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12366 dnk201e dynamics - sample problems 3 Problem 1: (15p) The tennis ball is struck with a horizontal velocity π£βπ΄ , strikes the smooth ground at B, and bounces upward at π = 30°. Determine the initial velocity π£βπ΄ , the final velocity π£βπ΅ , and the coefficient of restitution between the ball and the ground. 2.28 m 6m Problem 2: (25p) In the speed-governing mechanism π = 0.2ππ, πΏ = 10 ππ (rods AB have negligible mass), π = 500 πππ, π = 40°, πΜ = 0. Determine πΜ 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 ππΆπ· = 6 πππ/π , 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 ππ ladder is assumed to be a uniform bar which starts sliding without frictional resistance. Determine the forces at A and B. (πΌπ΄ = πΌπ΅ = ππ 2β3 , πΌπΆ = ππ 2 β12) ππΆπ· = 6 πππ/π 5m D Kinematics and Kinetics of a Particle: Coordinates Position (πβ) Rectangular ββ π₯πβ + π¦πβ + π§π Polar ππββπ Cylindrical ββ ππββπ + π§π Spherical ππββπ Velocity (π£β) ββ π₯Μ πβ + π¦Μ πβ + π§Μ π Μ πΜ πββπ + πππββπ ββ πΜ πββπ + ππΜπββπ + π§Μ π πΜ πββπ + ππΜπββπ + ππΜ sin π πββπ Acceleration (πβ) ββ π₯Μ πβ + π¦Μ πβ + π§Μ π 2 Μ (πΜ β ππ )πββπ + (ππΜ + 2πΜ πΜ)πββπ ββ (πΜ β ππΜ 2 )πββπ + (ππΜ + 2πΜ πΜ)πββπ + π§Μ π (πΜ β ππΜ 2 β ππΜ 2 sin2 π)πββπ +(ππΜ + 2πΜ πΜ β ππΜ 2 sin π cos π)πββπ +(ππΜ sin π + 2πΜ πΜ sin π + 2ππΜπΜ cos π)πββπ Tangential and normal components: π£β = π£πββπ‘ , πβ = π£Μ πββπ‘ + (π£ 2 βπ)πββπ 3/2 Radius of curvature: π = [1+(ππ¦βππ₯ )2 ] π2 π¦/ππ₯ 2 Work-energy principle: π12 = π2 β π1 2 Work of a force: π12 = β«1 πΉβ β ππβ ππ Power: = πΉβ β π£β ππ‘ ββV, Conservation of mechanical energy: π1 + π1 = π2 + π2 Conservative force: πΉβ = ββ π Time rate of change of linear momentum: β πΉβ = πΊβΜ = ππ‘ (ππ£β) ββΜ0 = π (πβ × ππ£β) = πβ × πΉβ , β π βββ0 = π» ββΜ0 Time rate of change of angular momentum: π» ππ‘ β ππ₯ = π»Μπ₯ β ππ¦ = π»Μπ¦ β ππ§ = π»Μπ§ Angular impulse: π‘ βββπ ππ‘ = π» ββπ β π» ββπ = βππ=1(πβ × ππ π£βπ )2 β βππ=1(πβ × ππ π£βπ )1 βππ=1 β«π‘ 2 π π 2 1 1 Kinematics of Rigid Bodies: Velocities in general plane motion: π£βπ΅ = π£βπ΄ + π ββ × πβπ΅/π΄ Accelerations in general plane motion: πβπ΅ = πβπ΄ + πΌβ × πβπ΅/π΄ + π ββ × (π ββ × πβπ΅/π΄ ) Kinetics of Rigid Bodies in Plane Motion: β πΉπ₯ = πππΆ,π₯ β πΉπ¦ = πππΆ,π¦ β ππΆ = πΌπΆ πΌ