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.
(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 the principle of conservation of mechanical energy. The total mechanical energy of the system is conserved and is the sum of the potential energy (PE) and kinetic energy (KE) of the bead.
(a) To find the speed v, we need to find the potential energy and kinetic energy when θ = 90∘ and equate them.
First, let's consider the potential energy of the bead when θ = 90∘. At this point, the height of the bead from the lowest point on the hoop is given by h = R - R = 0. Since the equilibrium length of the spring is R, the displacement (x) of the bead from its equilibrium position is also zero. Therefore, the potential energy (PE) at θ = 90∘ is zero.
Next, let's consider the kinetic energy of the bead at θ = 90∘. The total mechanical energy at any point is given by the sum of the kinetic energy and potential energy. At θ = 90∘, all the mechanical energy of the bead is in the form of kinetic energy. Assuming the speed of the bead at θ = 90∘ is v, the kinetic energy (KE) is given by KE = (1/2)mv², where m is the mass of the bead.
Since mechanical energy is conserved, we can equate the potential energy and kinetic energy:
0 = (1/2)mv²
Solving for v, we have:
v = 0
Therefore, the speed of the bead when θ = 90∘ is zero.
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we can consider the forces acting on the bead at that point.
The forces acting on the bead are gravity (mg) and the spring force (F_spring). The spring force always acts towards the equilibrium position, which is the bottom of the hoop. At θ = 90∘, the bead has moved away from the equilibrium position and the spring is extended. Therefore, the spring force is directed towards the bottom of the hoop.
Since the bead is not accelerating vertically at θ = 90∘, the net force in the vertical direction is zero. Therefore, the magnitudes of the two forces must be equal:
F_spring = mg
The force exerted by the hoop on the bead is equal to the spring force, so the magnitude of the force the hoop exerts on the bead when θ = 90∘ is equal to mg.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is mg.