# 经典力学 Classical Mechanics MATH228

0

$$v=c \widehat{\boldsymbol{r}}+(c t) \Omega \widehat{\boldsymbol{\theta}}=c(\widehat{\boldsymbol{r}}+\Omega t \hat{\boldsymbol{\theta}})$$
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.

## MATH228COURSE 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 $P$.