Today's lecture was a review for the second midterm exam, covering everything since the first midterm. As before, most of the lecture was spent going over problems, but I'm reproducing my condensed set of formulas here (not guaranteed to be 100% complete.)
Net external force acts on the center of mass (CM) like a point particle,
\[ \begin{aligned} \vec{F}_{\rm net, ext} = M \ddot{\vec{R}}. \end{aligned} \]
Total mass:
\[ \begin{aligned} M = \sum_\alpha m_\alpha \end{aligned} \]
or continuous version with density function \( \rho(\vec{r}) \):
\[ \begin{aligned} M = \int dm = \int \rho(\vec{r}) dV. \end{aligned} \]
CM position:
\[ \begin{aligned} \vec{R} = \frac{1}{M} \sum_\alpha m_\alpha \vec{r}_\alpha \end{aligned} \]
or continuous version,
\[ \begin{aligned} \vec{R} = \frac{1}{M} \int \vec{r} dm = \int \rho(\vec{r}) \vec{r}\ dV. \end{aligned} \]
Reflection symmetry: contributions by equal masses on opposite sides of a coordinate axis, e.g. \( (x,y,z) \) and \( (-x,y,z) \), cancel out for CM position. Complete reflection symmetry in a direction --> that CM coordinate is zero.
Rocket equation of motion:
\[ \begin{aligned} m \frac{dv}{dt} = F_{\rm thrust},\ \ F_{\rm thrust} = -|v_{\rm exh}| \frac{dm}{dt} \end{aligned} \]
with fuel exhaust speed \( |v_{\rm exh}| \) (always positive) and mass ejection rate \( dm/dt \) (always negative!) If other forces are present, add alongside \( \vec{F}_{\rm thrust} \) as normal.
Speed-mass relationship:
\[ \begin{aligned} v(m) = v_0 + |v_{\rm exh}| \ln(m_0/m) \end{aligned} \]
If initial mass \( m_0 = m_E + m_F \), where \( m_F \) is fuel mass and \( m_E \) is empty mass, then max delta-v from burning all fuel is
\[ \begin{aligned} \Delta v_{\rm max} = |v_{\rm exh}| \ln \left( 1 + \frac{m_F}{m_E} \right). \end{aligned} \]
Definition of work done by force \( \vec{F} \) from \( \vec{r}_1 \) to \( \vec{r}_2 \):
\[ \begin{aligned} W = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r} \end{aligned} \]
(note: depends on the path taken, in general!)
Work-KE theorem:
\[ \begin{aligned} \Delta T = W_{\rm net} \end{aligned} \]
Kinetic energy: \( T = \frac{1}{2} mv^2 \) Potential energy (only if \( \vec{F} \) is conservative):
\[ \begin{aligned} U(\vec{r}) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F} \cdot d\vec{r} \end{aligned} \]
Conservation of energy: \( E = T + U \) is time-independent if only conservative forces are present. If we have non-conservative forces, then
\[ \begin{aligned} \Delta E = W_{\rm non-cons} \end{aligned} \]
Force from potential:
\[ \begin{aligned} \vec{F} = -\vec{\nabla} U \end{aligned} \]
A force is conservative if \( \vec{\nabla} \times \vec{F} = 0 \) everywhere.
"1-d system": whenever a single coordinate \( x \) is enough to describe the motion.
Total energy \( E = T + U(x) \). Turning points where \( E=U(x) \), and thus \( T=0 \). "Forbidden regions" if \( T < 0 \) - system can't reach these points with the given \( E \).
Equilibrium points where \( dU/dx = 0 \). Taylor expand at equilibrium point \( x_0 \) to test stability:
Stable equilibrium: \( F(x) \approx -k(x-x_0) \) or \( U(x) \approx +\frac{1}{2} k(x-x_0)^2 \) - oscillating solutions.
Unstable equilibrium: \( U(x) \approx +k(x-x_0) \) or \( U(x) \approx -\frac{1}{2} k(x-x_0)^2 \) - exponential solutions.
Definition of central force, with source at origin:
\[ \begin{aligned} \vec{F}(\vec{r}) = f(\vec{r}) \hat{r} \end{aligned} \]
A conservative, central force has \( f(\vec{r}) = f(r) \), i.e. force only depends on distance.
Gravitational force, on mass \( m \) due to source \( M \) at origin:
\[ \begin{aligned} \vec{F}_g = -\frac{GMm}{r^2} \hat{r} \end{aligned} \]
Potential energy:
\[ \begin{aligned} U(r) = -\frac{GMm}{r} \end{aligned} \]
Gravitational field:
\[ \begin{aligned} \vec{g} = \frac{\vec{F}_g}{m} = -\frac{GM}{r^2} \hat{r} \end{aligned} \]
Gravitational potential:
\[ \begin{aligned} \Phi(r) = \frac{U(r)}{m} = -\frac{GM}{r} \end{aligned} \]
Dealing with extended sources: gravitational field at \( \vec{r} \) due to source at \( \vec{r}' \) (overall source mass density \( \rho(\vec{r}') \)) is
\[ \begin{aligned} \vec{g} = -G \int \frac{dV' \rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}') \\ = -G \int \frac{dm}{s^2} \hat{s} \end{aligned} \]
where \( dm = \rho(\vec{r}') dV' \) and \( \vec{s} = \vec{r} - \vec{r}' \).
Or, calculate gravitational potential directly:
\[ \begin{aligned} \Phi(\vec{r}) = -G \int \frac{dm}{s} = -G \int \frac{dV' \rho(\vec{r}')}{|\vec{r} - \vec{r}'|}. \end{aligned} \]
Gauss' law: given a volume of space \( V \) with closed boundary surface \( \partial V \),
\[ \begin{aligned} \oint_{\partial V} \vec{g} \cdot d\vec{A} = -4\pi G \int_V \rho(\vec{r}) dV \\ = -4\pi G M_{\rm enc} \end{aligned} \]
(where \( M_{\rm enc} \) is the enclosed mass inside of \( V \).)
To use in practice, use symmetry to find a Gaussian surface over which \( \vec{g} \) points in the same direction as \( d\vec{A} \), the normal vector to the surface. Then we pull \( \vec{g} \) out of the integral on the left-hand side, leaving the surface area of \( \partial V \).
Finally, here's the solution to the one practice problem that isn't from Taylor, the gravity problem I assigned:
Consider a thin rod of uniform density with mass \( M \) and length \( L \). Find the gravitational force exerted on a point mass \( m \) located a distance \( D \) away from the end of the rod, along the direction of its axis. (In other words, if the rod is on the \( x \)-axis from \( x=0 \) to \( x=L \), the point mass is at \( x=L+D \).)
Answer: Since this is a one-dimensional problem, I'll just use the gravitational force integral directly:
\[ \begin{aligned} \vec{g} = -G \int \frac{dm}{s^2} \hat{s} \end{aligned} \]
As given, the point mass is at \( x=L+D \), which means that the vector pointing to it is \( \vec{r} = (L+D) \hat{x} \). Meanwhile, the vector \( \vec{r}' = x' \hat{x} \) will point to \( dm \) along the rod. The difference is \( \vec{s} = \vec{r} - \vec{r}' = [(L+D) - x'] \hat{x} \). (You can draw a diagram for this if you like, I'm skipping it since it's just in one dimension.)
Normally it would be a pain to find \( \hat{s} \) from \( \vec{s} \), but in one dimension the answer is nice and simple: \( \hat{s} = \hat{x} \). Finally, for \( dm \) along the rod we have \( dm = \lambda dx' = (M/L) dx' \). Thus, we have for our integral
\[ \begin{aligned} \vec{g} = -G \int_0^L \frac{(M/L) dx'}{((L+D)-x')^2} \hat{x} \\ = -\frac{GM}{L} \hat{x} \int_{L+D}^{D} \frac{-du}{u^2} \end{aligned} \]
using \( u = (L+D) - x' \) and thus \( du = -dx' \). Doing the integral,
\[ \begin{aligned} \vec{g} = \frac{GM}{L} \hat{x} \left. \frac{1}{u} \right|_{D}^{L+D} \\ = \frac{GM}{L} \hat{x} \left[ \frac{1}{L+D} - \frac{1}{D} \right] \\ = \frac{GM}{L} \hat{x} \left[ \frac{D - (L+D)}{D(L+D)} \right] \\ = -\frac{GM}{D(L+D)} \hat{x}. \end{aligned} \]
We got the expected minus sign out at the very end, but notice there were a lot of sign flips in the middle there! Finally, to get the gravitational force, we add the other mass in:
\[ \begin{aligned} \vec{F}_g = -\frac{GmM}{D(L+D)} \hat{x}. \end{aligned} \]
As a check, if \( D \gg L \), then \( L+D \) is approximately the same as \( D \), and we recover the expected force proportional to \( GmM/D^2 \).