Question

The three masses shown in FIGURE EX12.7 are connected by massless, rigid rods. What are the coordinates of the center of mass?

Short answer

The coordinates of the center of mass are$(8\text{cm},5\text{cm})$

Step-by-step solution

Step 1. Given information

Let us assume that the mass $300\text{g}$be at origin.

The co-ordinates of ball A $200\text{g}$is $(0,0)$

The co-ordinates of ball B $300\text{g}$is$(12\text{\hspace{0.17em}cm\hspace{0.17em}},10\text{\hspace{0.17em}cm})$

The co-ordinates of ball C$100\text{g}$ is$(12\text{cm},0)$

Step 2. Explanation

The x - xoordinate of center of mass is

$\begin{array}{l}{x}_{cm}=\frac{m_A{x}_A+m_B{x}_B+m_C{x}_C}{m_A+m_B+m_C}\\ x_{\text{cm}}=\frac{(200\text{g})(0\text{cm})+(300\text{g})(12\text{cm})+(100\text{g})(12\text{cm})}{(200\text{g})+(300\text{g})+(100\text{g})}\\ x_{\text{cm}}=8\text{cm}\end{array}$

The y- coordinate of center of mass is

$\begin{array}{l}{y}_{cm}=\frac{m_A{y}_A+m_B{y}_B+m_C{y}_C}{m_A+m_B+m_C}\\ y_{\text{cm}}=\frac{(200\text{g})(0\text{cm})+(300\text{g})(10\text{cm})+(100\text{g})(0\text{cm})}{(200\text{g})+(300\text{g})+(100\text{g})}\\ y_{\text{cm}}=5\text{cm}\end{array}$

Therefore, the coordinates of the center of mass are$(8\text{cm},5\text{cm})$