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Question 2 (20 Marks) Ball bearings leave the horizontal trough with a velocity of magnitude u and fall through the 70-mm diameter hole as shown. Calculate the permissible range of u which will enable the balls to enter the hole. Take the dashed positions to represent the limiting conditions. - 120 mm 20 mm 80 mm mm

Answer 1

120mm Date somm o 70 mm Consider Point A, the horizontal displacement from end of the truff is calculated as 120 - (Radius of hole) + (Radius of ball bearing) Sx Sx 120 - 35 +10 95 mm Si Now coneider the kinematic Equation in vertical direction. Initially, velocity of ball bearing along Yoaxis direction is Therefore, uy = 0 zero Sy : Uyt + I at² 2 lly al: 9800 mm/s? Sy : 80 mm Sy : + 1 x 9800xt? 2 2 --- 9800t 1x980012 2 160 9800

It 0.27 Sec Now, consider kinematical equations along x-axis Sx = uxt + I at² 2 where sx = 95mm t:0.127 sec, a : 0 tx 38 x 0.1.27 to Sx Ux * 0.127 95 Ux x0.127 Ux 95 0.127 = 768.03 mm Is UK This is the minimum velocity of ball bearing Point C, the horizontal displacement from the end of the truff calculated as Consider can be Sa =120 + Radius of hole - Radius of ball bearing Sx - 10 = 120 + 35 145 mm S2

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Now, time will be same as before Coneider kinematic equatics along x-axis. = Uxt + I at² S 2 Sx = 145 mm 145 mm + 2 t = 0.1275 a=0 145 = Ux * 0.127 to ux 145 0.127 1,141,73 mms ux This is the maximum velocity of ball bearing So the permissible range of velocity is 748.03 mmls to 1,141.73 mmts.