1. **Problem statement:** A bead slides without friction on a vertical circular hoop of radius $r=0.100$ m rotating at $f=4.00$ rev/s about a vertical diameter. (a) Find the angle $\beta$ where the bead is in vertical equilibrium. (b) Determine if the bead can be at the same elevation as the hoop's center. 2. **Formulas and concepts:** - Angular velocity $\omega = 2\pi f$ rad/s. - Forces on bead: gravity $mg$ downward, and normal force from hoop. - At equilibrium, the bead experiences a balance of forces along the hoop. - Centrifugal force component along the hoop balances gravity component. 3. **Calculate angular velocity:** $$\omega = 2\pi \times 4.00 = 8\pi \approx 25.13 \text{ rad/s}$$ 4. **Set up equilibrium condition:** Let $\beta$ be the angle from the bottom of the hoop to the bead. - Gravity component along hoop: $mg \sin \beta$ - Centrifugal force component along hoop: $m \omega^2 r \sin \beta \cos \beta$ At equilibrium, these balance: $$mg \sin \beta = m \omega^2 r \sin \beta \cos \beta$$ 5. **Simplify equation:** Cancel $m \sin \beta$ (assuming $\sin \beta \neq 0$): $$g = \omega^2 r \cos \beta$$ 6. **Solve for $\beta$:** $$\cos \beta = \frac{g}{\omega^2 r}$$ Calculate denominator: $$\omega^2 r = (25.13)^2 \times 0.100 = 631.7$$ Calculate ratio: $$\frac{9.81}{631.7} \approx 0.01553$$ Therefore: $$\beta = \arccos(0.01553) \approx 89.11^\circ$$ 7. **Answer for (a):** The bead is in vertical equilibrium at approximately $\boxed{89.1^\circ}$. 8. **For (b), can bead be at hoop center elevation?** - Center elevation corresponds to $\beta = 90^\circ$. - At $\beta=90^\circ$, $\sin \beta = 1$, so centrifugal force component along hoop is zero. - Gravity acts downward, no balancing centrifugal force. - Hence, equilibrium is not possible at center elevation. **Answer:** No, the bead cannot ride at the same elevation as the center of the hoop.