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

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?

• I would highly suggest to type in your answer to point 2 in order for the reader to get used to your notations. Additionally for the point 3, the principles of dynamics hold if the mass distribution is not uniform. But it a bit odd to have a non null acceleration at rest, more over I thing that you forgot a Jacobian inversion in your formula. – N. Staub Oct 23 '17 at 15:33
• If an extenal force is applied at rest, we have an instantaneous acceleration and we can compute its components neglecting the velocity contribution I guess – B.junior Oct 23 '17 at 16:25
• A small Google search shows us, that the question is from an robotics exam, held by the DLR in 2004, source The robot in the image is the Justin-robot, which is a high-developed telerobotics project which includes a dexterous hand and a high-level-pathplanner. Without knowing the software (which is proprietary and not documented in Google Scholar) it is impossible to answer the question in detail. So perhaps, the OP can give us detailed information about Justin so that we can answer the question? – Manuel Rodriguez Feb 21 '18 at 14:55

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.