Tag Archives: Classical Mechanics代写

经典力学 Classical Mechanics PHYS470

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这是一份liverpool利物浦大学PHYS470的成功案例

经典力学 Classical Mechanics PHYS470


$$
\omega=\frac{u}{b},
$$
and on using this value of $\omega$ in we find that the velocity of the typical particle $P$ is given by
$$
v^{P}=u\left(1+\frac{\rho}{b} \cos \theta\right) i-u\left(\frac{\rho}{b} \sin \theta\right) j
$$
When $P$ lies on the circumference of the wheel, this formula simplifies to
$$
v^{P}=u(1+\cos \theta) i-u \sin \theta \boldsymbol{j}
$$
in which case the speed of $P$ is given by
$$
\left|v^{P}\right|=2 u \cos (\theta / 2), \quad(-\pi \leq \theta \leq \pi)
$$

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PHYS470 COURSE NOTES ：

$$
v=\left(\frac{d \boldsymbol{r}^{\prime}}{d t}\right){\mathcal{F}}+\boldsymbol{V}, $$ where $v$ is the velocity of $P$ observed in $\mathcal{F}$ and $\boldsymbol{V}$ is the velocity of $\mathcal{F}^{\prime}$ relative to $\mathcal{F}$. Now when two different reference frames are used to observe the same vector, the observed rates of change of that vector will generally be different. In particular, it is not generally true that $$ \left(\frac{d \boldsymbol{r}^{\prime}}{d t}\right){\mathcal{F}}=\left(\frac{d \boldsymbol{r}^{\prime}}{d t}\right){\mathcal{F}^{\prime}} . $$ However, as we will show in Chapter 17, these two rates of change are equal if the frame $\mathcal{F}$ does not rotate relative to $\mathcal{F}$. Hence, in our case, we do have $$ \left(\frac{d \boldsymbol{r}^{\prime}}{d t}\right){\mathcal{F}}=\left(\frac{d \boldsymbol{r}^{\prime}}{d t}\right)_{\mathcal{F}^{\prime}}=v^{\prime},
$$
where $v^{\prime}$ is the velocity of $P$ observed in $\mathcal{F}^{\prime}$.










经典力学 Classical Mechanics MATH228

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这是一份liverpool利物浦大学MATH228的成功案例

经典力学 Classical Mechanics MATH228

$$
v=c \widehat{\boldsymbol{r}}+(c t) \Omega \widehat{\boldsymbol{\theta}}=c(\widehat{\boldsymbol{r}}+\Omega t \hat{\boldsymbol{\theta}})
$$
and
$$
\boldsymbol{a}=\left(0-(c t) \Omega^{2}\right) \widehat{\boldsymbol{r}}+(0+2 c \Omega) \widehat{\boldsymbol{\theta}}=c \Omega(-\Omega t \widehat{\boldsymbol{r}}+2 \widehat{\boldsymbol{\theta}}) .
$$
The speed of the particle at time $t$ is thus given by $|v|=c\left(1+\Omega^{2} t^{2}\right)^{1 / 2}$. To find the angle between $v$ and $a$, consider
$$
\begin{aligned}
v \cdot a &=c^{2} \Omega(-\Omega t+2 \Omega t)=c^{2} \Omega^{2} t \
&>0
\end{aligned}
$$
for $t>0$. Hence, for $t>0$, the angle between $v$ and $a$ is acute.

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MATH228 COURSE NOTES ：

Suppose a particle $P$ moves in any manner around the circle $r=b$, where $r, \theta$ are plane polar coordinates. Then the velocity and acceleration vectors of $P$ are given by
$$
\begin{aligned}
&v=v \widehat{\theta} \
&a=-\left(\frac{v^{2}}{b}\right) \widehat{\boldsymbol{r}}+\dot{v} \widehat{\boldsymbol{\theta}}
\end{aligned}
$$
where $v(=b \dot{\theta})$ is the circumferential velocity of $P$.








经典力学|Classical Mechanics代写PHAS00010代考

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这是一份UCL伦敦大学学院PHAS00010作业代写的成功案

经典力学|Classical Mechanics代写PHAS00010代考
问题 1.


In vector notation, the gradient of $V$ is written as either grad $V$ or $\vec{\nabla} V$. Then
$$
\vec{E}=-\vec{\nabla} V
$$
In general, the potential function can depend on $x, y$, and $z$. The Cartesian components of the electric field are related to the partial derivatives of the potential with respect to $x, y$, or $z$. For example, the $x$ component of the electric field is given by
$$
E_{x}=-\frac{\partial V}{\partial x}
$$

partial V}{\partial x}
$$


证明 .

Similarly, the $y$ and $z$ components of the electric field are related to the potential by
$$
E_{y}=-\frac{\partial V}{\partial y}
$$
and
$$
E_{z}=-\frac{\partial V}{\partial z}
$$
Thus, Equation 23-15 in Cartesian coordinates is written
$$
\vec{E}=-\vec{\nabla} V=-\left(\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}\right) \quad 23-17
$$

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PHAS00010 COURSE NOTES ：

Outside the spherical shell, the electric field is radial and is the same as if all the charge $Q$ were a point charge at the origin:
$$
\overrightarrow{\boldsymbol{E}}=\frac{k Q}{r^{2}} \hat{\boldsymbol{r}}
$$
where $\hat{\boldsymbol{r}}$ is a unit vector directed away from the center of the sphere. The change in the potential for some displacement $d \vec{\ell}$ outside the shell is then
$$
d V=-\overrightarrow{\boldsymbol{E}} \cdot d \vec{\ell}=-\frac{k Q}{r^{2}} \hat{\boldsymbol{r}} \cdot d \vec{\ell}=-\frac{k Q}{r^{2}} d r
$$
where the product $\hat{\boldsymbol{r}} \cdot d \overrightarrow{\boldsymbol{\ell}}$ is equal to $d r$ (the component of $d \vec{\ell}$ in the direction of $\hat{\boldsymbol{r}}$ ). Integrating along a path from the reference point at infinity, we obtain
$$
V_{p}=-\int_{\infty}^{\vec{r}{P}} \overrightarrow{\boldsymbol{E}} \cdot d \vec{\ell}=-\int{\infty}^{r_{p}} \frac{k Q}{r^{2}} d r=-k Q \int_{\infty}^{r_{p}} r^{-2} d r=\frac{k Q}{r_{p}}
$$