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  • Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass(d)

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)

best_answer

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}

Posted by

Devendra Khairwa

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