357. A stick inside a hemisphere. A stick of length $d$ is placed in a hemispherical basin with a diameter of $d$. Neglecting the thickness of the stick and assuming that friction is absent, determine the angle that the stick will form with the diameter in the equilibrium position.
357. I. If we are inventive enough and draw the additional lines shown in Fig. 1, we immediately obtain the following relationships:

$$
\begin{gathered}
y=u \sin \theta \\
v=2 a \sin \theta \\
v^{2}+(u+a)^{2}=4 a^{2}
\end{gathered}
$$

By eliminating \( u \) and \( v \) from these equations, we arrive at the equation

$$
y=a(\sin 2 \theta - \sin \theta)
$$

If \( P \) is the center of gravity of our rod, then the equilibrium position will be achieved when \( P \) is in the lowest possible position. Therefore, the angle we are looking for is the angle at which \( y \) reaches its maximum value. Setting \( y'(\theta) \) to zero, we obtain the equation

$$
4 \cos^2 \theta - \cos \theta - 2 = 0
$$

from which we find

$$
\cos \theta = \frac{1 \pm \sqrt{33}}{8}
$$

But the angle we are looking for is acute. Therefore, the only solution to the problem is \( \theta \approx 32^\circ 23' \).

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-213.jpg?height=280&width=506&top_left_y=1596&top_left_x=775)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-214.jpg?height=317&width=326&top_left_y=321&top_left_x=865)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-214.jpg?height=343&width=414&top_left_y=708&top_left_x=821)

Fig. 3

II. Introduce polar coordinates as shown in Fig. 2. Then the trajectory of the center of gravity \( P \) (Fig. 3) will be given by the equation

$$
r = 2a \cos \theta - a, \quad -\frac{\pi}{2} \leq \theta \leq 0
$$

To find the lowest position of \( P \), recall the expression for the slope of the tangent to a curve given in polar coordinates

$$
k = \frac{r + r' \tan \theta}{r' - r \tan \theta}
$$

and set it to zero. Substituting the expression \( r(\theta) \) into the equation

$$
r + r' \tan \theta = 0
$$

we obtain the equation

$$
4 \cos^2 \theta - \cos \theta - 2 = 0
$$

Solving this, we find \( \theta \approx -32^\circ 32' \). The negative sign is related to the range of \( \theta \) in polar coordinates.

The second solution has the advantage that it does not require additional constructions, and it allows us to find the trajectory of the center of gravity and provide an interpretation for the second value of \( \cos \theta \).

[J. Stab, M. M., 45, 286 (May 1972).]