Съдържание

IPhO 1972, Теория и експеримент — Задача 1. Problem 1 (Mechanics)

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Всички материали от IPhO 1972

Problems of the 6th International Physics Olympiad (Bucharest, 1972) · 10 т.

Romulus Pop, Civil Engineering University, Physics Department, Bucharest, Romania

The sixth IPhO was held in Bucharest and the participants were: Bulgaria, Czechoslovakia, Cuba, France, German Democratic Republic, Hungary, Poland, Romania, and Soviet Union. It was an important event because it was the first time when a non-European country and a western country participated (Cuba), and Sweden sent one observer.

The International Board selected four theoretical problems and an experimental problem. Each theoretical problem was scored from 0 to 10 and the maximum score for the experimental problem was 20. The highest score corresponding to actual marking system was 47,5 points. Each team consisted in six students. Four students obtained the first prize, seven students obtained the second prize, ten students obtained the third prize, thirteen students had got honorable mentions, and two special prizes were awarded too.

The article contains the competition problems given at the 6th International Physics Olympiad (Bucharest, 1972) and their solutions. The problems were translated from the book published in Romania concerning the first nine International Physics Olympiads², because I couldn’t find the original English version.

¹ E-mail: popr@tvet.ro

² Marius Gall and Anatolie Hristev, Probleme date la Olimpiadele de Fizica, Editura Didactica si Pedagogica – Bucuresti, 1978

Условие

Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, μ, is known and has the same value for all the cylinders. The first cylinder is empty (tube) , the second is homogeneous filled, and the third has a cavity exactly like the first, but closed with two negligible mass lids and filled with a liquid with the same density like the cylinder’s walls. The friction between the liquid and the cylinder wall is considered negligible. The density of the material of the first cylinder is n times greater than that of the second or of the third cylinder.

Determine:

a) The linear acceleration of the cylinders in the non-sliding case. Compare all the accelerations.

b) Condition for angle α of the inclined plane so that no cylinders is sliding.

c) The reciprocal ratios of the angular accelerations in the case of roll over with sliding of all the three cylinders. Make a comparison between these accelerations.

d) The interaction force between the liquid and the walls of the cylinder in the case of sliding of this cylinder, knowing that the liquid mass is m $_l$ .

Решение

The inertia moments of the three cylinders are:

I_1 = \frac{1}{2}\rho_1\pi\left(R^4 - r^4\right)h, \quad I_2 = \frac{1}{2}\rho_2\pi R^4 h = \frac{1}{2}mR^2 \quad , \quad I3 = \frac{1}{2}\rho_2\pi\left(R^4 - r^4\right)h, \qquad (1)

Because the three cylinders have the same mass :

m = \rho_1\pi\left(R^2 - r^2\right)h = \rho_2\pi R^2 h \qquad (2)

it results:

r^2 = R^2\left(1 - \frac{\rho_2}{\rho_1}\right) = R^2\left(1 - \frac{1}{n}\right), n = \frac{\rho_1}{\rho_2} \qquad (3)

The inertia moments can be written:

I_1 = I_2\left(2 - \frac{1}{n}\right)\rangle I_2 \ , \quad I_3 = I_2\left(2 - \frac{1}{n}\right)\cdot\frac{1}{n} = \frac{I_1}{n} \qquad (4)

In the expression of the inertia momentum $I_3$ the sum of the two factors is constant:

\left(2 - \frac{1}{n}\right) + \frac{1}{n} = 2

independent of n, so that their products are maximum when these factors are equal: $2 - \frac{1}{n} = \frac{1}{n}$ ; it results n = 1, and the products $\left(2 - \frac{1}{n}\right)\cdot\frac{1}{n} = 1$ . In fact $n > 1$ , so that the products is les than 1. It results:

I_1 > I_2 > I_3 \qquad (5)

For a cylinder rolling over freely on the inclined plane (fig. 1.1) we can write the equations:

mg\sin\alpha - F_f = ma \qquad (6)

N - mg\cos\alpha = 0

F_f R = I\varepsilon \qquad (7)

where ε is the angular acceleration. If the cylinder doesn’t slide we have the condition:

a = \varepsilon R \qquad (8)

Solving the equation system (6-8) we find:

a = \frac{g\sin\alpha}{1 + \frac{I}{mR^2}} \ , \quad F_f = \frac{mg\sin\alpha}{1 + \frac{mR^2}{I}} \qquad (9)

The condition of non-sliding is:

F_f < \mu N = \mu mg\sin\alpha

tg\alpha < \mu\left(1 + \frac{mR^2}{I_1}\right) \qquad (10)

Inclined plane at angle α with three cylinders rolling down it: the upper one hatched with a dotted inner liquid core (labels ρ2, r), the middle one homogeneous hatched (ρ2, radius R), the lower one a hatched tube with an empty central hole (ρ1, r); each has forces N, mg and friction force Ff3, Ff2, Ff1 drawn.

In the case of the cylinders from this problem, the condition necessary so that none of them slides is obtained for maximum I:

tg\alpha\langle\mu\left(1 + \frac{mR^2}{I_1}\right) = \mu\frac{4n - 1}{2n - 1} \qquad (11)

The accelerations of the cylinders are:

a_1 = \frac{2g\sin\alpha}{3 + (1 - \frac{1}{n})} \ , \quad a_2 = \frac{2g\sin\alpha}{3} \ , \quad a_3 = \frac{2g\sin\alpha}{3 - (1 - \frac{1}{n})^2}. \qquad (12)

The relation between accelerations:

a_1 < a_2 < a_3 \qquad (13)

In the case than all the three cylinders slide:

F_f = \mu N = \mu mg\cos\alpha \qquad (14)

and from (7) results:

\varepsilon = \frac{R}{I}\mu mg\cos\alpha \qquad (15)

for the cylinders of the problem:

\varepsilon_1 : \varepsilon_2 : \varepsilon_3 = \frac{1}{I_1} : \frac{1}{I_2} : \frac{1}{I_3} = 1 : \left(1 - \frac{1}{n}\right) : n

\varepsilon_1 < \varepsilon_2 < \varepsilon_3 \qquad (16)

In the case that one of the cylinders is sliding:

mg\sin\alpha - F_f = ma \ , \quad F_f = \mu mg\cos\alpha \ , \qquad (17)

a = g\left(\sin\alpha - \mu\cos\alpha\right) \qquad (18)

Let $\vec{F}$ be the total force acting on the liquid mass m $_l$ inside the cylinder (fig.1.2), we can write:

F_x + m_l g\sin\alpha = m_l a = m_l g\left(\sin\alpha - \mu\cos\alpha\right), \quad F_y - m_l g\cos\alpha = 0 \qquad (19)

F = \sqrt{F_x^2 + F_y^2} = m_l g\cos\alpha\cdot\sqrt{1 + \mu^2} = m_l g\frac{\cos\alpha}{\cos\phi} \qquad (20)

where $\phi$ is the friction angle $\left(tg\phi = \mu\right)$ .

Cylinder filled with liquid on an inclined plane at angle α; the liquid core is dotted and marked ρ2, the wall hatched; force components Fy (perpendicular to the plane) and Fx (along the plane) and weight m_l g are drawn. — Fig. 1.2

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