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 11.10 g, 15.10 g, and 18.90 g, and their respective coordinates after one particular throw are (0.3150 m,−0.4990 m), (−0.4050 m,0.4850 m), and (−0.1150 m,−0.1850 m).
Requried:
What are the resulting coordinates of the center of mass of the dice, xcm and ycm?

Answer 1

Answer:

The resulting coordinate is   $(x,y) = (-0.170 \ m , -0.038 \ m )$

Explanation:

From the question we are told that  

     The  mass of the first dice is  $m_1 = 11.10 \ g$

      The  mass of the second dice is  $m_2 = 15.10 \ g$

     The mass of the third dice is  $m_3 = 18.90 \ g$

     The coordinate of the first dice is $(x_1, y_1) = (0.3150\ m , -0.4990 \ m )$

     The coordinate of the second dice is  $(x_2,y_2) = (-0.4050 \ m , 0.4850 \ m )$

      The coordinate of the third dice is  $(x_3 , y_3) = (-0.1150 \ m , -0.1850\ m )$

Generally the resulting coordinate of the center of mass of the dice in the x-axis is mathematically evaluated as

        $x\ cm = \frac{m_1 * x_1 + m_2 * x_2 + m_3 * x_3}{m_1 + m_2 +m_3 }$

i.e the summation of the moments about their x-axis divided by the magnitude of their masses

    substituting values

         $x\ cm = \frac{11.0 * 0.3150 + 15.10 * (-0.4050) + 18.90 * (-0.1150)}{11.100 + 15.10 +18.90 }$

        $x\ cm = -0.170 \ m$

Generally the resulting coordinate of the center of mass of the dice in the y-axis is mathematically evaluated as

     $y\ cm = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3}{m_1 + m_2 +m_3 }$

     $y\ cm = \frac{ 11.10 * (-0.4990) + (15.10) * (0.4850) + (18.90) * (-0.1850)}{ 11.10 + 15.10 +18.90 }$

     $y\ cm = -0.038 \ m$

Thus the resulting coordinate is   $(x,y) = (-0.170 \ m , -0.038 \ m )$