On a straight country road, a car travels at a constant speed $v$. Its motion is observed by a radar located at a distance $h$ from the road. Determine the angular velocity and angular acceleration of the radar's rotating antenna as a function of time and angular displacement. Plot the functions and determine the angular displacement at which the angular acceleration is maximal.
Let's set the time when the car passes the perpendicular point drawn from the locator to the road as 0! We measure the angular displacement of the locator from this perpendicular!

![](https://cdn.mathpix.com/cropped/2024_05_02_271f369abba9e95c27d9g-1.jpg?height=355&width=326&top_left_y=278&top_left_x=886)

Figure 1

Based on Figure 1:

$$
\operatorname{tg} \alpha=\frac{v \cdot t}{h} ; \quad t=\frac{h}{v} \operatorname{tg} \alpha
$$

Thus,

$$
\alpha=\operatorname{arctg}(v \cdot t / h)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_271f369abba9e95c27d9g-1.jpg?height=371&width=539&top_left_y=1105&top_left_x=782)

Figure 2

The angular velocity of the locator (Figure 2):

$$
\omega(t)=\frac{\mathrm{d} \alpha}{\mathrm{d} t}=\frac{v}{h} \cdot \frac{1}{1+\frac{v^{2} t^{2}}{h^{2}}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_271f369abba9e95c27d9g-1.jpg?height=358&width=458&top_left_y=1813&top_left_x=817)

Figure 3

Eliminating \( t \) using (1) gives us \( \omega \) as a function of \( \alpha \) (Figure 3):

$$
\omega(\alpha)=\frac{v}{h} \cos ^{2} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_271f369abba9e95c27d9g-2.jpg?height=447&width=588&top_left_y=129&top_left_x=758)

Figure 4

The angular acceleration is the time derivative of the angular velocity (Figure 4):

$$
\beta(t)=\frac{\mathrm{d} \omega}{\mathrm{d} t}=-2\left(\frac{v}{h}\right)^{2} \frac{v t}{h} \cdot \frac{1}{1+\left(\frac{v^{2} t^{2}}{h^{2}}\right)^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_271f369abba9e95c27d9g-2.jpg?height=521&width=726&top_left_y=978&top_left_x=686)

Using (1) (Figure 5):

$$
\beta(\alpha)=-2\left(\frac{v}{h}\right)^{2} \frac{\operatorname{tg} \alpha}{\left(1+\operatorname{tg}^{2} \alpha\right)^{2}}=-2\left(\frac{v}{h}\right)^{2} \cdot \sin \alpha \cos ^{3} \alpha
$$

The angular acceleration has a local maximum where its derivative is zero:

$$
\frac{\mathrm{d} \beta}{\mathrm{d} \alpha}=-2\left(\frac{v}{h}\right)^{2} \cos ^{2} \alpha\left(\cos ^{2} \alpha-3 \sin ^{2} \alpha\right)
$$

The interesting zeros of this are at \( \alpha = \pm 90^{\circ} \) and \( \alpha = \pm 30^{\circ} \). Among these, the maximum occurs at \( \alpha = -30^{\circ} \), where the derivative changes from positive to negative.

Schmidt József (Esztergom, Dobó K. Gymnasium, 4th grade)