Question

a 1.5 m thin brass rod is bent in half to form a right - angled l,
as shown in figure 1. the corner of the l is located at the
origin. the rod has a mass per unit length of 470 g/m.
determine the position \\( \vec { r } _ { \mathrm { cm } } \\) of the l - rods center of mass.
enter your answer using ij unit vector notation. round
numerical values to two significant figures.
\\( \vec { r } _ { \mathrm { cm } } = \\) m

Explanation:

Step1: Calculate the length of each segment

The total length of the rod is \(L = 1.5\space m\). Since it is bent into two equal - length segments at a right - angle, the length of each segment \(l=\frac{1.5}{2}=0.75\space m\)

Step2: Find the mass of each segment

The mass per unit length \(\lambda=470\space g/m = 0.47\space kg/m\). The mass of each segment \(m=\lambda l\). So \(m = 0.47\times0.75=0.3525\space kg\)

Step3: Determine the position of the center of mass of each segment

For the horizontal segment: Let the rod be along the \(x\) - axis. The center of mass of the horizontal segment \(\vec{r}_1=(0.375\hat{i}+0\hat{j})\space m\) (since the center of mass of a uniform rod of length \(l\) is at \(\frac{l}{2}\) from the end). For the vertical segment: Let the rod be along the \(y\) - axis. The center of mass of the vertical segment \(\vec{r}_2=(0\hat{i}+0.375\hat{j})\space m\)

Step4: Use the formula for the center of mass \(\vec{r}_{cm}=\frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}\)

Since \(m_1 = m_2=m\), \(\vec{r}_{cm}=\frac{m(0.375\hat{i}+0\hat{j})+m(0\hat{i}+0.375\hat{j})}{m + m}\)

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Answer:

\(\vec{r}_{cm}=(0.19\hat{i}+0.19\hat{j})\space m\)