1.3.15. A ball is launched along the inner surface of a smooth vertical cylinder of radius $R$ at an angle $\alpha$ to the vertical. What initial velocity must be given to it so that it returns to the starting point?
Solution
We will divide the movement into vertical and in the plane of the cross section of the cylinder
$x = vt \sin\alpha$
$$y = vt \cos\alpha - \frac{gt^2}{2}$$
We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0:
$$vt \cos\alpha - \frac{gt^2}{2} = 0$$
$$vt \sin\alpha = 2\pi n R,\,n\in \mathbb{N}$$ We find time $t$ as $$t = \frac{2\pi n R}{v \sin\alpha}$$$$2v \cos\alpha = g\frac{2\pi n R}{v \sin\alpha}$$
From here we express $v$:
$$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$
Answer
$v=\sqrt{2\pi Rgn/\sin2\alpha}$, where $n$ is any natural number; when $\alpha = 0$, the speed can be any in absolute value.
