Final_paper_solution

# Final_paper_solution - Statics and Dynamics Me 35b(Winter...

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Statics and Dynamics - Me 35b (Winter 2009) Final Examination Solutions Problem 1 (15 points): Member ABC (shown in the Figure) has a weight of 9 kg and is attached to a pin support at midpoint B . A 2 kg sphere D strikes the end C of member ABC with a vertical velocity v 1 of 12 m/s. The half-length of member ABC is h = 2 m and the coe ffi cient of restitution between the sphere and the member ABC is e = 0 . 5. Determine the angular velocity of member ABC and the velocity of the sphere right after the impact. Figure 1: Solution 1: For the member ABC ¯ I = 1 12 m ABC L 2 Figure 2: Kinematics: v = L ω (1) 1

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Condition of Impact: v C - v D = ev 1 or v D = h ω - ev 1 (2) Moments about B : m D v 1 h + 0 = ¯ I ω + m D v D h = ¯ I ω + m D ( h ω - ev 1 ) h (1 + e ) m D hv 1 = ( ¯ I + m D h 2 ) ω (3) Thus the angular velocity of ABC can be given to be: ω = (1 + e ) m D hv 1 ¯ I + m D h 2 . Using the given data: L = 2 h = 4m, e = 0 . 5, v 1 = 12m/s, m ABC = 9kg, m D = 2kg. We thus obtain that the angular velocity of the member ABC immediately after impact is ω = 3 . 6rad/s. Problem 2 (10 points): A 73-kg gymnast is executing a series of full-circle swings on the horizontal bar. In the position shown he has a small and negligible clockwise angular velocity and will maintain his body straight and rigid as he swings downward. Assuming that during the swing the centroidal radius of gyration of his body is 457mm, determine the angular velocity and the force exerted on his hands after he has rotated through 90 o . Figure 3: Solution 2: Position 1. (Directly above the bar). Elevation: h 1 = 1 . 1m Potential Energy: V 1 = mgh 1 = (73)(9 . 81)(1 . 1) = 787 . 74J Speeds: ω 1 = 0, ¯ v 1 = 0 Kinetic Energy: T 1 = 0 Position 2: (Body at level of bar after rotating 90 o ). 2
Elevation: H 2 = 0m Potential Energy: V 2 = 0 Speeds: ¯ v 2 = 1 . 1 ω 2 Kinetic Energy: T 2 = 1 2 m ¯ v 2 2 + 1 2 mk 2 ω 2 2 T 2 = 1 2 (73)(1 . 1 ω 2 ) 2 + 1 2 (73)(0 . 457) 2 ω 2 2 = 51 . 788 ω 2 By Principle of conservation of energy: T 1 + V 1 = T 2 + V 2 0 + 787 . 74 = 51 . 788 ω 2 2 Thus we have the angular velocity to be: ω 2 = 3 . 9rad/s Kinematics: ¯ α t = 1 . 1 α ¯ a n = 1 . 1 ω 2 2 = (1 . 1)(3 . 9) 2 = 16 . 732 m/s 2 Σ M o = Σ ( M o ) eff (1 . 1)(73)(9 . 81) = (73)(1 . 1)(1 . 1 α ) + (73)(0 . 457) 2 α (4) Thus we have the angular acceleration to be: α = 7 . 6055 rad/s 2 and ¯ a t = 8 . 3660 m/ s 2 Σ F x = ma n : R x = (73)(16 . 732) = 1221 . 4 N Σ F y = ma t : R y - (73)(9 . 81) = - (73)(8 . 366) R y = 105 . 4 N (5) Thus force exerted on his hands after he has rotated is 1 . 226 kN at 4 . 93 o Problem 3 (5 points per part, 35 points total): Please indicate whether the state- ments below are true or false and explain why.

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