Physlib.ClassicalMechanics.RigidBody.KineticEnergy

For a rigid body $R$ in a three-dimensional Euclidean space rotating with an angular velocity $\omega \in \mathbb{R}^3$, the rotational kinetic energy $T$ is defined as: $$T = \frac{1}{2} \omega \cdot (I \omega)$$ where $I$ is the inertia tensor of the rigid body $R$, and the expression represents the contraction of the angular velocity vector with the inertia tensor.

For a three-dimensional rigid body $R$ rotating with angular velocity $\boldsymbol{\omega} \in \mathbb{R}^3$, the rotational kinetic energy $T$ is equal to half the dot product (or contraction) of the angular velocity $\boldsymbol{\omega}$ and the angular momentum $\mathbf{L}$: $$T = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L}$$ where $\mathbf{L}$ is the angular momentum of the rigid body corresponding to the angular velocity $\boldsymbol{\omega}$.

For a three-dimensional rigid body $R$ rotating with an angular velocity $\boldsymbol{\omega} \in \mathbb{R}^3$, the rotational kinetic energy $T$ is equal to half the integral over the body's mass of the squared magnitude of the local velocity $\boldsymbol{\omega} \times \mathbf{r}$: $$T = \frac{1}{2} \int_R |\boldsymbol{\omega} \times \mathbf{r}|^2 \, dm$$ where $\mathbf{r}$ is the position vector of a point in the body relative to the reference point, $\times$ denotes the vector cross product, and $dm$ is the mass distribution of the rigid body.

For a rigid body motion $M$ in $d$-dimensional space, the total kinetic energy $T$ at time $t$ is defined as half the integral of the squared speed of its points with respect to its mass distribution: $$T = \frac{1}{2} \int \|\mathbf{v}(t, y)\|^2 \, dm(y)$$ where $\mathbf{v}(t, y)$ is the velocity of the point $y$ at time $t$ (using the `velocityClosedForm`) and $m$ is the mass distribution of the rigid body.

For a rigid body motion $M$ in $d$-dimensional space, if the body's orientation $\mathbf{R}(t)$ and its center of mass trajectory are differentiable at time $t$, then the total kinetic energy $T$ is given by half the integral of the squared speed of its material points with respect to its mass distribution $m$: $$T = \frac{1}{2} \int \|\mathbf{v}(t, y)\|^2 \, dm(y)$$ where $\mathbf{v}(t, y)$ is the velocity of the material point $y$ at time $t$.

For a rigid body motion $M$ in $d$-dimensional space at a given time $t$, the squared speed of a point $y$ in the body frame, $\|v(t, y)\|^2$, is decomposed into three terms: $$\|v(t, y)\|^2 = \|\dot{\mathbf{R}}(t) (y - \mathbf{c})\|^2 + \sum_{j=0}^{d-1} 2 \left( \mathbf{V}(t)^\top \dot{\mathbf{R}}(t) \right)_j (y_j - c_j) + \|\mathbf{V}(t)\|^2$$ where: - $v(t, y)$ is the velocity of point $y$ at time $t$. - $\dot{\mathbf{R}}(t)$ is the time derivative of the body's orientation matrix (rotation matrix). - $\mathbf{c}$ is the center of mass of the rigid body, with $c_j$ being its $j$-th component. - $\mathbf{V}(t)$ is the velocity of the center of mass. - $y_j$ is the $j$-th component of the body-frame coordinate $y$. - $\|\cdot\|$ denotes the Euclidean norm and $\top$ denotes the transpose. The three terms correspond respectively to the squared rotational speed, a term linear in the displacement from the center of mass, and the squared translational speed of the center of mass.

For a rigid body motion in $d$-dimensional space with non-zero total mass $M$, the total kinetic energy $T$ at time $t$ is the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass: $$T = \frac{1}{2} M \|\mathbf{V}\|^2 + \frac{1}{2} \int \|\dot{\mathbf{R}}(\mathbf{y} - \mathbf{c})\|^2 \, dm(\mathbf{y})$$ where: - $\mathbf{V}$ is the velocity of the center of mass at time $t$. - $\dot{\mathbf{R}}$ is the time derivative of the orientation (rotation) matrix at time $t$. - $\mathbf{c}$ is the center of mass of the rigid body in the body-fixed reference frame. - The integral is taken over the body's mass distribution $m$ with respect to the body-frame coordinates $\mathbf{y}$.

**König's theorem** in three dimensions: For a rigid body motion $M$ in three-dimensional space at time $t$, assuming the total mass $M_{\text{total}}$ is non-zero and the body's orientation is differentiable at $t$, the total kinetic energy $T$ is the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass: $$T = \frac{1}{2} M_{\text{total}} \|\mathbf{V}(t)\|^2 + \frac{1}{2} \int \|\boldsymbol{\omega}(t) \times \mathbf{r}(t, y)\|^2 \, dm(y)$$ where: - $\mathbf{V}(t)$ is the velocity of the center of mass. - $\boldsymbol{\omega}(t)$ is the angular velocity vector. - $\mathbf{r}(t, y) = \text{displacement}(t, y) - \mathbf{R}_{\text{com}}(t)$ is the position of point $y$ relative to the center of mass in the inertial frame. - $m$ is the mass distribution of the rigid body.

**König's theorem** in the body frame: For a rigid body motion $M$ in three-dimensional space at time $t$, assume that the total mass $M_{\text{total}}$ is non-zero, the orientation of the body is differentiable at $t$, and the center of mass is located at the origin in the body-fixed reference frame. Then the total kinetic energy $T$ is the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass, evaluated using the body-frame angular velocity $\boldsymbol{\omega}_{\text{body}}(t)$: $$T = \frac{1}{2} M_{\text{total}} \|\mathbf{V}(t)\|^2 + \frac{1}{2} \boldsymbol{\omega}_{\text{body}}(t) \cdot (I \boldsymbol{\omega}_{\text{body}}(t))$$ where: - $\mathbf{V}(t)$ is the velocity of the center of mass in the lab frame. - $\boldsymbol{\omega}_{\text{body}}(t)$ is the angular velocity vector in the body-fixed frame. - $I$ is the inertia tensor of the rigid body.