USAPhO 1999, Четвъртфинал — Задача 4

• $D = \dfrac{\pi^2}{12}L$

4. Imagine the door to be a compound pendululm in a gravitational field whose “g” equals the car's acceleration. We are interested in finding 1/4 of the period of the oscillation, the time it takes the door to go from a maximum small amplitude to its equilibrium position. The torque and rotational inertia are:

Diagram similar to accompanying [5]

$\tau = -Ma\left(L/2\right)\sin\theta$ [3]

$I = \left(1/3\right)ML^2$ [4]

$\left(1/3\right)ML^2\alpha = -Ma\left(L/2\right)\sin\theta$ [3]

For small angles we have $\quad \alpha \equiv -\left(3/2\right)\left(a/L\right)\theta$ [1]

$\omega^2 = \dfrac{3a}{2L}$ [3]

$t = \dfrac{T}{4} = \dfrac{1}{4}\left(\dfrac{2\pi}{\omega}\right) = \dfrac{\pi}{2}\sqrt{\dfrac{2L}{3a}}$ [2]

The distance the car travels is $\quad D = \dfrac{1}{2}at^2 = \dfrac{1}{2}a\dfrac{\pi^2}{4}\left(\dfrac{2L}{3a}\right) = \dfrac{\pi^2}{12}L$ [4]