# Q7.33     Separation of Motion of a system of particles into the motion of the centre of mass and motion about the               centre of mass :              (b) show$K=K^{'}+1/2MV^{2}$           where $K$ is the total kinetic energy of the system of particles, $K^{'}$is the total kinetic energy of the          system when the particle velocities are taken with respect to the centre of mass and $MV^{2}/2$  is the          kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the           system). The result has been used in Sec. 7.14.

From the first part we can write :

$\sum mv_i'\ =\ \sum mv_i\ -\ \sum mv$

or                                                     $\sum mv_i\ =\ \sum mv_i'\ +\ \sum mv$

Squaring both sides (in vector form taking dot products with itself), we get :

$\sum mv_i. \sum mv_i\ =\ \sum m(v_i'\ +\ v)\ .\ \sum m(v_i'\ +\ v)$

or                                                      $M^2 \sum v_i^2\ =\ M^2 \sum v_i'^2\ +\ M^2v^2$

or                                                      $\frac{1}{2}M \sum v_i^2\ =\ \frac{1}{2}M \sum v_i'^2\ +\ \frac{1}{2}Mv^2$

Hence                                                                 $K\ =\ K'\ +\ \frac{1}{2}Mv^2$

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