physics question about centripetal force.?

Question

Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.91 m long (shoulder to fist) and the distance from the handle to the surface of the water is 17.0 cm, what minimum speed is required?

Answer 1

The centripetal acceleration must balance the gravitational force of the water:

mv^2/r = mg

Therefore, v = sqrt(gr)

r is the total distance from the axis of rotation (your shoulder) to the surface of the water, or 1.08 m.

v = sqrt(1.08*9.81) = 3.26 m/s

Answer 2

I will tack a wack at it. The pail when at the top of its rotation must have sufficient speed that the centrifugal force balances out the gravitational pull. The water at the "bottom" of the pail is furthest away and must spin the fastest to cancel out gravity.

g = v^2/r => v = 3.254 m/sec

where g = 9.8 m/sec^2
r = 0.91m+0.17m

Answer 3

The centripetal force always acts perpendicular to the direction of motion of the body. The force is directed inward, toward the center of the circle.

People like to say CENTRIFICAL FORCE however its
Centrifugal Force this force is outwards from the center
and the force that would keep water in the bucket

Answer 4

♦ the critical situation is, when the pail is upside-down, that’s centripetal F must be >= m*v^2/R >= mg, where mg is weight of water, m is mass of water, R=0.91+0.17 = 1.08m; thus
♣ m*v^2/R = mg; → v^2/R = g; → v=√(R*g) = √(1.08*9.81) =3.26 m/s;

Answer 5

V^2 = Rg

R = 0.91 + 0.17 = 1.08m

V^2 = 1.08 * 9.8 = 10.584

V = 3.25 m/s