## THE RELATION BETWEEN COLLISION DYNAMICS OF SLIDER AND PI 1)

LI Kaiwei,2)

Guangdong Polytechnic College, Zhaoqing 526100, Guangdong, China

 基金资助: 1)广东理工学院科技项目资助.  2020GKJZD003

Abstract

The collision of slider seems not relevant to Pi, but there is a certain connection between the two things really. In this paper, the collision dynamics of slider is studied. By using MATLAB, the connection between the number of collision of slider and Pi is established. The calculation results are proved theoretically.

Keywords： collision of slider ; Pi ; number of collision

LI Kaiwei. THE RELATION BETWEEN COLLISION DYNAMICS OF SLIDER AND PI 1). MECHANICS IN ENGINEERING[J], 2021, 43(1): 108-111 DOI:10.6052/1000-0879-20-151

## 1 问题来源

### 图1

$\begin{eqnarray} \left. {{\begin{array}{l} {mu_{1} +Mv_{1} =Mv_{0} }\\ {\dfrac{1}{2}mu_{1}^{2} +\dfrac{1}{2}Mv_{1}^{2} =\dfrac{1}{2}Mv_{0}^{2} }\\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left. {{\begin{array}{l} {u_{1} =\dfrac{2k}{k+1}v_{0} } \\[2mm] {v_{1} =\dfrac{k-1}{k+1}v_{0} } \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left. {{\begin{array}{l} {u_{2} =-u_{1} } \\ {v_{2} =v_{1} } \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left. {{\begin{array}{*{20}c} {mu_{3} +Mv_{3} =mu_{2} +Mv_{2} } \\ {\dfrac{1}{2}mu_{3}^{2} +\dfrac{1}{2}Mv_{3}^{2} =\dfrac{1}{2}mu_{2}^{2} +\dfrac{1}{2}Mv_{2}^{2} } \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left. {{\begin{array}{*{20}c} {u_{3} =\dfrac{(1-k)u_{2} +2kv_{2} }{k+1}} \\[2mm] {v_{3} =\dfrac{2u_{2} +(k-1)v_{2} }{k+1}} \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left\{ {{\begin{array}{*{20}c} {u_{2} } \\ {v_{2} } \\ \end{array} }} \right\}=\left[ {{\begin{array}{*{20}c} {-1} & 0 \\ 0 & 1 \\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} {u_{1} } \\ {v_{1} } \\ \end{array} }} \right\} \end{eqnarray}$
$\begin{eqnarray} \left\{ {{\begin{array}{*{20}c} {u_{3} } \\ {v_{3} } \\ \end{array} }} \right\}=\frac{1}{k+1}\left[ {{\begin{array}{*{20}c} {1-k} & {2k} \\ 2 & {k-1} \\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} {u_{2} } \\ {v_{2} } \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \left\{ {{\begin{array}{*{20}c} {u_{1} } \\ {v_{1} } \\ \end{array} }} \right\}=\frac{1}{k+1}\left[ {{\begin{array}{*{20}c} {1-k} & {2k} \\ 2 & {k-1} \\ \end{array} }} \right]\left\{ {{\begin{array}{*{20}c} 0 \\ {v_{0} } \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \frac{1}{k+1}\left[ {{\begin{array}{*{20}c} {1-k} & {2k} \\ 2 & {k-1} \\ \end{array} }} \right],\ \ \left[ {{\begin{array}{*{20}c} {-1} & 0 \\ 0 & 1 \\ \end{array} }} \right] \end{eqnarray}$

## 2 理论证明

$\begin{eqnarray} 0\leqslant x(t)\leqslant y(t) \end{eqnarray}$

$\begin{eqnarray} \varphi =\psi \end{eqnarray}$

### 图3

$\begin{eqnarray} \theta_{1} =45^{\circ},\ \ \tan \varphi =u_{1} /v_{1},\ \ \theta_{2}=45^{\circ}+\varphi, \cdots \end{eqnarray}$

$\begin{eqnarray} \left( {X\ \ Y} \right)=(x\ \ y)\left[ {{\begin{array}{*{20}c} {\sqrt m } & 0 \\ 0 & {\sqrt M } \\ \end{array} }} \right] \end{eqnarray}$

$\begin{eqnarray} \tan \alpha =\sqrt {\frac{m}{M}} \end{eqnarray}$

### 图4

$\begin{eqnarray} \left. {{\begin{array}{l} r\cdot w=r\cdot {w}'={\rm cons1}\\ |w|=|w'| ={\rm cons2} \\ \end{array} }} \right\} \end{eqnarray}$

$\begin{eqnarray} \varphi_{1}=\varphi_{2} \end{eqnarray}$

$\begin{eqnarray} \left.\begin{array}{l} \alpha_{1} =\alpha_{2} =\alpha\\ \beta_{1}=\alpha +\alpha_{2} =2\alpha\\ \alpha_{3}=\alpha +\beta_{2} =3\alpha\\ \qquad \quad \vdots \end{array}\right\} \end{eqnarray}$

$\begin{eqnarray} n\alpha <\pi \end{eqnarray}$

$\begin{eqnarray} n=\left[ {\frac{\pi }{\alpha }} \right] \end{eqnarray}$

$\begin{eqnarray} n=\left[ {\frac{\pi }{\arctan \sqrt {{m}/{M}} }} \right]=\left[ {\frac{\pi }{\arctan \sqrt {{1}/{k}} }} \right] \end{eqnarray}$

$\begin{eqnarray} n=\left[ {\frac{\pi }{\arctan (10^{-N})}} \right] \end{eqnarray}$

$\begin{eqnarray} \arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots \end{eqnarray}$

$\begin{eqnarray} &&\dfrac{1}{\arctan x}-\dfrac{1}{x}=\dfrac{x-\arctan x}{\arctan x\cdot x}=\\&&\qquad \dfrac{\dfrac{1}{3}x^{3}-\dfrac{1}{5}x^{5}+\dfrac{1}{7}x^{7}-\cdots}{x\left( {x-\dfrac{1}{3}x^{3}+\dfrac{1}{5}x^{5}-\cdots} \right)}= \\&&\qquad x\dfrac{\dfrac{1}{3}-\dfrac{1}{5}x^{2}+\dfrac{1}{7}x^{4}-\cdots}{1\mbox{-}x^{2}\left( {\dfrac{1}{3}-\dfrac{1}{5}x^{2}+\cdots} \right)} \end{eqnarray}$

$\begin{eqnarray*} \lim\limits_{x\to 0} \left( {\frac{1}{\arctan x}-\frac{1}{x}} \right)=0,\ \ 0<\frac{1}{\arctan x}-\frac{1}{x}<x \end{eqnarray*}$

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Ju Shi'er, Zhang Yijie.

Did Liu Hui and Zu Chongzhi compute the approximation of $\pi$

The Chinese Journal for the History of Science and Technology, 2019(4):389-401 (in Chinese)

Mikami Y. The Development of Mathematics in China and Japan. New York: Chelsea Publishing Company, 1913

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