7.33     Separation of Motion of a system of particles into motion of the centre of mass and motion about the
           centre of mass :

        (d) Show \frac{dL^{'}}{dt}=\sum r_{i}^{'}\times \frac{dp^{'}}{dt}
             Further, show that

             \frac{dL^{'}}{dt}=\tau ^{'}_{ext}
          where \tau ^{'}_{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.)


Answers (1)

Since we know that :

                                           L'\ =\ \sum r_i' \times p_i'

Differentiating the equation with respect to time, we obtain :

                                           \frac{dL'}{dt}\ =\ d\frac {\left ( \sum r_i' \times p_i' \right )}{dt}

or                                                  =\ \frac{d\left ( \sum m_ir_i' \right )}{dt}\times v_i'\ +\ \sum r_i'\ \times \frac{d}{dt}p_i'

or                                          \frac {dL'}{dt} =\ \sum r_i'\ \times m_i \frac{d}{dt}v_i'                                                                      \left ( \because \sum m_ir_i\ =\ 0 \right )

Now using Newton's law of motion we can write : 

                                             \sum r_i'\ \times m_i \frac{d}{dt}v_i'\ =\ \tau _{ext}'

Thus                                         \frac{dL^{'}}{dt}=\tau ^{'}_{ext}