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106 lines
2.2 KiB
106 lines
2.2 KiB
3 years ago
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\documentclass{article}
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\usepackage[margin=2cm]{geometry}
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\usepackage{titlesec}
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\usepackage{titling}
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\usepackage[hidelinks]{hyperref}
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\usepackage{multicol}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{braket}
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\titleformat{\section}
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{\Large\bfseries}
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{}
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{0em}
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{}[\titlerule]
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\titleformat{\subsection}
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{\large\bfseries}
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{}
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{0em}
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{}
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\titlespacing{\subsection}
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{0em}{2em}{.4em}
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\titleformat{\subsubsection}[runin]
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{\bfseries}
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{}
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\titlespacing{\subsubsection}
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{0em}{2em}{1em}
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\renewcommand{\maketitle}{
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\begin{center}
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{\huge\bfseries\thetitle}\\
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\vspace{1em}
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{\Large\theauthor} \\
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\vspace{1em}
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elalmqvist@gmail.com --- \url{https://wych.dev}
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\end{center}
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}
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\begin{document}
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\title{Anteckningar 2022-05-05}
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\author{Elias Almqvist}
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\maketitle
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\newpage
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\section{Fallande kropp i atmosfär}
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Något som faller har tyngdaccelerationen $g$ i $\frac{m}{s^2}$ och en massa $m$ i kg. $v$ är dess hastighet i $\frac{m}{s}$ och $F$ är kraften som verkas på den i newton. Den har dessutom en luftmotståndsfunktion: $\gamma(v)$. Låt positiva ($v > 0$) gå uppåt:
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$$
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F_g = -mg
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$$
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$$
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F_\gamma = \gamma(v)
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$$
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$$
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\implies F = \Delta F = F_g - F_\gamma = -mg - \gamma(v)
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$$
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För att få den resulterande accelerationen $a$ dividerar vi med $m$:
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$$
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a = \frac{F}{m} \because F = ma
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$$
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$$
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a = \frac{-mg - \gamma(v)}{m} = -\left(g + \frac{\gamma(v)}{m}\right)
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$$
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Vi kan också skriva om $a$:
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$$
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a = \frac{dv}{dt} = \dot{v} = -\left(g + \frac{\gamma(v)}{m}\right), \quad +\frac{\gamma(v)}{m}
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$$
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$$
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\therefore \dot{v} + \frac{1}{m} \gamma(v) = -g
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$$
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Nu varierar ekvationen beroende på vad $\gamma(v)$ är för något. Men vi kan anta att $F_\gamma \propto v$ vilket ger:
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$$
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\dot{v} + \frac{1}{m} kv = -g \quad | \quad k \in \mathbb{R}
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$$
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$$
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\implies v_h = Ce^{-\frac{k}{m} t} \quad | \quad C \in \mathbb{R}
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$$
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$$
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\implies v_p = \frac{-mg}{k}
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$$
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$$
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\therefore v_a = v_p + v_h = Ce^{-\frac{k}{m} t} - \frac{mg}{k}
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$$
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Vi har nu den en generell lösning. Behöver bara lösa vad $C$ är givet ett villkor. etc etc. \\
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Exempel:
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$$
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v(0) = 0 \implies C - \frac{mg}{k} = 0 \implies C = \frac{mg}{k}
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$$
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$$
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\therefore v = \frac{mg}{k} \left( e^{\frac{-k}{m}t} - 1 \right)
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$$
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\end{document}
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