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.

sqrt(( (2*sqrt(2)-2)*k*R^2)/m+2*g*R)

(2*(sqrt(2)-1)*k*R)+(2*m*g)-((sqrt(2)-1)*(k*R)*(1/sqrt(2)))

To solve this problem, we need to analyze the forces acting on the bead at different angles θ and use the principles of Newtonian mechanics.

(a) The speed v of the bead when θ = 90∘ can be determined using conservation of mechanical energy.

At θ = 0, the bead has gravitational potential energy and no kinetic energy. At θ = 90∘, the bead has kinetic energy and no gravitational potential energy.

The total mechanical energy is given by:
E = PE + KE

At θ = 0:
E = mgh = m(0)(R) = 0

At θ = 90∘:
E = 1/2 mv^2

Since we assume no friction, the energy is conserved, so E = 0.

Setting the equations equal to each other, we have:
0 = 1/2 mv^2

Solving for v:
v = 0

Therefore, the speed v of the bead when θ = 90∘ is 0.

(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we need to consider the net force acting on the bead at this point.

The net force is the sum of the gravitational force, the force from the spring, and the normal force from the hoop.

At θ = 90∘, the gravitational force and the normal force cancel each other out since the bead is in equilibrium. Therefore, the net force is only due to the force from the spring.

Using Hooke's Law, the force of the spring is given by:
F = -kΔx

Since the equilibrium (relaxed) length of the spring is R and the bead is at the bottom of the hoop when θ = 90∘, the displacement Δx of the bead with respect to the equilibrium position of the spring is Δx = R.

Substituting these values into the equation, we have:
F = -kR

The magnitude of the force the hoop exerts on the bead is equal to the magnitude of the force from the spring. Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is |F| = kR.

In summary, the speed v of the bead when θ = 90∘ is 0, and the magnitude of the force the hoop exerts on the bead when θ = 90∘ is kR.