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 answer parts (a) and (b), we need to consider the energy and forces acting on the bead as it moves on the vertical hoop.
(a) To find the speed v of the bead when θ = 90∘, we can use the principle of conservation of mechanical energy. At angle θ = 0, the bead is released from rest, so it has potential energy U = mgh, where h = 2R is the height from the bottom of the hoop to the top. At θ = 90∘, all the potential energy is converted into kinetic energy.
The potential energy at θ = 0 is U = mgh, and the kinetic energy at θ = 90∘ is K = 0. Therefore, we have:
U = K
mgh = (1/2)mv^2
Simplifying the equation, we can find the speed v:
v = √(2gh)
Therefore, the speed v of the bead when θ = 90∘ is √(2gh), where g is the acceleration due to gravity and h = 2R is the height.
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we can use Newton's second law of motion. At θ = 90∘, the bead experiences both gravity and the force from the spring.
The force from the spring is given by Hooke's Law:
F = -kx
where k is the spring constant and x is the displacement from the equilibrium position (which is R in this case).
At θ = 90∘, the bead is in equilibrium, so the force from the spring balances the force due to gravity. The force due to gravity is given by Fg = mg.
Therefore, we have:
mg = kx
Simplifying the equation, we can find the magnitude of the force:
F = mg/k
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is mg/k, where m is the mass, g is the acceleration due to gravity, and k is the spring constant.