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Question #e35ee

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Answer 1 · 2014-04-23T21:49:52

The moment of inertia is calculated from the sum

$I = m_{1} \times r_{1}^{2} + m_{2} \times r_{2}^{2} + m_{3} \times r_{3}^{2}$ $I=m_1timesr_1^2 + m_2timesr_2^2 + m_3timesr_3^2$

where $r_{i}$ $r_i$ is the distance of point mass $i$ $i$ from the center of mass of the three points. The square of each distance is calculated in Cartesian coordinates as

$r_{i}^{2} = {(x_{i} - x_{c m})}_{i}^{2} + {(y_{i} - y_{c m})}_{i}^{2}$ $r_i^2=(x_i-x_(cm))^2+(y_i-y_(cm))^2$

The center-of-mass coordinates ( $x_{c m}, y_{c m})$ $x_(cm),y_(cm))$ can be found from the simple formula below if you know the masses of the three points and their coordinates in any (x,y) plane.

$x_{c m} = \frac{m_{1} \times x_{1} + m_{2} \times x_{2} + m_{3} \times x_{3}}{m_{1} + m_{2} + m_{3}}$ $x_(cm) = (m_1 timesx_1 + m_2 times x_2 + m_3 times x_3)/(m_1+m_2+m_3$

$y_{c m} = \frac{m_{1} \times y_{1} + m_{2} \times y_{2} + m_{3} \times y_{3}}{m_{1} + m_{2} + m_{3}}$ $y_(cm) = (m_1 timesy_1 + m_2 times y_2 + m_3 times y_3)/(m_1+m_2+m_3$

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