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# 数学代写|数学建模代写Mathematical Modeling代考|Components of Velocity and Acceleration Vectors along Radial and Transverse Directions

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## 数学代写|数学建模代写Mathematical Modeling代考|Components of Velocity and Acceleration Vectors along Radial and Transverse Directions

As the particle moves from $P$ to $Q$, the displacement along the radius vector
$$=O N-O P=(r+\Delta r) \cos \Delta \theta-r$$
and the radial component $u$ of velocity is
\begin{aligned} u & =\operatorname{Lt}_{\Delta t \rightarrow 0} \frac{(r+\Delta r) \cos \Delta \theta-r}{\Delta t} \ & =\underset{\Delta t \rightarrow 0}{\operatorname{Lt}} \frac{\Delta r}{\Delta t}=\frac{d r}{d t} \end{aligned}
Similarly the displacement perpendicular to the radius vector
$$=(r+\Delta r) \sin \Delta \theta$$
and the transverse component $v$ of the velocity is given by
$$v=\operatorname{Lt}{\Delta t \rightarrow 0} \frac{(r+\Delta r) \sin \Delta \theta}{\Delta t}=\operatorname{Lt}{\Delta t \rightarrow 0} r \frac{\sin \Delta \theta}{\Delta \theta} \frac{\Delta \theta}{\Delta t}=r \frac{d \theta}{d t}$$
As such the velocity components in polar coordinates are
$$u=\frac{d r}{d t}=r^{\prime} \text { and } v=r \frac{d \theta}{d t}=r \theta^{\prime}$$

## 数学代写|数学建模代写Mathematical Modeling代考|Motion Under a Central Force

Let the force acting on a particle of mass $m$ be $m F(r)$ and let it be directed toward the origin, then the equations of motion are
$$\begin{gathered} m\left(r^{\prime \prime}-r^{\theta^{\prime 2}}\right)=-m F(r) \ \frac{m}{r} \frac{d}{d t}\left(r^2 \theta^{\prime}\right)=0 \end{gathered}$$
From Eqn. (11)
$$r^2 \theta^{\prime}=\text { Constant }=h$$
then Eqn. (10) gives
$$r^{\prime \prime}-r^{\theta^{\prime 2}}=-F(r)$$
We can eliminate $t$ between Eqns. (12) and (13) to get a differential equation between $r$ and $\theta$. We find it convenient to use $u=1 / r$ instead of $r$, so that making use of Eqn. (12), we get
$$r^{\prime}=\frac{d r}{d t}=\frac{d r}{d u} \frac{d u}{d \theta} \frac{d \theta}{d t}=-\frac{1}{u^2} \frac{d u}{d \theta} \frac{h}{r^2}=-h \frac{d u}{d \theta}$$
and
\begin{aligned} r^{\prime \prime} & =\frac{d}{d t}\left(-h \frac{d u}{d t}\right)=\frac{d}{d \theta}\left(-h \frac{d u}{d \theta}\right) \frac{d \theta}{d t} \ & =-h \frac{d^2 u}{d \theta^2} h u^2=-h^2 u^2 \frac{d^2 u}{d \theta^2} \end{aligned}
From Eqns. (12), (13), and (15)
$$\begin{gathered} -F(r)=-h^2 u^2 \frac{d^2 u}{d \theta^2}-\frac{1}{u} h^2 u^4=-h^2 u^2\left(\frac{d^2 u}{d \theta^2}+u\right) \ \frac{d^2 u}{d \theta^2}+u=\frac{F}{h^2 u^2} \end{gathered}$$
where $F$ can be easily expressed as a function of $u$. This is the differential equation of the second order whose integration will give the relation between $u$ and $\theta$ or between $r$ and $\theta$, i.e., the equation of the path described by a particle moving under a central force $F$ per unit mass.

## 数学代写|数学建模代写Mathematical Modeling代考|Components of Velocity and Acceleration Vectors along Radial and Transverse Directions

$$=O N-O P=(r+\Delta r) \cos \Delta \theta-r$$

\begin{aligned} u & =\operatorname{Lt}_{\Delta t \rightarrow 0} \frac{(r+\Delta r) \cos \Delta \theta-r}{\Delta t} \ & =\underset{\Delta t \rightarrow 0}{\operatorname{Lt}} \frac{\Delta r}{\Delta t}=\frac{d r}{d t} \end{aligned}

$$=(r+\Delta r) \sin \Delta \theta$$

$$v=\operatorname{Lt}{\Delta t \rightarrow 0} \frac{(r+\Delta r) \sin \Delta \theta}{\Delta t}=\operatorname{Lt}{\Delta t \rightarrow 0} r \frac{\sin \Delta \theta}{\Delta \theta} \frac{\Delta \theta}{\Delta t}=r \frac{d \theta}{d t}$$

$$u=\frac{d r}{d t}=r^{\prime} \text { and } v=r \frac{d \theta}{d t}=r \theta^{\prime}$$

## 数学代写|数学建模代写Mathematical Modeling代考|Motion Under a Central Force

$$\begin{gathered} m\left(r^{\prime \prime}-r^{\theta^{\prime 2}}\right)=-m F(r) \ \frac{m}{r} \frac{d}{d t}\left(r^2 \theta^{\prime}\right)=0 \end{gathered}$$

$$r^2 \theta^{\prime}=\text { Constant }=h$$

$$r^{\prime \prime}-r^{\theta^{\prime 2}}=-F(r)$$

$$r^{\prime}=\frac{d r}{d t}=\frac{d r}{d u} \frac{d u}{d \theta} \frac{d \theta}{d t}=-\frac{1}{u^2} \frac{d u}{d \theta} \frac{h}{r^2}=-h \frac{d u}{d \theta}$$

\begin{aligned} r^{\prime \prime} & =\frac{d}{d t}\left(-h \frac{d u}{d t}\right)=\frac{d}{d \theta}\left(-h \frac{d u}{d \theta}\right) \frac{d \theta}{d t} \ & =-h \frac{d^2 u}{d \theta^2} h u^2=-h^2 u^2 \frac{d^2 u}{d \theta^2} \end{aligned}

$$\begin{gathered} -F(r)=-h^2 u^2 \frac{d^2 u}{d \theta^2}-\frac{1}{u} h^2 u^4=-h^2 u^2\left(\frac{d^2 u}{d \theta^2}+u\right) \ \frac{d^2 u}{d \theta^2}+u=\frac{F}{h^2 u^2} \end{gathered}$$

## MATLAB代写

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