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 this problem, we will apply the principles of conservation of mechanical energy and Hooke's law.
First, let's consider the potential energy and kinetic energy of the system at different positions of the bead.
(a) When θ = 90∘, the bead is at the bottom of the circle. At this position, all the potential energy is converted into kinetic energy.
The potential energy at θ = 0 is given by:
PE = mgh = mgR(1 - cosθ)
At the bottom of the circle, the potential energy is zero, so we can equate the potential energy at θ = 0 to the kinetic energy at θ = 90∘:
mgh = (1/2)mv^2
Simplifying the equation, we get:
mgR = (1/2)mv^2
Solving for v, we have:
v = √(2gR)
Therefore, the speed of the bead when θ = 90∘ is v = √(2gR).
(b) The magnitude of the force the hoop exerts on the bead when θ = 90∘ can be found by considering the net force acting on the bead at that position.
The net force acting on the bead is the sum of the gravitational force and the spring force.
The gravitational force is given by:
Fg = mg
The spring force is given by Hooke's law:
Fs = -kx
At θ = 90∘, the bead is at the relaxed length of the spring, and x = 0. Therefore, the spring force is zero.
The magnitude of the force the hoop exerts on the bead is equal to the gravitational force:
F = Fg = mg
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is F = mg.
Note: In this problem, the bead is assumed to be in static equilibrium at all positions, so the net force acting on the bead is always zero. The force exerted by the hoop is the force required to maintain this equilibrium.
To solve these problems, we will need to analyze the forces acting on the bead at different positions on the hoop.
(a) To find the speed v of the bead when θ = 90∘, we can use the principle of conservation of mechanical energy. At the top of the hoop, the total mechanical energy is the sum of the potential energy and the elastic potential energy of the spring.
The potential energy at the top of the hoop can be calculated as the gravitational potential energy:
PE_top = m * g * 2R
The elastic potential energy of the spring at the top is zero because the spring is at its equilibrium (relaxed) length.
At the position where θ = 90∘, the total mechanical energy is given by:
E = PE_top + 0 = m * g * 2R
At any point on the hoop, the total mechanical energy is the sum of potential energy and kinetic energy:
E = m * g * 2R = (1/2) * m * v^2
Solving for v, we get:
v = sqrt(2 * g * R)
Therefore, the speed v of the bead when θ = 90∘ is sqrt(2 * g * R).
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we can consider the net force acting on the bead at that position.
At θ = 90∘, the gravitational force is acting downward with a magnitude of m * g. The force exerted by the spring can be calculated using Hooke's law:
F_spring = -k * (x - R)
Since the bead is at its equilibrium position when θ = 90∘, the displacement x of the spring from its equilibrium length R is zero, and the force exerted by the spring is also zero.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is equal to the gravitational force:
F_hoop = m * g
So the magnitude of the force the hoop exerts on the bead when θ = 90∘ is m * g.