A horizontal plane train wheel has a diameter of 1 meter. The connecting rod is attached at half the length of the crank. The connection point describes a so-called stretched cycloid. What is the radius of the circle that best approximates the track curve near the highest and lowest points of this point's path?
At the upper point of the cycloid described by the connection point of the driving wheel, the centripetal acceleration in the coordinate system fixed to the train is

$$
a=\omega^{2} \cdot r / 2
$$

where $\omega$ is the angular velocity of the wheel, and $r$ is its radius. If the track at this point is best approximated by a circle of radius $R$, then the centripetal acceleration in the coordinate system fixed to the track is

$$
a^{\prime}=\frac{[r \omega+r / 2(\omega)]^{2}}{R_{1}}
$$

Since the values of the accelerations are the same in any inertial system, $a=a^{\prime}$, and from this

$$
R_{1}=4.5 r=2.25 \mathrm{~m}
$$

By the same reasoning, the following equation is valid at the lower point

$$
\begin{gathered}
\omega^{2} \cdot(r / 2)=\frac{[\omega r-(r / 2) \omega]^{2}}{R_{2}}, \quad \text { hence } \\
R_{2}=0.5 r=0.25 \mathrm{~m}
\end{gathered}
$$

Imre Bártfai (Bonyhád, S. Petőfi Gymnasium, 3rd grade)

Note. The radius of the circle of curvature can also be determined from the condition that the function value, as well as the first and second derivatives of the equation of the best approximating circle, coincide with the corresponding values of the curve. This approach, after a lengthy calculation, leads to the result.

Many used the formula found in mathematical literature

$$
R=\left|\frac{\sqrt{\left[1+f^{\prime 2}(x)\right]^{3}}}{f^{\prime \prime}(x)}\right|
$$

or the more general form written in parametric form

$$
R=\left|\frac{\dot{r}(t)^{3}}{\dot{\mathbf{r}}(t) \times \ddot{\mathbf{r}}(t)}\right|
$$

which can be derived from the previous considerations.