The Center of Mass — Question 3

Question 3

Removing mass from a plate

A uniform rectangular plate spans $0\le x\le6.0\,\mathrm m$ and $0\le y\le4.0\,\mathrm m$. A $2.0\,\mathrm m$ by $2.0\,\mathrm m$ square occupying $4\le x\le6$, $2\le y\le4$ is removed.

1. Find the center of mass of the remaining plate.

2. Explain the direction of its shift from the original center.

Question 3 - Solution

Removing mass from a plate

(a) Derivation and calculation

Use common areal density $\sigma$ and subtract the removed first moment: $$M=\sigma(24-4)=20\sigma,\qquad M\vec R=\sigma\left[24(3,2)-4(5,3)\right].$$ $$\boxed{\vec R=\left(\frac{72-20}{20},\frac{48-12}{20}\right) =(2.60,1.80)\ \mathrm m}.$$

(b) Result and interpretation

The shift relative to the original center is $$\boxed{\Delta\vec R=(2.60,1.80)-(3,2)=(-0.40,-0.20)\ \mathrm m}.$$ Subtracting upper-right material reduces both first moments. The negative-mass device is bookkeeping; all retained mass is positive.

Physical check

The shift points away from the center of the removed square.