- 1 A 5 Kg Collar Is On A Rod That Is Rotating In A Horizontal Plane Around A Vertical Pole As Shown In The Diagram The 1 (213.6 KiB) Viewed 51 times
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1. A 5 kg collar is on a rod that is rotating in a horizontal plane around a vertical pole, as shown in the diagram. There is no friction between the rod and the pole. At the instant shown, the collar is located at a distance r = 0.2 meters from the axis of rotation. A spring is attached to the collar and to the end of the rod. The undeformed (relaxed) length of the spring is 0.25 meters, and the spring constant is k = 150 N/m. The rod has zero mass. Friction between the rod and collar is determined by: µk=0.25, µs = 0.3. (a) Assume that the collar is initially not sliding along the rod. Find the maximum value of 0 of the rod without causing the collar to start sliding at the position shown in the diagram. (b) Suppose that the value of is initially equal to 3.25 radians/second, r = 0.2 m, and r = -0.3 m/s at the instant shown in the diagram. What will be the radial component of the collar's acceleration, a, at that instant? 10 (c) Suppose that the value of is initially equal to 3.25 radians/second, r = 0.2 m, and r = -0.3 m/s at the instant shown in the diagram. Determine the velocity of the collar in (r, 0) coordinates when the collar reaches the position r = 0.13m. 25 marks for question #1 rod pole collar g= 9.81 m/s² 10 5 TTTII 0.2 m 0.18 m
(C) 0 =3.25 rad/s (no normal velucity) {Fy=N-mg=o N Fs=0 ≤Fx = man = Ff ↓ y ал x < EFX = 12.2625= man = 38.46 v ² = V= 12.2625 38.46 -то 9=0 = N = (5) (9.81) = 49.05N F₁ = MK N = (0,25) (49.05) = 12.2635 № mv² = (5) v² Y (0,13) 38.46 v2 Fs = k ox= (150) 0,2-0,13 +0,18-0.25) = ON 0.5046 m/