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?


1 Answer 1


$m$ 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: $$\int_B\vec{R_c}\times\dot{\vec{r}}\left(m'\right)dm'=\int_B\left[\frac{1}{M}\int_B\vec{R}\left(m\right)dm\right]\times\dot{\vec{r}}\left(m'\right)dm'=\left[\frac{1}{M}\int_B\vec{R}\left(m\right)dm\right]\times\int_B\dot{\vec{r}}\left(m'\right)dm'=\vec{R_c}\times\int_B\dot{\vec{r}}\left(m'\right)dm'$$


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