Question

The rod CD is rotating with an angular velocity 0CD = 2 rad/s. Determine: 1) The velocity of point and show its direction in the figure below i1) What is the angular velocity of link BC? 1) 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 30° wcd=2 rad/s

Answer 1

solution: WBC E co C WAB 0.4m Wed J 30 AD F AGT30 CD -BF [BF = 0.4 m) In A ABF : BE sino = AB 12 0.4m AB A B = 0,8 m] [AB by Graphical method: Velocity analysis by b 60° ad c Velocity of point a 2 Vc = cd Vc = WCDX CD Vc = (2 rued/see) xo um CSScanned with CamScanner

[VC ca= 0.8 misee] + By bca i tan 60 = bc ca ba bc 0.8 6c = 1.39 misee] $ Cos (о ac ab 0,8 ret ay lis [04 - 1.6 misee] velocity of point c = vc Vc = ac Vc ='0.8 misee] direction! towards left ve с CSScanned with CamScanner 40

(1) bc = WBC & Be 1.39 = Wec X 0.4 Bc. [WBC = 3.48 rad) see] Angular velocity of the link (in cw direction) velocity of Point B (ii) VB V2 VB = ab B VE - 1.6 m/s ] AA VE NEY VE, x² velocity of point E = VE Ves Tlves?+(VE.a)? (VC)*+(W.BC X CE? J63-654830-23 [ ve = 1:06 m/s] E (iv) aba WAB X AB 1.6 - WAR X 0.8 Angular velocity of [WAB WAB = 2 rad /see cow] link AB CSScanned with CamScanner