Question

A Mй Given: The collar M is being moved at constant velocity v = 68cm/s along a straight track AB. Ring K with mass m 0.1 kg can move without friction along a curved wire CD with radius R - 1.9 m. The ring and the collar are connected with an inextensible cable with length 1 = 5R/3. The entire system is in horizontal plane (remember what this means from a problem we did?). At the instant shown, the angle is the angle between the cable and the collar motion direction, and angle B is the angle between the cable and the motion direction of ring K. You know that cos a 15/17 and cos B 4/5. Calculate: 1. The velocity of the ring K at the given instant (4 points) 2. The velocity of the ring with respect to the collar at the given instant (4 points) 3. Tension within the cable (4 points)

Answer 1

Given that velocuty of Collar =Vc = 68cm/s Ring maes - Me = 0.1 kg ; Radius = Yw=loquy length of the cable = 5 ; cosa=15; Cosp=4 / 2 I @ Let us take VR as velocity of the ring 1. VR & ve along string will be equal I . At the given instant, the a -/velocity of ringk = 75cy JURY las follows due cosa = up cosB > Vo = Vo.Cosa - Cosp = 68 x 15/17 - 4/5 TUR = 75 cm/s 6 velocity of the ring with respect to Collar at gives instant = UKM = UK - ve :. Horizontal component of VK Vc are equal ... Perpendicular " " " " will play role VRing wrt = siuß + Vesing 1 = 75 Vi-cosB +68 1-cosa P = 15/1-(415)2 + 68 [1- (15/17ge 1 [UR, E = 17 auls C. Ring performing circular motion wot Collar -) Radial acceleratioy will produce in the cable - T = wel .. w. (VRN) 2 It string length = 04 & 77² x 10 573 X1.9 0; i81 - S2