1. **Problem statement:** A bucket filled with water is tied to a string of length 50 cm and rotated in a vertical circle. We need to find the minimum angular speed at the highest point so that no water spills. 2. **Key concept:** At the highest point, the water will not spill if the centripetal force is at least equal to the gravitational force acting on the water. 3. **Formula:** The centripetal force required is given by $$F_c = m\omega^2 r$$ where $\omega$ is the angular speed in radians per second and $r$ is the radius (length of the string). 4. At the highest point, the tension in the string can be zero for minimum angular speed, so the only force providing centripetal acceleration is gravity: $$mg = m\omega^2 r$$ 5. Simplify to find angular speed: $$g = \omega^2 r$$ $$\omega = \sqrt{\frac{g}{r}}$$ 6. Given: $$r = 50\text{ cm} = 0.5\text{ m}$$ $$g = 9.8\text{ m/s}^2$$ 7. Calculate $\omega$: $$\omega = \sqrt{\frac{9.8}{0.5}} = \sqrt{19.6} \approx 4.427\text{ rad/s}$$ 8. Convert angular speed to revolutions per second: $$1\text{ revolution} = 2\pi\text{ radians}$$ $$f = \frac{\omega}{2\pi} = \frac{4.427}{2\pi} \approx 0.704\text{ rev/s}$$ **Final answer:** The minimum angular speed at the highest point is approximately **0.704 revolutions per second**.