Question

a tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. to add an extra degree of randomness, the coordinates of the center of mass of the three dice are used as well. the masses of the three dice are 11.90 g, 13.50 g, and 17.50 g, and their respective coordinates after one particular throw are (0.1170 m, -0.3270 m), (-0.4970 m, 0.2010 m), and (-0.2130 m, -0.1250 m). what are the resulting coordinates of the center of mass of the dice, ( x_{cm} ) and ( y_{cm} )?

Explanation:

Step1: Calculate the total mass

The formula for the total mass \(M\) of the three - dice system is \(M = m_1 + m_2+m_3\).
Given \(m_1 = 11.90\space g\), \(m_2 = 13.50\space g\), \(m_3=17.50\space g\)
\(M=(11.90 + 13.50+17.50)\space g=42.9\space g\)

Step2: Calculate the \(x\) - coordinate of the center of mass

The formula for the \(x\) - coordinate of the center of mass \(x_{cm}=\frac{m_1x_1 + m_2x_2+m_3x_3}{M}\)
where \(x_1 = 0.1170\space m\), \(x_2=- 0.4970\space m\), \(x_3=-0.2130\space m\)
\(m_1x_1 + m_2x_2+m_3x_3=(11.90\times0.1170)+(13.50\times(- 0.4970))+(17.50\times(-0.2130))\)
\(=1.3923-6.7095 - 3.7275=-9.0447\space g\cdot m\)
\(x_{cm}=\frac{-9.0447}{42.9}\space m=- 0.211\space m\)

Step3: Calculate the \(y\) - coordinate of the center of mass

The formula for the \(y\) - coordinate of the center of mass \(y_{cm}=\frac{m_1y_1 + m_2y_2+m_3y_3}{M}\)
where \(y_1=-0.3270\space m\), \(y_2 = 0.2010\space m\), \(y_3=-0.1250\space m\)
\(m_1y_1 + m_2y_2+m_3y_3=(11.90\times(-0.3270))+(13.50\times0.2010)+(17.50\times(-0.1250))\)
\(=-3.9913+2.7135-2.1875=-3.4653\space g\cdot m\)
\(y_{cm}=\frac{-3.4653}{42.9}\space m=-0.0808\space m\)

Answer:

\(x_{cm}=-0.211\space m\), \(y_{cm}=-0.0808\space m\)