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) showK=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.

Answers (1)

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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