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Description
Flashcards by Hayden Tornabene, updated more than 1 year ago
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Created by Hayden Tornabene
about 8 years ago
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| Question | Answer |
| Blocks on Ramps: Force Parallel and Force Perpendicular | \[F_{g,par} = mg \sin \theta\] \[F_{g,perp} = mg \cos \theta\] |
| Force of Friction (Normal Force) | \[F_{f} = \mu mg \cos \theta\] \[N = mg \cos \theta\] |
| Kinematic Equations | \[x(t) = v_{0x}t + x_0\] \[y(t) = -\frac{1}{2}gt^2 + v_{0y}t + y_0\] Only use kinematics if you need to know the explicit time dependence of a system. Energy considerations are faster otherwise. |
| Initial and Final Velocity Equation | \[v_f^2-v_i^2 = 2a\Delta x\] |
| Radial Acceleration for Uniform Circular Motion | \[a = \frac{v^2}{r}\] |
| Centripetal Force for Uniform Circular Motion | \[F = \frac{mv^2}{r}\] Does not tell you what kind of force acting on body, just that a body moving in uniform circular motion has this force. |
| Types of Energy | Translational Kinetic: \[\frac{1}{2}mv^2\] Rotational Kinetic: \[\frac{1}{2}I\omega^2\] Gravitational Potential: \[mgh\] Spring Potential: \[\frac{1}{2}kx^2\] |
| Potential Energy Line Integral | \[\Delta U = - \int_{a}^{b} \textbf{F} \cdot d\textbf{l}\] For any conservative force - Integral is indirection of the force vector. |
| Gravitational Force and Gravitational Potential | \[\textbf{F}_{grav} = \frac{Gm_1m_2}{r^2} \hat{r}\] \[U(r) = \frac{GmM}{r}\] |
| Potential Energy and Force Relationship | \[\textbf{F} = -\nabla U\] |
| Linear Velocity of Rolling without Slipping | \[v = R\omega\] Questions regarding which arrives with slower linear velocity are NOT questions of which arrive first. Need kinematics. Ex. energy expression for a sphere with mass m and radius r. \[mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)\omega^2\] REMEMBER: \[\omega = \frac{v}{r}\] |
| Work Energy Theorem - In terms of Kinetic Energy - In terms of Force | \[E_{initial} + W_{other} = E_{final}\] \[W = \Delta KE\] \[W = \int \textbf{F} \cdot d\textbf{l}\] |
| Linear Collisions Conservation of Momentum Two balls collide, M on m with M scattering at \(\theta\) with initial velocity V with the final velocity equal for both at v. What is the scattering angle of \(\phi\) of m? | \[MV = Mv\cos\theta + mv\cos\phi\] (parallel to initial) \[0 = Mv\sin\theta + mv\sin\phi\] (perpendicular to initial) \[\phi = \sin^{-1}(\frac{-M}{m}\sin\theta)\] Using limiting cases, triaging is essential. Finding phi analytically takes time. |
| Angular momentum of a point particle and for an extended body. | \[\vec{L} = \vec{r} x \vec{p}\] \[\vec{L} = I \vec{\omega}\] |
| Torque (Angular Force) | \[\vec{\tau} = \vec{r} x \vec{F}\] |
| Scalar Analogues of \(\vec{p} = m\vec{v}\) and \(\vec{F} = \frac{d\vec{p}}{dt}\) | Angular Momentum: \(L = I\omega\) Torque: \(\tau = \frac{dL}{dt}\) The vector angular momentum \(\vec{L}\) is generally parallel to \(\vec{\omega}\) with the angular momentum determined by the right hand rule. |
| Moment of Inertia (General and Extended Objects) | \[I = mr^2\] \[I = \int r^2 dm\] for dm = \(\rho dV\) and \(\rho = Ar^3\) Compute the integral to find the constant A and then insert into the extended objects integral. Ex. \(\rho(x) = Ax^2\) \[M = \int_{0}^{L} \rho(x)dx = \frac{1}{3} AL^3 \rightarrow A = \frac{3M}{L^3}\] |
| Parallel Axis Theorem | \[I = I_{CM} + Mr^2\] For penny through center, \(I = \frac{1}{2}MR^2\) For penny axis through edge, \(I = \frac{1}{2}MR^2 + MR^2\) |
| Center of Mass Displacement from Origin | \[\vec{r}_{CM} = \frac{\int \vec{r} dm}{M}\] Must solve three integrals, one for each dimension. Remember to select a volume integral and corresponding coordinate system that capitalizes on symmetry. For a system of point masses: \[\vec{r}_{CM} = \frac{\Sigma_i \vec{r_i} m_i}{M}\] |
| Lagrangian Function | Scalar Function \[L(q, \dot{q}, t) = T - U\] for T kinetic energy, U potential energy, and q the generalized coordinate describing degrees of freedom. |
| How do you compute the correct expression for T in the Lagrangian? | 1. Write down expressions for Cartesian coordinates in terms of chosen coordinates q. 2. Differentiate x,y,z to get \(\dot{x}, \dot{y}, \dot{z}\) 3. Form kinetic cartesian energy, \(T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\) for point particle, \(T = \frac{1}{2}I\omega^2\) for an extended object. For extended object, must express \(\omega\) in terms of \(\dot{q}\), but this is easy as the two are generally equal. See page 20 in notebook for excellent example. |
| Euler-Lagrange Equations | \[\frac{d}{dt}\frac{\delta L}{\delta\dot{q}} = \frac{\delta L}{\delta q}\] One equation for each degree of freedom, q. Remember that \(\frac{d}{dt}\) is a total time derivative, not a partial derivative. I.e. \(\frac{d}{dt}m\dot{x} = m\ddot{x}\) |
| Momentum Conjugate to q (Lagrangian) | \[\frac{\delta L}{\delta \dot{q}}\] Iff the Lagrangian is independent of a coordinate q, the corresponding conjugate momentum quantity, \(\frac{\delta L}{\delta \dot{q}}\), is conserved. Quantities whose time derivatives are zero are conserved quantities. |
| Hamiltonian | \[H(p,q) = \Sigma_i p_i \dot{q_i} - L\] for \(p_i = \frac{\delta L}{\delta \dot{q_i}}\) If the potential energy is not explicitly dependent on velocities of time, H = T + U Iff the Hamiltonian is independent of a coordinate 1, the corresponding conjugate momentum p is conserved. |
| Hamiltonian for Particle Moving in One Dimension | \[H = T + U = \frac{p_x^2}{2m} + U(x)\] |
| Hamiltonian for Particle Moving in 2D Polar | \[H = \frac{p_r^2}{2m} + \frac{p_{\theta}^2}{2mr^2}\] |
| Hamilton's Equations | Scalar function encoding equations of motion: \[\dot{p} = -\frac{\delta H}{\delta q}\] \[\dot{q} = \frac{\delta H}{\delta p}\] |
| Lagrangian for a Simple Pendulum | Take corresponding derivatives - e.g. \[\dot{x} = l\cos\theta\dot{\theta}\cos{\phi} - l\sin\theta\sin\phi\dot{\phi}\] Plug the \(\dot{x},\dot{y},\dot{z}\) into the expression for \[T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\] Simplify and put into L = T - U for \(U = mgz = - mgl\cos\theta\) |
| Conservation of Conjugate Momentum Example | For the Lagrangian independent of \(\phi\), \(p_{\phi}\) is conserved. Therefore the following quantity is conserved: \[p_{\phi} = \frac{\delta L}{\delta \dot{\phi}}\] |
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