a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top meaning n equals zero there. What is the speed of the water at the top

mv^2 / r = mg

2.35 kg*

gravity is providing centripetal acceleration ... v^2 / r = g

2.84 To find the speed of the water at the top of the swing, we can use conservation of mechanical energy. At the top of the swing, the water bucket has its maximum potential energy and zero kinetic energy. As the water bucket swings downward, its potential energy decreases and is converted into kinetic energy. At the very bottom of the swing, the potential energy is at its minimum, and the entire energy has been converted into kinetic energy.

Since we know the radius of the circle (0.824 m), we can calculate the potential energy at the top of the swing using the formula:

Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)

In this case, the height (h) is equal to the radius (0.824 m). The mass (m) of the water is given as 2.35 kg, and the acceleration due to gravity (g) is approximately 9.81 m/s^2.

PE = 2.35 kg * 9.81 m/s^2 * 0.824 m

Next, we can equate the potential energy to the kinetic energy at the top of the swing, since they must be equal:

PE = KE

So, at the top of the swing:

m * g * h = (1/2) * m * v^2

Where v is the speed of the water at the top of the swing.

Simplifying the equation, we have:

v^2 = 2 * g * h

v^2 = 2 * 9.81 m/s^2 * 0.824 m

v^2 = 16.073

Taking the square root of both sides:

v ≈ 4.008 m/s

Therefore, the speed of the water at the top of the swing is approximately 4.008 m/s.