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 Smáx S B 7T 57 4 27 @

Answer 1

PAGE NO.: DATE: Solution 5 E toom Smox af PT. of outstroke angular rebcity of w is sz cam B Sin r return stroke. E - F 2T Given AB so Const acceleration or to Comest velocity Dog 60° BC てい DE Const deceleration Dwell Crest) cycloidal o=45 EF For AB Time of outstroke to, وع کا (velaug wsi si to, (vo lang Vomox + Vomin 2 (Volmin - O at S1/2 2(Vo) ang No more = 2 wsi. Ne bosc Dome As is Const, V= utat from to 1:0 (0-0) t = toy 2 (No mox + ao to 4 W²s, 0.2 1 2 REDMI NOTE 8 PRO AL QUAD CAMERA 202017131 00:28

DA Ts2-5) W (v.), For BC S2-S toz oor (Volmox in uniform velocity (Vulang Tacol For co CD is similar to AB portion, En lang . z ustrone of ways. Cloudy - A = 4 w2s mox - Sz) 2 since, jollower retarding VDE & = a DE =0 followes at rest for EF / Return stroke) Rr S العيونك sin 9. 2T on Ver= w dzer cos cos ( - 200 Vq = do or 2ws, sin rol On 09 2 Wsmax og mo (Vg) mox REDMI NOTE 8 PRO AI QUAD CAMERA 202017131 00:29

aq = 2x was sin ( 20 ) Or (almon 27 w²smos 2 Og