A cord is tied to a pail of water and the pail is swung in a vertical circle of 1.2 m. What is the minimum velocity the pail must have at the top of the circle if no water spills?

2 Answers
Dec 4, 2015

Answer:

#3.43"m/s"#

Explanation:

At the top of the circle the weight is balanced by the centripetal force so:

#cancel(m)g=(cancel(m)v^2)/r#

#:.v=sqrt(gr)#

#v=sqrt(9.8xx1.2)=3.43"m/s"#

Dec 4, 2015

Answer:

#3.4 "m"/"s"#

Explanation:

Let us examine the forces acting not on the pail, but on the water inside the pail at the top of the circular path.

Imagine the pail being swung at a fairly high speed. At the top of the path, the pail exerts a normal force on the water, directed downwards. This combined with the force of gravity gives us a centripetal force that keeps the entire pail-water system moving in a circle.

Here's an appropriate free-body diagram I found online: (ignore the "bottom of circle" portion)
http://www.physicsclassroom.com/calcpad/circgrav

Now, if we lower the speed enough there is a certain value for which the normal force becomes zero at the exact instant in which the pail is at the top of the path. And, any speed lower than this value would result in the water falling out of the pail - the water would simply not have enough momentum to resist the force of gravity.

Recall that for an object undergoing circular motion at a constant speed we have

#a = v^2/r#

Since #F_"net" = F_"grav" + F_"norm"#, If we let #F_"norm" = 0# then we have #F_"net" = F_"grav"#.

So we can apply Newton's second law and obtain

#F_"net" = ma = mv^2/r#

Since we know that #F_"net" = F_"grav"#, and #F_"grav" = mg#, we can substitute this into the above equation to find

#mg = mv^2/r#

Cancelling #m#'s and solving for #v# yields

#v = sqrt(rg)#

Assuming that #g = 9.81 "m"/"s"# we have

#v = 3.4 "m"/"s"#