Question

Separating motion of a system of particles into motion of the centre of mass and motion about the centre of mass, Show that dL'/dt = Σ ri' x dp'/dt. Further show that dL'/dt = ζext where ζext 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 Newton's third law. Assume the internal forces between any two particles act along the line joining the particles]

Answer 1

We have , $L'=\Sigma r'_i\times P'_i$

Differentiating both sides with respect to time,
$\frac{dL'}{dt}=\frac{d}{dt}\left(\Sigma r'_i\times p'_i\right)\\\\=p'_i\times\Sigma\frac{dr'_i}{dt}+\Sigma r'_i\times\frac{dp'_i}{dt}\\\\=\Sigma m_ir'_i\times v'_i+\Sigma r'_i\times\frac{dp'_i}{dt}$

Where $r'_i$ is the position vector with respect to centre of mass of system of particles.

But from definition of centre of mass,
$\Sigma m_ir'_i=0$

so, $\frac{dL'}{dt}=\Sigma r'_i\times\frac{dp'_i}{dt}$ [ hence, proved]

We know, $\bf{\tau=r\times\frac{dv}{dt}}$

So, $\tau_{ext}=\Sigma r'_i\times\frac{dp'_i}{dt}$

And hence, $\frac{dL'}{dt}=\tau_{ext}$

Answer 2

Answer:

$We have , L'=\Sigma r'_i\times P'_iL′=Σri′×Pi′ </p><p></p><p>Differentiating both sides with respect to time, </p><p>\begin{gathered}\frac{dL'}{dt}=\frac{d}{dt}\left(\Sigma r'_i\times p'_i\right)\\\\=p'_i\times\Sigma\frac{dr'_i}{dt}+\Sigma r'_i\times\frac{dp'_i}{dt}\\\\=\Sigma m_ir'_i\times v'_i+\Sigma r'_i\times\frac{dp'_i}{dt}\end{gathered}dtdL′=dtd(Σri′×pi′)=pi′×Σdtdri′+Σri′×dtdpi′=Σmiri′×vi′+Σri′×dtdpi′ </p><p></p><p>Where r'_iri′ is the position vector with respect to centre of mass of system of particles. </p><p></p><p>But from definition of centre of mass, </p><p>\Sigma m_ir'_i=0Σmiri′=0 </p><p></p><p>so, \frac{dL'}{dt}=\Sigma r'_i\times\frac{dp'_i}{dt}dtdL′=Σri′×dtdpi′ [ hence, proved]</p><p></p><p>We know, \bf{\tau=r\times\frac{dv}{dt}}τ=r×dtdv </p><p></p><p>So, \tau_{ext}=\Sigma r'_i\times\frac{dp'_i}{dt}τext=Σri′×dtdpi′  \\ \\ </p><p></p><p>And hence, \frac{dL'}{dt}=\tau_{ext}dtdL′=τext</p><p>$