Engr 240 - Dynamics - Sample Test 2 - Solution NAME: ________________________ 1. Two wires AC and BC are tied to a 15-lb sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that θ1 = 50° and θ 2 = 25° and that d = 4 ft, determine the range of values of v for which both wires are taut. θ 3 = θ1 − θ 2 = 50° − 25° = 25° l1 sin θ 2 = d sin θ 3 ρ = l1 sin θ1 = = l1 = or d sin θ 2 sin θ 3 d sin θ 2 sin θ1 sin θ 3 ( 4 )( sin 25° )( sin 50° ) sin 25° = 3.0642 ft + ΣFy = 0: TAC cosθ 2 + TBC cosθ1 − W = 0 (1) 2 + ΣF = ma : T sin θ + T sin θ = Wv 2 1 x x AC BC (2) gρ Case 1: TBC = 0. TAC cosθ 2 − W = 0 TAC = or TAC sin θ 2 = W tan θ 2 = W cosθ 2 Wv 2 gρ v 2 = g ρ tan θ 2 = ( 32.2 )( 3.0642 ) tan 25° = 46.01 ft 2 /s 2 v = 6.78 ft/s Case 2: TAC = 0. TBC cosθ1 − W = 0 TBC = or TBC sin θ1 = W tan θ1 = W cosθ1 Wv 2 gρ v 2 = g ρ tan θ1 = ( 9.81)( 3.0642 ) tan 50° = 117.59 ft 2 /s 2 v = 10.84 ft/s 6.78 ft/s ≤ v ≤ 10.84 ft/s W 2. The slotted arm revolves in the horizontal plane about the fixed vertical axis through point O. The 3-lb slider C is drawn toward O at the constant rate of 2 in/sec by pulling cord S. At the instant for which r =9 in, the arm has a counterclockwise angular velocity of 6 rad/sec and is slowing down at the rate of 2 rad/sec2. For this instant, determine the tension in the cord and the force exerted on the slider by the sides of the smooth radial slot. Ans.: T = 2.52 lb, N = 0.326 lb 3. The 0.75-kg bob of a pendulum is fired from rest at position A. If the spring is compressed 50 mm and released, determine (a) its stiffness k so that the speed of the bob is 2 m/s as it passes position B, (b) the minimum value of its stiffness k if the bob is to pass position C and move along a vertical circular arc. Ans: a) k=4731.6 N/m b) k=14126 N/m Solution: a) Apply work energy from A to B. Velocities (and hence kinetic energies) at A and B are known. Work done by gravity is –mg(.6), and spring force is ½ k (.050)2. Solve for k. b) Find minimum velocity at C, using Newton’s Second Law. Minimum velocity is obtained from a summation of vectors in the normal (vertical) direction: -mg = - m vC2/R (since Tension is zero.) 4. The 2-kg block A is connected to a spring of constant k = 20 N/m, and by an inextensible rope to the 8-kg block B. The smooth pulley is of negligible mass, and the coefficient of friction between the horizontal surface and block A is μk=0.2. If the system is released from rest when the spring is initially undeformed, find a) the velocity of block B after it has moved 0.5 m. b) the position at which the velocity of A is maximum. Ans: 3.7278 m 4. Two identical spheres A and B each of mass m =1 kg collide while moving with velocities vA = 4 m/s, and vB = 4 m/s as shown. If the coefficient of restitution is e = 1, find a. the velocities of A and B after the collision, and b. the average impulsive force between the spheres during collision if the spheres were in contact 0.01 s. Ans.: r r v A' = −2iˆ + 2 ˆj , v B' = 4 cos 30 o iˆ + 4 sin 60 o ˆj ; P = 546.4 N 5. A 300-g block is released from rest after a spring of constant k = 600 N/m has been compressed 160 mm. Determine the force exerted by the loop ABCD on the block as the block passes through (a) point A, (b) point B, (c) point C. Assume no friction. 6. The 5-kg sphere A is attached to a 2-m inextensible cable. It is released from rest at the position shown, swings along a vertical plane, and strikes the 8-kg block B. Block B is initially at rest and is attached to a spring of constant k=20 N/m that is initially undeformed. If the coefficient of kinetic friction between B and the horizontal floor is μk=0.2, and the coefficient of restitution between A and B is e=0.8, determine a) the tension in the cable immediately after the collision, and b) the maximum deformation of the spring. Answer:T=49.62 N; x=1.307 m 8. Two small spheres A and B, weighing 5 lb and 2 lb, respectively, are connected by a rigid 7-in rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity vo = 10 ft/s i. Determine a) the linear momentum of the system and its angular momentum about its mass center G, b) the velocities of A and B after the rod AB has rotated through 90°. Answer: r r v A' = 7.143iˆ − 2.857 ˆj , vB' = 7.143iˆ + 7.143 ˆj