Rigid Body Rotation and the Tensor of Inertia
Angular Momentum and the Tensor of Inertia
In example 8.1, we analyzed the angular momentum of a rotating skew rod, but we found that the angular velocity and angular momentum weren’t point in the same direction. Therefore, the equation L ⃗ = I 0 ω ⃗ \vec{L} = I_0 \vec{\omega} L = I 0 ω
Let’s start with the definition of angular momentum and go from there:
L ⃗ = ∫ M r ⃗ × p ⃗ d m L ⃗ = ∫ M r ⃗ × m v ⃗ d m L ⃗ = ∫ M r ⃗ × m ( ω ⃗ × r ⃗ ) d m \begin{align*}
\vec{L} &= \int_M \vec{r} \times \vec{p} \,dm \\
\vec{L} &= \int_M \vec{r} \times m \vec{v} \,dm \\
\vec{L} &= \int_M \vec{r} \times m (\vec{\omega} \times \vec{r}) \,dm
\end{align*} L L L = ∫ M r × p d m = ∫ M r × m v d m = ∫ M r × m ( ω × r ) d m You may fear the double cross product. I fear the double cross product. Let’s use a cross product identity:
A ⃗ × ( B ⃗ × C ⃗ ) = ( A ⃗ ⋅ C ⃗ ) B ⃗ − ( A ⃗ ⋅ B ⃗ ) C ⃗ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C} A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C blah blah blah blah blah
( L x L y L z ) = ( ∫ M y 2 + z 2 d m − ∫ M x y d m − ∫ M x z d m − ∫ M x y d m ∫ M x 2 + z 2 d m − ∫ M y z d m − ∫ M x z d m − ∫ M y z d m ∫ M x 2 + y 2 d m ) ( ω x ω y ω z ) L ⃗ = I ~ ω ⃗ \begin{align*}
\begin{pmatrix}
L_x \\
L_y \\
L_z
\end{pmatrix} &=
\begin{pmatrix}
\int_M y^2 + z^2 \,dm & -\int_M xy \,dm & -\int_M xz \,dm \\
-\int_M xy \,dm & \int_M x^2 + z^2 \,dm & -\int_M yz \,dm \\
-\int_M xz \,dm & -\int_M yz \,dm & \int_M x^2 + y^2 \,dm
\end{pmatrix}
\begin{pmatrix}
\omega_x \\
\omega_y \\
\omega_z
\end{pmatrix} \\
\vec{L} &= \tilde{I} \vec{\omega}
\end{align*} L x L y L z L = ∫ M y 2 + z 2 d m − ∫ M x y d m − ∫ M x z d m − ∫ M x y d m ∫ M x 2 + z 2 d m − ∫ M yz d m − ∫ M x z d m − ∫ M yz d m ∫ M x 2 + y 2 d m ω x ω y ω z = I ~ ω And the inertia tensor said softly, “I am your worst nightmare.”
Principal Axes
In every rotation, you can chose your axes so the inertia tensor simplifies to
I ~ = ( I x x 0 0 0 I y y 0 0 0 I z z ) \tilde{I} =
\begin{pmatrix}
I_{xx} & 0 & 0 \\
0 & I_{yy} & 0 \\
0 & 0 & I_{zz}
\end{pmatrix} I ~ = I xx 0 0 0 I yy 0 0 0 I zz Which is a lot better! If we label these principal axes 1, 2, and 3, then we can calculate our angular momentum in a simpler way:
L 1 = I 1 ω 1 L 2 = I 2 ω 2 L 3 = I 3 ω 3 \begin{align*}
L_1 &= I_1 \omega_1 \\
L_2 &= I_2 \omega_2 \\
L_3 &= I_3 \omega_3
\end{align*} L 1 L 2 L 3 = I 1 ω 1 = I 2 ω 2 = I 3 ω 3 Rotational Kinetic Energy of a Rigid Body