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.

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(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.

(a) To find the speed of the bead when θ = 90°, we need to consider the forces acting on the bead at that point. At θ = 90°, the bead is at the highest point of the hoop.

The forces acting on the bead are gravity and the spring force. The gravitational force is given by F_gravity = mg, where m is the mass of the bead and g is the acceleration due to gravity.

The spring force is given by Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the equilibrium position of the spring is at the bottom of the hoop, so the displacement is R. The spring force is given by F_spring = -kx, where k is the spring constant and x is the displacement.

At θ = 90°, the net force on the bead is zero, since it is at equilibrium. So we have:

F_net = F_gravity + F_spring = 0

mg - kR = 0

mg = kR

Now, let's solve for the speed v of the bead. The gravitational potential energy at θ = 0 is zero, and at θ = 90° it is mgR (since the bead is at the highest point). The spring potential energy at θ = 0 is also zero, and at θ = 90° it is 1/2 kR^2 (since the spring is stretched by R).

The total mechanical energy of the system is conserved, so we have:

0 + 0 = mgR + 1/2 kR^2

Simplifying the equation, we get:

mgR = 1/2 kR^2

Cancelling out R on both sides, we have:

mg = 1/2 kR

Now, let's solve for the speed v. The mechanical energy E is equal to the sum of the kinetic energy and potential energy:

E = 1/2 mv^2 + mgR + 1/2 kR^2

Since the bead is released from rest, the initial speed v0 is zero. Therefore, the equation becomes:

0 = 1/2 mv^2 + mgR + 1/2 kR^2

Rearranging the equation, we get:

1/2 mv^2 = -mgR - 1/2 kR^2

Multiplying both sides by 2/m, we have:

v^2 = -2gR - kR^2/m

Taking the square root of both sides, we get:

v = √(-2gR - kR^2/m)

Therefore, the speed v of the bead when θ = 90° is given by:

v = √(-2gR - kR^2/m)

(b) The magnitude of the force the hoop exerts on the bead when θ = 90° can be found using the centripetal force.

At θ = 90°, the bead is at the highest point of the hoop and is moving in a circular motion. The force the hoop exerts on the bead is the centripetal force, which is directed towards the center of the circle.

The centripetal force is given by F_centripetal = m(v^2/R), where m is the mass of the bead, v is its speed, and R is the radius of the hoop.

From part (a), we found the speed v of the bead when θ = 90°. Substituting this into the equation, we have:

F_centripetal = m[(√(-2gR - kR^2/m))^2/R]

Simplifying the equation, we get:

F_centripetal = m(-2gR - kR^2/m)/R

Canceling out the "m" terms, we have:

F_centripetal = -2gR - kR^2/R

Simplifying further, we have:

F_centripetal = -2mg - kR

Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90° is given by:

|F_hoop| = |-2mg - kR| = 2mg + kR

Note that since we are only interested in the magnitude of the force, we take the absolute value.