A bead of mass m slides without friction on a vertical hoop of radius R . The bead moves under the combined action of gravity and a spring, with spring constant k , attached to the bottom of the hoop. Assume that the equilibrium (relaxed) length of the spring is R. The bead is released from rest at θ = 0 with a non-zero but negligible speed to the right. The bead starts on the top of the circle opposing gravitational pull of the earth

(a) What is the speed v of the bead when θ = 90∘ ? Express your answer in terms of m, R, k, and g.

(b) What is the magnitude of the force the hoop exerts on the bead when θ = 90∘ ? Express your answer in terms of m, R, k, and g.

To solve these problems, we need to analyze the forces acting on the bead as it moves on the vertical hoop.

Let's start with the free body diagram of the bead at θ = 90°:

1. (a) Speed of the bead (v) when θ = 90°:
At this point, the gravitational force and the spring force are the only forces acting on the bead.

Gravitational force (Fg): The gravitational force on the bead is given by Fg = mg, where m is the mass of the bead and g is the acceleration due to gravity.

Spring force (Fs): The spring force can be found using Hooke's law, which states that the force exerted by a spring is proportional to its displacement. The equilibrium length of the spring is R, so when θ = 90°, the displacement is R as well. Therefore, Fs = k(R - R) = 0.

Since there is no net force acting on the bead at θ = 90°, it will continue to move at the same speed (v = 0) or come to a momentary stop and change direction.

2. (b) Magnitude of the force the hoop exerts on the bead when θ = 90°:
The force exerted by the hoop on the bead is the sum of the gravitational force and the spring force because these are the only forces acting on the bead at this point.

Therefore, the magnitude of the force the hoop exerts on the bead is given by:
Fhoop = Fg + Fs = mg + 0 = mg.

So, when θ = 90°, the magnitude of the force the hoop exerts on the bead is equal to the gravitational force, Fhoop = mg.