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IPhO 2007, Теория — Задача 1.4

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Всички материали от IPhO 2007

ORANGE — 38th International Physics Olympiad, Isfahan-Iran, July 2007

Условие

4 We would like to adjust the parameters in the problem such that the air bag will not be activated in normal braking but opens fast enough during a collision to prevent the driver’s head from colliding with the windshield or the steering wheel. As you have seen in Part 2, the force exerted on the movable plate by the springs and the electrical charges can be represented as that of a spring with an effective spring constant $k_{eff}$ . The whole capacitor complex is similar to a mass and spring system of mass $M$ and spring constant $k_{eff}$ under the influence of a constant acceleration $a$ , which in this problem is the acceleration of the automobile.

Note: In this part of the problem, the assumption that the mass and spring are in equilibrium under a constant acceleration and hence are fixed relative to the automobile, no longer holds.

Ignore friction and consider the following numerical values for the parameters of the problem:

$d = 1.0\,\mathrm{cm}$ , $A = 2.5\times 10^{-2}\,\mathrm{m^2}$ , $k = 4.2\times 10^{3}\,\mathrm{N/m}$ , $\varepsilon_0 = 8.85\times 10^{-12}\,\mathrm{C^2/Nm^2}$ ,
$V = 12\,\mathrm{V}$ , $M = 0.15\,\mathrm{kg}$ .

4.1 Using this data, find the ratio of the electrical force you calculated in section 2.3 to the force of the springs and show that one can ignore the electrical forces compared to the spring forces. [0,6 т.]

Although we did not calculate the electrical forces for the case when the switch is in the state $\beta$ , it can be shown that in this situation, quite similarly, the electrical forces are as small and can be ignored.

4.2 If the automobile while traveling with a constant velocity, suddenly brakes with a constant acceleration $a$ , what is the maximum displacement of the movable plate? Give your answer in parameter. [0,6 т.]

Assume that the switch is in state $\beta$ and the system has been designed such that when the electrical voltage across the capacitor reaches $V_S = 0.15\,\mathrm{V}$ , the air bag is activated. We would like the air bag not to be activated during normal braking when the automobile’s acceleration is less than the acceleration of gravity $g = 9.8\,m/s^2$ , but be activated otherwise.

4.3 How much should $C_S$ be for this purpose? [0,6 т.]

We would like to find out if the air bag will be activated fast enough to prevent the driver’s head from hitting the windshield or the steering wheel. Assume that as a result of collision, the automobile experiences a deceleration equal to $g$ but the driver’s head keeps moving at a constant speed.

4.4 By estimating the distance between the driver’s head and the steering wheel, find the time $t_1$ it takes before the driver’s head hits the steering wheel. [0,8 т.]

4.5 Find the time $t_2$ before the air bag is activated and compare it to $t_1$ . Is the air bag activated in time? Assume that airbag opens instantaneously. [0,9 т.]

Отговори

Решение

4.1)

The ratio of the electrical force to the mechanical (spring) force is

\frac{F_E}{F_m} = \frac{\varepsilon_0 A V^2}{k\,d^3},

Putting the numerical values:

\frac{F_E}{F_m} = 7.6\times 10^{-9}.

((0.2) + (0.2) + (0.2): (0.2) for order of magnitude, (0.2) for two significant digits and (0.2) for correct answer (7.6 or 7.5)).

As it is clear from this result, we can ignore the electrical forces compared to the electric force.

4.2)

As seen in the previous section, one may assume that the only force acting on the moving plate is due to springs:

F = 2k\,x.

(The concept of equilibrium (0.2))

Hence in mechanical equilibrium, the displacement of the moving plate is

x = \frac{ma}{2k}.

The maximum displacement is twice this amount, like the mass spring system in a gravitational force field, when the mass is let to fall.

x_{\max} = 2\,x

(0.2)

x_{\max} = \frac{m\,a}{k}

(0.2)

4.3)

At the acceleration

a = g,

(0.2)

The maximum displacement is

x_{\max} = \frac{m\,g}{k}.

Moreover, from the result obtained in 3.2, we have

V_S = V\,\frac{2\varepsilon_0 A\,x_{\max}}{d^2 C_S + 2\varepsilon_0 A\,d}

This should be the same value given in the problem, $0.15\,\mathrm{V}$ .

\Rightarrow \qquad C_S = \frac{2\varepsilon_0 A}{d}\left(\frac{V\,x_{\max}}{V_S\,d} - 1\right)

(0.2)

\Rightarrow \qquad C_S = 8.0\times 10^{-11}\,\mathrm{F}

(0.2)

4.4)

Let $\ell$ be the distance between the driver’s head and the steering wheel. It can be estimated to be about

\ell = 0.4\,m - 1\,m.

(0.2)

Just at the time the acceleration begins, the relative velocity of the driver’s head with respect to the automobile is zero.

\Delta v(t = 0) = 0,

(0.2)

then

\ell = \frac{1}{2} g\,{t_1}^2 \qquad \Rightarrow \qquad t_1 = \sqrt{\frac{2\,\ell}{g}}

(0.2)

t_1 = 0.3 - 0.5\ s

(0.2)

4.5)

The time $t_2$ is half of period of the harmonic oscillator, hence

t_2 = \frac{T}{2},

(0.3)

The period of harmonic oscillator is simply given by

T = 2\pi\sqrt{\frac{m}{2k}},

(0.2)

therefore,

t_2 = 0.013\ s.

(0.2)

As $t_1 > t_2$ , the airbag activates in time.

(0.2)

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf