Question

a different rod with the same mass per unit length, 470 g/m. is bent into an asymmetric z - shape, as shown in the figure. the lengths of the segments are ( a = 1.4 , m ), ( b = 0.65 , m ), and ( c = 0.45 , m ). determine the position ( vec{r}_{cm} ) of the z - rods center of mass. enter your answer using ij unit vector notation. round numerical values to two significant figures. ( vec{r}_{cm}= )

Explanation:

Step1: Calculate the mass of each segment

Let the mass per unit length be \(\lambda = 470\space g/m=0.47\space kg/m\).
The mass of segment \(a\), \(m_a=\lambda a = 0.47\times1.4 = 0.658\space kg\).
The mass of segment \(b\), \(m_b=\lambda b=0.47\times0.65 = 0.3055\space kg\).
The mass of segment \(c\), \(m_c=\lambda c = 0.47\times0.45=0.2115\space kg\).

Step2: Find the coordinates of the center - of - mass of each segment

Assume the left - most end of the vertical part (not shown in the problem's coordinate setup, but we can set a coordinate system where the intersection of the axes is a reference).
For segment \(a\): Let's assume a coordinate system. If we consider the \(x - y\) coordinate system, for segment \(a\) (horizontal segment above), if we assume the left - hand end of \(a\) is at \((x_1,y_1)\), and since it is a horizontal segment, \(y_1\) is non - zero (say \(y = 1.4/2\) in terms of its own length's center - of - mass in a proper coordinate setup. But if we assume the vertical axis is the \(y\) - axis and the horizontal axis is the \(x\) - axis.
The center - of - mass of segment \(a\): \(x_a=\frac{a}{2}\), \(y_a = b+\frac{a}{2}\) (assuming the vertical drop from the top of the \(Z\) to the middle of \(a\) is \(b\) plus half of \(a\)'s vertical position relative to the intersection of the axes).
The center - of - mass of segment \(b\): \(x_b=a\), \(y_b=\frac{b}{2}\)
The center - of - mass of segment \(c\): \(x_c=a + b+\frac{c}{2}\), \(y_c = 0\)

Step3: Use the center - of - mass formula \(\vec{r}_{cm}=\frac{\sum_{i = 1}^{n}m_i\vec{r}_i}{\sum_{i = 1}^{n}m_i}\)

The \(x\) - coordinate of the center of mass:

$$
LATEXBLOCK0
$$

The \(y\) - coordinate of the center of mass:

$$
LATEXBLOCK1
$$

Answer:

\(\vec{r}_{cm}=(1.2\hat{i}+0.84\hat{j})\space m\)