Правка раздела «Solution»
en/1.3.15.md
+1 −1
| ### Statement | |||
| $1.3.15.$ 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 $$ t = \frac{2\pi n R}{v \sin\alpha} $$ $$ 2v \cos\alpha = g\frac{2\pi n R}{v \sin\alpha} $$ | |||
| @@ -18,7 +18,7 @@Solution | |||
| 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. | |||
| ещё строк без изменений 3 | |||
| ### Statement | ### Statement | ||
| $1.3.15.$ 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? | $1.3.15.$ 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 | ### Solution | ||
| We will divide the movement into vertical and in the plane of the cross section of the cylinder | We will divide the movement into vertical and in the plane of the cross section of the cylinder | ||
| $x = vt \sin\alpha$ | $x = vt \sin\alpha$ | ||
| $$ y = vt \cos\alpha - \frac{gt^2}{2} $$ | $$ y = vt \cos\alpha - \frac{gt^2}{2} $$ | ||
| We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0: | We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0: | ||
| $$ vt \cos\alpha - \frac{gt^2}{2} = 0 $$ | $$ vt \cos\alpha - \frac{gt^2}{2} = 0 $$ | ||
| $$vt \sin\alpha = 2\pi n R,\,n\in \mathbb{N}$$ We find time $t$ as $$ t = \frac{2\pi n R}{v \sin\alpha} $$ $$ 2v \cos\alpha = g\frac{2\pi n R}{v \sin\alpha} $$ | $$vt \sin\alpha = 2\pi n R,\,n\in \mathbb{N}$$ We find time $t$ as $$ t = \frac{2\pi n R}{v \sin\alpha} $$ $$ 2v \cos\alpha = g\frac{2\pi n R}{v \sin\alpha} $$ | ||
| @@ -18,7 +18,7 @@Solution | |||
| From here we express $v$: | From here we express $v$: | ||
|  | ||
| $$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$ | $$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$ | ||
| #### Answer | #### 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. | $v=\sqrt{2\pi Rgn/\sin2\alpha}$, where $n$ is any natural number; when $\alpha = 0$, the speed can be any in absolute value. | ||
| ещё строк без изменений 3 | |||