Съдържание

EuPhO 2018, Теория — Задача 1. Three balls

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Всички материали от EuPhO 2018

EuPhO 2018, Moscow — Theory

Условие

Three small identical balls (denoted as $A$ , $B$ , and $C$ ) of mass $m$ each are connected with two massless rods of length $\ell$ so that one of the rods connects the balls $A$ and $B$ , and the other rod connects the balls $B$ and $C$ . The connection at the ball $B$ is hinged, and the angle between the rods can change effortlessly. The system rests in weightlessness so that all the balls lie on one line. The ball $A$ is given instantaneously a velocity perpendicular to the rods.

Find the minimal distance $d$ between the balls $A$ and $C$ during the subsequent motion of the system. Any friction is to be neglected.

Отговори

Решение

Let the ball $A$ is given an initial velocity $v$ along the axis $Y$ , which is perpendicular to the rod. The total momentum of the system conserves, therefore the center-of-mass (CM) of the system moves with a constant velocity:

v_{\mathrm{CM}} = \frac{mv}{3m} = \frac{v}{3}

along $Y$ . In what follows, we will work in the CM frame of reference, which is an inertial system of reference. Therefore, in the CM frame the laws of conservation of energy, momentum, and the angular momentum hold true. The initial velocities of the three balls along $Y$ are:

v_A = \frac{2v}{3}, \qquad v_B = v_C = -\frac{v}{3}.

Correspondingly, the total kinetic energy of the balls is:

E = \frac{mv^2}{2}\left(\frac{4}{9} + \frac{1}{9} + \frac{1}{9}\right) = \frac{mv^2}{3}

and the total angular momentum with respect to the CM equals:

L = m\frac{2v}{3}\ell - m\frac{v}{3}(-\ell) = mv\ell.

In any moment the three balls form an isosceles triangle with an angle $2\varphi$ at the top vertex. The distance between $A$ and $C$ is minimal when either $\varphi = 0$ , or $\dot\varphi = 0$ . For $\varphi = 0$ , however, the laws of conservation are not consistent with the rigidity of the rods. Therefore, at the minimal distance $\dot\varphi = 0$ , and in this particular instance the system behaves as a rigid body whose moment of inertia with respect to the CM can be obtained through:

I = \frac{L^2}{2E} = \frac{3}{2}ml^2. \qquad (1)

On the other hand, the moment of inertia $I$ could be found independently from geometric considerations. Although, $I$ could be found using the distances from the balls to the CM (medicentre of a triangle), it is more convenient to use this relatively unknown formula for the moment of inertia of a collection of point masses with respect to CM:

I = \frac{\sum_{i,j} m_i m_j (\mathbf{r}_i - \mathbf{r}_j)^2}{\sum_i m_i}.

In our case:

I = \frac{m^2}{3m}\left(AB^2 + BC^2 + AC^2\right) = \frac{m}{3}\left(2\ell^2 + d^2\right). \qquad (2)

From (1) and (2), we obtain the minimal distance:

d = \ell\sqrt{\frac{5}{2}} \approx 1.58\,\ell

Оригинал в Архива: th_pr.pdf · официални решения: th_sol.pdf