1. The problem statement, all variables and given/known data To explain why water falls out of a bucket when it is not being swung fast enough 2. Relevant equations a = mv^2/r 3. The attempt at a solution I have had a look on previous threads and nearly get it but still a bit confused. When the bucket is upside down and the required centripetal force is 9.81m/s^2 then the water will stay in the bucket and the tension in the string will be zero and the normal force from the bottom of the bucket on the water will also be zero. If the bucket was swung where the required centripetal acceleration was 5m/s^2 then gravity would provide too much and the water would fall out of the bucket. It's the description in my book that confuses me: "In order for the water to stay in the bucket when it is upside down the water needs to be accelerated (pulled downwards) faster than what gravity would move the water". If water is being pulled downwards faster than gravity then wouldn't it fall out the bucket even faster not stay in the bucket? Also if the required centripetal acceleration is 5m/s^2 then surely the bucket would also fall so since they are both falling wouldn't the water still stay in the bucket until it hit the ground? Any help is much appreciated!