Question

Problem 3. A flat surface follower behaves in the following manner. There is a constant acceleration ascent for 60°, followed by 60 of constant velocity, and finally constant deceleration for 60" The follower rests for the next 45 before beginning its descent, which is cycloidal for the rest of its trajectory. Determine the equations for each section. Take Smax as 100 cm S Smax В O 57 4 27

Answer 1

Solution Given Qo, = 60° ) 2 600 AB - Const accelesation BC – Const velocity Oo, CD Const deceleration , Ou; = 60° DE Dwell Crest) 0 - 45° Cycloidal ST EF - E Smax Smax = 100cm O outstoke Sz Angulas velocity of cam B Si return stroke A 7 SA 57 27 0 For AB Time of ows fruke (tor) - 00 و در (No lang Oui S, to, (Vo) max & (Volmin Sordo Nomin=0 Navy at Sy 3 (Volmax = 2 ous, 2 Volary - > 2 @o) a ix constant V= utat * from t=0 (0:0) to to &- to (o= e Volmax auto, 4 w²s, ao L ot と

For BC (vola S₂-S, ta (52-5) a Ou (Nolaug - (Volmax in uniform Velocity aro For CD AB Postion is Co similar to (No Imaxz 20 € Smax - S₂ con (No lavdi Volmax 2 4we (sman-se Steuerberaten Sie eing ave sign indicate follower is setanding 3 aug oo For DE follower at rest ADE 50, VDE For EF (Return Starke) 2TO (sin :)] s dx = 27 [. - cos (20)] Ew dxx as Or - 2ws, Sin2 Or => Vr or 2ws, 2 > No) max

ax 27 os Og sin ( 20 ) 2 W Smax (Armax