Introductory Classical Mechanics Equations

Introduction to Classical Mechanics

EQUATIONS

INFORMATION

$$ \large \rho = \frac{m}{V} $$

$\\ {\rho} = density \hspace{4 pt} (kg \hspace{2 pt} m^{-3}) \\ \textbf{m} = mass \hspace{4 pt} (kg) \\ \textbf{V} = volume \hspace{4 pt} (m^3)$

Displacement

$$ \large d = x_f - x_i $$

$\\ \textbf{d} = displacment \hspace{4 pt} (m) \\ {x_f} = final \hspace{4 pt} position \hspace{4 pt} (m) \\ {x_i} = initial \hspace{4 pt} position \hspace{4 pt} (m)$

Speed

$$ \large v = \frac{d}{t} $$

$\\ \textbf{v} = speed \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{d} = distance \hspace{4 pt} traveled \hspace{4 pt} (m) \\ \textbf{t} = time \hspace{4 pt} (s)$

Average Velocity

$$ \large \vec{v}_{avg} = \frac{\vec{d}}{t} $$

$\\ \vec{v}_{avg} = average \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \vec{d} = displacement \hspace{4 pt} (m) \\ \textbf{t} = time \hspace{4 pt} (s)$

Average Acceleration

$$ \large \bar{a} = \frac{\Delta \vec{v}}{\Delta t} $$

$\\ {\bar{a}} = average \hspace{4 pt} acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ {\Delta} \vec{v} = change \hspace{4 pt} in \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ {\Delta t} = duration \hspace{4 pt} (s) $

Kinematic Equations for Linear Motion (For Constant Acceleration)

$$ \large \vec{v_f} = \vec{v_i} + \vec{a}t \\ {v_f}^2 = {v_i}^2 + 2 \vec{a} \cdot (\vec{r} - \vec{r_0}) \\ \vec{r} = \vec{r_0} + \vec{v_i} t + \frac{1}{2}\vec{a} t^2 \\ \vec{r} = \vec{r}_0 + \frac{1}{2} (\vec{v_f} + \vec{v_i}) t \\ \vec{r} = \vec{r}_0 + \vec{v}_ft - \frac{1}{2} \vec{a} t^2 $$

$\\ \vec{v_f} = final \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1})\\ \vec{v_i} = initial \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1})\\ \vec{a} = acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2})\\ \textbf{t} = time \hspace{4 pt} (s)\\ \vec{r} = final \hspace{4 pt} position \hspace{4 pt} (m) \\ \vec{r_0} = initial \hspace{4 pt} position \hspace{4 pt} (m)$

Momentum

$$ \large \vec{p} = m\vec{v} $$

$\\ \vec{p} = momentum \hspace{4 pt} (kg \hspace{2 pt} m \hspace{2 pt} s^{-1}) \\ \textbf{m} = mass \hspace{4 pt} (kg)\\ \vec{v} = velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1})$

Conservation of Momentum

$$ \large \sum {\vec{p}_{init, \hspace{2 pt} i}} = \sum{\vec{p}_{final, \hspace{2 pt} i}} $$

$\\ \vec{p}_{init, \hspace{2 pt} i} = initital \hspace{4 pt} momentum \hspace{4 pt} contribution \\ \hspace{23 pt} (kg \hspace{2 pt} m \hspace{2 pt} s^{-1}) \\ \vec{p}_{final, \hspace{2 pt} i} = final \hspace{4 pt} momentum \hspace{4 pt} contribution \\ \hspace{30 pt} (kg \hspace{2 pt} m \hspace{2 pt} s^{-1}) \\$

Force (Newton's Second Law)

$$ \large \vec{F} = m\vec{a} $$

$ \vec{F} = force \hspace{4 pt} (N)\\ \textbf{m} = mass \hspace{4 pt} (kg)\\ \vec{a} = acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) $

Newton's Third Law

$$ \large \vec{F}_{12} = -\vec{F}_{21} $$

$ \vec{F}_{12} = force \hspace{4 pt} exerted \hspace{4 pt} by \hspace{4 pt} 1 \hspace{4 pt} on \hspace{4 pt} 2 \hspace{4 pt} (N) \\ \vec{F}_{21} = force \hspace{4 pt} exerted \hspace{4 pt} by \hspace{4 pt} 2 \hspace{4 pt} on \hspace{4 pt} 1 \hspace{4 pt} (N) $

Impulse

$$ \large J = F_{net} \hspace{2 pt} t = \Delta ps $$

$\\ \textbf{J} = impulse \hspace{4 pt} (N \hspace{2 pt} s) \\ {F_{net}} = net \hspace{4 pt} force \hspace{4 pt} (N) \\ \textbf{t} = time \hspace{4 pt} of \hspace{4 pt} collision \hspace{4 pt} (s) \\ {\Delta p} = change \hspace{4 pt} in \hspace{4 pt} momentum \hspace{4 pt} (N \hspace{2 pt} s)$

Work Done by a Constant Force (One Direction)

$$ \large W = Fd $$

$\\ \textbf{W} = work \hspace{4 pt} (J) \\ \textbf{F} = force \hspace{4 pt} (N) \\ \textbf{d} = distance \hspace{4 pt} (m)$

Work Done by a Constant Force (Generalized)

$$ \large W = \vec{F} \cdot \vec{d} = \left \| \vec{F} \right \| \left \| \vec{d} \hspace{2 pt} \right \| cos \hspace{4 pt}\theta $$

$\\ \textbf{W} = work \hspace{4 pt} (J) \\ \textbf{F} = varying \hspace{4 pt} force \hspace{4 pt} (N) \\ \textbf{s} = distance \hspace{4 pt} (m)\\ \textbf{i} = initial \hspace{4 pt} location \hspace{4 pt} (m)\\ \textbf{f} = final \hspace{4 pt} location \hspace{4 pt} (m)$

Kinetic Energy

$$ \large K = \frac{1}{2} m{\bar{v}}^2 $$

$\\ \textbf{K} = kinetic \hspace{4 pt} energy \hspace{4 pt} (J) \\ \textbf{m} = mass \hspace{4 pt} (kg) \\ {\bar{v}} = average \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1})$

Potential Energy

$$ \large U = mgh $$

$\\ \textbf{U} = potential \hspace{4 pt} energy \hspace{4 pt} (J) \\ \textbf{m} = mass \hspace{4 pt} (kg) \\ \vec{g} = acceleration \hspace{4 pt} due \hspace{4 pt} to \hspace{4 pt} gravity \\ \hspace{4 pt} (9.81 \hspace{2 pt} m \hspace{2 pt} s^{-2}) \\ \textbf{h} = height \hspace{4 pt} (m)$

Potential Energy Stored in a Spring

$$ \large U = \frac{1}{2} k x^2 $$

Newton's Law of Universal Gravitation

$$ \large F = G \left( \frac{m_1 m_2}{r^2} \right ) $$

$ \textbf{F} = gravitational \hspace{4 pt} force \hspace{4 pt} between \\ \hspace{6 pt} masses \hspace{4 pt} (N)\\ \textbf{G} = gravitational \hspace{4 pt} constant \\ \hspace{7 pt} (6.67384 x 10^{-11} \hspace{2 pt} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) \\ {m_\#} = mass \hspace{4 pt} of \hspace{4 pt} object \hspace{4 pt} \# \hspace{4 pt} (kg)\\ \textbf{r} = distance \hspace{4 pt} between \hspace{4 pt} objects \hspace{4 pt} (m)$

Newton's Form of Kepler's Third Law

$$ \large T^2 = \left( \frac{4 \pi ^2}{G (m_1 + m_2)} \right) $$

$ \textbf{T} = orbital \hspace{4 pt} period \hspace{4 pt} (s) \\ \textbf{G} = gravitational \hspace{4 pt} constant \\ \hspace{8 pt} (6.67384 \hspace{2 pt} x \hspace{2 pt} 10^{-11} \hspace{2 pt} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) $

Gravitational Field

$$ \large g = \frac{F_g}{m_0} = \frac{GM}{r^2} $$
$$ \large = \frac{GM}{(R + h)^2} $$

$ \textbf{g} = gravitational \hspace{4 pt} acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ F_g = gravitational \hspace{4 pt} force \hspace{4 pt} (N) \\ m_0 = mass \hspace{4 pt} of \hspace{4 pt} test \hspace{4 pt} particle \hspace{4 pt} (kg) \\ \textbf{G} = gravitational \hspace{4 pt} constant \\ \hspace{8 pt} (6.674 \hspace{2 pt} x \hspace{2 pt} 10^{-11} \hspace{4 pt} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} planet \hspace{4 pt} (kg) \\ \textbf{r} = distance v between \hspace{4 pt} the \hspace{4 pt} centers \hspace{4 pt} (m) \\ \textbf{R} = radius \hspace{4 pt} of \hspace{4 pt} planet \hspace{4 pt} to \hspace{4 pt} surface \hspace{4 pt} (m) \\ \textbf{h} = distance \hspace{4 pt} above \hspace{4 pt} planet's \hspace{4 pt} surface \hspace{4 pt} (m) $

Gravitational Potential Energy

$$ \large U = -\frac{GMm}{r} $$

$\\ \textbf{U} = gravitational \hspace{4 pt} potential \hspace{4 pt} energy (J) \\ \textbf{G} = gravitational \hspace{4 pt} constant \\ \hspace{8 pt} (6.67384 x 10^{-11} \hspace{2 pt} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} planet \hspace{4 pt} (kg) \\ \textbf{m} = mass \hspace{4 pt} of \hspace{4 pt} object \hspace{4 pt} (kg) \\ \textbf{r} = distance \hspace{4 pt} between \hspace{4 pt} the \hspace{4 pt} center \\ \hspace{3 pt} of \hspace{4 pt} masses \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} objects \hspace{4 pt} (m)$

Total Mechanical Energy

$$ \large {{ME}}_{total} = K + U $$

$\\ {{{ME}}_{total}} = total \hspace{4 pt} mechanical \hspace{4 pt} energy \hspace{4 pt} (J) \\ \textbf{K} = total \hspace{4 pt} kinetic \hspace{4 pt} energy \hspace{4 pt} (J) \\ \textbf{U} = total \hspace{4 pt} potential \hspace{4 pt} energy \hspace{4 pt} (J)$

Conservation of Energy

$$ \large \Delta {ME}_{total} = 0 $$

$\\ {{{ME}}_{total}} = total \hspace{4 pt} mechanical \hspace{4 pt} energy \hspace{4 pt} (J)$

Total Energy (for Objects in Orbit)

$$ \large E = - \frac{GMm}{2a} $$

$\\ \textbf{E} = total \hspace{4 pt} energy \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} system \hspace{4 pt} (J) \\ \textbf{G} = gravitational \hspace{4 pt} constant \\ \hspace{8 pt} (6.674 \times 10^{-11} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} massive \hspace{4 pt} object \hspace{4 pt} (kg) \\ \textbf{m} = mass \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} less \\ \hspace{9 pt} massive \hspace{4 pt} object \hspace{4 pt} (kg) \\ \textbf{a} = semi-major \hspace{4 pt} axis \hspace{4 pt} (m)$

Average Power

$$ \large P_{avg} = \frac{W}{t} $$

$ P_{avg} = average \hspace{4 pt} power \hspace{4 pt} (J \hspace{2 pt} s^{-1}) \\ \textbf{W} = work \hspace{4 pt} (J) \\ \textbf{t} = time \hspace{4 pt} (s) $

Moment of Inertia

$$ \large I = \sum_{i} m_i r_i ^2 $$

$\\ \textbf{I} = moment \hspace{4 pt} of \hspace{4 pt} inertia \hspace{4 pt} (kg \hspace{2 pt} m^2) \\ \textbf{m} = mass \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} component \hspace{4 pt} (kg) \\ \textbf{r} = distance \hspace{4 pt} between \hspace{4 pt} the \hspace{4 pt} component \\ \hspace{4 pt} and \hspace{4 pt} the \hspace{4 pt} rotation \hspace{4 pt} axis \hspace{4 pt} (m) $

Moment of Inertia of a Thin Cylindrical Shell

$$ \large I = MR^2 $$

Moment of Inertia of a Hollow Cylinder

$$ \large I = \frac{1}{2} \hspace{2 pt} M(R_1 ^2 + R_2 ^2 ) $$

$\\ \textbf{I} = moment \hspace{4 pt} of \hspace{4 pt} inertia \hspace{4 pt} (kg \hspace{2 pt} m^2) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} shell \hspace{4 pt} (kg)\\ {R_1} = distance \hspace{4 pt} to \hspace{4 pt} inner \hspace{4 pt} shell \hspace{4 pt} (m) \\ {R_2} = distance \hspace{4 pt} to \hspace{4 pt} outer \hspace{4 pt} shell \hspace{4 pt} (m) \\$

Moment of Inertia of a Solid Cylinder

$$ \large I = \frac{1}{2} \hspace{2 pt} MR^2 $$

Moment of Inertia of a Rectangular Plate

$$ \large I = \frac{1}{12} \hspace{2 pt} M \hspace{2 pt} (a^2 + b^2) $$

$\\ \textbf{I} = moment \hspace{4 pt} of \hspace{4 pt} inertia \hspace{4 pt} (kg \hspace{2 pt} m^2) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} plate \hspace{4 pt} (kg)\\ \textbf{a} = width \hspace{4 pt} of \hspace{4 pt} plate \hspace{4 pt} (m) \\ \textbf{b} = length \hspace{4 pt} of \hspace{4 pt} plate \hspace{4 pt} shell \hspace{4 pt} (m) \\$

Angle and Solid Angle

$$ \large \theta = \frac{S}{r} \\ $$ $$ \large \Omega = \frac{A}{r^2} $$

$\\ \theta = angle \hspace{4 pt} subtended \hspace{4 pt} by \hspace{4 pt} arc \hspace{4 pt} S \hspace{4 pt} (deg) \\ \textbf{S} = arc \hspace{4 pt} length \hspace{4 pt} (m) \\ \textbf{r} = radius \hspace{4 pt} of \hspace{4 pt} circle \hspace{4 pt} (m) \\ \Omega = solid \hspace{4 pt} angle \hspace{4 pt} subtended \hspace{4 pt} by \\ \hspace{4 pt} surface \hspace{4 pt} area \hspace{4 pt} A \hspace{4 pt} (sr) \\ \textbf{A} = surface \hspace{4 pt} area \hspace{4 pt} (m)$

Average Angular Speed

$$ \large \bar{\omega} = \frac{\theta}{t} $$

$ \bar{\omega} = average \hspace{4 pt} angular \hspace{4 pt} velocity \hspace{4 pt} (deg) \\ \theta = angle \hspace{4 pt} (deg) \\ \textbf{t} = time \hspace{4 pt} (s) $

Average Angular Acceleration

$$ \large \alpha = \frac{a_t}{r} $$

$\\ \alpha = angular \hspace{4 pt} acceleration \hspace{4 pt} (deg \hspace{2 pt} s^{-2}) \\ {a_t} = translational \hspace{4 pt} acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ \textbf{r} = radius \hspace{4 pt} (m)$

Kinematic Equations for Angular Motion

$$ \large \omega_f = \omega_i + \alpha t \\ $$ $$ \large \omega_f ^2 = \omega_i^2 + 2 \alpha \cdot (\theta - \theta_0) \\ $$ $$ \large \theta = \theta_0 + \omega_i t + \frac{1}{2} \alpha t^2 \\ $$ $$ \large \theta = \theta_0 + \omega_i t + \frac{1}{2} (\omega_f \omega_i) t \\ $$ $$ \large \theta = \theta_0 + \omega_f t - \frac{1}{2} \alpha t^2 $$

$ \omega_f = final \hspace{4 pt} angular \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1} \\ \omega_i = initial \hspace{4 pt} angular \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \alpha = angular \hspace{4 pt} acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ \textbf{t} = time \hspace{4 pt} (s) \\ \theta = final \hspace{4 pt} angular \hspace{4 pt} position \hspace{4 pt} (m) \\ \theta_0 = initial \hspace{4 pt} angular \hspace{4 pt} position \hspace{4 pt} (m) $

Angular Frequency (for Circular Motion) / Rotational Speed

$$ \large \omega = \frac{v}{r} $$

$\\ {\omega} = angular \hspace{4 pt} frequency \hspace{2 pt} / \hspace{2 pt} rotational \\ \hspace{2 pt}\hspace{4 pt} speed \hspace{4 pt} (rad \hspace{4 pt} s^{-1}) \\ \textbf{v} = tangential \hspace{4 pt} speed \hspace{4 pt} (m \hspace{2 pt} s^{-1})\\ \textbf{r} = radial \hspace{4 pt} distance \hspace{4 pt} (m)$

Angular Acceleration (for Circular Motion) / Rotational Acceleration

$$ \large \alpha = \frac{a_t}{r} $$

$ \alpha = angular \hspace{4 pt} acceleration \hspace{4 pt} (deg \hspace{2 pt} s^{-2}) \\ a_t = translational \hspace{4 pt} acceleration \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ \textbf{r} = radius \hspace{4 pt} (m) $

Normal Force

$$ \large \vec{F_N} = m \vec{g} \hspace{2 pt} cos \hspace{2 pt} \theta $$

$ \vec{F_N} = normal \hspace{4 pt} force, \hspace{4 pt} perpendicular \\ \hspace{11 pt} to \hspace{4 pt} the \hspace{4 pt} surface \hspace{4 pt} (N) \\ \textbf{m} = mass \hspace{4 pt} (kg) \\ \vec{g} = acceleration \hspace{4 pt} due \hspace{4 pt} to \hspace{4 pt} gravity, \\ \hspace{4 pt}(9.81 \hspace{2 pt} m \hspace{2 pt} s^{-2}) \\ {\theta} = angle \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} inclined \\ \hspace{4 pt} measured \hspace{4 pt} from \hspace{4 pt} the \hspace{4 pt} horizontal \hspace{4 pt} (deg)$

Coulomb Friction

$$ \large \vec{F_f} \leq \mu \vec{F_N} $$

$\\ \vec{F_f} = force \hspace{4 pt} due \hspace{4 pt} to \hspace{4 pt} friction \hspace{4 pt} exerted \\ \hspace{11 pt} by \hspace{4 pt} each \hspace{4 pt} object\hspace{4 pt} on \hspace{4 pt} the \hspace{4 pt} other \\ \hspace{11 pt} in \hspace{4 pt} the \hspace{4 pt} direction \hspace{4 pt} that \hspace{4 pt} opposes \\ \hspace{11 pt} the \hspace{4 pt} direction \hspace{4 pt} of \hspace{4 pt}the \hspace{4 pt} net \\ \hspace{11 pt} applied \hspace{4 pt} force \hspace{4 pt}(N) \\ {\mu} = coefficient \hspace{4 pt} of \hspace{4 pt} friction \\ \vec{F_N} = normal \hspace{4 pt} force, \hspace{4 pt} perpendicular \\ \hspace{11 pt} to \hspace{4 pt} the \hspace{4 pt} force \hspace{4 pt} of \hspace{4 pt} friction \hspace{4 pt} (N)$

Tangential Velocity

$$ \large v_T = \frac{2 \pi r}{T} $$

Centripetal Acceleration

$$ \large \vec{F_c} = m\vec{a_c} = \frac{m\vec{v}^{\hspace{2 pt} 2}}{r} $$

$\\ \vec{a_c} = centripetal \hspace{4 pt} acceleration, \\ \hspace{7 pt} pointed \hspace{4 pt} towards \hspace{4 pt} the \hspace{4 pt} center \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ \vec{v} = tangential \hspace{4 pt} velocity, \hspace{4 pt} perpendicular \\ \hspace{4 pt} to \hspace{4 pt} the \hspace{4 pt} acceleration \hspace{4 pt} vector \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{r} = radial \hspace{4 pt} distance \hspace{4 pt} (m)$

Centripetal Force

$$ \large \vec{F_c} = m\vec{a_c} = \frac{m\vec{v}^{\hspace{2 pt} 2}}{r} $$

$\\ \vec{F_c} = centripetal \hspace{4 pt} force, \hspace{4 pt} pointed \\ \hspace{9 pt} towards \hspace{4 pt} the \hspace{4 pt} center \hspace{4 pt} (N) \\ \textbf{m} = mass \hspace{4 pt} of \hspace{4 pt} object \hspace{4 pt} in \hspace{4 pt} motion \hspace{4 pt} (kg) \\ \vec{a_c} = centripetal \hspace{4 pt} acceleration, \\ \hspace{7 pt} pointed \hspace{4 pt} towards \hspace{4 pt} the \hspace{4 pt} center \hspace{4 pt} (m \hspace{2 pt} s^{-2}) \\ \textbf{r} = radial \hspace{4 pt} distance \hspace{4 pt} (m)$

Torque

$$ \large \vec{\tau} = \vec{r} \times \vec{F} $$

$\\ \vec{\tau} = torque \hspace{4 pt} (N \hspace{2 pt} m) \\ \vec{r} = distance \hspace{4 pt} between \hspace{4 pt} where \hspace{4 pt} force \\ \hspace{4 pt} is \hspace{4 pt} applied \hspace{4 pt} and \hspace{4 pt} where \hspace{4 pt} torque \\ \hspace{4 pt} is \hspace{4 pt} measured \hspace{4 pt} (m) \\ \vec{F} = force \hspace{4 pt} applied \hspace{4 pt} (N) $

Rigid Object Under a Net Torque

$$ \large \tau_{net} = I \alpha $$

$\\ {\tau_{net}} = net \hspace{4 pt} external \hspace{4 pt} torque \hspace{4 pt} (N \hspace{2 pt} m) \\ \textbf{I} = moment \hspace{4 pt} of \hspace{4 pt} inertia \hspace{4 pt} (kg \hspace{2 pt} m^2) \\ {\alpha} = angular \hspace{4 pt} acceleration \hspace{4 pt} (deg \hspace{2 pt} s^{-2})$

Angular Momentum of a Rotating Rigid Object

$$ \large \tau_{net} = I \alpha $$

Conservation of Angular Momentum

$$ \large \Delta \vec{L} = 0 $$

$\\ {\Delta} \vec{L} = change \hspace{4 pt} in \hspace{4 pt} the \hspace{4 pt} net \hspace{4 pt} angular \\ \hspace{16 pt} momentum \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} system \hspace{4 pt} (kg \hspace{2 pt} m^2 \hspace{2 pt} s^{-1})$

Escape Velocity

$$ \large \Delta \vec{L} = 0 $$

Terminal Velocity (in a Liquid or Gas)

$$ \large v_{esc} = \sqrt{\frac{2GM}{r}} $$

$\\ {v_{esc}} = escape \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{G} = gravitational \hspace{4 pt} constant \hspace{4 pt} \\ \hspace{8 pt} (6.67 \hspace{2 pt} x \hspace{2 pt} 10^{-11} \hspace{2 pt} m^3 \hspace{2 pt} kg^{-1} \hspace{2 pt} s^{-2}) \\ \textbf{M} = mass \hspace{4 pt} of \hspace{4 pt} planet \hspace{4 pt} (kg) \\ \textbf{r} = distance \hspace{4 pt} from \hspace{4 pt} center \hspace{4 pt} of \\ \hspace{3 pt} gravity \hspace{4 pt} (m)$

Velocity (in Terms of the Terminal Velocity)

$$ \large v_T = \frac{mg}{b} = \tau g $$

$\\ {v_T} = terminal \hspace{4 pt} velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{m} = mass \hspace{4 pt} of \hspace{4 pt} object \hspace{4 pt} (kg) \\ \textbf{g} = acceleration \hspace{4 pt} due \hspace{4 pt} to \hspace{4 pt} gravity \\ \hspace{4 pt} (9.81 \hspace{4 pt} m \hspace{2 pt} s^{-2}) \\ \textbf{b} = constant \hspace{4 pt} dependent \hspace{4 pt} on \hspace{4 pt} the \\ \hspace{4 pt} medium \hspace{4 pt} (kg \hspace{4 pt} s^{-1}) \\ {\tau} = time \hspace{4 pt} constant \hspace{4 pt} (s)$

Terminal Velocity (for Objects Travelling Through Air at High Speeds)

$$ \large v_T = \frac{mg}{b} = \tau g $$

General Form of an Elastic Modulus

$$ \large E_M = \frac{s_s}{s_n} $$

$ E_m = elastic \hspace{4 pt} modulus \hspace{4 pt} (Pa) \\ s_s = stress \hspace{4 pt} (Pa) \\ s_n = strain $

Young's Modulus

$$ \large E = \frac{\sigma}{\varepsilon} = \frac{FL_0}{A_0 \Delta L} $$

$\\ \textbf{E} = Young's \hspace{4 pt} modulus \hspace{4 pt} (Pa) \\ {\sigma} = tensile \hspace{4 pt} stress \hspace{4 pt} (Pa) \\ {\varepsilon} = extensional \hspace{4 pt} strain \\ \textbf{F} = force \hspace{4 pt} exerted \hspace{4 pt} on \hspace{4 pt} an \hspace{4 pt} object \hspace{4 pt} under \\ \hspace{4 pt} tension \hspace{4 pt} (N) \\ {L_0} = initial \hspace{4 pt} length \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} object \hspace{4 pt} (m)\\ {A_0} = initial \hspace{4 pt} cross-sectional \hspace{4 pt} area \\ \hspace{14 pt} through \hspace{4 pt} which \hspace{4 pt} the \hspace{4 pt} force \hspace{4 pt} is \hspace{4 pt} applied \\ \hspace{8 pt} (m^2) \\ {\Delta L} = change \hspace{4 pt} in \hspace{4 pt} the \hspace{4 pt} length \hspace{4 pt} of \hspace{4 pt} the \\ \hspace{8 pt} object \hspace{4 pt} (m)$

Shear Modulus

$$ \large G = \frac{ F l}{A \Delta x} $$

$ \textbf{G} = shear \hspace{4 pt} modulus \hspace{4 pt} (Pa) \\ \textbf{F} = force \hspace{4 pt} (N) \\ \textbf{l} = original \hspace{4 pt} length (m) \\ \textbf{A} = area \hspace{4 pt} that \hspace{4 pt} force \hspace{4 pt} is \hspace{4 pt} applied \\ \hspace{8 pt} to \hspace{4 pt} (m^2) \\ \Delta x = displacement \hspace{4 pt} (m) $

Pascal's Law

$$ \large P = P_0 + \rho g h $$

$ \textbf{P} = pressure \hspace{4 pt} in \hspace{4 pt} the \hspace{4 pt} fluid \hspace{4 pt} (Pa) \\ P_0 = pressure \hspace{4 pt} at \hspace{4 pt} the \hspace{4 pt} surface \hspace{4 pt} (Pa) \\ \rho = density \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} fluid \hspace{4 pt} (kg \hspace{2 pt} m^{-3}) \\ \textbf{g} = gravitational \hspace{4 pt} acceleration \\ \hspace{8 pt} (9.81 \hspace{4 pt} m \hspace{2 pt} s^{-2}) \\ \textbf{h} = depth \hspace{4 pt} (m) $

Archimedes' Principle

$$ \large B = g \hspace{2 pt} V_{disp} \hspace{2 pt} \rho_{fluid} $$

$\\ \textbf{B} = buoyant \hspace{4 pt} force \hspace{4 pt} (N) \\ \textbf{g} = gravitational \hspace{4 pt} acceleration \\ \hspace{3 pt} (9.81 \hspace{2 pt} m \hspace{2 pt} s^{-2}) \\ {V_{disp}} = volume \hspace{4 pt} of \hspace{4 pt} fluid \hspace{4 pt} displaced \hspace{4 pt} (m^3) \\ {\rho \hspace{2 pt} _{fluid}} = density \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} fluid \hspace{4 pt} (kg \hspace{2 pt} m^{-3}) $

Continuity Equation

$$ \large A_1 v_1 = A_2 v_2 = k $$

$ A_{\#} = cross - sectional \hspace{4 pt} area \hspace{4 pt} at \\ \hspace{8 pt} point \hspace{4 pt} \# \hspace{4 pt} (m^2) \\ v_{\#} = velocity \hspace{4 pt} at \hspace{4 pt} point \hspace{4 pt} \# \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{k} = constant \hspace{4 pt} (m^3 \hspace{2 pt} s^{-1}) $

Bernoulli's Equation

$$ \large P = \frac{1}{2} \rho v^2 + \rho g y = k $$

$\\ \textbf{P} = pressure \hspace{4 pt} (Pa) \\ {\rho} = density \hspace{4 pt} of \hspace{4 pt} the \hspace{4 pt} liquid \hspace{4 pt} (kg \hspace{2 pt} m^{-3}) \\ \textbf{v} = velocity \hspace{4 pt} (m \hspace{2 pt} s^{-1}) \\ \textbf{g} = gravitational \hspace{4 pt} acceleration \\ \hspace{4 pt} (9.81 \hspace{2 pt} m \hspace{2 pt} s^{-2}) \\ \textbf{y} = height \hspace{4 pt} (m)\\ \textbf{k} = constant \hspace{4 pt} (Pa)$

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