The Center of Mass — Question 9

Question 9

Moving one load without moving the center

Two loads of $3.0$ and $5.0\,\mathrm{kg}$ are rearranged on a fixed light platform. The first moves by $(0.40,-0.20)\,\mathrm m$. No mass is added or removed.

1. Find the second displacement that keeps the combined center unchanged.

2. If the second load can move only horizontally, find the smallest possible center displacement magnitude.

Question 9 - Solution

Moving one load without moving the center

(a) Derivation and calculation

At fixed total mass, zero center displacement requires zero change in first moment: $$8\Delta\vec R=3\Delta\vec r_1+5\Delta\vec r_2=\vec0.$$ $$\boxed{\Delta\vec r_2=-\frac35(0.40,-0.20)=(-0.24,0.12)\ \mathrm m}.$$

(b) Result and interpretation

For horizontal-only motion of load 2, write $\Delta\vec r_2=(d,0)$: $$\Delta\vec R=\left(\frac{1.20+5d}{8},-\frac{0.60}{8}\right).$$ $$|\Delta\vec R|^2=\left(\frac{1.20+5d}{8}\right)^2+0.075^2 \Rightarrow\boxed{d=-0.24\ \mathrm m,\quad |\Delta\vec R|_{\min}=0.075\ \mathrm m}.$$

Physical check

The permitted horizontal adjustment cancels the horizontal moment but cannot cancel the vertical one.