question 1
Consider the mass A and B as a system
State 1
(Before contact)
State 2
(Mass B stops sliding)
Ƹ𝑖
Ratio of final to the initial K.E of the system
Kinetic energy of the system at state 1 and 2
𝐾. 𝐸1 =
1
2
𝑚 𝐵 𝑉𝐵
2
= 37.5𝐽
𝐾. 𝐸2 =
1
2
(𝑚 𝐴 + 𝑚 𝐵)𝑉𝑓
2
= 10.7𝐽
𝐾. 𝐸2
𝐾. 𝐸1
=
10.7
37.5
= 0.29
This shows that kinetic energy of the system is not conserved
in an inelastic collision

157. 157
Question 1
Consider the mass A and B as a system
State 1
(Before contact)
State 2
(Mass B stops sliding)
Ƹ𝑖
Ratio of final to the initial K.E of the system
Kinetic energy of the system at state 1 and 2
𝐾. 𝐸1 =
1
2
𝑚 𝐵 𝑉𝐵
2
= 37.5𝐽
𝐾. 𝐸2 =
1
2
(𝑚 𝐴 + 𝑚 𝐵)𝑉𝑓
2
= 10.7𝐽
𝐾. 𝐸2
𝐾. 𝐸1
=
10.7
37.5
= 0.29
This shows that kinetic energy of the system is not considered
in an inelastic collision

158. 158
Question 2
Apply conservation of energy principal between position 1 and 2
1
2
Datum
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
0 + 0 =
1
2
𝑚𝑉2
− 𝑚(9.81)(1.2)
𝑉 = )2(9.81)(1.2 = 4.85 Τ𝑚 𝑠
3
4
5
−
4
5
Ƹ𝑖 +
3
5
Ƹ𝑗
3
5
Ƹ𝑖 +
4
5
Ƹ𝑗
4
5
Ƹ𝑖 −
3
5
Ƹ𝑗
Unit vectors
X’
Y’
𝑉2,𝑥′ =
3
5
(4.85) = 2.91m/s
𝑉2,𝑦′ =
4
5
𝑒(4.85) = 3.1m/s
+→ 𝑠 = 𝑠0 + 𝑉0 𝑡
+↓ 𝑠 = 𝑠0 + 𝑉0 𝑡 +
1
2
𝑔𝑡2
𝑑 = 0 + 𝑉2,𝑥′t
4/5𝑑 = 0 + 𝑉2,𝑦′ 𝑡 +
1
2
9.81𝑡2
𝑑 = 𝑉2,𝑥′ 𝑡
(4/5)𝑉2,𝑥′ 𝑡 = 𝑉2,𝑦′ 𝑡 +
1
2
(9.81)(
4
5
)𝑡2
−1.35𝑡 + 3.96𝑡2
= 0
𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑑 =
4
5
Ƹ𝑖 −
3
5
Ƹ𝑗

159. 159
Question 2
Apply conservation of energy principal between position 1 and 2
1
2
Datum
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
0 + 0 =
1
2
𝑚𝑉2
− 𝑚(9.81)(1.2)
𝑉 = )2(9.81)(1.2 = 4.85 Τ𝑚 𝑠
3
4
5
−
4
5
Ƹ𝑖 +
3
5
Ƹ𝑗
3
5
Ƹ𝑖 +
4
5
Ƹ𝑗
4
5
Ƹ𝑖 −
3
5
Ƹ𝑗
Unit vectors
X’
Y’
𝑉2,𝑥′ =
3
5
(4.85) = 2.91m/s
𝑉2,𝑦′ =
4
5
𝑒(4.85) = 3.1m/s
+→ 𝑠 = 𝑠0 + 𝑉0 𝑡
+↓ 𝑠 = 𝑠0 + 𝑉0 𝑡 +
1
2
𝑔𝑡2
𝑑 = 0 + 𝑉2,𝑥′t
3
5
𝑑 = 0 + 𝑉2,𝑦′ 𝑡 +
1
2
9.81𝑡2
𝑑 = 𝑉2,𝑥′ 𝑡
3
5
⋅
5
4
𝑉2,𝑥′ 𝑡 = 𝑉2,𝑦′ 𝑡 +
1
2
(9.81)𝑡2
0.918𝑡 + 4.91𝑡2
= 0

160. 160
Question 3
Plane of contact
(line drawn tangent to circles)
Line of impact-Momentum is conserved in both directions
-Coefficient of restitution applies only along line
of impact
t
n
Conservation of momentum of balls in t direction
Ball A: t-dir 𝑚𝑣 𝑜sin𝜃 = 𝑚𝑣 𝐴𝑡
′
⇒ 𝑣 𝐴𝑡
′
= 𝑣 𝑜sin𝜃
Ball B: t-dir 0 = 𝑚𝑣 𝐵𝑡
′
⇒ 𝑣 𝐵𝑡
′
= 0
Conservation of momentum of balls in n direction
Ball A+B: n-dir 𝑚𝑣 𝑜cos𝜃 + 0 = 𝑚𝑣 𝐴𝑛
′
+ 𝑚𝑣 𝐵𝑛
′
∴ 𝑣 𝑜cos𝜃 = 𝑣 𝐴𝑛
′
+ 𝑣 𝐵𝑛
′
Apply coefficient of restitution along line of impact
൯𝑣 𝐵𝑛
′
− 𝑣 𝐴𝑛
′
= 𝑒(𝑣 𝐴𝑛 − 𝑣 𝐵𝑛
)𝑣 𝐵𝑛
′
− 𝑣 𝐴𝑛
′
= 𝑒(𝑣0cos𝜃 − 0
Solving highlighted equations along n direction
𝑣 𝐴𝑛
′
= 𝑣0
1 − 𝑒
2
cos𝜃 𝑣 𝐵𝑛
′
= 𝑣0
1 + 𝑒
2
cos𝜃

161. 161
Conservation of momentum of balls in t direction
Ball A: t-dir 𝑚𝑣 𝑜sin𝜃 = 𝑚𝑣 𝐴𝑡
′
⇒ 𝑣 𝐴𝑡
′
= 𝑣 𝑜sin𝜃
Ball B: t-dir 0 = 𝑚𝑣 𝐵𝑡
′
⇒ 𝑣 𝐵𝑡
′
= 0
Conservation of momentum of balls in n direction
Ball A+B: n-dir 𝑚𝑣 𝑜cos𝜃 + 0 = 𝑚𝑣 𝐴𝑛
′
+ 𝑚𝑣 𝐵𝑛
′
∴ 𝑣 𝑜cos𝜃 = 𝑣 𝐴𝑛
′
+ 𝑣 𝐵𝑛
′
Apply coefficient of restitution along line of impact
൯𝑣 𝐵𝑛
′
− 𝑣 𝐴𝑛
′
= 𝑒(𝑣 𝐴𝑛 − 𝑣 𝐵𝑛
)𝑣 𝐵𝑛
′
− 𝑣 𝐴𝑛
′
= 𝑒(𝑣0cos𝜃 − 0
Solving highlighted equations along n direction
𝑣 𝐴𝑛
′
= 𝑣0
1 − 𝑒
2
cos𝜃 𝑣 𝐵𝑛
′
= 𝑣0
1 + 𝑒
2
cos𝜃
Question 3
Substitute the given parameters
𝑣 𝐴𝑛
′
= 𝑣0
1 − 0.8
2
cos45 = 0.070𝑣0
𝑣 𝐵𝑛
′
= 𝑣0
1 + 0.8
2
cos45 = 0.6364𝑣0
𝑣 𝐴𝑡
′
= 𝑣0sin45 = 0.707𝑣0
𝑣 𝐵𝑡
′
= 0
Velocity magnitude and direction of each ball after impact is…
|𝑣 𝐴
′
| = 0.707𝑣0
2 + 0.070𝑣0
2 Τ1 2 = 0.711𝑣0
𝛽 = tan−1
0.0707
0.707
= 5.70
𝜃 = 450
− 5.70
= 39.30
Ԧ𝑣 𝐴
′
= 0.711𝑣0∠39.30
Ԧ𝑣 𝐵
′
= 0.636𝑣0∠ −450

162. 162
Question 5
t
n
-Momentum is conserved in both directions (n and t dir.)
-Coefficient of restitution applies only along line
of impact (n-dir.)
1
2
𝑚 𝐴 𝑣 𝐴0
2
= 𝑚 𝐴 𝑔(𝑙 − 𝑙cos450
) → 𝑣 𝐴0 = 0.7654 𝑔𝑙
Conservation of energy equation to find impact velocity of A
Conservation of linear momentum along n-direction
𝑚 𝐴 𝑣 𝐴0 + 0 = −𝑚 𝐴 𝑣 𝐴1 + 𝑚 𝐵 𝑣 𝐵1
Apply restitution along n-direction
)−𝑣 𝐴1 − 𝑣 𝐵1 = 𝑒(−𝑣 𝐴0 − 0
൯𝑣 𝐴1 + 𝑣 𝐵1 = 0.75(0.7654 𝑔𝑙
1.5(0.7654 𝑔𝑙) = −1.5𝑣 𝐴1 + 3𝑣 𝐵1
Solving highlighted equations simultaneously
𝑣 𝐴1 = 0.1276 𝑔𝑙 𝑣 𝐵1 = 0.4465 𝑔𝑙
Apply conservation of energy after impact to find the max height
𝑚 𝐴 𝑔ℎ 𝐴 =
1
2
𝑚 𝐴 𝑣 𝐴1
2
→ ℎ 𝐴 = 0.00814𝑙
𝑚 𝐴 𝑔ℎ 𝐴 =
1
2
𝑚 𝐴 𝑣 𝐵1
2
→ ℎ 𝐵 = 0.09968𝑙
𝜃 = cos−1
𝑙 − ℎ
𝑙
𝜃 𝐴 = 7.30
𝜃 𝐵 = 25.80

163. 163
Question 6
-Momentum is conserved in both directions
-Coefficient of restitution applies only along line
of impact ( Ƹ𝑒 𝑛-dir.)
Ƹ𝑖
Ƹ𝑗
Ƹ𝑒 𝑛 = 1∠300
Ƹ𝑒𝑡 = 1∠1200
)𝑣 𝐴0 = −15 Ƹ𝑖( Τ𝑚 𝑠
Express initial velocity of sphere in terms of n-t direction
𝑣 𝐴0 = −15 Ƹ𝑖( Τ𝑚 𝑠) = 7.5 Ƹ𝑒𝑡 − 12.99 Ƹ𝑒 𝑛
Express final velocity of block and sphere in terms of n-t direction
𝑣 𝐵1 = −𝑣 𝐵1 Ƹ𝑖 = 0.5𝑣 𝐵1 Ƹ𝑒𝑡 − 0.866𝑣 𝐵1 Ƹ𝑒 𝑛
𝑣 𝐴1 = 7.5 Ƹ𝑒𝑡 + 𝑣 𝐴1 Ƹ𝑒 𝑛
Note that momentum is conserved an t-dir. That’s why velocity of
the sphere is remains unchanged after impact
Conservation of linear momentum of the system along i direction is
𝑚 𝐴 Ԧ𝑣 𝐴0 = 𝑚 𝐴 Ԧ𝑣 𝐴1 + 𝑚 𝐵 Ԧ𝑣 𝐵1
1.5(−15) = 1.5(7.5 Ƹ𝑒𝑡 + 𝑣 𝐴1 Ƹ𝑒 𝑛) − 4𝑣 𝐵1
−22.5 = 11.25cos1200
+ 1.5𝑣 𝐴1cos300
− 4𝑣 𝐵1

164. 164
Question 6
-Momentum is conserved in both directions
-Coefficient of restitution applies only along line
of impact ( Ƹ𝑒 𝑛-dir.)
Ƹ𝑖
Ƹ𝑗
Apply restitution along line of impact
𝑣𝐴1
Ƹ𝑒 𝑛
− 𝑣 𝐵1
Ƹ𝑒 𝑛
) = −𝑒(𝑣𝐴0
Ƹ𝑒 𝑛
− 𝑣 𝐵0
Ƹ𝑒 𝑛
ቀ𝑣𝐴1
Ƹ𝑒 𝑛
+ 0.866𝑣 𝐵1) = 9.7425
Solving highlighted equations…
𝑣 𝐵1 = 5.762 Τ𝑚 𝑠
1
2
𝑘𝑥2
=
1
2
𝑚 𝐵 𝑣 𝐵1
2
→ 𝑥 = 0.163𝑚
Use conservation of energy equations to find spring deflection
𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Since system isn’t subjected to any external force, such a friction,
workdone is zero.
𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸 = 0
P.E before impact from sphere is zero and K.E is zero at max spring deflection
−𝐾. 𝐸1 + 𝑃. 𝐸2 = 0
−𝐾. 𝐸1 = −𝑃. 𝐸2

165. Tutorial 11
Kinetics of rigid body(Newton’s Second Law)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

166. 166
Question 1
Find the absolute acceleration of point B, by fixing
a rotating reference frame at A. Since…
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐴 + Ԧ𝛼 × Ԧ𝑟𝐵𝐴 + 𝜔 × (𝜔 × Ԧ𝑟𝐵𝐴) + 2𝜔 × Ԧ𝑣 𝑟𝑒𝑙 + Ԧ𝑎 𝑟𝑒𝑙
)∴ Ԧ𝑎 𝐵 = 𝜔 × (𝜔 × Ԧ𝑟𝐵𝐴
- Point A does not translate
- Disk rotates with constant angular velocity
- Point me does not translated w.r.t rotating reference frame
൯𝜔 = 18.85෠𝑘(𝑟𝑎 Τ𝑑 𝑠
Ԧ𝑟𝐵𝐴 = 0.2𝑚∠600
Ԧ𝑎 𝐵 = −35.53 Ƹ𝑖 − 61.54 Ƹ𝑗
Since rod BC is in translational (pin-pin end boundary condition)
Ԧ𝑎 𝐵 = Ԧ𝑎 𝐺
G
Ԧ𝐹 = 𝑚 Ԧ𝑎 𝐺
Ԧ𝐹 = 7(−35.53 Ƹ𝑖 − 61.54 Ƹ𝑗) 𝑁
Since pin-pin end boundary condition does not resists moment
𝑀 𝐺 = 0

167. 167
Question 1
G
Free body diagram of the
road with constrains at both
ends
෍ Ԧ𝐹 = 𝑚 Ԧ𝑎
൯𝐹𝐵
𝑥
Ƹ𝑖 + 𝐹𝐵
𝑦
Ƹ𝑗 − 𝑚𝑔 Ƹ𝑗 + 𝐹𝐶
𝑥
Ƹ𝑖 + 𝐹𝐶
𝑦
Ƹ𝑗 = 7(−35.53 Ƹ𝑖 − 61.54 Ƹ𝑗
Equating Ƹ𝑖 and Ƹ𝑗 components
𝐹𝐵
𝑥
+ 𝐹𝐶
𝑥
= −248.71
𝐹𝐵
𝑦
+ 𝐹𝐶
𝑦
= −362.11
Since moment about C.G is zero… 𝐹𝐵
𝑦
= 𝐹𝐶
𝑦
𝑀 𝐺 = 0
𝐹𝐵
𝑦
= 𝐹𝐶
𝑦
= −181.1𝑁

168. 168
Question 2
𝐼𝐴 = 𝐼 𝐺 + 𝑚 𝑟𝑜𝑑
𝐿
2
2
𝐼𝐴 =
1
12
𝑚 𝑟𝑜𝑑 𝐿2 + 𝑚 𝑟𝑜𝑑
𝐿
2
2
= 0.375𝑘𝑔 ⋅ 𝑚2
+𝑐𝑐𝑤 ෍ 𝑀 = 𝐼𝐴 Ԧ𝛼
𝑇 ⋅ 𝑟 = 𝐼 𝐺 Ԧ𝛼
12 ⋅ 𝐿 = (0.375) Ԧ𝛼
Ԧ𝛼 = 24𝑟𝑎 Τ𝑑 𝑠2
𝑎 𝐺 = Ԧ𝛼 ⋅
𝐿
2
= −9 Ƹ𝑖
Apply Equation of motion at point G
൯−2𝑔 Ƹ𝑗 − 12 Ƹ𝑖 + 𝐹𝐴
𝑥
Ƹ𝑖 + 𝐹𝐴
𝑦
Ƹ𝑗 = 2(−9 Ƹ𝑖
Ƹ𝑖
Ƹ𝑗
Equation i and j components
𝐹𝐴
𝑥
= −6𝑁
𝐹𝐴
𝑦
= 19.63𝑁

169. 169
Question 3
Ƹ𝑖
Ƹ𝑗
𝐼 𝐺 =
1
2
𝑚𝑟2
= 0.5(7) 0.125 2
𝑘𝑔 ⋅ 𝑚2
+𝑐𝑤 ෍ 𝑀 = 𝐼 𝐺 𝛼
4(0.125) = 0.5(0.7) 0.125 2
Ԧ𝛼
Ԧ𝛼 = 9.143𝒌(𝑟𝑎 Τ𝑑 𝑠2)
Ԧ𝐹 = 𝑚 Ԧ𝑎 𝐺
4 = 7 Ԧ𝑎 𝐺
Ԧ𝑎 𝐺 = 0.57𝐢 Τ(𝑚 𝑠2)
Ԧ𝑎 Τ𝐴 𝐺 = Ԧ𝛼 × Ԧ𝑟𝐴𝐺
൯Ԧ𝑎 Τ𝐴 𝐺 = 9.143෠𝑘 × −0.125 Ƹ𝑗 = 1.14 Ƹ𝑖( Τ𝑚 𝑠
𝑠 =
1
2
𝑎 Τ𝐴 𝐺 𝑡2 = 2.28𝑚

170. 170
Question 5
Ԧ𝑎 𝑃 = Ԧ𝑎 𝐴 + Ԧ𝑎 Τ𝑃 𝐴
Ԧ𝑎 𝐴 = 𝑎 𝐴 Ƹ𝑖
Ԧ𝑎 Τ𝑃 𝐴 = −𝑎 Τ𝑃 𝐴 Ƹ𝑒𝑡
Ƹ𝑒𝑡 = 1∠300
Ƹ𝑒 𝑛 = 1∠1200
𝛼 =
𝑎 Τ𝑃 𝐵
𝑟
Angular acceleration of the pipe is
Linear equation of motion of the ring
൯𝑁 Ƹ𝑒 𝑛 + 𝐹 Ƹ𝑒𝑡 − 500𝑔 Ƹ𝑗 = 500(𝑎 𝐴 Ƹ𝑖 − 𝑎 Τ𝑃 𝐴 Ƹ𝑒𝑡
𝑁𝑜𝑡𝑒:
Ƹ𝑖 ⋅ Ƹ𝑖 = 1
Ƹ𝑖 ⋅ Ƹ𝑗 = 0
Ƹ𝑗 ⋅ Ƹ𝑖 = 0
Ƹ𝑗 ⋅ Ƹ𝑗 = 1
)Ƹ𝑖 ⋅ Ƹ𝑒𝑡 = cos(𝜃
Dot Equation of motion of pipe by Ƹ𝑒 𝑛
)𝑁 − 500𝑔(0.866) = 500𝑎 𝐴(−0.5
Dot Equation of motion of pipe by Ƹ𝑒𝑡
𝐹 − 500𝑔(0.5) = 500𝑎 𝐴(0.866) − 500𝑎 Τ𝑃 𝐴
+𝑐𝑐𝑤 ෍ 𝑀 = 𝐼 𝑝 𝛼
𝐹𝑟 = 𝐼 𝑝 𝛼 = 𝐼 𝑝
𝑎 Τ𝑃 𝐵
𝑟
𝐹 = 500𝑎 Τ𝑃 𝐴
Rotational equation of motion of pipe
Where theta is the angle
between the unit vectors.

171. 171
Question 5
Ԧ𝑎 𝐴 = 𝑎 𝐴 Ƹ𝑖
Equation of motion of the ramp
−𝑁 Ƹ𝑒 𝑛 − 𝐹 Ƹ𝑒𝑡 − 300𝑔 Ƹ𝑗 + 𝑅 Ƹ𝑗 = 300𝑎 𝐴 Ƹ𝑖
Dot Ƹ𝑖
𝑁(0.5) − 𝐹(0.866) = 300𝑎 𝑎
Since…
𝐹 = 500𝑎 Τ𝑃 𝐴
∴ 𝑁 = 866𝑎 Τ𝑃 𝐴 + 600𝑎 𝐴
)𝑁 − 500𝑔(0.866) = 500𝑎 𝐴(−0.5
𝐹 − 500𝑔(0.5) = 500𝑎 𝐴(0.866) − 500𝑎 Τ𝑃 𝐴
Linear equation of motion of ring is…
Where…
𝐹 = 500𝑎 Τ𝑃 𝐴
𝑁 = 866𝑎 Τ𝑃 𝐴 + 600𝑎 𝐴
866𝑎 Τ𝑃 𝐴 + 850𝑎 𝐴 = 433𝑔
Therefore…
1000𝑎 Τ𝑃 𝐴 − 433𝑎 𝐴 = 250𝑔
Solving…
𝑎 Τ𝑃 𝐴 = 3.203 Τ𝑚 𝑠2 𝑎 𝐴 = 1.73 Τ𝑚 𝑠2

172. 172
Question 6
Length of the rope has to be constant
2𝑥 𝐵 + 𝑥 𝐷 = constant
2 ሶ𝑥 𝐵 + ሶ𝑥 𝐷 = 0
Velocity relationship
Acceleration relationship
2 ሷ𝑥 𝐵 + ሷ𝑥 𝐷 = 0
Acceleration relationship
ሷ𝑥 𝐺 = 𝛼𝑟 = 0.15𝛼
ሷ𝑥 𝐷 = 𝛼(2𝑟) = 0.3𝛼
ሷ𝑥 𝐵 = −0.5 ሷ𝑥 𝐷 = −0.15𝛼
+↓ ෍ 𝐹 = 𝑚𝑎
20 − 2𝑇 =
20
𝑔
ሷ𝑥 𝐵 =
20
𝑔
(−0.15𝛼)
Linear equation of motion for weight
Linear equation and rotational EOM for disc
+→ ෍ 𝐹 = 𝑚𝑎
𝑇 + 𝐹 =
−50
𝑔
(0.15𝛼)

173. 173
Question 6
Linear EOM for disc
+→ ෍ 𝐹 = 𝑚𝑎
𝑇 + 𝐹 =
−50
𝑔
(0.15𝛼)
Rotational EOM for disc
+𝑐𝑐𝑤 ෍ 𝑀 = 𝐼 𝐺 𝛼
( 𝐹 − 𝑇)𝑟 = 𝐼 𝐺 𝛼
(𝐹 − 𝑇)(0.15) =
1
2
50
𝑔
0.15 2 𝛼
Simultaneously solving highlighted EOM gives
𝛼 = −13.77𝑟𝑎 Τ𝑑 𝑠2
𝐹 = 2.6315𝑁
𝑇 = 7.89𝑁

174. Tutorial 12
Kinetics of rigid body (Auxiliary Principles)
Dr N SATHEESH Kumar
Research Fellow,SMRT-NTU SmartUrban Rail Corporate Laboratory
50 Nanyang Avenue, S2.1-B3-01, Singapore 639798
T 65-98466232 F65-6790-9313 nsatheesh@ntu.edu.sg www.ntu.edu.sg
http://tiny.cc/smlssy

175. 175
Question 1
𝑟𝐴 𝜃 = 𝑠 𝐴 = 900𝑚𝑚
𝑟𝐵 𝜃 = 𝑠 𝐵
𝜃 =
𝑠 𝐴
𝑟 𝐴
=7.5 rad
𝑟𝐵 ⋅
𝑠 𝐴
𝑟𝐴
= 𝑠 𝐵
180 ⋅
900
120
= 𝑠 𝐵
𝑠 𝐵 = 1.35𝑚
Apply Conservation of Energy Principal to System
Placing datum at state 1 position, P.E and K.E is…
𝐾. 𝐸1 = 0
𝑃. 𝐸1 = 0
State 2
𝐾. 𝐸2 =
1
2
𝐼 𝐺 𝜔2 +
1
2
𝑚 𝐴 𝑉𝐴
2
+
1
2
𝑚 𝐵 𝑉𝐵
2
𝐾. 𝐸2 = 0.5(0.1176)𝜔2 + 0.5(3) 0.12 2 𝜔2 + 0.5(3)(0.18)2 𝜔2
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
𝐼 𝐺 = 𝑚𝑘2
= 6 ⋅ 0.142
= 0.1176
𝜔 = 8.5𝑟𝑎 Τ𝑑 𝑠
𝑣 = 𝜔 ⋅ 𝑟 = 8.5(0.12) = 1.03 Τ𝑚 𝑠
𝐾. 𝐸1 = 𝑃. 𝐸1 = 0
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = −0.5 ⋅ 7.5 = −3.75𝐽
𝑃. 𝐸2 = −𝑚𝑔𝑠 𝐵 + 𝑚𝑔𝑠 𝐴
)𝑃. 𝐸2 = −3𝑔(1.35) + 3𝑔(0.9 = -13.24J
∴ −3.75 = −13.24 + 0.13𝜔2

176. 176
Question 2
𝜔 𝐴 𝑟𝐴 = 𝜔 𝐵 𝑟𝐵
)𝜔 𝐴(150) = 𝜔 𝐵(100
𝜔 𝐴 =
𝑣 𝐴
𝑟𝐴
= 10𝑣 𝐴
∴ 𝜔 𝐵 = 10𝑣 𝐴
150
100
= 15𝑣 𝐴
Likewise…
𝛼 𝐴 𝑟𝐴 = 𝑎
𝛼 𝐴 =
𝑎
𝑟𝐴
= 10𝑎
𝛼 𝐴 𝑟𝐴 = 𝛼 𝐵 𝑟𝐵
𝛼 𝐵 = 15𝑎
Apply Conservation of Energy Principal to System
Placing datum at state 1 position, P.E and K.E is…
𝑃. 𝐸1 = 0
𝐾. 𝐸1 = 0
State 2
𝑃. 𝐸2 = −𝑚 𝐶 𝑔𝑠
𝐾. 𝐸2 =
1
2
𝑚 𝐶 𝑉𝐶
2
+
1
2
𝐼 𝐺 𝜔 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐵
2
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸=0 (no external force)
∴ 𝑃. 𝐸2 = 𝐾. 𝐸2
10𝑔𝑠 = 45.625𝑣 𝑎
2
)10𝑔𝑠 = 45.625(2𝑎𝑠
∴ 𝑎 =
10𝑔
2 ⋅ 45.625
= 1.075 Τ𝑚 𝑠2
Velocity after 3s is…
𝑣 = 𝑎𝑡 = 1.075(3) = 3.23 Τ𝑚 𝑠

177. 177
Question 2
+𝑐𝑐𝑤 ෍ 𝑀 = 𝐼 𝐺 𝛼 𝐵
𝑇𝐵(0.1) = 𝐼 𝐺 𝛼 𝐵 → 𝑇𝐵 = 40.32𝑁
𝑇𝐴(0.1) − 𝑇𝐵(0.15) = 𝐼 𝐺 𝛼 𝐴 → 𝑇𝐴 = 87.35𝑁
Apply equation of motion to pulley A and pulley B

178. 178
Question 3
𝑣 𝐺 =
𝐿
2
ሶ𝜃 =
𝐿
2
𝜔
Making use of IC
𝑇 =
1
2
𝑚𝑣 𝐺
2
+
1
2
𝐼 𝐺 𝜔2 =
1
6
𝑚𝐿2 𝜔2
K.E of the rod AB can be expressed as…
P.E of the rod AB can be expressed as…
𝑉 = 𝑚𝑔𝑦 +
1
2
𝑘𝛿2
𝑉 = 𝑚𝑔(−0.5𝐿sin𝜃) +
1
2
𝑘 𝐿sin𝜃 2
Since the rod is not subjected to any external force
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 0
0 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Since, the K.E and P.E of the rod is zero when theta
is equal to zero
𝐾. 𝐸2 = −𝑃. 𝐸2
1
6
𝑚𝐿2
𝜔2
= 𝑚𝑔(0.5𝐿sin𝜃) −
1
2
𝑘 𝐿sin𝜃 2
𝜔 = ± ቇ3(
𝑔
𝐿
sin𝜃 −
𝑘
𝑚
sin2 𝜃
𝑣 𝐴 = 𝐿cos(𝜃)𝜔
𝑣 𝐴 = ±𝐿cos𝜃 ቇ3(
𝑔
𝐿
sin𝜃 −
𝑘
𝑚
sin2 𝜃
Velocity of point A can be found using IC
C.G of the rod

179. 179
Question 4
Using the instant center for the rolling without
slipping, we have…
𝑣 𝐴 = 𝜔 ⋅ 𝑟𝐴
𝑣 𝐶 = 𝜔 ⋅ 2𝑟𝐶
𝑣 𝐵 = 𝜔 ⋅ 𝑟𝐵
𝐵𝑢𝑡 𝑟𝐶 = 2𝑟𝐴 = 2𝑟𝐵
𝑣 𝐴 = 𝑣 𝐵 = 0.5𝑣 𝐶
Therefore…
𝜔 𝐴 = 𝜔 𝐵 = 𝜔 =
0.5𝑣
𝑟
The wheel rotates without slipping, work is
force times distance. If there is no slip, the
force of friction acts over a distance of 0.
Therefore frictional force does no work.
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Since the cart starts from rest…
𝐹 ⋅ 𝑠 = 𝐾. 𝐸2 + 𝑃. 𝐸2
The cart has translational K.E while the
wheels have both translational and
rotational K.E
10 ⋅ 𝑠 =
1
2
(6)𝑣2
+ 2[
1
2
(4) 0.5𝑣 2
+
1
2
𝐼 𝐺 𝜔2
]
𝐼 𝐺 =
1
2
(4)𝑟2
10𝑠 = 4.5𝑣2

180. 180
Question 4
The wheel rotates without slipping, work is
force times distance. If there is no slip, the
force of friction acts over a distance of 0.
Therefore frictional force does no work.
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 = 𝛥𝐾. 𝐸 + 𝛥𝑃. 𝐸
Since the cart starts from rest…
𝐹 ⋅ 𝑠 = 𝐾. 𝐸2 + 𝑃. 𝐸2
The cart has translational K.E while the
wheels have both translational and
rotational K.E
10 ⋅ 𝑠 =
1
2
(6)𝑣2
+ 2[
1
2
(4) 0.5𝑣 2
+
1
2
𝐼 𝐺 𝜔2
]
𝐼 𝐺 =
1
(4)𝑟2
Recall constant acceleration formulas…
10𝑠 = 4.5𝑣2
)∴ 10𝑠 = 4.5(2𝑎𝑠
𝑎 = 1.11 Τ𝑚 𝑠2
𝑣 = 𝑎𝑡
𝑣 = 1.11(2.5) = 2.78 Τ𝑚 𝑠

181. 181
Question 5
2𝑠 𝐴 − 𝑠 𝐷 = constant
Author Setup
Recommended
D
SD
SA
C1
𝑆𝐴 − 𝐶1
𝑆𝐴 − 𝑆 𝐷
( 𝑆𝐴 − 𝑆 𝐷) + (𝑆𝐴 − 𝐶1) = constant
2𝑆𝐴 − 𝑆 𝐷 = constant

182. 182
Question 5
2𝑠 𝐴 − 𝑠 𝐷 = constant
2𝑣 𝐴 = 𝑣 𝐷 𝑎𝑛𝑑 2𝑎 𝐴 = 𝑎 𝐷
𝜔 𝐵 =
𝑣 𝐷
𝑟
=
2𝑣 𝐴
𝑟
𝜔 𝐴 =
𝑣 𝐴
𝑟
𝛼 𝐴 =
𝑎 𝐴
𝑟
𝛼 𝐵 =
2𝑎 𝐴
𝑟
Apply Conservation of Energy Principal to System
𝐾. 𝐸1 = 0
𝑃. 𝐸1 = −𝑚𝑔𝑠 𝐴
Placing datum at s = 0 position, P.E and K.E is…
State 1
State 2
𝐾. 𝐸2 =
1
2
𝐼 𝐺 𝜔 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐵
2
+
1
2
𝑚𝑣 𝐴
2
𝑃. 𝐸2 = −𝑚𝑔𝑠𝑓𝑖𝑛𝑎𝑙
Work done by or on the system is zero
as there is no external force
𝑙𝑒𝑡 𝑠 = 𝑠𝑓𝑖𝑛𝑎𝑙 − 𝑠 𝑎
𝑚𝑔𝑠 =
1
2
𝑚𝑣 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐵
2
𝑔𝑠 =
7
4
𝑣2
Recall constant acceleration formulas…

183. 183
Question 5
𝑙𝑒𝑡 𝑠 = 𝑠𝑓𝑖𝑛𝑎𝑙 − 𝑠 𝑎
𝑚𝑔𝑠 =
1
2
𝑚𝑣 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐴
2
+
1
2
𝐼 𝐺 𝜔 𝐵
2
𝑔𝑠 =
7
4
𝑣2
Recall constant acceleration formulas…
2𝑎𝑠 = 𝑣2
Since 𝑣2 =
4
7
𝑔𝑠
𝑎 =
1
2𝑠
4
7
𝑔𝑠 =
2
7
𝑔 = 2.803 Τ𝑚 𝑠2
Speed after 2.5 seconds is…
𝑣 = 𝑎𝑡 = 2.803(2.5) = 7.01 Τ𝑚 𝑠
+𝑐𝑐𝑤 ෍ 𝑀 = 𝐼 𝐺 𝛼 𝐵
𝑇𝑟 = 𝐼 𝐺 𝛼 𝐵
𝑇𝑟 =
1
2
𝑚𝑟2
2𝑎
𝑟
𝑇 = 28.03𝑁