Discussion
Let \(L_{x}, L_{y}, L_{z}\) be the components of angular momentum of a rigid body with one point fixed. If \(L_{x}\) is of the form
\[ L_{x} = I_{xx}W_{x} + I_{xy}W_{y} + I_{xz}W_{z}, \]
where \(I_{xx}\) is the moment of inertia about x-axis, then :
Answer: C
The moment of inertia about the x-axis is \(I_{xx} = \sum_{i} m_{i}(y_{i}^{2} + z_{i}^{2})\).
The moment of inertia about an axis uses the squared distance of each particle from that axis. The distance of \((x_{i}, y_{i}, z_{i})\) from the x-axis is \(\sqrt{y_{i}^{2} + z_{i}^{2}}\), which gives option C.
\(\sum m_{i}(x_{i}^{2} + y_{i}^{2})\) is \(I_{zz}\) and \(\sum m_{i}(x_{i}^{2} + z_{i}^{2})\) is \(I_{yy}\). A negative sum like \(-\sum m_{i} x_{i}^{2}\) resembles a product of inertia such as \(I_{xy} = -\sum m_{i} x_{i}y_{i}\), which carries the minus sign in this convention, but a moment of inertia is never negative.
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