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IPhO 1989, Теория и експеримент — Задача 2. Problem 2

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Problems of the 20th International Physics Olympiad (Warsaw, 1989) · 16 юли 1989 г.

Abstract

The article contains problems given at the 20th International Physics Olympiad (1989) and their solutions. The 20th IPhO was the third IPhO organized in Warsaw, Poland.

¹ This article has been sent for publication in Physics Competitions in October 2003

² e-mail: gorzk@ifpan.edu.pl

Logo

The emblem of the XX International Physics Olympiad contains a picture that is a historical record of the first hypernuclear event observed and interpreted in Warsaw by M. Danysz and J. Pniewski. The collision of a high-energy particle with a heavy nucleus was registered in nuclear emulsion. Tracks of the secondary particles emitted in the event, seen in the picture (upper star), consist of tracks due to fast pions (“thin tracks”) and to much slower fragments of the target nucleus (“black tracks”). The “black track” connecting the upper star (greater) with the lower star (smaller) in the figure is due to a hypernuclear fragment, in this case due to a part of the primary nucleus containing an unstable hyperon Λ instead of a nucleon. Hyperfragments (hypernuclei) are a new kind of matter in which the nuclei contain not only protons and neutrons but also some other heavy particles.

In the event observed above the hyperon Λ, bound with nucleon, decays like a free particle through a week (slow) process only. This fact strongly suggested the existence of a new quantum number that could explain suppression of the decay, even in presence of nucleons. Indeed, this was one of the observations that, 30 months later, led to the concept of strangeness.

M. Danysz and J. Pniewski, Bull. Acad. Polon. Sci., 3(1) 42 (1952) and Phil. Mag., 44, 348 (1953). Later the same physicists, Danysz and Pniewski, discovered the first case of a nucleus with two hyperons (double hyperfragment).

Увод

Theoretical problems (including solutions and marking schemes) were prepared especially for the 20th IPhO by Waldemar Gorzkowski. The experimental problem (including the solution and marking scheme) was prepared especially for this Olympiad by Andrzej Kotlicki. The problems were refereed independently (and many times) by at least two persons after any change was made in the text to avoid unexpected difficulties at the competition. This work was done by:

First Problem:
Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz Ungier

Second Problem:
Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz Ungier, Stanisław Woronowicz

Third Problem:
Andrzej Rajca, Andrzej Szymacha, Włodzimierz Ungier

Experimental Problem:
Krzysztof Korona, Anna Lipniacka, Jerzy Łusakowski, Bruno Sikora

Several English versions of the texts of the problems were given to the English-speaking students. As far as I know it happened for the first time (at present it is typical). The original English version was accepted (as a version for the students) by the leaders of the Australian delegation only. The other English-speaking delegations translated the English originals into English used in their countries. The net result was that there were at least four English versions. Of course, physics contained in them was exactly the same, while wording and spelling were somewhat different (the difference, however, were not too great).

This article is based on the materials quoted at the end of the article and on personal notes of the author.

Условие

Three non-collinear points $P_1$ , $P_2$ and $P_3$ , with known masses $m_1$ , $m_2$ and $m_3$ , interact with one another through their mutual gravitational forces only; they are isolated in free space and do not interact with any other bodies. Let $\sigma$ denote the axis going through the center-of-mass of the three masses, and perpendicular to the triangle $P_1P_2P_3$ . What conditions should the angular velocities $\omega$ of the system (around the axis $\sigma$ ) and the distances:

P_1P_2 = a_{12}, \qquad P_2P_3 = a_{23}, \qquad P_1P_3 = a_{13},

fulfill to allow the shape and size of the triangle $P_1P_2P_3$ unchanged during the motion of the system, i.e. under what conditions does the system rotate around the axis $\sigma$ as a rigid body?

Отговори

Решение

As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points P₁, P₂ and P₃ with the masses $m_1$ , $m_2$ and $m_3$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points (P₁,P₂), (P₂,P₃) and (P₁,P₃). It depends only on the distances $a_{12}$ , $a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\sigma$ depends only on the distances from the points P₁, P₂ and P₃ to the axis $\sigma$ which, for fixed $a_{12}$ , $a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:

\omega = \mathrm{const}. \qquad (1)

This is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.

Let the positions of the points P₁, P₂ and P₃ with the masses $m_1$ , $m_2$ and $m_3$ be given by the vectors $\mathbf{r}_1$ , $\mathbf{r}_2$ and $\mathbf{r}_3$ . For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points P₁, P₂ and P₃ with the masses $m_1$ , $m_2$ and $m_3$ and that all the vectors $\mathbf{r}_1$ , $\mathbf{r}_2$ and $\mathbf{r}_3$ are in the same coordinate plane, e.g. in the plane $(x,y)$ . Then the axis $\sigma$ is the axis $z$ .

In this coordinate system, according to the definition of the center of mass, we have:

m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + m_3\mathbf{r}_2 = 0 \qquad (2)

Now we will find the second condition by using several methods.

FIRST METHOD

Consider the point P₁ with the mass $m_1$ . The points P₂ and P₃ act on it with the forces:

\mathbf{F}_{21} = G\frac{m_1m_2}{a_{12}^3}(\mathbf{r}_2 - \mathbf{r}_1), \qquad (3)

\mathbf{F}_{31} = G\frac{m_1m_3}{a_{13}^3}(\mathbf{r}_3 - \mathbf{r}_1). \qquad (4)

where $G$ denotes the gravitational constant.

In the inertial frame the sum of these forces is the centripetal force

\mathbf{F}_{r1} = -m_1\omega^2\mathbf{r}_1,

which causes the movement of the point P₁ along a circle with the angular velocity $\omega$ . (The moment of this force with respect to the axis $\sigma$ is equal to zero.) Thus, we have:

\mathbf{F}_{21} + \mathbf{F}_{31} = \mathbf{F}_{r1}. \qquad (5)

In the non-inertial frame, rotating around the axis $\sigma$ with the angular velocity $\omega$ , the sum of the forces (3), (4) and the centrifugal force

\mathbf{F}'_{r1} = m_1\omega^2\mathbf{r}_1

should be equal to zero:

\mathbf{F}_{21} + \mathbf{F}_{31} + \mathbf{F}'_{r1} = 0. \qquad (6)

(The moment of this sum with respect to any axis equals to zero.)

The conditions (5) and (6) are equivalent. They give the same vector equality:

G\frac{m_1m_2}{a_{12}^3}(\mathbf{r}_2 - \mathbf{r}_1) + G\frac{m_1m_3}{a_{13}^3}(\mathbf{r}_3 - \mathbf{r}_1) + m_1\omega^2\mathbf{r}_1 = 0, \qquad (7')

G\frac{m_1}{a_{12}^3}m_2\mathbf{r}_2 + G\frac{m_1}{a_{13}^3}m_3\mathbf{r}_3 + m_1\mathbf{r}_1\left(\omega^2 - \frac{Gm_2}{a_{12}^3} - \frac{Gm_3}{a_{13}^3}\right) = 0 \qquad (7'')

From the formula (2), we get:

m_2\mathbf{r}_2 = -m_1\mathbf{r}_1 - m_3\mathbf{r}_3 \qquad (8)

Using this relation, we write the formula (7) in the following form:

G\frac{m_1}{a_{12}^3}(-m_1\mathbf{r}_1 - m_3\mathbf{r}_3) + G\frac{m_1}{a_{13}^3}m_3\mathbf{r}_3 + m_1\mathbf{r}_1\left(\omega^2 - \frac{Gm_2}{a_{12}^3} - \frac{Gm_3}{a_{13}^3}\right) = 0,

i.e.

\mathbf{r}_1m_1\left(\omega^2 - \frac{Gm_2}{a_{12}^3} - \frac{Gm_3}{a_{13}^3} - \frac{Gm_1}{a_{12}^3}\right) + \mathbf{r}_3\left(\frac{1}{a_{13}^3} - \frac{1}{a_{12}^3}\right)Gm_1m_3 = 0.

The vectors $\mathbf{r}_1$ and $\mathbf{r}_3$ are non-col1inear. Therefore, the coefficients in the last formula must be equal to zero:

\left(\frac{1}{a_{13}^3} - \frac{1}{a_{12}^3}\right)Gm_1m_3 = 0,

m_1\left(\omega^2 - \frac{Gm_2}{a_{12}^3} - \frac{Gm_3}{a_{13}^3} - \frac{Gm_1}{a_{12}^3}\right) = 0.

The first equality leads to:

\frac{1}{a_{13}^3} = \frac{1}{a_{12}^3}

and hence,

a_{13} = a_{12}.

Let $a_{13} = a_{12} = a$ . Then the second equality gives:

\omega^2a^3 = GM \qquad (9)

where

M = m_1 + m_2 + m_3 \qquad (10)

denotes the total mass of the system.

In the same way, for the points P₂ and P₃, one gets the relations:

a) the point P₂:

a_{23} = a_{12}; \qquad \omega^2a^3 = GM

b) the point P₃:

a_{13} = a_{23}; \qquad \omega^2a^3 = GM

Summarizing, the system can rotate as a rigid body if all the distances between the masses are equal:

a_{12} = a_{23} = a_{13} = a, \qquad (11)

the angular velocity $\omega$ is constant and the relation (9) holds.

SECOND METHOD

At the beginning we find the moment of inertia $I$ of the system with respect to the axis $\sigma$ . Using the relation (2), we can write:

0 = (m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + m_3\mathbf{r}_3)^2 = m_1^2\mathbf{r}_1^2 + m_2^2\mathbf{r}_2^2 + m_3^2\mathbf{r}_3^2 + 2m_1m_2\mathbf{r}_1\mathbf{r}_2 + 2m_1m_3\mathbf{r}_1\mathbf{r}_3 + 2m_3m_2\mathbf{r}_3\mathbf{r}_2.

Of course,

\mathbf{r}_i^2 = r_i^2 \qquad i = 1, 2, 3

The quantities $2\mathbf{r}_i\mathbf{r}_j$ ( $i, j$ = 1, 2, 3) can be determined from the following obvious relation:

a_{ij}^2 = \left|\mathbf{r}_i - \mathbf{r}_j\right|^2 = (\mathbf{r}_i - \mathbf{r}_j)^2 = \mathbf{r}_i^2 + \mathbf{r}_j^2 - 2\mathbf{r}_i\mathbf{r}_j = r_i^2 + r_j^2 - 2\mathbf{r}_i\mathbf{r}_j.

We get:

2\mathbf{r}_i\mathbf{r}_j = r_i^2 + r_j^2 - a_{ij}^2.

With help of this relation, after simple transformations, we obtain:

0 = (m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + m_3\mathbf{r}_3)^2 = (m_1 + m_2 + m_3)(m_1r_1^2 + m_2r_2^2 + m_3r_3^2) - \sum_{i<j}m_im_ja_{ij}^2.

The moment of inertia $I$ of the system with respect to the axis $\sigma$ , according to the definition of this quantity, is equal to

I = m_1r_1^2 + m_2r_2^2 + m_3r_3^2.

The last two formulae lead to the following expression:

I = \frac{1}{M}\sum_{i<j}m_im_ja_{ij}^2

where $M$ (the total mass of the system) is defined by the formula (10).

In the non-inertial frame, rotating around the axis $\sigma$ with the angular velocity $\omega$ , the total potential energy $V_{tot}$ is the sum of the gravitational potential energies

V_{ij} = -G\frac{m_im_j}{a_{ij}}; \qquad i, j = 1, 2, 3;\ i < j

of all the masses and the potential energies

V_i = -\tfrac{1}{2}\omega^2m_ir_i^2; \qquad i = 1, 2, 3

of the masses $m_i$ ( $i$ = 1, 2, 3) in the field of the centrifugal force:

V_{tot} = G\sum_{i<j}\frac{m_im_j}{a_{ij}} - \tfrac{1}{2}\omega^2\sum_{i=1}^{3}m_ir_i^2 = G\sum_{i<j}\frac{m_im_j}{a_{ij}} - \tfrac{1}{2}\omega^2I = G\sum_{i<j}\frac{m_im_j}{a_{ij}} - \tfrac{1}{2}\omega^2\frac{1}{M}\sum_{i<j}m_im_ja_{ij}^2 =

= -\sum_{i<j}m_im_j\left(\frac{\omega^2}{2M}a_{ij}^2 + \frac{G}{a_{ij}}\right)

i.e.

V_{tot} = -\sum_{i<j}m_im_j\left(\frac{\omega^2}{2M}a_{ij}^2 + \frac{G}{a_{ij}}\right).

A mechanical system is in equilibrium if its total potential energy has an extremum. In our case the total potential energy $V_{tot}$ is a sum of three terms. Each of them is proportional to:

f(a) = \frac{\omega^2}{2M}a^2 + \frac{G}{a}.

The extrema of this function can be found by taking its derivative with respect to $a$ and requiring this derivative to be zero. We get:

\frac{\omega^2}{M}a - \frac{G}{a^2} = 0.

It leads to:

\omega^2a^3 = GM \qquad \text{or} \qquad \omega^2a^3 = G(m_1 + m_2 + m_3).

We see that all the terms in $V_{tot}$ have extrema at the same values of $a_{ij} = a$ . (In addition, the values of $a$ and $\omega$ should obey the relation written above.) It is easy to show that it is a maximum. Thus, the quantity $V_{tot}$ has a maximum at $a_{ij} = a$ .

This means that our three masses can remain in fixed distances only if these distances are equal to each other:

a_{12} = a_{23} = a_{13} = a

and if the relation

\omega^2a^3 = GM,

where $M$ the total mass of the system, holds. We have obtained the conditions (9) and (11) again.

THIRD METHOD

Let us consider again the point P₁ with the mass $m_1$ and the forces $\mathbf{F}_{21}$ and $\mathbf{F}_{31}$ given by the formulae (3) and (4). It follows from the text of the problem that the total moment (with respect to any fixed point or with respect to the mass center) of the forces acting on the point P₁ must be equal to zero. Thus, we have:

\mathbf{F}_{21} \times \mathbf{r}_1 + \mathbf{F}_{31} \times \mathbf{r}_1 = 0

where the symbol $\times$ denotes the vector product. Therefore

G\frac{m_1m_2}{a_{12}^3}(\mathbf{r}_2 - \mathbf{r}_1) \times \mathbf{r}_1 + G\frac{m_1m_3}{a_{13}^3}(\mathbf{r}_3 - \mathbf{r}_1) \times \mathbf{r}_1 = 0.

But

\mathbf{r}_1 \times \mathbf{r}_1 = 0.

Thus:

\frac{m_2}{a_{12}^3}\mathbf{r}_2 \times \mathbf{r}_1 + \frac{m_3}{a_{13}^3}\mathbf{r}_3 \times \mathbf{r}_1 = 0.

Using the formula (8), the last relation can be written as follows:

\frac{1}{a_{12}^3}(-m_1\mathbf{r}_1 - m_3\mathbf{r}_3) \times \mathbf{r}_1 + \frac{m_3}{a_{13}^3}\mathbf{r}_3 \times \mathbf{r}_1 = 0,

-\frac{m_3}{a_{12}^3}\mathbf{r}_3 \times \mathbf{r}_1 + \frac{m_3}{a_{13}^3}\mathbf{r}_3 \times \mathbf{r}_1 = 0,

\left(\frac{1}{a_{13}^3} - \frac{1}{a_{12}^3}\right)\mathbf{r}_3 \times \mathbf{r}_1 = 0.

The vectors $\mathbf{r}_1$ and $\mathbf{r}_3$ are non-col1inear (and different from 0). Therefore

\mathbf{r}_3 \times \mathbf{r}_1 \neq 0

and

\frac{1}{a_{13}^3} - \frac{1}{a_{12}^3} = 0,

hence,

a_{12} = a_{13}.

Similarly, one gets:

a_{12} = a_{23} \ (= a).

We have re-derived the condition (11).

Taking into account that all the distances $a_{ij}$ have the same value $a$ , from the equation (7) concerning the point P₁, using the relation (2) we obtain:

G\frac{m_1m_2}{a^3}(\mathbf{r}_2 - \mathbf{r}_1) + G\frac{m_1m_3}{a^3}(\mathbf{r}_3 - \mathbf{r}_1) + m_1\omega^2\mathbf{r}_1 = 0,

-\left(G\frac{m_1}{a^3} + G\frac{m_2}{a^3}G\frac{m_3}{a^3}\right)m_1\mathbf{r}_1 + m_1\omega^2\mathbf{r}_1 = 0,

\frac{GM}{a^3} = \omega^2.

This is the condition (9). The same condition is got in result of similar calculations for the points P₂ and P₃.

The method described here does not differ essentially from the first method. In fact they are slight modifications of each other. However, it is interesting to notice how application of a proper mathematical language, e.g. the vector product, simplifies the calculations.

The relation (9) can be called a “generalized Kepler’s law” as, in fact, it is very similar to the Kepler’s law but with respect to the many-body system. As far as I know this generalized Kepler’s law was presented for the first time right at the 20th IPhO.

Marking scheme

the proof that $\omega = \mathrm{const}$ — 1 point
the conditions at the equilibrium (conditions for the forces and their moments or extremum of the total potential energy) — 3 points
the proof of the relation $a_{ij} = a$ — 4 points
the proof of the relation $\omega^2a^3 = GM$ — 2 points

Remarks and typical mistakes in the pupils' solutions

No type of error was observed as predominant in the pupils' solutions. Practically all the mistakes can be put down to the students' scant experience in calculations and general lack of skill. Several students misunderstood the text of the problem and attempted to prove that the three masses should be equal. Of course, this was impossible. Moreover, it was pointless, since the masses were given. Almost all the participants tried to solve the problem by analyzing equilibrium of forces and/or their moments. Only one student tried to solve the problem by looking for a minimum of the total potential energy (unfortunately, his solution was not fully correct). Several participants solved the problem using a convenient reference system: one mass in the origin and one mass on the $x$ -axis. One of them received a special prize.

Acknowledgement

I would like to thank very warmly to Prof. Jan Mostowski and Dr. Andrzej Wysmołek for reading the text of this article and for valuable critical remarks. I express special thanks to Dr. Andrzej Kotlicki for critical reviewing the experimental part of the article and for a number of very important improvements.

Literature

[1] Waldemar Gorzkowski and Andrzej Kotlicki, XX Międzynarodowa Olimpiada Fizyczna - cz. I, Fizyka w Szkole nr 1/90, pp. 34 - 39

[2] Waldemar Gorzkowski, XX Międzynarodowa Olimpiada Fizyczna - cz. II, Fizyka w Szkole nr 2/3-90, pp. 23 - 32

[3] XX International Physics Olympiad - Proceedings of the XX International Physics Olympiad, Warsaw (Poland), July 16 - 24, 1989, ed. by W. Gorzkowski, World Scientific Publishing Company, Singapore 1990 [ISBN 981-02-0084-6]

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