Question

8. In the Figure below, bar AB rotates with a clockwise angular velocity of 5 rad/s. Points A and D are fixed (VA= Vp=0). Find: a) The velocity of point B b) The angular velocity of BC and DC c) The vertical velocity VR of the rack in the rack-and-pinion gear (35 points) B 12 in 10 in 10 rad/s muna 6 in [VR -- 6 in- 16 in 4-6 in

Answer 1

Given to AB - 5 radly Let the origin be at 'A' AB rotades dockwise ФА = -5k rad/ vedor AB TAC = ( 61 +125) in - WAB TAG gaxda ) SEX (6i +125) (601 - 305) in / 160* + (-30) The magnitude og veloudy of VG 67.08 20 iols JB = (boi - 3oj) in /s (or) VB = 67.0820 inis Let the velocity at point 't be vo Ve= VB + uclo uclo and VC= Vot We FDOU and Hs magnitude Be WBC hat the anguka velocity of Be be Фес and direchon TO Bo = - WBCF .! uclo ūc JB + (60i - 30J) + wpcx Yeo - 1601 - 305) + -WOC Ex (167 -25) Vc = 60ī -301 26 OCT 16WBCJ leo tarehe Clara rce = Trc-ro C16 +6)+105 -(61 +125) Tel : (160 - aj) in Vc = 601-200 Bc ł - 301 - 160 BC 1 het the angulou Velocity of DC be Woc and Its magnitude be wpc and divedón be to .. W DC= - ODCk The vector Telo & (-61 +105) in Vc = Jo + volo - Otwocx ico EX (-6i +10) uc 10 lopci + Go De 1 -WOC

Eq 0 9 ea both represent velowy at to equating them weget boi - 20 BCT - 301 - 16 WRCJ 10 copeł +6WDC5 equaling i, J Coeffeaents we get 60 awBo = 100DC 106pc + 2WBC = 60 y solve them -30 -16WBC = 60 DC 6wbc + 16 WBC =-30 Soling them we get WDC = 6.8918 radla - 4.4594 radls WBC - WBC ve Indicates ou so med direchon -It is wrong so wBC in te directon Woc = 4.4594 e rad's Angular Velout of BC Co.c.w) WDC = 6.8918 F rad is → Angular velour of Dc (c.w) velours VR of rack The velocity at the tip of gear is same as UR veluty at tip of geal = WDC (6) 6.8918 (6) 41.3508 in /s - Velocty of rack VR = 41.3508 in 3