WoPhO 2011, Теория — Задача 1.3

• (a) $\mathbf{v}(t) = \left(v_0 + \frac{g\sin\alpha}{\omega_0}(\cos(\omega_0 t) - 1)\right)\begin{pmatrix}\cos(\omega_0 t)\\ \sin(\omega_0 t)\end{pmatrix}$
• (b) $\frac{v_0\omega_0}{2g\sin\alpha} \le 1$
• (c) $y_{\max} = \frac{v_0^2}{2g\sin\alpha}\ \text{if}\ \frac{v_0\omega_0}{2g\sin\alpha} \le 1;\quad y_{\max} = \frac{2(v_0\omega_0 - g\sin\alpha)}{\omega_0^2}\ \text{if}\ \frac{v_0\omega_0}{2g\sin\alpha} > 1$

C) a) Equations (1) and (2) are still valid, we only have to fit them to the initial conditions:

b) The velocity of the center of the rod is zero iff

Since $-1 \le \cos(\omega_0 t) \le 1$, the condition of the stopping of the center of the rod reads

c) We have to distinguish the following two cases: the center of the rod stops for a moment during the motion or not. In the first case, the conservation of mechanical energy tells us that the maximal displacement along the $y$ direction is

In the second case (when $\frac{v_0\omega_0}{2g\sin\alpha} > 1$), the center of the rod has a finite velocity even at the highest point of its orbit. From the expression of $\mathbf{v}(t)$ one can see, that at the highest point of the orbit the velocity of the center of the rod is

so the conservation of mechanical energy reads

from which one can get the result