Kvant Physics Problem 70
The physical system is a cyclist moving along a road.
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m15s
Source on kvant.digital
Problem
F70. A cyclist can easily develop a tractive force of 10 kgf. The friction force does not exceed 5 kgf. It would seem that in a few hours the cyclist could reach the second cosmic velocity. However, no one has ever managed to do this. Why?
Moscow Institute of Electronic Engineering (MIEM) Physics and Mathematics Olympiad, 1969.
Setup and Assumptions
The physical system is a cyclist moving along a road. The cyclist can exert a tractive force
$$F_t = 10\ \text{kgf},$$
while the total resistive force due to rolling friction and similar mechanical losses does not exceed
$$F_f = 5\ \text{kgf}.$$
The unknown is whether the cyclist can accelerate to the second cosmic velocity, approximately
$$v_2 \approx 11.2\ \text{km/s}.$$
The analysis is performed in the reference frame of the Earth. The road is assumed horizontal. Mechanical friction in the bicycle is neglected except for the stated resistive force. The crucial physical effect to be examined is the interaction with the surrounding air.
The unit kilogram-force is converted to SI units through
$$1\ \text{kgf} \approx 9.8\ \text{N}.$$
Thus
$$F_t \approx 98\ \text{N}, \qquad F_f \approx 49\ \text{N}.$$
Physical Principles
The motion of the cyclist is governed by Newton's second law,
$$m\frac{dv}{dt}=F_{\text{net}},$$
where the net force equals the tractive force minus all resistive forces.
At low speeds,
$$F_{\text{net}}=F_t-F_f.$$
For motion through air, the aerodynamic drag force is approximately
$$F_d=\frac12 C_d\rho A v^2,$$
where $C_d$ is the drag coefficient, $\rho$ is the air density, and $A$ is the effective frontal area.
The cyclist can continue accelerating only while
$$F_t>F_f+F_d.$$
The maximum attainable speed is reached when the net force becomes zero,
$$F_t=F_f+F_d.$$
Derivation
If air resistance did not exist, the net force would remain constant:
$$F_{\text{net}}=10\ \text{kgf}-5\ \text{kgf}=5\ \text{kgf}.$$
A constant positive force would produce continuous acceleration, and after a sufficiently long time the speed could become arbitrarily large. This reasoning leads to the apparent paradox mentioned in the problem.
The missing factor is aerodynamic drag. As the cyclist's speed increases, the drag force grows approximately as the square of the speed. The equation of motion becomes
$$m\frac{dv}{dt}=F_t-F_f-\frac12 C_d\rho A v^2.$$
At first, when $v$ is small, the drag force is negligible and the cyclist accelerates. As $v$ grows, the drag force increases rapidly. Eventually a speed is reached at which
$$\frac12 C_d\rho A v^2=F_t-F_f.$$
Beyond this point the net force would become negative, so further acceleration is impossible.
For a typical cyclist,
$$C_dA \approx 0.5\ \text{m}^2, \qquad \rho \approx 1.2\ \text{kg/m}^3.$$
The available force for overcoming air resistance is
$$F_t-F_f = (10-5),\text{kgf} \approx 49\ \text{N}.$$
Substituting into the balance equation gives
$$49 = \frac12(1.2)(0.5)v^2.$$
Hence
$$49 = 0.3,v^2,$$
so
$$v^2 \approx 163,$$
and
$$v \approx 12.8\ \text{m/s}.$$
This corresponds to
$$v \approx 46\ \text{km/h}.$$
The second cosmic velocity is
$$v_2=11.2\times10^3\ \text{m/s},$$
which is about
$$\frac{11.2\times10^3}{12.8} \approx 8.8\times10^2$$
times larger than the attainable cycling speed. At such a velocity the aerodynamic drag would exceed the cyclist's available force by many orders of magnitude.
Result
The decisive factor is air resistance. The drag force increases approximately as $v^2$, while the cyclist can supply only a limited tractive force.
The limiting speed is determined by
$$F_t-F_f=\frac12 C_d\rho A v_{\max}^2.$$
Using representative values,
$$v_{\max} = \sqrt{\frac{2(F_t-F_f)}{C_d\rho A}} = \sqrt{\frac{2(49\ \text{N})}{(1.2\ \text{kg/m}^3)(0.5\ \text{m}^2}} \approx 12.8\ \text{m/s}.$$
Thus
$$\boxed{v_{\max}\approx 13\ \text{m/s}\approx 46\ \text{km/h}}.$$
A cyclist cannot approach the second cosmic velocity because aerodynamic drag grows enormously with speed and balances the available tractive force at an ordinary cycling speed.
Sanity Checks
The equation
$$F_d=\frac12 C_d\rho A v^2$$
has units
$$(\text{kg/m}^3)(\text{m}^2)(\text{m}^2/\text{s}^2) = \text{kg}\cdot\text{m}/\text{s}^2 = \text{N},$$
which matches the unit of force.
If the air density were reduced toward zero, then $F_d$ would vanish and the limiting speed would increase without bound. This reproduces the naive argument and shows exactly which assumption fails under real terrestrial conditions.
For the second cosmic velocity, the drag force would scale by roughly
$$\left(\frac{11.2\times10^3}{12.8}\right)^2 \approx 7.7\times10^5,$$
relative to the drag at the cyclist's top speed. The required force would be hundreds of thousands of times larger than the cyclist can generate.
The most common source of error is treating the resistive force as a constant $5\ \text{kgf}$ at all speeds. The central physical point is that air resistance is not constant; it increases approximately as $v^2$ and ultimately prevents further acceleration.