Dynamic model of a two-mass electric drive taking into account the Gear Ratio

Question

Two-mass rotational system has the following form and is represented in following structural diagram.

where $\tau_e$, $\omega_1$ and $J_m$ - motor torque, angular velocity and moment of inertia

$\tau_s$, $\tau_s$, $\omega_2$ and $J_d$ - shaft torque, load torque, angular velocity and load moment of inertia;

$K_{md}$ - shaft stiffness

Problem: how to include a gear ratio $N=\frac{\omega_1}{\omega_2}$ in equation of motion and in in a block diagram respectively?

I wrote down the Lagrangian for each mass in terms of the gear ratio, but in the usual case, I just compiled a system of differential equations according to the Lagrange equation, but I cannot understand how now using the TWO Lagrangian to get one block diagram with a gear ratio.

Answer 1

Assuming that the gear box is on the left end of the shaft (i.e. no flexible shaft between motor and gearbox).

1. The angular velocity on the left end of the gear box is $\omega_1$.
2. The angular velocity of the shaft side of the gear box is assumed as $\omega_1' = \frac{\omega_1}{N}$.
3. The angular velocity on the right end of the shaft is $\omega_2$. So the torque on the shaft is $\pm K_m (\frac{\phi_1}{N} - \phi_2)$. (sign to be checked).
4. Because of the way I described the gearbox, $\omega_1' < \omega_1 $. so the torque on the shaft when acting on the motor through the gearbox is $\frac{1}{N}$. This can be seen in the below derivation.
5. Since I have assumed that shaft is directly connected to the load, the torque in the shaft is made available 1:1. This can also be seen in below derivation.

(Below derivation to be verified independently by OP) $$ L = \frac{J_m \omega_1^2}{2} + \frac{J_d \omega_2^2}{2} + \frac{Km (\frac{1}{N} \phi_1 - \phi_2)^2}{2} $$

$$ \frac{d}{dt} \frac{\partial L}{\partial \omega_1} = \frac{d}{dt} J_m \omega_1 = J_m \frac{d \omega_1}{dt} $$

$$ \frac{\partial L}{\partial \phi_1} = \frac{K_m}{\color{red}{N}} (\frac{1}{N} \phi_1 - \phi_2) $$

Similarly for the other body also (exercise left to you).

$$ \frac{\partial L}{\partial \phi_2} = -K_m (\frac{1}{N} \phi_1 - \phi_2) $$

I have not considered the input torque. It can be added to this result.