$\Sigma F, a$ towards centre, $v$ tangential
- $F_{centrip} = {{mv^2} \over r} = {{4 \pi^2 rm} \over T^2}$
+ $\Sigma F = F_{centrip} = {{mv^2} \over r} = {{4 \pi^2 rm} \over T^2}=T \sin \theta = mg \tan \theta$
\includegraphics[height=4cm]{graphics/circ-forces.png}
% -----------------------
\subsection*{Vertical circular motion}
- $T =$ tension, e.g. circular pendulum
+ % $T =$ tension, e.g. circular pendulum
- $T+mg = {{mv^2}\over r}$ at highest point
+ $T+mg = {{mv^2}\over r}, v = \sqrt{rg}$ (top)
- $T-mg = {{mv^2} \over r}$ at lowest point
+ $T-mg = {{mv^2} \over r}, v = \sqrt{2rg}$ (bottom)
$E_K_{\text{bottom}}=E_K_{\text{top}}+mgh$
\item{Force-time: $A=\Delta \rho$}
\item{Force-disp: $A=W$}
\item{Force-ext: $m=k,\quad A=E_{spr}$}
- \item{Force-dist: $A=\Delta \operatorname{gpe}$}
+ \item{$F_g$-dist: $A=\Delta \operatorname{gpe}$}
\item{Field-dist: $A=\Delta \operatorname{gpe} / \operatorname{kg}$}
\end{itemize}
\item closer field lines means larger force
\item dot: out of page, cross: into page
\item +ve corresponds to N pole
+ \item Inv. sq. ${E_1 \over E_2} = ({r_2 \over r_1})^2$
\end{itemize}
\includegraphics[height=2cm]{graphics/field-lines.png}
\[{V_p \over V_s}={N_p \over N_s}={I_s \over I_p} \tag{xfmr coil ratios} \]
\textbf{Lenz's law:} $I_{\operatorname{emf}}$ opposes $\Delta \Phi$ \\
- (emf creates $I$ with associated field that opposes $\Delta \phi$)
+ (emf creates $I$ with associated field that opposes $\Delta \Phi$)
\textbf{Eddy currents:} counter movement within a field
\subsection*{Power transmission}
% \begin{align*}
- \[V_{\operatorname{rms}}={V_{\operatorname{p\rightarrow p}}\over \sqrt{2}} \]
+ \[V_{\operatorname{rms}}={V_{\operatorname{p}}\over \sqrt{2}}={V_{\operatorname{p\rightarrow p}}\over {2 \sqrt{2}}} \]
\[P_{\operatorname{loss}} = \Delta V I = I^2 R = {{\Delta V^2} \over R} \]
\[V_{\operatorname{loss}}=IR \]
% \end{align*}
\includegraphics[height=4cm]{graphics/dc-motor-2.png}
\includegraphics[height=3cm]{graphics/ac-motor.png} \\
- Force on current-carying wire, not copper \\
+ Force on I-carying wire, not Cu \\
$F=0$ for front & back of coil (parallel) \\
Any angle $> 0$ will produce force \\
% \end{wrapfigure}
% \(\Delta x\) = fringe spacing \\
\(l\) = distance from source to observer\\
\(d\) = separation between each wave source (e.g. slit) \(=S_1-S_2\)
- \item diffraction $\propto {\lambda \over d}$
+ \item diffraction $\propto {\lambda \over d} \propto$ fringe spacing
+ \item $d(|\overrightarrow{S_1W}|-|\overrightarrow{S_2W}|)=d \Delta x = \lambda l$
\item significant diffraction when ${\lambda \over \Delta x} \ge 1$
- \item diffraction creates distortion (electron $>$ optical microscopes)
+ \item diffraction creates distortion (electron $\gt$ optical microscopes)
\end{itemize}
\subsection*{Refraction}
\includegraphics[height=3.5cm]{graphics/refraction.png}
- When a medium changes character, light is \emph{reflected}, \emph{absorbed}, and \emph{transmitted}
+ When a medium changes character, light is \emph{reflected}, \emph{absorbed}, and \emph{transmitted}. $\lambda$ changes, not $f$.
angle of incidence $\theta_i =$ angle of reflection $\theta_r$
\subsection*{Uncertainty principle}
- measuring location of an e- requires hitting it with a photon, but this causes $\rho$ to be transferred to electron, moving it.
+ $\Delta x \approx {\text{slit width} \over 2$}
+
+ measurement: $\rho$ transferred to e-\\ slit: possibility of diff. before slit
\subsection*{Wave-particle duality}