Tangential Speed Physics Help?

Question

A 0.700 kg ball is on the end of a rope that is 3.50 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. The rope makes an angle of 70.0° down to the right. What is the tangential speed of the ball?
____m/s

Answer 1

EDITED: see corrected result

The speed must be such that the centrifugal force (acting radially outward, i.e. horizontally) creates a vertical component equal to the force of gravity (weight) of the ball.

Centrifugal force = Fc = m*v^2/r, where v is the tangential velocity and r the (horizontal) radius of motion. Here, r = L*sinø, where L = rope length and ø is the angle of the rope with the pole. The centrifugal force creates a tension in the rope of T = Fc/sinø, which in turn produces a vertical component of Fv = T*cosø. Combining these,

Fv = Fc*cosø/sinø = (m*v^2/r)*cosø/sinø = (m*v^2*cosø*sinø)/(L*sinø) = (m*v^2*cosø)/(L*sin^2ø)

This must be equal to m*g, so

m*g = (m*v^2*cosø) / (L*sin^2ø):

v^2 = (g*L*sin^2ø) / cosø

v = √[L*g/cosø] * sinø


EDIT: see the diagram http://img165.imageshack.us/img165/4140/ropeanglevx3.jpg