1. **Problem statement:** A bead slides without friction on a circular hoop of radius $r=0.100$ m rotating at $f=4.00$ rev/s about a vertical diameter. We need to find the angle $\beta$ where the bead is in vertical equilibrium and check if it can be at the same elevation as the hoop's center. 2. **Formula and concepts:** The hoop rotates with angular velocity $\omega = 2\pi f = 2\pi \times 4.00 = 8\pi$ rad/s. The bead is in equilibrium when the component of gravitational force balances the centripetal force due to rotation. The equilibrium condition is: $$\tan(\beta) = \frac{\omega^2 r}{g}$$ where $g=9.8$ m/s$^2$ is acceleration due to gravity. 3. **Calculate $\tan(\beta)$:** $$\tan(\beta) = \frac{(8\pi)^2 \times 0.100}{9.8} = \frac{64\pi^2 \times 0.100}{9.8} = \frac{64 \times 9.8696 \times 0.100}{9.8}$$ $$= \frac{63.165}{9.8} = 6.447$$ 4. **Find $\beta$:** $$\beta = \arctan(6.447) = 81.1^\circ$$ 5. **Check if bead can be at hoop center elevation:** At the center elevation, $\beta = 90^\circ$. Since $\tan(90^\circ)$ is undefined (infinite), and our $\tan(\beta)$ is finite, the bead cannot be in equilibrium at the center elevation. **Final answers:** - $\beta = 81.1^\circ$ - No, the bead cannot ride at the same elevation as the center of the hoop.