Балкански 2010, Теория — Задача 5

Concours Général de Physique “Minko Balkanski” · 8 май 2010 г.

Указания

The main problem and the three exercises are completely independent and can be solved in any order.

All the answers must be given in English or French. The clarity and precision of expression will be taken into account in the attribution of the final note.

The exam is 4 hours long! The calculators are authorized.

Physical constants

Physical constants:

  • Mass of the Earth: ME=6,0.1024kgM_E = 6{,}0.10^{24}kg
  • Mass of the Sun: MS=2,0.1030kgM_S = 2{,}0.10^{30}kg
  • Average Earth radius: RE=6,4.106mR_E = 6{,}4.10^{6}m
  • Average Sun radius: RS=7,0.108mR_S = 7{,}0.10^{8}m
  • Average radius of the Earth’s orbit around the Sun: L=1,5.1011mL = 1{,}5.10^{11}m
  • Time needed for one rotation of the Sun around its axe: θS=2,6.106s\theta_S = 2{,}6.10^{6}s
  • Universal gravitational constant: G=6,67.10−11N.m2.kg−2G = 6{,}67.10^{-11}N.m^2.kg^{-2}
  • Perfect gaz constant: ℜ=8,31J.K−1.mol−1\Re = 8{,}31J.K^{-1}.mol^{-1}
  • Avogadro’s number: Na=6,02.1023mol−1Na = 6{,}02.10^{23}mol^{-1}
  • Hydrogen molar mass: MH=1,0.10−3kg.mol−1M_H = 1{,}0.10^{-3}kg.mol^{-1}
  • Speed of light: c=3,0.108m.s2c = 3{,}0.10^{8}m.s^2

Условие

A vertical ladder falls down without initial speed. The contact with the ground is frictionless.

A ladder drawn as a thick blue line: vertically on the horizontal ground, labelled “t=0”, and inclined in a position labelled “intermediate position”, its lower end sliding on the ground to the right; the angle θ between the inclined ladder and the ground is marked.

(a) Find the speed VGV_G (of the center of masses) as a function of Ω\Omega (the angular speed) and θ\theta. Calculate the speed VGV_G when the ladder touches the ground entirely.

(b) Write down the equation of movement of the center of masses in the very first moment of the falling. We will note α=π/2−θ\alpha = \pi/2 - \theta. Show that the angle α\alpha is increasing under an exponential law with a characteristic time τ=L6g\tau = \sqrt{\frac{L}{6g}}.

Решение

Внимание: Непълно решение

В архива няма официално решение на тази задача.

Оригинал в Архива: CGP_2010_en.pdf