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12366 dnk201e dynamics - sample problems 3
Problem 1: (15p) The tennis ball is struck with a
horizontal velocity 𝑣⃗𝐴 , strikes the smooth ground at B, and
bounces upward at πœƒ = 30°. Determine the initial velocity
𝑣⃗𝐴 , the final velocity 𝑣⃗𝐡 , and the coefficient of restitution
between the ball and the ground.
2.28 m
6m
Problem 2: (25p) In the speed-governing mechanism π‘š = 0.2π‘˜π‘”, 𝐿 =
10 π‘π‘š (rods AB have negligible mass), πœ” = 500 π‘Ÿπ‘π‘š, πœƒ = 40°, πœƒΜ‡ =
0. Determine πœƒΜˆ at that instant and the tension in the rod.
0.3 m 0.3 m
C
B
Problem 3: (30p) If link CD has an angular velocity of
πœ”πΆπ· = 6 π‘Ÿπ‘Žπ‘‘/𝑠,
E
(a) determine the velocity of point B,
(b) and determine the acceleration of point B.
0.6 m
A
30°
Problem 4: (30p) A 27 π‘˜π‘” ladder is assumed to be a uniform bar which
starts sliding without frictional resistance. Determine the forces at A and B.
(𝐼𝐴 = 𝐼𝐡 = π‘šπ‘™ 2⁄3 , 𝐼𝐢 = π‘šπ‘™ 2 ⁄12)
πœ”πΆπ· = 6 π‘Ÿπ‘Žπ‘‘/𝑠
5m
D
Kinematics and Kinetics of a Particle:
Coordinates Position (π‘Ÿβƒ—)
Rectangular
βƒ—βƒ—
π‘₯𝑖⃗ + 𝑦𝑗⃗ + π‘§π‘˜
Polar
π‘Ÿπ‘›βƒ—βƒ—π‘Ÿ
Cylindrical
βƒ—βƒ—
π‘Ÿπ‘›βƒ—βƒ—π‘Ÿ + π‘§π‘˜
Spherical
π‘Ÿπ‘›βƒ—βƒ—π‘Ÿ
Velocity (𝑣⃗)
βƒ—βƒ—
π‘₯Μ‡ 𝑖⃗ + 𝑦̇ 𝑗⃗ + 𝑧̇ π‘˜
Μ‡
π‘ŸΜ‡ π‘›βƒ—βƒ—π‘Ÿ + π‘Ÿπœƒπ‘›βƒ—βƒ—πœƒ
βƒ—βƒ—
π‘ŸΜ‡ π‘›βƒ—βƒ—π‘Ÿ + π‘ŸπœƒΜ‡π‘›βƒ—βƒ—πœƒ + 𝑧̇ π‘˜
π‘ŸΜ‡ π‘›βƒ—βƒ—π‘Ÿ + π‘Ÿπœ™Μ‡π‘›βƒ—βƒ—πœ™
+ π‘ŸπœƒΜ‡ sin πœ™ π‘›βƒ—βƒ—πœƒ
Acceleration (π‘Žβƒ—)
βƒ—βƒ—
π‘₯̈ 𝑖⃗ + π‘¦Μˆ 𝑗⃗ + π‘§Μˆ π‘˜
2
Μ‡
(π‘ŸΜˆ βˆ’ π‘Ÿπœƒ )π‘›βƒ—βƒ—π‘Ÿ + (π‘ŸπœƒΜˆ + 2π‘ŸΜ‡ πœƒΜ‡)π‘›βƒ—βƒ—πœƒ
βƒ—βƒ—
(π‘ŸΜˆ βˆ’ π‘ŸπœƒΜ‡ 2 )π‘›βƒ—βƒ—π‘Ÿ + (π‘ŸπœƒΜˆ + 2π‘ŸΜ‡ πœƒΜ‡)π‘›βƒ—βƒ—πœƒ + π‘§Μˆ π‘˜
(π‘ŸΜˆ βˆ’ π‘Ÿπœ™Μ‡ 2 βˆ’ π‘ŸπœƒΜ‡ 2 sin2 πœ™)π‘›βƒ—βƒ—π‘Ÿ
+(π‘Ÿπœ™Μˆ + 2π‘ŸΜ‡ πœ™Μ‡ βˆ’ π‘ŸπœƒΜ‡ 2 sin πœ™ cos πœ™)π‘›βƒ—βƒ—πœ™
+(π‘ŸπœƒΜˆ sin πœ™ + 2π‘ŸΜ‡ πœƒΜ‡ sin πœ™ + 2π‘Ÿπœ™Μ‡πœƒΜ‡ cos πœ™)π‘›βƒ—βƒ—πœƒ
Tangential and normal components: 𝑣⃗ = 𝑣𝑛⃗⃗𝑑 , π‘Žβƒ— = 𝑣̇ 𝑛⃗⃗𝑑 + (𝑣 2 β„πœŒ)𝑛⃗⃗𝑛
3/2
Radius of curvature: 𝜌 =
[1+(𝑑𝑦⁄𝑑π‘₯ )2 ]
𝑑2 𝑦/𝑑π‘₯ 2
Work-energy principle: π‘ˆ12 = 𝑇2 βˆ’ 𝑇1
2
Work of a force: π‘ˆ12 = ∫1 𝐹⃗ βˆ™ π‘‘π‘Ÿβƒ—
π‘‘π‘ˆ
Power: = 𝐹⃗ βˆ™ 𝑣⃗
𝑑𝑑
βƒ—βƒ—V, Conservation of mechanical energy: 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
Conservative force: 𝐹⃗ = βˆ’βˆ‡
𝑑
Time rate of change of linear momentum: βˆ‘ 𝐹⃗ = 𝐺⃗̇ = 𝑑𝑑 (π‘šπ‘£βƒ—)
βƒ—βƒ—Μ‡0 = 𝑑 (π‘Ÿβƒ— × π‘šπ‘£βƒ—) = π‘Ÿβƒ— × πΉβƒ— , βˆ‘ 𝑀
βƒ—βƒ—βƒ—0 = 𝐻
βƒ—βƒ—Μ‡0
Time rate of change of angular momentum: 𝐻
𝑑𝑑
βˆ‘ 𝑀π‘₯ = 𝐻̇π‘₯ βˆ‘ 𝑀𝑦 = 𝐻̇𝑦 βˆ‘ 𝑀𝑧 = 𝐻̇𝑧
Angular impulse:
𝑑
⃗⃗⃗𝑖 𝑑𝑑 = 𝐻
⃗⃗𝑂 βˆ’ 𝐻
⃗⃗𝑂 = βˆ‘π‘›π‘–=1(π‘Ÿβƒ— × π‘šπ‘– 𝑣⃗𝑖 )2 βˆ’ βˆ‘π‘›π‘–=1(π‘Ÿβƒ— × π‘šπ‘– 𝑣⃗𝑖 )1
βˆ‘π‘›π‘–=1 βˆ«π‘‘ 2 𝑀
𝑂
2
1
1
Kinematics of Rigid Bodies:
Velocities in general plane motion: 𝑣⃗𝐡 = 𝑣⃗𝐴 + πœ”
βƒ—βƒ— × π‘Ÿβƒ—π΅/𝐴
Accelerations in general plane motion: π‘Žβƒ—π΅ = π‘Žβƒ—π΄ + 𝛼⃗ × π‘Ÿβƒ—π΅/𝐴 + πœ”
βƒ—βƒ— × (πœ”
βƒ—βƒ— × π‘Ÿβƒ—π΅/𝐴 )
Kinetics of Rigid Bodies in Plane Motion:
βˆ‘ 𝐹π‘₯ = π‘šπ‘ŽπΆ,π‘₯
βˆ‘ 𝐹𝑦 = π‘šπ‘ŽπΆ,𝑦
βˆ‘ 𝑀𝐢 = 𝐼𝐢 𝛼
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