Question

.A planning machine works in two strokes namely cutting stroke and non-cutting
stroke. The total time taken for both the strokes is 2 minutes. The displacement of
the cutting stroke of the machine is given by the expression x+2y+z; whereas x is
150 cm; y is 60% lesser than x; z is average of x and y. During the cutting stroke,
initially the planning table accelerates uniformly for a displacement equal to 75%
of x from rest; then the speed remains constant for a displacement equal to 30%
of cutting stroke and retards uniformly for the remaining stroke till it comes to
stop. Find the maximum cutting velocity of the machine in cm/s, if time taken for
cutting stroke is 60% of the total time taken.

Answer 1

solve for 3 different intervals and find the final solution

Given total time for both the strokes = 2 minutes Let displacement is d, le. d= x + 2y tz where x = 150 cm y = (2- 60 x x ) 100 = 1-0.6 x = 0.un = 0.4 (150 cm) = 60cm 2 = x+y = (150760)cm = 105 cm 2 i d = (150+ 2(60) + 105) cm = 375 cm → During Cutting staoke , initially planning table accelerates uniformly (let acch be a) to 75% of a :: 75% of x = 0.75(150) cm X = 112.5 cm also wsing equ of kinetics 1 = hot that? * {i Imehet veritas しか {x, I displacement during Cutting S stroke or 112.5 m = I at, 2 100 00 also Speed, 9 astes accelesation = Motat 29 = at, t time to reach constant velocity or i 112.5 m= 128) t, 1 5 75% of x ( -> Const. Speed for 30% of lutting stroke (re. 0.3x), velocity remains Const. (1.e. 2.) 100 $x2= 0.38 : 0.3%, = 2+2 - (2) x₂ = 0.3 x 112.5 m = 29tz m2 108 Remaining displacement to cover = d-x-X2 (r.e. 13) x3 = 375 cm - 112.5cm -0.3 (112.5) im = 228.75 cm * = 1 ht - ② Adding ear 0 ① & ② = x+x₂+x₂ = ļ zh ti + 2 tz + 1 29 t 3 - as X, +82 +83 = d = 375 cm and to tatt3 time for lutting stroke time for non-cutting stroke and as tttttz = 2 min = 120 sec. → also as given ti = 60 (titta + tz) 100 = 0.6 X 120 sec = 72 sec #here maximum relcity is e and from equ o x, = 112.5 cm = 1 et =) 2 (112.5 cm) = 24 72 sec » 2, = 2 max = 3,125 cm/s Vmax = 3.125 cmls Scanned with CamScanner