--- title: "Equivalent inertia and torque of geared loads" subject: "Utilization of Electrical Energy" chapter: 2 key: "semester_7/utilization_of_electrical_energy:geared-loads-equivalent-inertia-dynamics" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/geared-loads-equivalent-inertia-dynamics site: BCT Engineering --- # Equivalent inertia and torque of geared loads Utilization of Electrical Energy · Chapter 2: Electric Drive System - Asked in: 2072 Kartik - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/geared-loads-equivalent-inertia-dynamics - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2 ## Question A motor running at speed N rpm driving a rotational load L1 directly coupled to its shaft and another load L2 through a gear to reduce its speed by a factor K. The inertia of motor, loads L1, loads L2 are Jm, J1, J2 respectively. The load torque of L1, and L2 are T1 and T2. Find the expression for total inertia and T reflected to motor. If "Te" is the electrical torque, what will be the dynamic equation for the motor speed? ## Short answer Equivalent inertia is J = Jm + J1 + J2/K^2, and equivalent load torque is Tl = T1 + T2/K, or T2/(Kη) with gear efficiency. Dynamic equation: Te = Tl + J dωm/dt, where ωm = 2πN/60. ## Answer **Set-up.** Motor speed $\omega_m$ (rad/s, $\omega_m = 2\pi N/60$). Load $L_1$ directly coupled, so it runs at $\omega_m$. Load $L_2$ runs through a gear of ratio $K$, so $\omega_2 = \omega_m/K$. Gear assumed lossless (efficiency $\eta$ given in a remark). ```text J1, T1 gear 1:K J2, T2 [Motor]======[L1]======[G]=========[L2] Jm, Te w_m w_m / K ``` #### Equivalent inertia referred to motor shaft The kinetic energy must be the same: $$ \frac12 J\omega_m^2 = \frac12J_m\omega_m^2 + \frac12J_1\omega_m^2 + \frac12J_2\left(\frac{\omega_m}{K}\right)^2 $$ $$ J = J_m + J_1 + \frac{J_2}{K^2} $$ #### Equivalent load torque referred to motor shaft The power must be the same (lossless gear): $$ T_l\,\omega_m = T_1\omega_m + T_2\frac{\omega_m}{K} \;\Rightarrow\; T_l = T_1 + \frac{T_2}{K} $$ With gear efficiency $\eta$, $T_l = T_1 + \dfrac{T_2}{K\eta}$. So both inertia and torque of the low-speed load appear smaller at the motor shaft (by $K^2$ and $K$ respectively), which is why a geared high-speed motor is used for slow, heavy loads. #### Dynamic equation The net torque accelerates the total equivalent inertia: $$ T_e - T_l = J\frac{d\omega_m}{dt} $$ $$ T_e = \left(T_1 + \frac{T_2}{K}\right) + \left(J_m + J_1 + \frac{J_2}{K^2}\right)\frac{d\omega_m}{dt} $$ In terms of speed in rpm, $\dfrac{d\omega_m}{dt} = \dfrac{2\pi}{60}\dfrac{dN}{dt}$. - If $T_e > T_l$: the drive accelerates. - If $T_e < T_l$: the drive decelerates. - If $T_e = T_l$: steady speed (equilibrium). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the total inertia referred to the motor shaft? A. Jm + J1 + J2 B. Jm + J1 + J2/K^2 C. Jm + J1 + K^2 J2 D. Jm + J1 + J2/K Answer: B. Equating kinetic energy, L2 runs at wm/K, so its inertia reflects as J2/K^2. (Key point: Equivalent inertia) 2. What is the load torque referred to the motor shaft (lossless gear)? A. T1 + T2/K B. T1 + T2/K^2 C. T1 + K T2 D. T1 + T2 Answer: A. Equating power gives Tl wm = T1 wm + T2 wm/K, so Tl = T1 + T2/K. (Key point: Equivalent torque) 3. Which is the motor dynamic equation? A. Te = Tl - J dw/dt B. Te + Tl = J dw/dt C. Te - Tl = J dw/dt with J the equivalent inertia D. Te = J dw/dt only Answer: C. Net torque accelerates the total equivalent inertia. (Key point: Dynamic equation) 4. What happens when Te = Tl? A. The drive accelerates B. The drive decelerates C. Speed is steady D. The motor stops Answer: C. With zero net torque, dw/dt = 0 and speed is constant. 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