# Integration of center of mass cross position vector of non-rigid body

Suppose we have a coordinate frame with origin in the center of mass, on which each point mass of a continuum has a position vector $$\overrightarrow{r}$$, suppose with respect to an inertia frame origin, the center of mass of the continuum is $$\overrightarrow{R_c}$$. Now if we have the integration $$\int_B \overrightarrow{R_c} \times \dot{\overrightarrow{r}} dm$$,

for rigid body, $$\int_B \overrightarrow{R_c} \times \dot{\overrightarrow{r}} dm = \overrightarrow{R_c} \times \int_B \dot{\overrightarrow{r}} dm =0$$, according to the definition of center of mass, where $$B$$ is the volume of the continuum, $$dm$$ is an infinitesimal mass.

My question is, why can we take $$\overrightarrow{R_c}$$ out from the integration? $$\overrightarrow{R_c}$$ actually also depend on $$dm$$? Although we could take $$R_c = 1/M \int_B \overrightarrow{R}dm$$, where $$\overrightarrow{R}$$ is the position of $$dm$$ w.r.t. the origin of the inertia frame.

This seems to be true for rigid body but why? And for general continuum also?

$$m$$ is the dummy variable for integration and so the calculation can be done without ambiguity by using two dummy variables for the integration:
$$\vec{R_c}\equiv\frac{1}{M}\int_B\vec{R}\left(m\right)dm$$ and the integral of interest expressed as: $$\int_B\vec{R_c}\times\dot{\vec{r}}\left(m'\right)dm'=\int_B\left[\frac{1}{M}\int_B\vec{R}\left(m\right)dm\right]\times\dot{\vec{r}}\left(m'\right)dm'=\left[\frac{1}{M}\int_B\vec{R}\left(m\right)dm\right]\times\int_B\dot{\vec{r}}\left(m'\right)dm'=\vec{R_c}\times\int_B\dot{\vec{r}}\left(m'\right)dm'$$