9.5 KiB
Chapter 4 - Energy
Section 4.1 - Kinetic Energy and Work
Theorem. Work-Energy Theorem. \frac{dT}{dt} = \frac{1}{2} m \frac{d}{dT}v^2 = \frac{1}{2} m \frac{d}{dT}(\mathbf{v} \cdot \mathbf{v}) = \frac{1}{2} m (\dot{\mathbf{v}} \cdot \mathbf{v} + \mathbf{v} \cdot \dot{\mathbf{v}}) = m \dot{\mathbf{v}} \cdot \mathbf{v} = \mathbf{F} \cdot \mathbf{v}. Then, we can see that dT = \mathbf{F} \cdot d\mathbf{r}, where \mathbf{F} \cdot d\mathbf{r} is the work done by force \mathbf{F} over a small displacement d\mathbf{r}.
Section 4.2 - Potential Energy and Conservative Forces
Theorem. For a force to be conservative, it must be able to be written as some \mathbf{F}(\mathbf{r}) = f(r) \hat{\mathbf{r}}.
Theorem. For a conservative force (and thus a conservative vector field), we can apply the Fundamental Theorem of Line Integrals to see that \int_1^2 \mathbf{F} \cdot d\mathbf{r} = F(2) - F(1), where F is the antiderivative of the integral. That is, the work done by said force is independent of the path taken.
Theorem. For a conservative force, we can find some function U(r), where U(r) is the potential of said force. This function will make the mechanical energy of a system E = T + U a constant.
In the presence of a nonconservative force, we see that W = W_{cons} + W_{nc}. We can then rewrite \Delta E = \Delta(T + U) = W_{nc}. Mechanical energy is no longer conserved, but the change in energy is equal to the work done by nonconservative forces.
Section 4.3 - Force as the Gradient of Potential Energy
We know that W = \mathbf{F} \cdot d\mathbf{r} = F_x dx + F_y dy + F_z dz.
Additionally, W = -dU = -[U(\mathbf{r} + d\mathbf{r}) - U(\mathbf{r})] = -[U(x + dx, y + dy, z + dz) - U(x, y, z)] = -[\frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y} dy + \frac{\partial U}{\partial z} dz].
We can match terms to see that F_x = -\frac{\partial U}{\partial x} and so on, so that \mathbf{F} = -[\hat{\mathbf{x}} \frac{\partial U}{\partial x} + \hat{\mathbf{y}} \frac{\partial U}{\partial y} + \hat{\mathbf{z}} \frac{\partial U}{\partial z}] = -\nabla U
Section 4.4 - The Second Condition that F be Conservative
An easier way to prove a force is conservative is to show that \nabla \times \mathbf{F} = 0.
If this is true, then \oint_\Gamma (\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} = 0. We can split the curve \Gamma into \mathbf{r}_1 and \mathbf{r}_2, to show that 0 = \int_1^2 \mathbf{F} \cdot d\mathbf{r}_1 + \int_2^1 \mathbf{F} \cdot d\mathbf{r}_2, so then \int_1^2 \mathbf{F} \cdot d\mathbf{r}_1 = \int_1^2 \mathbf{F} \cdot d\mathbf{r}_2 and thus the integral is path independent.
Section 4.5 - Time-Dependent Potential Energy
Sometimes, \mathbf{F} = \mathbf{F}(\mathbf{r}, t). Then, if the curl of the force is zero, the work integral is now path-independent. We can define the potential as U(\mathbf{r}, t) = -\int_{\mathbf{r}_0}^\mathbf{r} \mathbf{F}(\mathbf{r}', t) \cdot d\mathbf{r}'.
In this case, it is still true that \mathbf{F}(\mathbf{r}, t) = -\nabla U(\mathbf{r}, t). Again, dT = \frac{dT}{dt} dt = (m \hat{\mathbf{v}} \cdot \mathbf{v}) dt = \mathbf{F} \cdot d\mathbf{r}.
However, we see that dU = \frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y} dy + \frac{\partial U}{\partial z} + \frac{\partial U}{\partial t} dt. Then, we see that dU = -\mathbf{F} \cdot d\mathbf{r} + \frac{\partial U}{\partial t} dt.
Then, dE = d(T + U) = \frac{\partial U}{\partial t} dt. So, energy is only conserved if \frac{\partial U}{\partial t} = 0.
Section 4.6 - Energy for Linear One-Dimensional Systems
For a particle only moving in the $x$-direction, we can define U(x) = -\int_{x_0}^x F_x(x') dx'. If we let F_x = -kx (by Hooke's law), we see that U = \frac{1}{2} kx^2.
We can plot the potential energy as a function of x, and visually see what position an object will tend towards. Notably, where dU/dx = 0 and U is at a minimum or maximum, net force is zero.
Definition. When d^2 U/dx^2 > 0 and U(x) is at a minimum, the particle is said to be at stable equilibrium. That is, a small displacement will lead to a corrective force. When d^2 U / dx^2 < 0 and U(x) is at a maximum, the particle is said to be unstable. In this case, a small displacement will result in a force moving the particle further from equilibrium.
Additionally, in one-dimensional systems, E = T + U(x) and T(x) = \frac{1}{2} m \dot{x} = E - U(x), so we can solve for \dot(x) = \pm \sqrt{\frac{2}{m}} \sqrt{E - U(x)}
If we then let dt = \frac{dx}{\dot{x}}, so that t = \sqrt{\frac{m}{2}} \int_{x_0}^x \frac{dx'}{\sqrt{E - U(x')}}, we can then perform the integral and solve for x.
Section 4.7 - Curvilinear One-Dimensional Systems
Consider the direction of motion s such that T = \frac{1}{2} m \dot{s}, in which the direction of motion is not in a fixed direction. Notably, the normal force is constraining the object to a fixed path, yet doesn't move the object and thus does no work. Thus, it is the force tangential to the path that does work. We can see then that
F_{\perp} = m \ddot{s}
Here, we can define a potential U(s) such that F_{tang} = -dU/ds and the total mechanical energy E = T + U(s) is constant.
Definition. In an Atwood machine, there are two masses of mass m_1 and m_2$, suspended with an inextensible massless string over a pulley. The system can be constrained by a single parameter x, where x is the vertical distance from the center of the pulley and the center of mass of m_1.
Then, we can see that \Delta T_1 + \Delta U_1 = W_1^{tension}, and respectively \Delta T_2 + \Delta U_2 = W_2^{tension}. Then, we can see that \W_1^{tension} = -W_2^{tension}, so \Delta(T_1 + U_1 + T_2 + U_2) = 0. That is, E = T_1 + U_1 + T_2 + U_2, which is conserved.
Notably, if all forces are conservative, we can define a potential U_\alpha for each particle \alpha such that
E = \sum_\alpha^N (T_\alpha U_\alpha)
Section 4.8 - Central Forces
Definition. A central force is some force such that \mathbf{F}(\mathbf{r}) = f(\mathbf{r}) \hat{\mathbf{r}}..
Definition. A spherically symmetric or rotationally invariant force is a force such that f(\mathbf{r}) = f(r). Notably, this is equivalent to the force being conservative.
This book uses spherical polar coordinates.
Definition. \phi, the azimuth, is the angle between the $x$-axis and the projection of the vector \mathbf{r} on the $x$-y plane.
This book uses the convention of r, \theta, and \phi, where x = r \sin \theta \cos \phi, y = r \sin \theta \sin \phi, and z = r \cos \theta. That is, \theta is the only angle extending into the $z$-direction.
Recall that the dot product is defined as normal.
Additionally, we see that d\mathbf{r} = dr \hat{\mathbf{r}} + r d\theta \hat{\mathbf{\theta}} + r \sin \theta d \phi. Then, as df = \nabla f \cdot d\mathbf{r}, we see that df = (\nabla f)_r dr + (\nabla f)_\theta r d\theta + (\nabla f)_\phi r \sin \theta d\phi.
Notably, (\nabla f)_r = \frac{\partial f}{\partial r}, (\nabla f)_\theta = \frac{1}{r} \frac{\partial f}{\partial \theta}, and (\nabla f)_\phi = \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi}.
An important detail is that for \mathbf{F}(\mathbf{r}) = -\nabla U = f(r) \hat{\mathbf{r}}, this forces U = U(r).
Section 4.9 - Energy Interaction of Two Particles
Consider two particles, with forces \mathbf{F}_{12} the force on particle 1 by particle 2 and the equal and opposite force \mathbf{F}_{21}. Notably, we can write \mathbf{F}_{12} = f(r) \hat{\mathbf{r}} = f(r) \frac{\mathbf{r}}{r}.
For two forces not at the origin, we see that \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2 for \mathbf{F}_{12}. As a consequence, we tend to write \mathbf{F}_{12} = \mathbf{F}_{12} (\mathbf{r}_1 - \mathbf{r}_2).
We can extend this to the potential. If we fix \mathbf{r}_2, we see that \mathbf{F}_{12} = -\nabla_1 U(\mathbf{r}_1), and for an unfixed second particle, \mathbf{F}_{12} = -\nabla_1 U(\mathbf{r}_1 - \mathbf{r}_2).
An important note is that \nabla_1 U(\mathbf{r}_1 - \mathbf{r}_2) = -\nabla_2 U(\mathbf{r}_1 - \mathbf{r}_2). We can then say that \mathbf{F}_{21} - \nabla_2 U(\mathbf{r}_1 - \mathbf{r}_2).
Now, we can generalize this to see that W_{tot} = d\mathbf{r}_1 \cdot \mathbf{F}_{12} + d\mathbf{r}_2 \cdot \mathbf{F}_{21} = (d\mathbf{r}_1 - d\mathbf{r}_2) \cdot \mathbf{F}_{12} = d(\mathbf{r}_1 - \mathbf{r}_2) \cdot [-\nabla_1 U(\mathbf{r}_1 - \mathbf{r}_2)] = -dU. We then see that d(T + U) = 0. That is, E = T_1 + T_2 + U.
Recall elastic collisions, in which T_{in} = T_{fin}.
Section 4.10 - The Energy of a Multiparticle System
Consider a system with N particles, each donated $\alpha. Then, we see that
T = \sum_\alpha T_\alpha = \sum_\alpha \frac{1}{2}m_\alpha v_\alpha^2
Assuming all forces are conservative, for each pair of particles \alpha \beta, there exists a potential energy U_{\alpha\beta} between the two. Then, total potential energy can be written as
U = U^{int}+U^{ext} = \sum_\alpha \sum_{\beta > \alpha} U_{\alpha \beta} + \sum_\alpha U_\alpha
With this, the net force on any particle is given by -\nabla_\alpha U, and total energy is conserved given no external potential energy.
For a rigid body, we can they state U^{int} = \sum_\alpha \sum_{\beta > \alpha} U_{\alpha \beta}(\mathbf{r}_\alpha - \mathbf{r}_\beta). In the case that the forces are central, U^{int} = \sum_\alpha \sum_{\beta > \alpha} U_{\alpha \beta}(|\mathbf{r}_\alpha - \mathbf{r}_\beta|).