Question

7. The rod CD is rotating with an angular velocity OCD = 2 rad/s. Determine: i) The velocity of point C 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? 0.2 m 0.2 m В. E 0.4 m cd= 2 rad/s 30° D

Answer 1

GIVEN B C = 0.4m, co= 0.4m. EC =BE= 0.2m. Weo= 2rad's CA Velocity Analysis Velo - Vevo O is fixed loco Velo = Ve Ve WCO X CD V= 2 ( x 014 0:4 (T) = 0.81-9) Vele = Vp-Ve Vele Wocx BC {ret ange rotate clockwise. со вс FÅ) x x 0-4 sin 30:04 AB UB/C = 0.4 Woch 0.4 -> AB- 0.8m = Ve A Ve/A - VB - VA P { let won be in ccw VRE WOA XBA. direction.

Vge Wan (Alx (0.8 car 30 ( + 0.8 sin 30 (1) Vg = 0.693 CUBA () +0.4 WeA Vote VB - Vc 0 4 Woc (ſ) = 0.693 WBA (f ) +04 WBA ti 08 fî) Equating terms in same direction : (^ O = -0.4 WBA +0.8 = WBA 2 rad/s WBA 2 (f) rad's 02 0.4 WBC = 0.693 WBA 0:4 Woc 0.693x2 تولید 3.465 (-k) radle

VE - Ve Vele Vele WoeX EC VELE - 3.465 FÅ) x 0.2 (-2) VELE - 0-693) Vele VE-Ve 0.69 3 ) = VE- 0.8 ( => VE 0:69.3 CD +0.869) IC B VC C 31 Wco D Tot stw 8.0 -- (1) link BC Angular velocity of WBE 3-465 rad/s ER) [clock wise

(ii) Velocity of point B & E 0.693 WEA 9 +0.4 WBA t2 Vo = 0.623x2 Q2 +0.4x2 692 1-386 Ï ) + 0-869) ITA! / 1.386² +0.82 1.6 m/s TE = 0.693) +0.8622 1270-693° +0.82 = 1.058 m/s (iv) Angular velocity of link AB WAB 2 rad/s (e) [Counter Clock wise].