- Two Spur Gears In A Simple Train Shown In Figure Q1 Are Having The Same Module M 2 And Their Respective Number Of Teet 1 (88.55 KiB) Viewed 29 times
- Two Spur Gears In A Simple Train Shown In Figure Q1 Are Having The Same Module M 2 And Their Respective Number Of Teet 2 (13.46 KiB) Viewed 29 times
X=1
Y=4
Im=0.1
Ig=0.4
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 Nīm = 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 kgmand 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. (b) (c) (4 marks) Considering acceleration of load at 2 ms-?, evaluate the torque of the motor to overcome gears inertia. (9 marks) Considering the 2 kg load, examine the torque required to accelerate the load through gear system. (9 marks) Suggest the minimum torque of the motor to accelerate the 2 kg load at 2 ms. (3 marks) (d) 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
NTG = 45 r=30 mm Nam = 30 Ta driver a = 2 ms? 2 kg Figure Q1