A bead can slide without friction on a circular hoop with a radius of .100 m in a vertical plane. The hoop rotates at a constant rate of 3.00 rev/s about a vertical diameter.
a) find the angle θ at which the bead is a vertical equilibrium.
b) is it possible for the bead to "ride" at the same elevation as the center of the Hoop?
c) what will happen if the hoop rotates at 1 rev/s?© BrainMass Inc. brainmass.com February 24, 2021, 2:14 pm ad1c9bdddf
Radius of hoop (R) = 1.00 m (I think .100m is typo, but any way! I have solved for both values.)
angular rotation of hoop (w)= 2*pi*f
pi = 22/7, f = rev/sec = 3.00, w == omega (angular frequency)
w = 2*pi*3 = 18.85 rad/sec
see attached file.
because, radius of circular motion of bead = R*sin(Q)
here Q == theta.
The solution explains the problem and shows the workings to arrive at the answers.