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BPO 2020, Теория — Задача 1. City car

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Всички материали от BPO 2020

Balkan Physics Olympiad – BPO 2, online contest, 24 October 2020, Craiova, ROMANIA · 24 октомври 2020 г.

Условие

A typical city car of a mass $m = 1000\ \mathrm{kg}$ is equipped with a motor of a maximum output power $P_{max} = 80\ \mathrm{kW}$ . Only the front wheels are leading, i.e. propelled by the motor, while the rear wheels rotate freely, with negligible friction. The center of mass of the car is midway between the front and the rear wheels and midway between the left and the right wheels. The coefficient of static friction between the wheel tires and the road is $\mu = 0.8$ . Assume that the acceleration due to gravity is $g = 10\ \mathrm{m/s^2}$ . In all parts of the problem the car moves on a horizontal road in a straight line.

a) What is the maximum acceleration $a_{\mathrm{max}}$ of the car, when starting from rest? [4 т.]

b) What is the minimum time $t_{\mathrm{min}}$ necessary to accelerate the car from rest to a velocity $v_{\mathrm{f}} = 30\ \mathrm{m/s}$ ? [14 т.]

c) As the velocity surpasses $v_{\mathrm{f}}$ , the further motion of the car is affected significantly by the air drag force $F_{\mathrm{d}}$ , which was neglected in points a) and b). Its magnitude is proportional to the square of the velocity $v$ of the car:

F_{\mathrm{d}} = Dv^2

where $D = 0.5\ \mathrm{kg/m}$ is a drag coefficient. By using this information find the maximum velocity $v_{\mathrm{max}}$ , which could be attained by the car. [7 т.]

Отговори

Решение

Solution

a) The propulsion force, which accelerates the car, is due to the friction between the leading wheels and the road – explicit statement or a proper drawing. (1 pt)

The center of mass of the car is in the middle between the front and the rear wheels. Therefore, the total reaction force on the leading wheels is half of the weight of the car:

R = \frac{mg}{2} \qquad (1)

(1 pt)

Correspondingly, the maximum friction force on the leading wheels, and the maximum propulsion force, thereof, is:

f_{\mathrm{max}} = \mu R = \frac{\mu m g}{2} = 4000\ \mathrm{N} \qquad (2)

(1 pt)

There is no air drag force on the car, when it is in rest. Therefore, the maximum acceleration at the start is due to the propulsion force only:

a_{\mathrm{max}} = \frac{f_{\mathrm{max}}}{m} = \frac{\mu g}{2} = 4\ \mathrm{m/s^2} \qquad (3)

(1 pt)

b) While accelerating from rest to $v_{\mathrm{f}}$ , the air drag force is neglected. Therefore, the acceleration of the car is a result of the propulsion force between the leading wheels and the road only. It means also that the work, done by the motor, is transformed completely into kinetic energy of the car. In order that the car can reach $v_{\mathrm{f}}$ for a minimum time, the propulsion force should be maintained at the maximum value compliant with the friction between the front wheels and the road (i.e. no tire slipping):

f \le \frac{\mu m g}{2} \qquad (4)

(1 pt)

and the maximum output power of the motor:

f v \le P_{\mathrm{max}} \qquad (5)

(2 pt)

Therefore, the motion of the car can be divided into two stages. During first stage the car starts from rest and maintains a constant propulsion force $f_{\mathrm{max}} = \mu m g/2$ , and a constant acceleration $a_{\mathrm{max}}$ , until the output power of the motor reaches $P_{\mathrm{max}}$ . During the second stage the motor maintains a constant power equal to $P_{\mathrm{max}}$ – explicit statement describing the two stages, or implicit understanding of the two stages evident from the subsequent solution. (1 pt)

In the first period the velocity of the car increases uniformly in time:

v = a_{\mathrm{max}} t \qquad (6)

(1 pt)

The uniformly accelerated motion continues until the velocity reaches specific value $v_1$ at which:

f_{\mathrm{max}} v_1 = P_{\mathrm{max}}

i.e.

v_1 = \frac{P_{\mathrm{max}}}{f_{\mathrm{max}}} = \frac{8.0 \times 10^4\ \mathrm{W}}{4.0 \times 10^3\ \mathrm{N}} = 20\ \mathrm{m/s} \qquad (7)

(2 pt)

Therefore, the uniformly accelerated stage of the motion lasts for a time:

t_1 = \frac{v_1}{a_{\mathrm{max}}} = \frac{20\ \mathrm{m/s}}{4.0\ \mathrm{m/s^2}} = 5\ \mathrm{s} \qquad (8)

(1 pt)

From that moment on, the motor is not able to maintain a constant propulsion force. The car still accelerates, but not uniformly, as the propulsion force decreases with the increase of the velocity according to Equation 5:

f v = P_{\mathrm{max}} = \mathrm{const}

Let the second stage of motion lasts for a period $t_2$ . The work done by the car engine during that time is:

W = P_{\mathrm{max}} t_2 \qquad (9)

(2 pt)

According to the work-energy theorem the increase of the kinetic energy of the car in this period is equal to the work done by the motor:

\frac{m v_{\mathrm{f}}^2}{2} - \frac{m v_1^2}{2} = W = P_{\mathrm{max}} t_2 \qquad (10)

(2 pt)

The time of acceleration from $v_1 = 20\ \mathrm{m/s}$ to $v_{\mathrm{f}} = 30\ \mathrm{m/s}$ is:

t_2 = \frac{m(v_{\mathrm{f}}^2 - v_1^2)}{2P_{\mathrm{max}}} = \frac{1000\ .\ 500}{2\ .\ 80\,000} \approx 3.1\ \mathrm{s} \qquad (11)

(1 pt)

Therefore, the minimum time of acceleration from rest to $v_{\mathrm{f}} = 30\ \mathrm{m/s}$ is:

t_{\mathrm{min}} = t_1 + t_2 \approx 8.1\ \mathrm{s} \qquad (12)

(1 pt)

(c) As the velocity surpasses $v_{\mathrm{f}}$ the air drag becomes more and more important, and cannot be neglected as far as we consider the maximum velocity achievable by the car. The car reaches a constant terminal velocity when the air drag equilibrates the propulsion force:

f = Dv^2 \qquad (13)

(1 pt)

During uniform motion, the work done by the motor does not transform into kinetic energy of the car but is dissipated completely into heat due to the work done by the air drag force. Therefore the output power of the motor satisfies the relation:

P - F_{\mathrm{d}} v = 0

P = Dv^3 \qquad (14)

(1 pt)

Evidently, the terminal velocity is restricted by the friction between the tires and the road:

Dv^2 \le \frac{\mu m g}{2} \qquad (15)

(1 pt)

and by the maximum power of the motor:

Dv^3 \le P_{\mathrm{max}} \qquad (16)

(1 pt)

It follows from (15) that:

v \le \sqrt{\frac{\mu m g}{2D}} = 89.4\ \mathrm{m/s} \qquad (17)

(1 pt)

and from (16) that:

v \le \sqrt[3]{\frac{P_{\mathrm{max}}}{D}} = 54.3\ \mathrm{m/s} \qquad (18)

(1 pt)

Therefore, for the given parameters, the maximum velocity of the car is restricted by the maximum power of the motor:

v_{\mathrm{max}} = 54.3\ \mathrm{m/s}\ \ (195\ \mathrm{km/h}) \qquad (19)

(1 pt)

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf