Moment of inertia
Scalar moment of inertia in a plane

The scalar moment of inertia, I L {\displaystyle I_{L}}, of a body about a specified axis whose direction is specified by the unit vector k ^ {\displaystyle \mathbf {\hat {k}} } and passes through the body at a point R {\displaystyle \mathbf {R} } is as follows: I L = k ^ ⋅ (− ∑ i = 1 N m i 2) k ^ = k ^ ⋅ I R k ^ = k ^ T I R k ^, {\displaystyle I_{L}=\mathbf {\hat {k}} \cdot \left(-\sum _{i=1}^{N}m_{i}\left^{2}\right)\mathbf {\hat {k}} =\mathbf {\hat {k}} \cdot \mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} =\mathbf {\hat {k}} ^{\mathsf {T}}\mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}},} where I R {\displaystyle \mathbf {I_{R}} } is the moment of inertia matrix of the system relative to the reference point R {\displaystyle \mathbf {R} }, and {\displaystyle } is the skew symmetric matrix obtained from the vector Δ r i = r i − R {\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} _{i}-\mathbf {R} }.

This is derived as follows. Let a rigid assembly of n {\displaystyle n} particles, P i, i = 1, …, n {\displaystyle P_{i},i=1,\dots,n}, have coordinates r i {\displaystyle \mathbf {r} _{i}}. Choose R {\displaystyle \mathbf {R} } as a reference point and compute the moment of inertia around a line L defined by the unit vector k ^ {\displaystyle \mathbf {\hat {k}} } through the reference point R {\displaystyle \mathbf {R} }, L (t) = R + t k ^ {\displaystyle \mathbf {L} (t)=\mathbf {R} +t\mathbf {\hat {k}} }. The perpendicular vector from this line to the particle P i {\displaystyle P_{i}} is obtained from Δ r i {\displaystyle \Delta \mathbf {r} _{i}} by removing the component that projects onto k ^ {\displaystyle \mathbf {\hat {k}} }. Δ r i ⊥ = Δ r i − (k ^ ⋅ Δ r i) k ^ = (E − k ^ k ^ T) Δ r i, {\displaystyle \Delta \mathbf {r} _{i}^{\perp }=\Delta \mathbf {r} _{i}-\left(\mathbf {\hat {k}} \cdot \Delta \mathbf {r} _{i}\right)\mathbf {\hat {k}} =\left(\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}\right)\Delta \mathbf {r} _{i},} where E {\displaystyle \mathbf {E} } is the identity matrix, so as to avoid confusion with the inertia matrix, and k ^ k ^ T {\displaystyle \mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}} is the outer product matrix formed from the unit vector k ^ {\displaystyle \mathbf {\hat {k}} } along the line L {\displaystyle L}.

To relate this scalar moment of inertia to the inertia matrix of the body, introduce the skew-symmetric matrix {\displaystyle \left} such that y = k ^ × y {\displaystyle \left\mathbf {y} =\mathbf {\hat {k}} \times \mathbf {y} }, then we have the identity − 2 ≡ | k ^ | 2 (E − k ^ k ^ T) = E − k ^ k ^ T, {\displaystyle -\left^{2}\equiv \left|\mathbf {\hat {k}} \right|^{2}\left(\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}}\right)=\mathbf {E} -\mathbf {\hat {k}} \mathbf {\hat {k}} ^{\mathsf {T}},} noting that k ^ {\displaystyle \mathbf {\hat {k}} } is a unit vector.

The magnitude squared of the perpendicular vector is | Δ r i ⊥ | 2 = (− 2 Δ r i) ⋅ (− 2 Δ r i) = (k ^ × (k ^ × Δ r i)) ⋅ (k ^ × (k ^ × Δ r i)) {\displaystyle {\begin{aligned}\left|\Delta \mathbf {r} _{i}^{\perp }\right|^{2}&=\left(-\left^{2}\Delta \mathbf {r} _{i}\right)\cdot \left(-\left^{2}\Delta \mathbf {r} _{i}\right)\\&=\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\end{aligned}}}

The simplification of this equation uses the triple scalar product identity (k ^ × (k ^ × Δ r i)) ⋅ (k ^ × (k ^ × Δ r i)) ≡ ((k ^ × (k ^ × Δ r i)) × k ^) ⋅ (k ^ × Δ r i), {\displaystyle \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\equiv \left(\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\times \mathbf {\hat {k}} \right)\cdot \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right),} where the dot and the cross products have been interchanged. Exchanging products, and simplifying by noting that Δ r i {\displaystyle \Delta \mathbf {r} _{i}} and k ^ {\displaystyle \mathbf {\hat {k}} } are orthogonal: (k ^ × (k ^ × Δ r i)) ⋅ (k ^ × (k ^ × Δ r i)) = ((k ^ × (k ^ × Δ r i)) × k ^) ⋅ (k ^ × Δ r i) = (k ^ × Δ r i) ⋅ (− Δ r i × k ^) = − k ^ ⋅ (Δ r i × Δ r i × k ^) = − k ^ ⋅ 2 k ^. {\displaystyle {\begin{aligned}&\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\cdot \left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\\={}&\left(\left(\mathbf {\hat {k}} \times \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\right)\times \mathbf {\hat {k}} \right)\cdot \left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\\={}&\left(\mathbf {\hat {k}} \times \Delta \mathbf {r} _{i}\right)\cdot \left(-\Delta \mathbf {r} _{i}\times \mathbf {\hat {k}} \right)\\={}&-\mathbf {\hat {k}} \cdot \left(\Delta \mathbf {r} _{i}\times \Delta \mathbf {r} _{i}\times \mathbf {\hat {k}} \right)\\={}&-\mathbf {\hat {k}} \cdot \left^{2}\mathbf {\hat {k}}.\end{aligned}}}

Thus, the moment of inertia around the line L {\displaystyle L} through R {\displaystyle \mathbf {R} } in the direction k ^ {\displaystyle \mathbf {\hat {k}} } is obtained from the calculation I L = ∑ i = 1 N m i | Δ r i ⊥ | 2 = − ∑ i = 1 N m i k ^ ⋅ 2 k ^ = k ^ ⋅ (− ∑ i = 1 N m i 2) k ^ = k ^ ⋅ I R k ^ = k ^ T I R k ^, {\displaystyle {\begin{aligned}I_{L}&=\sum _{i=1}^{N}m_{i}\left|\Delta \mathbf {r} _{i}^{\perp }\right|^{2}\\&=-\sum _{i=1}^{N}m_{i}\mathbf {\hat {k}} \cdot \left^{2}\mathbf {\hat {k}} =\mathbf {\hat {k}} \cdot \left(-\sum _{i=1}^{N}m_{i}\left^{2}\right)\mathbf {\hat {k}} \\&=\mathbf {\hat {k}} \cdot \mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}} =\mathbf {\hat {k}} ^{\mathsf {T}}\mathbf {I} _{\mathbf {R} }\mathbf {\hat {k}},\end{aligned}}} where I R {\displaystyle \mathbf {I_{R}} } is the moment of inertia matrix of the system relative to the reference point R {\displaystyle \mathbf {R} }.

This shows that the inertia matrix can be used to calculate the moment of inertia of a body around any specified rotation axis in the body.
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