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?
$L=V-P=J_m \frac{\omega_1^2}{2}+J_d \frac{\omega_2^2}{2}-\frac{K_{md}(\phi_1-\phi_2)^2}{2}$
$V$ – kinetic, and $P$ – potential energy
Here is the Lagrangian for the entire system. And I don’t understand how to insert the gear ratio here?
Best Answer
Directly plug in constant ratio between gear rotation angles which is the same ratio between their time derivative angular velocities. If
$$N=\dfrac{{\phi_1}}{{\phi_2}}=\dfrac{\dot{\phi_1}}{\dot{\phi_2}}=\dfrac{\omega_1}{\omega_2}$$
then the action is:
$$L=V-P=\frac{\omega_2^2}{2} \left(J_m N^2 +J_d \right)-\frac{K_{md}}{2}\;(N-1)^2\;\phi_2^2.$$