A bucket full of water (15kg) is swung in a vertical circle on a rope 1.5m long. What is the minimum speed it can be swung so that the water doesn't fall out at the top?
Which equations should I start with first?
To solve this problem, you can start by considering the forces acting on the bucket at the top of the swing. At the topmost point, the water in the bucket experiences a downward force due to gravity and an upward force due to the tension in the rope.
The equation that relates these forces at the top of the swing is:
Tension + Weight of Water = Centripetal Force
First, calculate the weight of the water using the equation:
Weight of Water = Mass of Water × Acceleration due to Gravity
Given that the mass of water is 15 kg and the acceleration due to gravity is approximately 9.8 m/s², you can calculate the weight of the water.
Next, consider the centripetal force acting on the bucket. At the top of the swing, the centripetal force is given by:
Centripetal Force = Mass of Water × (Velocity^2 / Radius)
In this case, the radius is given as 1.5 m.
Since we want to find the minimum speed at which the water won't fall out, we set the tension to zero. Therefore, the equation becomes:
Weight of Water = Centripetal Force
By substituting the expressions for weight of the water and centripetal force, and solving for velocity, you can find the minimum speed at which the water won't fall out.