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
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
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
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, \(U^{int} = \sum_\alpha \sum_{\beta > \alpha} U_{\alpha \beta}(|\mathbf{r}_\alpha - \mathbf{r}_\beta|)\).