Equivalent inertia and torque of geared loads

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).

         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).