Recent content by thisisbenbtw

Thanks for your help

Little r is the distance from the bead to the axis, correct? So that would be r = R sin θ ω2 r cos θ/ sin θ = g ω2 (R sin θ) cos θ/ sin θ = g cos θ = g / (ω2 R) and θ = arccos(g / (ω2 R))

So Fn = m v2 / (r sin θ) = m ω2 r / sin θ And solving Fn cos θ - m g = 0 : ω2 r cos θ/ sin θ - g = 0 ω2 r cos θ/ sin θ = g cot θ = g / (ω2 r) and θ = arccot(g / (ω2 r)) Is that right?

So there would be -mg and Fn cos θ in the vertical direction, and Fn sin θ = m v2 / r in the horizontal direction?

So the centripetal force would be perpendicular against the vertical axis, and point torwards it?

So there would be a downward force for gravity, a force towards the center of the hoop for centripetal force, and another force in the same direction for the normal force?

Homework Statement A bead of mass m is constrained to move on a hoop of radius R which is spinning with angular velocity ω. There is no friction. Determine the angle θ at which the bead does not move for angular velocity ω. Do not consider the solution θ = 0. Homework Equations...