$\require{begingroup}\begingroup \newcommand{\bbeta}{\boldsymbol{\beta}} \newcommand{\bomega}{\boldsymbol{\omega}} \newcommand{\bnu}{\boldsymbol{\nu}} \newcommand{\p}{\boldsymbol{p}} \newcommand{\R}{\boldsymbol{R}} \newcommand{\x}{\boldsymbol{x}} \newcommand{\f}{\boldsymbol{f}} \newcommand{\v}{\boldsymbol{v}} \newcommand{\w}{\boldsymbol{w}} \newcommand{\y}{\boldsymbol{y}} \newcommand{\s}{\boldsymbol{s}} \newcommand{\g}{\boldsymbol{g}} \renewcommand{\u}{\boldsymbol{u}} %% OLD MACROS (by IROS 2016 Paper) \newcommand{\calS}{\mathcal{S}} \newcommand{\calW}{\mathcal{W}} \newcommand{\calA}{\mathcal{A}} \newcommand{\calB}{\mathcal{B}} \newcommand{\bcalB}{\boldsymbol{\mathcal{B}}} % \newcommand{\graph}{\mathcal{G}} \newcommand{\calV}{\mathcal{V}} \newcommand{\calE}{\mathcal{E}} \newcommand{\calG}{\mathcal{G}} \newcommand{\calK}{\mathcal{K}} \newcommand{\calN}{\mathcal{N}} \newcommand{\calO}{\mathcal{O}} $
I have a rigid body $i$ in 3D space ($SE(3)=\mathbb{R}^3\times SO(3)$). To this rigid body is associated a body frame $\calB$. In addition to that we can define a world-frame with the letter $\calW$.
$\p_i$ is the position of the center of gravity of the rigid body expressed in the world frame $\R_i$ is the rotation matrix which allows to go from $\calB$ to $\calW$. Therefore if I apply a velocity $\v_i$ in $\calB$ the new position of our body in $\calW$ will be:
$\p_i+\R_i\v_i\Delta t$
where $\Delta t$ is the amount of time for which I applied the velocity $\v_i$
If, I want to apply also an angular velocity $\w_i$ (expressed in $\calB$) to the rigid body $i$ the new rotation matrix associated to the object $i$ will be
$\exp_{\R_i}(\w_i \Delta t)$
Does this make sense for you?
$\endgroup$