这是一份liverpool利物浦大学PHYS470的成功案例

$$
\omega=\frac{u}{b},
$$
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
$$
\left|v^{P}\right|=2 u \cos (\theta / 2), \quad(-\pi \leq \theta \leq \pi)
$$

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 $\mathcal{F}^{\prime}$.