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Conical Pendulum.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Conical Pendulum} \begin{align*} \text{Hang off a point:}\quad&\text{An object of mass $m$ suspended by a string of length $l$ is performing uniform circular motion}\\ &\text{with angular velocity $\omega$. The inclination of the string and the vertical axis is $\theta$. There is tension}\\ &\text{$T$ on the string. The plane of circular motion is $h$ under the suspension point $A$.}\\ \\ &\ddot{x}=r\omega^2\:,\quad\text{where }\:\:r=l\sin\theta\:.\\ &F_N=T\sin\theta=m\ddot{x}=mr\omega^2=ml\sin\theta\cdot\omega^2\:,\quad\boxed{T=ml\omega^2}\:.\\ &\text{To balance the weight:}\quad T\cos\theta=mg\:,\quad ml\omega^2\cos\theta=mg\:,\quad\boxed{\cos\theta=\frac{g}{l\omega^2}}\:.\\ &h=l\cos\theta=\frac{g}{\omega^2}\:,\quad\boxed{h=\frac{g}{\omega^2}}\:.\\ &\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{T\sin\theta}{T\cos\theta}=\frac{mr\omega^2}{mg}=\frac{r^2\omega^2}{rg}=\frac{v^2}{rg}\:,\quad\boxed{\tan\theta=\frac{v^2}{rg}}\:.\\ &\text{Frequency }F=\frac{\omega}{2\pi}=\frac{\sqrt{\omega^2}}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{\:g\:}{h}}\:,\quad\boxed{f=\frac{1}{2\pi}\sqrt{\frac{\:g\:}{h}}}\:.\\ &\text{Period }T=\frac{\:1\:}{f}\:,\quad\boxed{T=2\pi\sqrt{\frac{\:h\:}{g}}}\:.\\ \\ \text{Hang off a disc:}\quad&\text{An object of mass $m$ suspended by a string of length $l$ hanging off the rim of a disc of radius $R$}\\ &\text{is performing uniform circular motion with angular velocity $\omega$. The inclination of the string and}\\ &\text{the vertical axis is $\theta$. There is tension $T$ on the string. The plane of motion is $h$ under the disc.}\\ \\ &\text{Radius of the circular motion }r=R+l\sin\theta\:,\quad \ddot{x}=r\omega^2=(R+l\sin\theta)\omega^2\:,\quad h=l\cos\theta\:.\\ &F_N=T\sin\theta=m\ddot{x}=mr\omega^2=m(R+l\sin\theta)\omega^2\:,\quad\boxed{T=m\left(\frac{R}{\sin\theta}+l\right)\omega^2}\:.\\ &\text{To balance the weight:}\quad mg=T\cos\theta=m\left(\frac{R}{\sin\theta}+l\right)\omega^2\cos\theta\:,\quad \frac{\:g\:}{\omega^2}=\frac{R}{\tan\theta}+l\cos\theta=\frac{R}{\tan\theta}+h\\ &\boxed{\tan\theta=\frac{R\omega^2}{g-h\omega^2}}\:,\quad \boxed{\omega=\sqrt{\frac{g\tan\theta}{R+l\sin\theta}}}\:,\quad \boxed{h=\frac{g}{\omega^2}-\frac{R}{\tan\theta}}\:,\quad \\ % &\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{T\sin\theta}{T\cos\theta}=\frac{mr\omega^2}{mg}=\frac{r^2\omega^2}{rg}=\frac{v^2}{rg}\:,\quad\boxed{\tan\theta=\frac{v^2}{(R+l\sin\theta)g}}\:,\quad\boxed{v^2=\tan\theta(R+l\sin\theta)g}\:.\\ % &\text{Frequency }F=\frac{\omega}{2\pi}\:,\quad\boxed{f=\frac{1}{2\pi}\sqrt{\frac{g\tan\theta}{R+l\sin\theta}}}\:.\\ &\text{Period }T=\frac{\:1\:}{f}\:,\quad\boxed{T=2\pi\sqrt{\frac{R+l\sin\theta}{g\tan\theta}}}\:.\\ \\ &\text{Note: When $R=0$, the disc becomes a point, so the above is reduced to the case of ``hanging off a point''.}\\ \end{align*} \end{document}