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Equivalent inertia, speed profile and quadrants (Fig 2b)

Short answer

J_eq = 0.12 kg·m². Speed rises 0→30 rad/s (A), holds 30 (B), falls 30→0 (C), reverses 0→-30 (D), holds -30 (E), rises -30→0 (F). Quadrants: I, I, II, III, IV, IV.

[Figure 1a: a motor (electric torque Te, speed ωm) is directly coupled to Load 1 (load torque Tl1); the motor and Load 1 together have inertia J1. Load 1's shaft drives Load 2 (load torque Tl2, inertia J2, speed ωm2) through a gear pair with n1 teeth on the motor side and n2 teeth on the load side.] [Figure 2b: electric torque Te (N-m) against time t (s): Section A, 0 to 0.1 s: +41; Section B, 0.1 to 0.2 s: +5; Sections C and D, 0.2 to 0.3 s (C from 0.2 to 0.25, D from 0.25 to 0.3): −67; Section E, 0.3 to 0.4 s: +5; Section F, 0.4 to 0.5 s: +41; after 0.5 s: +5.] Consider a dc motor driving system (Figure 1a) with following parameter; J2 = 0.01 kg-m², n1/n2 = 2, J1 = 0.08. Determine the equivalent 'Jeq' of the system. If electric torque profile of figure 2b is applied to motor, construct the profile of the speed (Assume Load torque TL = 5 NM). Using the plots (T vs. t and ω vs. t), discuss the quadrant of operation in each section A to F as shown.

Assumptions: the drive starts from rest at t=0t = 0; the load torque TL=5T_L = 5 N-m is referred to the motor shaft and stays constant (as given) in every section.

Equivalent inertia

Load 2 runs at ωm2=n1n2ωm=2ωm\omega_{m2} = \dfrac{n_1}{n_2}\omega_m = 2\omega_m. Equating kinetic energies:

Jeq=J1+(n1n2)2J2=0.08+(2)2(0.01)=0.08+0.04=0.12 kg-m2\begin{aligned} J_{eq} &= J_1 + \left(\frac{n_1}{n_2}\right)^2J_2 = 0.08 + (2)^2(0.01) \\ &= 0.08 + 0.04 = 0.12\ \text{kg-m}^2 \end{aligned}

Speed profile

From Te−TL=JeqdωmdtT_e - T_L = J_{eq}\dfrac{d\omega_m}{dt}, in each section the acceleration is constant:

α=Te−50.12\alpha = \frac{T_e - 5}{0.12}
SectionTime (s)TeT_e (N-m)Te−TLT_e - T_Lα\alpha (rad/s²)Δω\Delta\omegaω\omega at end (rad/s)
A0 to 0.14136300+3030
B0.1 to 0.2500030
C0.2 to 0.25-67-72-600-300
D0.25 to 0.3-67-72-600-30-30
E0.3 to 0.45000-30
F0.4 to 0.54136300+300
after> 0.550000

Sample working, section A: ω=0+300×0.1=30\omega = 0 + 300\times0.1 = 30 rad/s; section C: ω=30−600×0.05=0\omega = 30 - 600\times0.05 = 0.

 Te (N-m)
  41 |___                    ___
   5 |   |____          ____|   |_____
   0 +---+----+----+----+---+---+-----> t (s)
 -67 |        |____|____|
     0  .1   .2  .25  .3   .4  .5
          A    B   C  D   E   F

 w (rad/s)
  30 |    /------\
     |   /        \
   0 +--/----------\-----------/------> t
     |              \         /
 -30 |               \-------/
     0  .1   .2  .25  .3   .4  .5

Quadrant of operation

SectionTeT_eω\omegaQuadrantOperation
A++ (rising)IForward motoring, accelerating
B++ (constant)IForward motoring, steady speed
C-+ (falling)IIForward braking (regenerative)
D-- (rising in reverse)IIIReverse motoring, accelerating
E+- (constant)IVReverse braking (motor torque opposes reverse motion, steady)
F+- (falling to 0)IVReverse braking, decelerating to rest

Answer: Jeq=0.12J_{eq} = 0.12 kg-m²; speed rises to 30 rad/s (A), holds (B), reverses to -30 rad/s (C, D), holds (E) and returns to zero at 0.5 s (F); quadrants I, I, II, III, IV, IV for A to F.

✓ Answer checked by Claude Opus 5.5 (high effort)

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