Angular momentum and General Plan Motion

Question

Let us consider a general body, having both translation and rotational motion. Let O be the origin, and C be the center of mass of body. Total mass of body is $M$

Let us take small point mass $m_i$ in the body. It has position vector wrt to origin as $\vec r_i$. The position vector of center of mass wrt origin is $\vec r_{cm}$. Hence according to vector law of addition: $$\vec r_i=\vec r_{i,c}+\vec r_{cm}$$ where $\vec r_{i,c}$ is position vector of $m_i$ wrt to center of mass. Similarly we have: $$\vec v_i=\vec v_{i,c}+\vec v_{cm}$$ Let the angular velocity of body b $\vec \omega$ So we get angular momentum of body about O be: $$\vec L=\sum[\vec r_i\times(m_i\vec v_i)]$$ $$\vec L=\sum(\vec r_{i,c}+\vec r_{cm})\times m_i\sum(\vec v_{i,c}+\vec v_{cm})$$ $$\vec L=\sum(\vec r_{cm}\times M\vec v_{cm})+\vec r_{cm}\times\sum(m_i\vec v_{i,c})+\sum(m_i\vec r_{i,c})\times\vec v_{cm}+\sum m_i(\vec r_{i,c}\times \vec v_{i,c})$$ Here second and third terms becomes zero, and this is my doubt here, that why second and third terms becomes zero. Simplifying fourth term, we get $$\sum m_i(\vec r_{i,c}\times(\vec \omega\times \vec r_{i,c})) $$ $$\sum m_i r^2_{i,c} \vec \omega $$ $\rightarrow$$I_{cm}\vec \omega$

Hence we get: $$\vec L=\vec r_{cm}\times M\vec v_{cm}+I_{cm}\vec \omega$$ Please help to clarify my doubt.

Any suggestions are massively appreciated.

Answer 1

For the second term, note that $m_i\vec{v}_{i, c}$ is the momentum of the unit of mass relative to the center of mass.

All momenta must sum to zero in the center-of-mass frame.

Let $P$ be the total momentum, $V = v_{cm}, R = r_{cm}$.

$$ \vec{P} = \sum m_i \vec{v}_i = \sum m_i (\vec{v}_{i,c} + V) = MV + \sum m_i \frac{d\vec{r}_{i,c}}{dt}$$

But, by definition,

$$\vec{P} = M\vec{V} = M \frac{d}{dt}\vec{R} = \sum m_i \frac{d\vec{r}_{i}}{dt}$$

$$ = \sum m_i \frac{d\vec{r}_{i,c}}{dt} + \sum m_i \frac{d\vec{R}}{dt}$$

$$ = \sum m_i \frac{d\vec{r}_{i,c}}{dt} + \vec{P}$$

$$\Rightarrow \sum m_i \frac{d\vec{r}_{i,c}}{dt} = \vec{0} $$

For the third term, note that $\frac{1}{M} \sum m_i \vec{r}_{i,c}$ is the position of the center of mass relative to the center of mass, i.e. it's zero:

$\vec{R} = \frac{1}{M} \sum m_i \vec{r}_{i} = \frac{1}{M} \sum m_i \vec{r}_{i, c} + \vec{R}$

$\Rightarrow \sum m_i \vec{r}_{i} =\vec{0}$