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Two spur gears in a simple train shown in Figure Q1 are having the same module, m = 2 and their respective number of teeth is NTm = 30 for the driver gear, and NTG = 45 for the driven gear. The inertia of the gear attached to the motor is Im = 0.X kgm2 and the inertia of the second gear is IG = 0.Y kgm2. The second gear is connected to a pulley with the radius of 30 mm. The pulley is hoisting up the 2kg load with acceleration of a = 2 ms-2. Given the gear transmission efficiency is 90%. (a) Based on the module and number of teeth given, examine the pitch diameter of each gear. (4 marks) (b) Considering acceleration of load at 2 ms-2, evaluate the torque of the motor to overcome gears inertia. (9 marks) (c) Considering the 2 kg load, examine the torque required to accelerate the load through gear system. (9 marks) (d) Suggest the minimum torque of the motor to accelerate the 2 kg load at 2 ms. (3 marks) Note: X is the second last digit of your matrics number, while Y is the last digit. For example, student with matric number of CD 190242, will have X = 4 making Im = 0.4, whilst Y = 2 for IG = 0.2. For those numbers ending up with 00, can take X = 1 and Y =1
Two spur gears in a simple train shown in Figure Q1 are having the same module, m = 2 and their respective number of tee
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Two spur gears in a simple train shown in Figure Q1 are having the same module, m = 2 and their respective number of tee
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