Physlib.ClassicalMechanics.Pendulum.SimplePendulum.Geometric.PhysicalSpace

For a simple pendulum $S$ with rod length $\ell$, this function maps a time-dependent angular lift $\theta : \text{Time} \to \mathbb{R}$ to the trajectory of the pendulum bob in 2D physical space $\mathbb{R}^2$. At any time $t$, the position of the bob is given by the vector $(\ell \sin(\theta(t)), -\ell \cos(\theta(t)))$, where the pivot is located at the origin.

For a simple pendulum $S$ with rod length $\ell$, let $\theta: \text{Time} \to \mathbb{R}$ be an angular lift representing the angle of the pendulum over time. The physical trajectory of the bob in the 2D plane, denoted by $S.\text{spaceTrajectory}(\theta)$, is given by the function $$t \mapsto (\ell \sin(\theta(t)), -\ell \cos(\theta(t))),$$ where the origin is the pivot of the pendulum.

For a simple pendulum $S$ with rod length $\ell$, let $\theta(t)$ be the time-dependent angular lift. The horizontal coordinate (the component with index $0$) of the pendulum bob's position in physical space at time $t$ is given by $\ell \sin(\theta(t))$.

Let $S$ be a simple pendulum with rod length $\ell$. For any time-dependent angular lift $\theta : \text{Time} \to \mathbb{R}$, the vertical position (the second coordinate) of the pendulum bob in 2D physical space at time $t$ is given by $$y(t) = -\ell \cos(\theta(t))$$ where the pivot is located at the origin $(0,0)$ and the second axis points upwards.

For a simple pendulum $S$ with rod length $\ell$, let $\theta : \text{Time} \to \mathbb{R}$ be a time-dependent angular lift. At any time $t$, the Euclidean norm of the bob's position in physical space, denoted as $\| S.\text{spaceTrajectory } \theta(t) \|$, is equal to the length of the rod $\ell$. This reflects the physical constraint that the bob remains at a fixed distance $\ell$ from the pivot.

Let $\theta : \text{Time} \to \mathbb{R}$ be a differentiable angular lift of a simple pendulum. At any time $t$, the time derivative of the sine of the angle is the product of the cosine of the angle and the angular velocity: $$\frac{d}{dt} \sin(\theta(t)) = \cos(\theta(t)) \cdot \dot{\theta}(t)$$ where $\dot{\theta}(t)$ denotes the time derivative of the angle $\theta$ at time $t$.

Let $\theta : \text{Time} \to \mathbb{R}$ be a differentiable function representing the angular lift of a pendulum. For any time $t$, the time derivative of the cosine of the angle, $\cos(\theta(t))$, is equal to the negative of the sine of the angle multiplied by the angular velocity: $$\frac{d}{dt} \cos(\theta(t)) = -\sin(\theta(t)) \dot{\theta}(t),$$ where $\dot{\theta}(t)$ denotes the time derivative of $\theta$ at time $t$.

Let $S$ be a simple pendulum. If the angular lift $\theta : \text{Time} \to \mathbb{R}$ is a differentiable function of time, then the physical trajectory of the pendulum bob in 2D space, denoted by $S.\text{spaceTrajectory}(\theta)$ (which maps time $t$ to the position $(\ell \sin(\theta(t)), -\ell \cos(\theta(t)))$), is also differentiable with respect to time.

Let $S$ be a simple pendulum with rod length $\ell$. If the angular lift $\theta : \text{Time} \to \mathbb{R}$ is a differentiable function of time, then the velocity of the pendulum bob in 2D physical space at time $t$ is the time derivative of its position $S.\text{spaceTrajectory}(\theta)$, given by the vector: $$\frac{d}{dt} (S.\text{spaceTrajectory}(\theta))(t) = \left( \ell \cos(\theta(t)) \dot{\theta}(t), \ell \sin(\theta(t)) \dot{\theta}(t) \right)$$ where $\dot{\theta}(t)$ denotes the angular velocity (the time derivative of $\theta$) at time $t$.

For a simple pendulum $S$ with rod length $\ell$, let $\theta: \text{Time} \to \mathbb{R}$ be a differentiable angular trajectory. At any time $t$, the square of the speed of the pendulum bob in physical space—defined as the squared norm of the time derivative of its position vector $\mathbf{r}(t)$—is equal to the product of the square of the rod length and the square of the angular velocity $\dot{\theta}(t)$: $$\|\partial_t (S.\text{spaceTrajectory } \theta)(t)\|^2 = \ell^2 \left( \frac{d\theta}{dt} \right)^2$$ where $S.\text{spaceTrajectory } \theta$ is the function mapping the angular lift to the bob's coordinates $(\ell \sin \theta, -\ell \cos \theta)$ in $\mathbb{R}^2$.

Consider a simple pendulum $S$ with mass $m$. Let $\theta: \text{Time} \to \mathbb{R}$ be a differentiable angular trajectory. At any time $t$, the chart kinetic energy of the pendulum $T(\theta, t)$ is equal to the kinetic energy of the bob in physical space, given by: $$T(\theta, t) = \frac{1}{2} m \left\| \frac{d}{dt} \mathbf{r}(t) \right\|^2$$ where $\mathbf{r}(t)$ is the position of the pendulum bob in 2D physical space at time $t$ (defined as $S.\text{spaceTrajectory}(\theta)(t)$), and $\left\| \frac{d}{dt} \mathbf{r}(t) \right\|$ is the speed of the bob.

For a simple pendulum $S$ with mass $m$, gravitational acceleration $g$, and rod length $\ell$, let $\theta(t)$ be a time-dependent angular trajectory. The chart potential energy $V(\theta(t))$ is equal to the gravitational potential energy in physical space, given by $$V(\theta(t)) = m g (y(t) + \ell)$$ where $y(t)$ is the vertical component (the second coordinate) of the bob's position $(x(t), y(t))$ in the plane. Here, $y(t) + \ell$ represents the height of the bob above the bottom of its swing.

Consider a simple pendulum $S$ with mass $m$, rod length $\ell$, and gravitational acceleration $g$. Let $\theta: \text{Time} \to \mathbb{R}$ be a differentiable angular trajectory. At any time $t$, the chart Lagrangian $L(t, \theta(t), \dot{\theta}(t))$ of the pendulum is equal to the kinetic energy of the bob minus its gravitational potential energy in 2D physical space: $$L(t, \theta(t), \dot{\theta}(t)) = \frac{1}{2} m \|\mathbf{v}(t)\|^2 - m g (y(t) + \ell)$$ where $\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)$ is the velocity of the bob in physical space, $\mathbf{r}(t)$ is its position vector, and $y(t)$ is its vertical coordinate (the second component of the position vector). The term $y(t) + \ell$ represents the height of the bob above the bottom of its swing.

Consider a simple pendulum $S$ with mass $m$, rod length $\ell$, and gravitational acceleration $g$. Let $\theta: \text{Time} \to \mathbb{R}$ be a differentiable angular trajectory. At any time $t$, the chart energy $E(t)$ of the pendulum is equal to the total energy of the bob in physical space, which is the sum of its kinetic energy and its gravitational potential energy: $$E(t) = \frac{1}{2} m \|\mathbf{v}(t)\|^2 + m g (y(t) + \ell)$$ where $\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)$ is the velocity of the bob in 2D physical space, $\mathbf{r}(t)$ is its position vector, and $y(t)$ is the vertical component (the second coordinate) of its position. The term $y(t) + \ell$ represents the height of the bob above the bottom of its swing.