(!) 7. An arc of a circle of radius $R$ contains $\alpha$ radians. Find the position of its center of gravity. Where is the center of gravity of a semicircle? (O. 117, P. 206.)
7. Let $O$ be the center of the circle (Fig. 166), $\widetilde{A B}$ be the given arc, $A B$ be the chord subtending it, $O C \perp A B$, and $M N \perp O C$.

The center of gravity $Z$ of the arc lies on the radius $O C$ (due to the symmetry of the arc relative to $O C$).

When rotating around the axis $M N$, the arc $\overline{A B}$ will describe a spherical zone with surface area $S$, where

$$
S=2 \pi R \cdot A B=2 \pi R \cdot 2 R \cdot \sin \frac{\alpha}{2}=4 \pi R^{2} \sin \frac{\alpha}{2}
$$

Since the length of the arc $\breve{A B}$ is $R \alpha$, by the first theorem of Guldin,

$$
S=2 \pi \cdot O Z \cdot R \cdot \alpha
$$

Comparing the two expressions for $S$, we find

$$
O Z=\frac{2 R}{\alpha} \sin \frac{\alpha}{2}
$$

If $\breve{A B}$ is a semicircle, then $\alpha=\pi$ and $O Z=\frac{2 R}{\pi}$ (compare with problem $2 \S 8$).

![](https://cdn.mathpix.com/cropped/2024_05_21_27f86caa754616304e40g-208.jpg?height=272&width=373&top_left_y=445&top_left_x=685)

Fig. 166.