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Equilibrium points and steady-state stability

Short answer

Equilibrium at T_m = T_L: ω_m = 1 rad/s and ω_m =0.25 rad/s. Stability using dT_L/dω - dT_m/dω >0: at ω=1, diff = 0.5 (>0, stable); at ω=0.25, diff = -1 (<0, unstable).

Obtain the equilibrium points and determine their steady-state stability when motor and load torque are Tm = −(1 + 2ωm) and TL = −3√ωm respectively.

Equilibrium points: where motor torque equals load torque, Tm=TLT_m = T_L:

−(1+2ωm)=−3ωm2ωm−3ωm+1=0\begin{aligned} -(1 + 2\omega_m) &= -3\sqrt{\omega_m} \\ 2\omega_m - 3\sqrt{\omega_m} + 1 &= 0 \end{aligned}

Let x=ωmx = \sqrt{\omega_m}: 2x2−3x+1=0⇒(2x−1)(x−1)=02x^2 - 3x + 1 = 0 \Rightarrow (2x-1)(x-1) = 0, so x=1x = 1 or x=0.5x = 0.5.

ωm=1 rad/sandωm=0.25 rad/s\omega_m = 1\ \text{rad/s} \quad \text{and} \quad \omega_m = 0.25\ \text{rad/s}

Stability criterion. An equilibrium point is steady-state stable if a small speed increase produces a net decelerating torque:

dTLdωm−dTmdωm>0\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} > 0

Derivatives:

dTmdωm=−2,dTLdωm=−32ωm\frac{dT_m}{d\omega_m} = -2, \qquad \frac{dT_L}{d\omega_m} = -\frac{3}{2\sqrt{\omega_m}}

At ωm=1\omega_m = 1:

dTLdωm−dTmdωm=−32+2=0.5>0  ⇒  stable\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2} + 2 = 0.5 > 0 \;\Rightarrow\; \text{stable}

At ωm=0.25\omega_m = 0.25:

dTLdωm−dTmdωm=−32×0.5+2=−3+2=−1<0  ⇒  unstable\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2\times0.5} + 2 = -3 + 2 = -1 < 0 \;\Rightarrow\; \text{unstable}

Check at ωm=0.25\omega_m = 0.25: torque Tm=TL=−1.5T_m = T_L = -1.5 N-m; at ωm=1\omega_m = 1: Tm=TL=−3T_m = T_L = -3 N-m.

Equilibrium speedTorque (N-m)dTL/dω−dTm/dωdT_L/d\omega - dT_m/d\omegaStability
1 rad/s-3+0.5Stable
0.25 rad/s-1.5-1Unstable

Answer: equilibrium points at ωm=1\omega_m = 1 rad/s (stable) and ωm=0.25\omega_m = 0.25 rad/s (unstable).

✓ Answer checked by Claude Opus 5.5 (high effort)

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