Chapter 10, Problem 10.1P

To determine

Show that the result of ${\displaystyle \sum m_{\alpha }r{\text{'}}_{\alpha }=0}$.

Answer to Problem 10.1P

The result of ${\displaystyle \sum m_{\alpha }r{\text{'}}_{\alpha }=0}$ is shown here.

Explanation of Solution

Write the expression for the position of the CM ,

    $r_{\alpha }=R+r{\text{'}}_{\alpha }$        (I)

Here, $r_{\alpha }$ is the position of the CM, $R$ is the position relative to the origin O and $r_{\alpha }\text{'}$ is the position.

Rewrite the above equation,

    $r_{\alpha }\text{'}=r_{\alpha }-R$        (II)

Write the expression for the force,

    ${\displaystyle \sum _{\alpha }{m}_{\alpha }{r}_{\alpha }\text{'}}=0$        (III)

Here, $m_{\alpha }$ is the mass of the object.

Substitute (III) in (II),

    $\begin{array}{c}{\displaystyle \sum _{\alpha }{m}_{\alpha }\left(r_{\alpha }-R\right)}=0\\ {\displaystyle \sum _{\alpha }{m}_{\alpha }{r}_{\alpha }}-{\displaystyle \sum _{\alpha }{m}_{\alpha }R}=0\\ \frac{M}{M}{\displaystyle \sum _{\alpha }{m}_{\alpha }{r}_{\alpha }}-R{\displaystyle \sum _{\alpha }{m}_{\alpha }}=0\end{array}$        (IV)

Substitute $\frac{1}{M}{\displaystyle \sum _{\alpha }{m}_{\alpha }{r}_{\alpha }}$ for $R$ in the above equation,

    $\begin{array}{c}{\displaystyle \sum _{\alpha }{m}_{\alpha }{r}_{\alpha }\text{'}}=MR-RM\\ =0\end{array}$        (V)

Therefore, the result of ${\displaystyle \sum m_{\alpha }r{\text{'}}_{\alpha }=0}$ is shown here.