The Center of Mass — Question 7

Question 7

An origin change cannot move a center

Point masses $1$, $2$ and $3\,\mathrm{kg}$ occupy $(0,0)$, $(3,0)$ and $(0,2)\,\mathrm m$. A second coordinate system has parallel axes and origin at $(1,-2)\,\mathrm m$ in the original frame.

1. Find the center in each coordinate system.

2. A student subtracts the origin displacement from the numerator before dividing by total mass. Identify the error.

Question 7 - Solution

An origin change cannot move a center

(a) Derivation and calculation

Compute the first moment in the original frame: $$M=1+2+3=6,\qquad M\vec R=1(0,0)+2(3,0)+3(0,2)=(6,6).$$ $$\boxed{\vec R=(1,1)\ \mathrm m},\qquad \boxed{\vec R^{\prime}=\vec R-(1,-2)=(0,3)\ \mathrm m}.$$

(b) Result and interpretation

Translate every particle before summing: $$\sum_i m_i\vec r_i^{\prime}=\sum_i m_i(\vec r_i-\vec a) =\sum_i m_i\vec r_i-M\vec a.$$ Subtract $M\vec a$ from the first moment, or subtract $\vec a$ after dividing by $M$. A position cannot be subtracted directly from a mass-times-position quantity.

Physical check

Changing origin changes coordinates but does not change the physical center location.