The Center of Mass — Question 2

Question 2

A center outside the material

A uniform thin wire consists of two straight arms joined at a right angle: a $4.0\,\mathrm m$ arm along $+x$ and a $2.0\,\mathrm m$ arm along $+y$. Its linear density is constant.

1. Find the center of mass.

2. Explain why the result can be in empty space and why averaging the three endpoints is invalid.

Question 2 - Solution

A center outside the material

(a) Derivation and calculation

Replace each uniform arm by its mass at its own midpoint: $$m_x=4\lambda,\quad\vec r_x=(2,0),\qquad m_y=2\lambda,\quad\vec r_y=(0,1).$$ $$\vec R=\frac{4\lambda(2,0)+2\lambda(0,1)}{6\lambda} =\boxed{\left(\frac43,\frac13\right)\ \mathrm m}.$$

(b) Result and interpretation

For the distributed wire, the defining expression is $$\vec R=\frac{\int\vec r\,\lambda\,ds}{\int\lambda\,ds}.$$ This weighted average can lie in empty space between the arms. Averaging endpoints instead would impose three fictitious point masses and would not reproduce either arm’s first moment.

Physical check

The center lies inside the convex region spanned by the wire even though it is not on either arm.