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 of the bead when θ = 90∘, we can use conservation of energy.
(a) First, let's consider the initial and final states of the system:
Initial state: The bead is at θ = 0 with no speed.
Final state: The bead is at θ = 90∘ with speed v.
The total mechanical energy of the system is conserved. It is the sum of the kinetic energy and potential energy:
Initial energy (Ei) = Final energy (Ef)
The initial energy consists of only potential energy:
Ei = mgh = mgR (where h is the height at θ = 0, which is R)
The final energy consists of both potential energy and kinetic energy:
Ef = mgh' + 1/2mv^2
At θ = 90∘, the height h' is 2R (the highest point of the circle).
Plugging in these values, we get:
mgR = mg(2R) + 1/2mv^2
Now, let's solve for v:
v^2 = 2gR
Therefore, the speed v of the bead when θ = 90∘ is given by:
v = √(2gR)
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we can use Newton's second law:
F = ma
The net force acting on the bead is the centripetal force, which is provided by the tension in the spring and the gravitational force.
Centripetal force = Tension force + Weight force
The weight force is given by mg.
Now, let's analyze the forces:
Tension force: The spring provides a force that is proportional to the change in length from the equilibrium position. In this case, the spring is stretched by a distance R (2R - R). So, the tension force is given by:
Tension force = k * (change in length) = kR
Weight force: mg
Centripetal force = Tension force + Weight force = kR + mg
At θ = 90∘, the centripetal force is equal to the mass times the centripetal acceleration:
mv^2/R = kR + mg
Solving for the magnitude of the force the hoop exerts, we get:
Magnitude of force (F) = mv^2/R - mg = (m/R)(v^2 - gR)