problem on circular motion

Question

You swing a bucket of water at constant speed in a vertical circle at arm's length(0.7m).What is the minimum number of revolutions per second you must maintain to keep the water from spilling out of the bucket?Please show all the calcuate steps.

Answer 1

let the velocity at the highest point be v
the velocity at the lowest point be V
the radius of the circular path be r
the angular speed be w
the acceleration due to gravity be g=9.8 ms^(-2)

in order to maintain a complete circular motion
at the highest point:
(mv^2)/r=mg
v^2=gr ------(i)

by conservation of energy:
(1/2)mV^2=(1/2)mv^2+mgh
(1/2)mV^2=(1/2)m(gr)+mg(2r) <----sub (i) into here
(1/2)V^2=(1/2)gr+2gr <----eliminate m
V^2=gr+4gr <----mutiply by 2
V=(5gr)^(1/2) <----take root for both sides
rw=(5gr)^(1/2) <----by V=rw
w=(1/r)(5gr)^(1/2)
w=(1/0.7)[(5)(9.8)(0.7)]^(1/2)
w=8.37 rad/s or 1.33 rev/s <----simply divided by 2 pi

so the answer is 1.33 rev/s

Answer 2

Let m be the mass of water in the bucket

When the bucket is at the highest point of the vertical circle, the weight mg of the water just provides the necessary centripetal force, then the water would not spill out.

thus, mg = mr.w^2
where r is the radius of the circle (=0.7 m)
w is the angular velocity of the bucket
g is the acceleration due to gravity

we get, w^2 = g/r
but w = 2.pi.f
where f is the frequency (no of revolutions per second) and pi = 3.14159.....

hence, f = (1/2.pi).sqrt(g/r)
[sqrt = square root]