A bucket of water is swung in a vertical plane at the end of a rope of length l= 6 m. The mass of the bucket plus water is 5 kg and the gravitational acceleration is g=10 m/s2. We assume that the mass of the rope can be neglected.
(a) What is the minimal speed of the bucket at its highest point in the circular motion, such that the water does not fall out? (in m/s)
(b) For this speed, what is the magnitude of the centripetal acceleration that the water in the bucket experiences at the highest point? (in m/s2)
ω = ω0 + α * t
θ = ω0 * t + 0.5 * α * t ^ 2
ω = ω0 ^ 2 + 2 * α * θ
The Attempt at a Solution
I am newbie in physics, I solved some problems in kinematics already, but I don't even know how to start in this case