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本帖最后由 hbghlyj 于 2022-2-5 07:24 编辑
力学讲义的39页
Example (Motion on a paraboloid): A particle moves under gravity on the smooth inside surface of the paraboloid $z=r^{2} / 4 a=H(r)$. Initially it is at a height $z=a$ and is projected horizontally with speed $v$ along the surface of the paraboloid. Show that the particle moves between two heights in the subsequent motion, and find them.
Solution: At $t=0, z=a$. Since $r^{2}=4 a z$ (the particle is on the paraboloid), initially $r=2 a$. Also
$$
\left.\begin{array}{rl}
r \dot{\theta} & =v \\
\dot{r} & =0 \\
\dot{z} & =0
\end{array}\right\} \text { at } t=0 .
$$
where Fig. 8b in particular highlights $r \dot{\theta}=v$. Thus
$$
h=r^{2} \dot{\theta}=r \times r \dot{\theta}=2 a v .\tag{5.31}
$$
Conservation of energy, on use of the initial conditions, gives
$$
\frac{1}{2} m \dot{\mathbf{r}}^{2}+m g z=\mathrm{constant}=\frac{1}{2} m v^{2}+m g a .\tag{5.32}\label{5.32}
$$
Thus
$$
\frac{1}{2}\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}+\dot{z}^{2}\right)+g z=\frac{1}{2} v^{2}+g a .\tag{5.33}
$$
Eliminate $\dot{\theta}$ and $r$ to get a first order differential equation for $z(t)$ only, as we are interested in heights of the motion.
We have
$$
r=2 \sqrt{a z}, \quad \dot{r}=\sqrt{\frac{a}{z}} \dot{z}\tag{5.34}
$$
Substituting into \eqref{5.32} using $\dot{\theta}=h / r^{2}$, and eliminating $r$ for $z$, gives
$$
\frac{1}{2}\left[\left(1+\frac{a}{z}\right) \dot{z}^{2}+\frac{4 a^{2} v^{2}}{4 a z}\right]+g z=\frac{1}{2} v^{2}+g a,\tag{5.35}
$$
and thus
$$
\frac{1}{2}\left(1+\frac{a}{z}\right) \dot{z}^{2}=\frac{1}{2} v^{2}\left(1-\frac{a}{z}\right)+g(a-z)=\frac{g}{z}(z-a)\left(\frac{v^{2}}{2 g}-z\right)\tag{5.36}
$$
Since $z>0$ and $\dot{z}^{2} \geq 0$ it follows that
$$
\left(\frac{v^{2}}{2 g}-z\right)(z-a) \geq 0 .\tag{5.37}
$$
Therefore the particle always stays between the two heights $z=a$ and $z=v^{2} / 2 g$, at which $\dot{z}=0$.
In particular the particle is confined to $z \geq a$ if $v^{2}>2 g a$, or to $z \leq a$ if $v^{2}<2 g a$, or to the horizontal circle $z=a$ if $v^{2}=2 g a$. |
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