you swing the pail in a vertical circle of radius 0.900m.What minimum speed must you give the pail at the highest point of the circle if no water is to spill from it?

well, centripetal force must be greater than weight

mg<=mv^2/r
v>=sqrt(g*r)
check that.

Consider the following equation in the y-direction.

Fy=T+Fg=ma

Or in this case...

T+Fg=(mv^2)/R

Tension equals to 0 because when the pail of water is at the highest point, tension is nonexistent.

Therefore...

mg=(mv^2)/R

Cancel out mass

g=(v^2)/R

(Rg)^(1/2)=v

Plug in values.

i don't know where to go from there

v is the minimum speed.

To find the minimum speed required for the pail at the highest point of the circle, we can use the concept of centripetal force and equate it to the force due to gravity.

At the highest point in the circle, the pail is upside down, and the water inside it is trying to spill out. The force that prevents the water from spilling out is the centripetal force.

The centripetal force is given by the equation:

F = m * v² / r

Where:
F = Centripetal force (in this case, the force to prevent the water from spilling)
m = Mass of the pail
v = Velocity of the pail
r = Radius of the circle

Since the pail is at the highest point of the circle, the centripetal force should be equal to the force due to gravity:

F = m * g

Where:
g = Acceleration due to gravity (approximately 9.8 m/s²)

Now we can equate the two equations:

m * v² / r = m * g

Canceling out the mass (m) on both sides of the equation, we get:

v² / r = g

Rearranging the equation to solve for v, we have:

v = √(g * r)

Substituting the given values:
r = 0.900 m
g = 9.8 m/s²

v = √(9.8 * 0.900)

v ≈ 3.0 m/s

Therefore, the pail must be given a minimum speed of approximately 3.0 m/s at the highest point of the circle to prevent the water from spilling out.