Question

7. The rod CD is rotating with an angular velocity OcD = 2 rad/s. Determine: i) The velocity of point and show its direction in the figure below ii) What is the angular velocity of link BC? iii)What is the velocity of points B and E? iv) What is the angular velocity of link AB? (30 points) 0.2 m 0.2 m В. E 0.4 m cd= 2 rad/s 130° А.

Answer 1

(3) Bo. 2m (13) Colm, P. (2) E (ly) 10.4m to.4m (de) ww A (4) 330 Step 1:- we can way to solve this solve it by different approaches easiest is instantaneous center Tethod no of instantaneous Centers = n(n+) caculate the Here n4: no. of Jocs ) 4(4-1). 6. 464-0 which are Iz 13. 114 I 23 I24 I3 4 Step 2 Mark those Instantaneous centers on diagram IB can be “D I24 22 3 obtained by extending I2, I23 and I 34,- I 14 can be obtained by extending I2, I34 and I14, 112 Given angular velocity of link 2 is wa W, = 2 rad/sec velocity of the point "c" is perpendicular to the link (de) and in the direction of we loot of I23 from the triangle 345 0.4 14 sinzoo → 94 singoo 10.4m 134 2 3 dz 0.4 Qy 14 0.8m 300 5 II 4

Page no 1

6 from ble 236 sinzora - 2 / 2 . COS30 = 0.4 20 / 2 (or) YE los30 0.4618m Tango° = x 0.4 a=0.4 (tanso) =0.2309m Now 9 velocity of point a Vc = (CD) x W2 = 0.8 mis c) (0.4) (2) Angulae velocity of link we know 3 تا distance b/w I 12 & 123 distance blw I 13 & 123 wz wz 0.4 0.2309 [from the above & 0.42 z = 3.464 rad (0.4)(043) 6.9309 il) velocity of Sec points B and E 10.2309 VE 30 0.2 Wzx (oist b/w I 12 & E) = 3.464x (0.3054) 1.0579 mis = (W) x (bist blw Ilze B) - (3) X(0.4610) = 3.464x604614) 1.5996 m/s. VB r-10.2² 0.2309² 0.3054 m

Page no 2

Page no 3

If you have any queries add them in comments

Thank you