Question

Gearing 1. A motor with armature inertia, J. = 400 g-cm, is coupled to the input shaft of a gear pair with tooth counts, Ni = 40 and N, = 400. A load moment of inertia, JL = 40 kg.cm², is coupled to the output shaft, to which a constant torque, TL = 1 Nom, is applied in the negative direction. If it is desired to rotate the load one revolution in 0.2 seconds, find the required motor torques, Tm1.2.3, for an equal-time trapezoidal velocity profile motion.

Answer 1

Motor Jer load Je = load inertia = 210kg :cm² N, 040 Ng=2400 Ja - Armatol inertia = 200 g.cme applied torque shaft TLE IN-m. revolution in o.2 sec ) SE= 360 Ben/ enia. Motor speed - Sma SGXN E56 x 400 240 - 360 (10) - 3600 pm Motor torque 400x40 -0.0025 (40) oil

Rejected load imeria 91 Io 0 =10 Total inertia at mater. If I + Jon = 40 +400 = 10 +400 Jt = 410 %