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.
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 find the magnitude of the force the hoop exerts on the bead when θ = 90°, we need to consider the forces acting on the bead when it reaches that position.
At θ = 90°, the bead is at the highest point of the hoop. We can break down the forces into two components: the gravitational force and the spring force.
1. Gravitational Force:
The gravitational force acting on the bead is given by the equation F_gravity = m * g, where m is the mass of the bead and g is the acceleration due to gravity.
2. Spring Force:
The spring force at equilibrium is zero when the spring is at its relaxed length. However, when the bead is displaced from equilibrium, the spring starts exerting a force to bring the bead back to its equilibrium position. The magnitude of the spring force can be calculated using Hooke's law: F_spring = -k * x, where k is the spring constant and x is the displacement from equilibrium.
At θ = 90°, the displacement of the bead from equilibrium is R - R = 0 because the bead is at equilibrium. Therefore, the spring force is zero.
Since the spring force is zero, the only force acting on the bead at θ = 90° is the gravitational force. Therefore, the magnitude of the force the hoop exerts on the bead at θ = 90° is equal to the gravitational force.
Answer: The magnitude of the force the hoop exerts on the bead when θ = 90° is F_gravity = m * g.