Question

Can you please help me with a detailed solution to this problem?

|The rod CD is rotating with an angular velocity CD = 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 B E 0.4 m WCD= 2 rad/s 30° D

Answer 1

Solution : 0.2 Wco = aradis 0-2 > 04 m 30° WCD = 2 A AMD (8) velocity of pointe IC Vc = WC D X VCD E 2 X 0.4-0.8 m/s ve 0.4 tano" Vc = 0.8mis 30 B 0.2 E 0:2 (i) angular velocity of link be Vc-(WBC) * YC-IC 2 * IC Instantaneous centre VC 0.8 =) WBC = rc-IC 0.4 tanzo 0.8 = 3.464 0.4 tano" IC = WBC = WBC = 3.464 radis TA 30 B (iii) velocity of point D (Un) 66330 = 0.4 B-IC Un = ( WBC) x VG-IC -> BIC: 0.4 Cosso = (3.464) ) = 1.599 mis VB = 1.599uis Aug CS Scanned with CamScanner

velouty a point E IVE) IC VE -(WBC) YE-IC 1- 0.4 tano E 0:2 C VE = 3.464 x 0 305 VE = 1.056 mis Tas liv) angula velouity of link AB IWAB) (E10) (0+014 ->(Ete) (62+6.4 tanse? -0.04 + 0.0533 ELC= 10.04 +0.0533 ECC = 0.305m (E102 VB = (WAN)X(700) 1.599 & WAB= VB JAB 0.4 Simo 110599x singo 0.4 WAB 2 0.4 WAB = 1.998 radis 1. Aug 30 A 04 Sinzo An TAB 2 0.4 sino CS Scanned with CamScanner