Let \(\bfp_i\) denote the position in the world frame of a reference point on the robot's \(i^{\textrm{th}}\) link. The center of mass (CoM) \(G\) is located at position \(\bfp_G\) in the world frame. Also, let \(m_i\) denote the mass of link \(i\), and \(m = \sum_i m_i\) the total mass of the robot.

The robot is subject to gravity and contact forces. Let \(\bfg\) denote the gravity vector. For a link \(i\) in contact with the environment, we write \(\bff_i\) the resultant force exerted on the link. Let \(\bfh_{ij}\) denote the internal force exerted by link \(i\) on link \(j\). We take the convention that \(\bfh_{ij} = 0\) if links \(i\) and \(j\) are not connected (similarly, \(\bff_i = 0\) if link \(i\) is not in contact). All force vectors are expressed in the world frame.

Equation

Newton's equation of motion links the resultant accelerations and forces:

\begin{equation*} \sum_{\textrm{link } i} m_i \bfpdd_i = \sum_{\textrm{link } i} m_i \bfg + \bff_i + \sum_{\textrm{link } j

eq i} \bfh_{ij} \end{equation*}

An interesting point here is that \(\bfh_{ij} = - \bfh_{ji}\) (Newton's third law of motion), so that internal forces vanish in this sum over all links: \(\sum_i \sum_{j

eq i} \bfh_{ij} = 0\). In concise form, Newton's equation then binds the acceleration of the center of mass (CoM) with the whole-body resultant force:

The linear momentum of the robot is defined by:

\begin{equation*} \bfP_G := m \bfpd_G \end{equation*}

Then, Newton's equation can be written in concise form as:

\begin{equation*} \dot{\bfP}_G = m \bfpdd_G = m \bfg + \sum_{\textrm{contact } i} \bff_i \end{equation*}

In other words, the rate of change of the linear momentum is equal to the resultant of external forces exerted on the robot.