Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass: Show $\frac{\text{dL}}{\text{dt}}=\sum\text{r}'_\text{i}\times\frac{\text{dp}'}{\text{dt}}$ Further, show that $\frac{\text{dL}'}{\text{dt}}=\tau'_\text{ext}$ where $\tau'_\text{ext}$ is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles)
(Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles)
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We have the relation, $\text{L}'=\sum\limits_\text{i}\text{r}'_\text{i}\times\text{p}'_\text{i }$
$\frac{\text{dL}'}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i }\text{r}'_\text{i}\times\text{p}'_\text{i}\Big)$
$=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i }\text{r}'_\text{i}\Big)\times\text{p}'_\text{i }+\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$
$=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}\Big)\times\text{v}'_\text{i}+\sum\limits\limits\text{r}'_\text{i} \times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$ Where, $r'_i$ is the position vector with respect to the centre of mass of the system of particles. $\therefore\ \sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}=0$
$\therefore\ \frac{\text{dL}'}{\text{dt}}=\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$ We have the relation, $\frac{\text{dL}'}{\text{dt}}=\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$
$=\sum_\limits{\text{i}}\text{r}'_\text{i}\times\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ Where, $\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ is the rate of change of velocity of the $i^{th}$^ particle with respect ot the centre of mass of the system. Therefore, according to Newton's third law of motion, we can write, $\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ = Extrenal force acting on the ith particle $=\sum\limits_\text{i}(\tau'_\text{i})_\text{ext}$ i.e., $\sum\limits_\text{i}\text{r}'\times\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})=\tau'_\text{ext}=$ External torque acting on the system as a whole, $\therefore\ \frac{\text{dL}'}{\text{dt}}=\tau'_\text{ext}$
$\frac{\text{dL}'}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i }\text{r}'_\text{i}\times\text{p}'_\text{i}\Big)$
$=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i }\text{r}'_\text{i}\Big)\times\text{p}'_\text{i }+\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$
$=\frac{\text{d}}{\text{dt}}\Big(\sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}\Big)\times\text{v}'_\text{i}+\sum\limits\limits\text{r}'_\text{i} \times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$ Where, $r'_i$ is the position vector with respect to the centre of mass of the system of particles. $\therefore\ \sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}=0$
$\therefore\ \frac{\text{dL}'}{\text{dt}}=\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$ We have the relation, $\frac{\text{dL}'}{\text{dt}}=\sum\limits_\text{i}\text{r}'_\text{i}\times\frac{\text{d}}{\text{dt}}(\text{p}'_\text{i})$
$=\sum_\limits{\text{i}}\text{r}'_\text{i}\times\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ Where, $\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ is the rate of change of velocity of the $i^{th}$^ particle with respect ot the centre of mass of the system. Therefore, according to Newton's third law of motion, we can write, $\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})$ = Extrenal force acting on the ith particle $=\sum\limits_\text{i}(\tau'_\text{i})_\text{ext}$ i.e., $\sum\limits_\text{i}\text{r}'\times\text{m}_\text{i}\frac{\text{d}}{\text{dt}}(\text{v}'_\text{i})=\tau'_\text{ext}=$ External torque acting on the system as a whole, $\therefore\ \frac{\text{dL}'}{\text{dt}}=\tau'_\text{ext}$
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