Dynamics of a robot manipulator with a force acting on the end-effector

Question

I have this exercise

The dynamic model is

$B(q)\ddot{q}+S(q,\dot{q})\dot{q}=\tau+\tau_k$

Where $B(q)$ is the inertia matrix, $S(q,\dot{q})\dot{q}$ are the centrifugal and Coriolis terms, $\tau$ is the actuator input and $\tau_k=J^T(q)F$ is the torque imposed by the external force.

So since at rest $\ddot{p}=J(q)\ddot{q}$, $S(q,\dot{q})\dot{q}=0$ and we have no actuator input, substituting $\ddot{q}$ from the dynamic model we have

$\ddot{p}=J(q)B^{-1}(q)J^T(q)F$

I assume that for the point 1, for the second principle of the dynamics, the end-effector accelerates as in the case B. Now, in the point 3, if we have non-uniform masses of the links, the second principle of the dynamics is still valid? I guess I can choose an inertia matrix $B(q)$ with proper centers of mass of the three links such that at the rest configuration $q_{eq}$

$\ddot{p}=J(q_{eq})B^{-1}(q_{eq})J^T(q_{eq})F=\begin{bmatrix} p_1 \\ 0 \\ 0\end{bmatrix}$

and so the end effector accelerates as the case D right?

Answer 1

If you consider a null velocity when you apply the force, then you cancel out the terms in your dynamics that are multiplied by the velocity, hence you can consider $S(q,\dot{q})q$ null and $\ddot{p} = J q$ as $\dot{J}\dot{q}$ is null leading to your expression of the instantaneous acceleration.

If you rewrite it in the form $\dot{p} = A f$ (prefer capital for matrix and lowercase for vector if you can), then what you want to prove is that $A \in \mathbb{R}^{3 \times 3}$ is such that it maps a vector of the form $\begin{bmatrix} 0 \\ f_2 \\ 0 \end{bmatrix}$ to an acceleration $\begin{bmatrix} \ddot{p}_1 \\ 0 \\ 0 \end{bmatrix}$. You known the analytical structure of A from the Jacobian and Inertia matrices, so you can see if it is possible to get a second column of A of the desired form.