Really having trouble with these types of questions so if someone would be able to explain in the simplest way possible that would be amazing.
ok so this is the example I have (but all the working has been missed out):
- rigid body, triangular shape
- consists of 3 pt particles of mass $m_1=m_2=m_3=m$
- relative positions of the particles are kept fixed by massless rigid rods
- body fixed coordinate system: the 3 particles have positions $x_1=(0,b,0)$, $x_2=(a,-b,0)$ and $x_3=(-a,-b,0)$.
Calculate the inertial tensor.
My vector calc is a bit rusty so a simple explanation of things would really be appreciated!
For a set of point-like masses $m_k$ at locations ${\bf r}_k$ The inertia tensor is defined as
$$ {\bf I} = \sum_k m_k ({\bf r}_k^T{\bf r}_k \Bbb{1} - {\bf r}_k{\bf r}_k^T) \tag{1} $$
where
$$ {\bf r}_k = \left(\begin{array}{c}x_k \\ y_k \\ z_k \end{array}\right) ~~~\mbox{and}~~~ {\bf r}_k^T = (x_k ~~ y_k ~~ z_k) \tag{2} $$
and $\Bbb{1}$ is the $3\times 3$ identity matrix. Replacing (2) in (1) we get
\begin{eqnarray} {\bf I} &=& \sum_k m_k\left\{(x_k ~~ y_k ~~ z_k) \left(\begin{array}{c}x_k \\ y_k \\ z_k \end{array}\right) \left(\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) - \left(\begin{array}{c}x_k \\ y_k \\ z_k \end{array}\right)(x_k ~~ y_k ~~ z_k) \right\} \\ &=& \sum m_k \left\{(x_k^2 + y_k^2 + z_k^2) \left(\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) - \left( \begin{array}{ccc} x_k^2 & x_k y_k & x_k z_k \\ x_k y_k & y_k^2 & y_k z_k \\ x_k z_k & y_k z_k & z_k^2 \end{array} \right) \right\} \\ &=& \sum_k m_k \left\{ \left( \begin{array}{ccc} x_k^2+y_k^2+z_k^2 & 0 & 0 \\ 0 & x_k^2+y_k^2+z_k^2 & 0 \\ 0 & 0 & x_k^2+y_k^2+z_k^2 \end{array} \right) - \left( \begin{array}{ccc} x_k^2 & x_k y_k & x_k z_k \\ x_k y_k & y_k^2 & y_k z_k \\ x_k z_k & y_k z_k & z_k^2 \end{array} \right)\right\} \\ &=& \sum_k m_k \left( \begin{array}{ccc} y_k^2+z_k^2 & -x_k y_k & -x_k z_k \\ -x_k y_k & x_k^2+z_k^2 & -y_k z_k \\ -x_k z_k & -y_k z_k & x_k^2+y_k^2 \end{array} \right) \tag{3} \end{eqnarray}
So, you only need to evaluate Eq. (3) with the values you have, the result is
$$ \bf{I} = \left( \begin{array}{ccc} 3 mb^2 & 0 & 0 \\ 0 & 2 m a^2 & 0 \\ 0 & 0 & 2 m a^2+3 mb^2 \end{array} \right) $$