数学代写|matlab仿真代写simulation代做|Planar Dynamics Analysis

MATLAB是一个编程和数值计算平台，被数百万工程师和科学家用来分析数据、开发算法和创建模型。

MATLAB主要用于数值运算，但利用为数众多的附加工具箱，它也适合不同领域的应用，例如控制系统设计与分析、影像处理、深度学习、信号处理与通讯、金融建模和分析等。另外还有配套软件包提供可视化开发环境，常用于系统模拟、动态嵌入式系统开发等方面。

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|matlab仿真代写simulation代做|Planar Dynamics Analysis

Consider an arbitrary rigid body in motion with the total mass $m$. The position of the mass center of the rigid body is $\mathbf{r}{C}$ and $\mathbf{a}{C}$ is the acceleration of the mass center $C$. The sum of the external forces, $\mathbf{F}$, acting on the system equals the product of the mass and the acceleration of the mass center
$$m \mathbf{a}{C}=\mathbf{F}$$ Equation (1.44) is Newton’s second law for a rigid body and is applicable to planar and three-dimensional motions. Resolving the sum of the external forces into Cartesian rectangular components $$\mathbf{F}=F{x} \mathbf{1}+F_{y} \mathbf{J}+F_{z} \mathbf{k}$$
and the position vector of the mass center
$$\mathbf{r}{C}=x{C}(t) \mathbf{1}+y_{C}(t) \mathbf{j}+z_{C}(t) \mathbf{k},$$
Newton’s second law for the rigid body is
$$m \ddot{\mathbf{r}}{C}=\mathbf{F}$$ or $$m \ddot{x}{C}=F_{x}, \quad m \ddot{y}{C}=F{y}, \quad m \ddot{z}{C}=F{z} .$$

Figure $1.14$ shows a rigid body moving in the $(x, y)$ plane with the origin at $O$. The mass center $C$ of the rigid body is in the plane of the motion. The axis $O z$ is perpendicular to the plane of motion of the rigid body, where $O$ is the origin. The axis $C z$ is perpendicular to the plane of motion through the mass center $C$. The angular velocity vector of the rigid body is $\omega=\omega \mathbf{k}$ and the angular acceleration is $\alpha=\dot{\omega}=\ddot{\theta} \mathbf{k}$, where the angle $\theta$ is the angular position of the rigid body about a fixed axis. The mass moment of inertia of the rigid body about the $z$-axis through $C$ is also the polar mass moment of inertia of the rigid body about $C, I_{C z}=I_{C}$.
The rotational equation (Euler’s equation) of motion for the rigid body is
$$I_{C} \boldsymbol{\alpha}=\sum \mathbf{M}{C},$$ where $\mathbf{M}{C}$ is the sum of the moments about $C$ due to external forces and couples.
Consider the case when the rigid body rotates about a fixed point $P$. The mass moment of inertia of the rigid body about the $z$-axis through $P$ is (parallel-axis theorem)
$$I_{P z}=I_{C z}+m(C P)^{2}$$
For the rigid body rotation about a fixed point $P$, Euler’s equation is
$$I_{P z} \boldsymbol{\alpha}=\sum \mathbf{M}{P}$$ where $\mathbf{M}{P}$ is the sum of the moments about the fixed point $P$ due to external forces and couples.

Abstract The planar motion of a mechanism with three moving links is analyzed. Symbolical and numerical MATLAB are used for the kinematics and dynamics of the system. The classical vectorial equations for velocity and acceleration of the rigid body are used. The joint reaction forces and the moment applied to the driver link are calculated for a given position with Newton-Euler equations for inverse dynamics.
The planar R-RTR mechanism considered is shown in Fig. 2.1. The driver link is the rigid link 1 (the link $A B)$. The dyad RTR $\left(B_{\mathrm{R}}, B_{\mathrm{T}}, C_{\mathrm{R}}\right)$ is composed of the slider 2 and the rocker 3 . The following numerical data are given: $A B=0.10 \mathrm{~m}$, $A C=0.05 \mathrm{~m}$, and $C D=0.15 \mathrm{~m}$. The angle of the driver link 1 with the horizontal axis is $\phi=30^{\circ}$. The constant angular speed of the driver link 1 is $-50 \mathrm{rpm}$.

Given an external moment $\mathbf{M}{c}=-100 \operatorname{sign}\left(\omega{3}\right) \mathbf{k} \mathrm{Nm}$ applied on the link 3 , calculate the motor moment $\mathbf{M}{m}$ required for the dynamic equilibrium of the mechanism. All three links are rectangular prisms with the depth $d=0.001 \mathrm{~m}$ and the mass density $\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}$. The height of the links 1 and 3 is $h=0.01 \mathrm{~m}$. The slider 2 has the height $h{S}=0.02 \mathrm{~m}$, and the width $w_{S}=0.04 \mathrm{~m}$. The center of mass location of the links $i=1,2,3$ are designated by $C_{i}\left(x_{C_{i}}, y_{C_{i}}, 0\right)$. The gravitational acceleration is $g=9.807 \mathrm{~m} / \mathrm{s}^{2}$.

数学代写|matlab仿真代写simulation代做|Position Analysis

A Cartesian reference frame $x y$ is selected. The joint $A$ is in the origin of the reference frame, that is, $A \equiv O$,
$$x_{A}=0 \text { and } y_{A}=0 .$$
The coordinates of the joint $C$ are
$$x_{C}=A C=0.05 \text { and } y_{C}=0 \mathrm{~m} .$$

Position of joint $B$
The unknowns are the coordinates of the joint $B, x_{B}$ and $y_{B}$. Because the joint $A$ is fixed and the angle $\phi$ is known, the coordinates of the joint $B$ are computed from the following expressions
\begin{aligned} &x_{B}=A B \cos \phi=0.10 \cos 30^{\circ}=0.087 \mathrm{~m} \ &y_{B}=A B \sin \phi=0.10 \sin 30^{\circ}=0.050 \mathrm{~m} \end{aligned}
The MATLAB commands for joints $A, C$, and $B$ are:
$A B=0.10 ; \quad$ 읳 $(\mathrm{m})$
$A C=0.05 ; \quad$ 읳 $(\mathrm{m})$
$C D=0.15 ;$ 앟 (m)
phi $=$ pi $/ 6 ;$ 웋 (rad)
$x A=0 ; y A=0$;
$r A_{-}=[x A y A l] ;$
$x C=A C ; y C=0$;
$r C_{-}=[x C y C 0] ;$
$\mathrm{xB}=A \mathrm{~B}^{} \cos ($ phi $)$; $y B=A B^{} \sin ($ phi $)$;
$r B_{-}=[x B y B 0] ;$
Angle $\phi_{2}$
The angle of link 2 (or link 3 ) with the horizontal axis is calculated from the slope of the straight line $B C$ :
$$\phi_{2}=\phi_{3}=\arctan \frac{y_{B}-y_{C}}{x_{B}-x_{C}}=\arctan \frac{0.050}{0.087-0.050}=0.939 \mathrm{rad}=53.794^{\circ}$$

数学代写|matlab仿真代写simulation代做|Planar Dynamics Analysis

rC=XC(吨)1+是C(吨)j+和C(吨)ķ,

数学代写|matlab仿真代写simulation代做|Position Analysis

X一种=0 和 是一种=0.

XC=一种C=0.05 和 是C=0 米.

X乙=一种乙因⁡φ=0.10因⁡30∘=0.087 米 是乙=一种乙罪⁡φ=0.10罪⁡30∘=0.050 米

CD=0.15;(m)

X一种=0;是一种=0;
r一种−=[X一种是一种l];
XC=一种C;是C=0;
rC−=[XC是C0];
$\mathrm{xB}=A\mathrm{~B}^{ } \cos(pH一世);和 B=AB^{ } \sin (pH一世);r B_{-}=[x B y B 0] ;一种nGl和\phi_{2}吨H和一种nGl和这Fl一世nķ2(这rl一世nķ3)在一世吨H吨H和H这r一世和这n吨一种l一种X一世s一世sC一种lC在l一种吨和dFr这米吨H和sl这p和这F吨H和s吨r一种一世GH吨l一世n和公元前:φ2=φ3=反正切⁡是乙−是CX乙−XC=反正切⁡0.0500.087−0.050=0.939r一种d=53.794∘$

有限元方法代写

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MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。