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.

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 find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we need to analyze the forces acting on the bead at that specific angle.

At θ = 90∘, the bead is at the highest point of its vertical circular motion. At this point, the net force acting on the bead should be equal to the centripetal force required to keep it moving in a circle.

Let's break down the forces acting on the bead at θ = 90∘:

1. Gravity (mg): The bead experiences a downward force due to gravity. This force can be expressed as mg, where m is the mass of the bead and g is the acceleration due to gravity.

2. Centripetal Force (Fc): The bead is undergoing circular motion, so it requires a centripetal force to keep it moving in a circle. The centripetal force at θ = 90∘ can be expressed as: Fc = m * v^2 / R, where m is the mass of the bead, v is the instantaneous velocity at that point, and R is the radius of the hoop.

3. Spring Force (Fs): The spring attached to the bottom of the hoop will exert a force on the bead. However, at the point θ = 90∘, the spring is completely compressed, and its force is zero. This is because the equilibrium length of the spring is R, and the bead is at R distance from the bottom of the hoop.

Now, to find the magnitude of the force the hoop exerts on the bead at θ = 90∘, we need to calculate the net force acting on the bead.

Net Force (Fnet) = Fc + Fs - mg

Since Fs is zero at θ = 90∘, the net force simplifies to:

Fnet = Fc - mg

Substituting the expression for Fc:

Fnet = (m * v^2 / R) - mg

Remember, we were given that the bead is released from rest at θ = 0 with a non-zero but negligible speed to the right. In this case, the magnitude of the velocity (v) at θ = 90∘ is negligible.

Thus, the magnitude of the force the hoop exerts on the bead at θ = 90∘ is:

F = -mg

So, the answer is -mg, expressed in terms of m, R, k, and g.