LOOP, SPRING AND BEAD (14 points possible)

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 v of the bead when θ = 90∘, we need to consider the energy of the system.

Let's go step by step:

Step 1: Find the total mechanical energy of the system.
- The mechanical energy is the sum of the kinetic energy and potential energy.
- At any point on the hoop, the gravitational potential energy is given by mgh, where h is the height of the bead from the lowest point of the hoop to the current point. Since the bead starts from rest at θ = 0, the initial potential energy is 0.
- The kinetic energy is given by (1/2)mv², where v is the speed of the bead.
- The potential energy due to the spring is given by (1/2)kx², where x is the displacement of the bead from the relaxed position. At θ = 0, the spring is neither stretched nor compressed, so the initial potential energy due to the spring is 0.
- Therefore, the initial mechanical energy of the system is given by E = 0.

Step 2: Find the mechanical energy at θ = 90∘.
- At θ = 90∘, the maximum potential energy due to gravity is mgh = mgR.
- The spring potential energy at θ = 90∘ is (1/2)kx², where x = R since the spring is stretched to its maximum at this point.
- Therefore, the mechanical energy at θ = 90∘ is E = mgR + (1/2)kR².

Step 3: Apply the principle of conservation of energy.
- According to the conservation of energy, the initial mechanical energy should equal the mechanical energy at θ = 90∘.
- Therefore, mgR + (1/2)kR² = (1/2)mv².

Step 4: Solve for v.
- Rearrange the equation to solve for v:
v = √[(2/m)(mgR + (1/2)kR²)].

So, the speed v of the bead when θ = 90∘ is given by the equation:
v = √[(2/m)(mgR + (1/2)kR²)].

Now, let's move on to part (b).

To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we need to consider the forces acting on the bead.

Step 1: Determine the net force acting on the bead.
- The net force is the sum of the spring force and the gravitational force.
- The spring force is given by Hooke's Law: Fs = -kx, where x = R is the displacement of the bead from the relaxed position.
- The gravitational force is given by Fg = mg.
- Therefore, the net force is Fnet = Fs + Fg = -kR + mg.

Step 2: Find the magnitude of the force the hoop exerts on the bead.
- The hoop exerts a centripetal force to keep the bead moving in a circular path.
- At θ = 90∘, the direction of the force exerted by the hoop is towards the center of the hoop.
- Therefore, the magnitude of the force the hoop exerts on the bead is Fhoop = kR - mg.

So, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is given by the equation:
Fhoop = kR - mg.