Homework 1 Problem 2

Task a. torque dynamics

$$\begin{align*} M\ddot{q} + C\dot{q} + G(q) + K(q - \theta) &= 0 \quad \text{(1) link dynamic} \\ J_m\ddot{\theta} + B_m\dot{\theta} - K(q - \theta) &= \tau_m \quad \text{(2) motor dynamic} \end{align*}$$

The transmitted elastic torque:

$$\tau_c = K(\theta - q)$$

So we simplify this system to:

$$\ddot{q} = M^{-1}(\tau_c - C\dot{q} - G(q))$$ $$\ddot{\theta} = J_m^{-1}(\tau_m - \tau_c - B_m\dot{\theta} )$$

Compute the second derivative of $\tau_c$:

$\dot{\tau}_c = K(\dot{\theta} - \dot{q}) \quad$ $\ddot{\tau}_c = K(\ddot{\theta} - \ddot{q})$

$$\ddot{\tau}_c = K \left[ J_m^{-1}(\tau_m - \tau_c - B_m\dot{\theta}) - M^{-1}(\tau_c - C\dot{q} - G(q)) \right]$$

link-side motion appears in the torque dynamics:

$$\ddot{\tau}_c + K(J_m^{-1} + M^{-1})\tau_c = K J_m^{-1}\tau_m - K J_m^{-1}B_m\dot{\theta} + K M^{-1}(C\dot{q} + G(q))$$

Task b. The influence of stiffness K from motor torque $\tau_m$ to elastic torque $\tau_c$

1. $K \rightarrow \infty$

$$\ddot{\tau}_c + K(J_m^{-1} + M^{-1})\tau_c = K J_m^{-1}\tau_m - K J_m^{-1}B_m\dot{\theta} + K M^{-1}(C\dot{q} + G(q))$$

so the function becomes:

$$(J_m^{-1} + M^{-1})\tau_c = J_m^{-1}\tau_m - J_m^{-1}B_m\dot{\theta} + M^{-1}(C\dot{q} + G(q))$$

at the same time, as there is no deformation, the movement of the motor and the link is completely synchronised $\theta = q, \dot{\theta} = \dot{q}, \ddot{\theta} = \ddot{q}$.

$$(M\ddot{q} + J_m\ddot{\theta}) + C\dot{q} + B_m\dot{\theta} + G(q) = \tau_m$$ $$\mathbf{(M + J_m)\ddot{q} + (C + B_m)\dot{q} + G(q) = \tau_m}$$

2. $K \rightarrow 0$

from the torque dynamics above, $K \rightarrow 0$, then

$$\ddot{\tau}_c = 0 \implies \tau_c = 0$$

From the :

$$\begin{align*} M\ddot{q} + C\dot{q} + G(q) &= 0 \quad \text{link side}\\ J_m\ddot{\theta} + B_m\dot{\theta} &= \tau_m \quad \text{motor side} \end{align*}$$

The motor and the connecting rod are completely decoupled. The torque cannot be transmitted to the connecting link side $(\tau_c = 0)$ because the intermediate medium is too soft. The connecting link will simply drop straight down under its own weight, whilst the motor spins idly at the other end.