IZhO 2006, Теория — Задача 1

• 1) $\alpha = 0,\quad \alpha = \pi,\quad \alpha = \arccos(\omega_0^2/\omega^2)\ (\omega^2 > \omega_0^2),\quad \omega_0^2 \equiv 3g/(2L)$
• 2) α = 0 is stable if ω² < ω0²; α = π is always unstable; α = arccos(ω0²/ω²) is stable for all ω² > ω0², where ω0² = 3g/(2L).
• 3) $T = \frac{2\pi}{\sqrt{\omega_0^2-\omega^2}}\ (\alpha = 0,\ \omega^2 < \omega_0^2);\qquad T = \frac{2\pi\omega}{\sqrt{\omega^4-\omega_0^4}}\ (\alpha = \arccos(\omega_0^2/\omega^2),\ \omega^2 > \omega_0^2)$

It is easier to solve the problem in noninertial frame of reference connected with the rotating axis, where the rod is in equilibrium under the action of forces including centrifugal inertial forces. Let as a result of the rotation the rod is deflected from the vertical by an angle. Consider an element of the rod at the distance $\ell$ from the point O with a length $d\ell$ and a mass $dm$. Given that the mass of the rod with a length $L$ is $m$, we have $dm = dl\cdot m/L$. The centrifugal force and the acting on that element is equal to (see fig.1.2)

(0,5 points)

(0,5 points)

The torques of these forces about the point O are

$dM_{cf} = \ell\cos\alpha\cdot dF_{cf}$ and $dM_{gr} = \ell\sin\alpha\cdot dF_{gr}$

Thus the torque of the force of gravity is

(0,5 points)

and the torque of the centrifugal force is equal to

(0,5 points)

At equilibrium we have $\sum\limits_i \vec{M}_i = 0$, i.e. $\left(\dfrac{m\omega^2L^2}{3}\cos\alpha - \dfrac{mgL}{2}\right)\sin\alpha = 0$, consequently

a1) $\sin\alpha = 0$, hence $\alpha = 0$ and $\alpha = \pi$; (1.0 point)

a2) $\cos\alpha = \omega_0^2/\omega^2$, where $\omega_0^2 \equiv 3g/(2L)$. Thus $\alpha = \arccos(\omega_0^2/\omega^2)$ with $\omega^2 > \omega_0^2$. (1.0 point)

The variation of the total momentum is defined by

The equilibrium state is stable provided $(\partial M/\partial\alpha) < 0$. Then

b1) For $\alpha = 0$ the derivative of the total torque is equal to

consequently that state is stable if $\omega^2 < \omega_0^2$; (1.0 point)

For $\alpha = \pi$ we have $\partial M/\partial\alpha = (mL^2/3)\cdot(\omega^2 + \omega_0^2) \ge 0$ hence that state is always unstable. (1.0 point)

b2) At $\cos\alpha = \omega_0^2/\omega^2$ the derivative of the torque is

(1.0 point)

hence the state is stable for all $\omega^2 > \omega_0^2$.

Thus for different values of the parameters of the systems oscillations are possible about the angles $\alpha = 0$ and $\alpha = \arccos(\omega_0^2/\omega^2)$. The equation for oscillatory motion has the form

(0,5 points)

where $I$ is the moment of inertia of the rod about the point O:

(0,5 points)

For small $\Delta\alpha$ the torque can be written as $\Delta M = (\partial M/\partial\alpha)\cdot\Delta\alpha$. Then the period of oscillations is equal to

(1.0 point)

From 1.6 and 1.8 we get for $\alpha = 0$ ($\omega^2 < \omega_0^2$)

(0,5 points)

For $\alpha = \arccos\left(\omega_0^2/\omega^2\right)$ ($\omega^2 > \omega_0^2$) from 1.7 and 1.8 the period of oscillations is equal to

(0,5 points)