4.8 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.