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 find the speed v of the bead when θ = 90∘, we need to consider the forces acting on the bead at that position.
Let's analyze the situation:
1. Gravity force (mg):
The bead experiences a downward force due to gravity, which can be represented as mg, where m is the mass of the bead and g is the acceleration due to gravity.
2. Spring force (Fs):
The spring attached to the bottom of the hoop is compressed, exerting a force on the bead in the upward direction. This force can be calculated using Hooke's Law: Fs = k * (R - d), where k is the spring constant and d is the displacement of the spring from its relaxed length (R).
At θ = 90∘, the bead is at the bottom of the hoop, and the displacement of the spring (d) is R, as it is stretched to its maximum length. Hence, Fs = k * (R - R) = 0.
Since there is no spring force acting on the bead at θ = 90∘, only gravity is responsible for the motion of the bead.
Applying the principle of conservation of energy, the initial potential energy of the bead is converted into kinetic energy when it reaches θ = 90∘.
The potential energy at the top of the hoop is given by mg * (2R), as the height is twice the radius.
The kinetic energy at θ = 90∘ is given by (1/2) * m * v^2, where v is the velocity of the bead.
Equating the potential energy at the top to the kinetic energy at θ = 90∘, we have:
mg * (2R) = (1/2) * m * v^2
Simplifying, we find:
v^2 = 4gR
Taking the square root of both sides, we get:
v = 2√(gR)
Therefore, the speed of the bead when θ = 90∘ is v = 2√(gR).
For part (b), we need to find the magnitude of the force the hoop exerts on the bead when θ = 90∘.
The force exerted by the hoop can be decomposed into two components:
1. Normal force (N):
The hoop exerts a normal force (N) on the bead in the direction perpendicular to the hoop's surface, which balances the gravitational force acting on the bead.
2. Centripetal force (Fc):
Since the bead is moving along a circular path, there must be a centripetal force that keeps it in this motion. In this case, it is provided by the normal force.
At θ = 90∘, the normal force (N) becomes equal in magnitude but opposite in direction to the gravitational force (mg) to maintain circular motion.
Hence, the magnitude of the force the hoop exerts on the bead at θ = 90∘ is given by:
F = N = mg
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is equal to mg.