Newton's laws only hold for inertial reference frames ### Linear Momentum $ \vec{f}_{E} = M \dot{\vec{v}}_{C} = \dot{\vec{p}}_{E} $ ### Angular Momentum Torque and angular momentum of particle $S$ about $Q$ is defined as $\vec{\tau}_{S/Q} = \vec{r}_{S/Q} \times \vec{f}_{S}$ $\vec{h}_{S/Q}= \vec{r_{S/Q}}\times \vec{p_{S}}$ This gives us $\vec{\tau}_{S/Q} = \dot{\vec{h}}_{S/Q} + \vec{v}_{Q}\times \vec{p}_{S}$ For a rigid body $E$ we have $ \vec{\tau}_{E} = \dot{\vec{H}}_{E/Q} + \vec{v}_{Q}\times \vec{p}_{E} $ and using $\vec{H}_{E/Q}=\vec{H}_{C}+\vec{r}_{C/Q}\times \vec{p}_{E}$ we get $ \vec{\tau}_{E/Q} = \dot{\vec{H}}_{E/C} + \vec{r}_{C/Q} \times \vec{p}_{C} $ when $Q = C$ this becomes simply $ \vec{\tau}_{E/C}=\dot{\vec{H}}_{E/C} $ Planar case: $H_{z}=I_{C}\omega_{z}$ #### Moment of Inertia For rigid planar bodies, we have $\vec{H}_{C}=I\vec{\omega}$ where $I = \int r^{2} \, dm$ | Geometry | Formula | | --------------------------------- | -------------------------------------------------- | | Rod through center, Perpendicular | $I = \frac{1}{12}ML^{2}$ | | Hoop about symmetry axis | ${} I=MR^{2}$ | | Cylinder through symmetry axis | ${} I = \frac{1}{2}MR^{2}$ | | Solid Sphere | ${} I = \frac{2}{5}MR^{2} {}$ | | Cylinder about diameter | ${} I = \frac{1}{4}MR^{2} + \frac{1}{12}ML^{2} {}$ | | Hoop about diameter | ${} I = \frac{1}{2}MR^{2} {}$ | | Rectangular Plate about CM | $I = \frac{1}{12}M(a^{2}+b^{2})$ | For 3D Bodies, $ \vec{H}_{CM}=\int \begin{bmatrix}y^{2}+z^{2} & -xy & -xz \\ -xy & x^{2}+z^{2} & -yz \\ -xz & -yz & x^{2}+y^{2}\end{bmatrix}\vec{\omega} $ which becomes $ \vec{H}_{A}=\begin{bmatrix}I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz}\end{bmatrix}\vec{\omega} $ The inertia tensor is symmetric positive definite, so there are three axes about which it is a diagonal matrix: $ I = Q I_\text{diag}Q^{-1} $ Symmetry rules: - Principal axis perpendicular to planes of symmetry - Principal axis on axis of symmetry - Principal axis on intersection of orthogonal planes of symmetry Superposition works! ##### Parallel Axis Theorem In-plane, we have $ I_{Q} = I_{C}+M||r_{Q/C}||^{2} $ in 3D: $ I_{Q}=I_{C}+ M \cdot \begin{bmatrix}\Delta y^{2}+\Delta z^{2} & -\Delta x\Delta y & -\Delta x\Delta z \\ -\Delta x\Delta y & \Delta x^{2}+\Delta z^{2} & -\Delta yz \\ -\Delta x\Delta z & -\Delta y\Delta z & \Delta x^{2}+\Delta y^{2}\end{bmatrix} $