The Center of Mass — Question 6

Question 6

An adjustable balance interval

A fixed $8.0\,\mathrm{kg}$ payload has center $x=0.50\,\mathrm m$. A $2.0\,\mathrm{kg}$ ballast can slide on a light rail over $-1.0\le x\le2.0\,\mathrm m$.

1. Find the entire attainable range of the combined center.

2. Can the center be placed at $x=0$? If not, find the smallest ballast mass that can achieve it on the same rail.

Question 6 - Solution

An adjustable balance interval

(a) Derivation and calculation

The original ballast produces $$X(x)=\frac{8(0.50)+2x}{10}=0.40+0.20x.$$ Because $dX/dx>0$, the rail endpoints give $$\boxed{0.20\le X\le0.80\ \mathrm m}.$$

(b) Result and interpretation

For a new ballast mass $m$, the target $X=0$ requires $$4+mx=0\Rightarrow x=-\frac4m.$$ The most negative permitted location is $x=-1$ m, so $$\boxed{m_{\min}=4.0\ \mathrm{kg}\quad\text{at}\quad x=-1.0\ \mathrm m}.$$ The original 2 kg ballast cannot supply a sufficiently negative first moment.

Physical check

A heavier ballast can reach zero from a less negative location, consistent with $x=-4/m$.