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 find the speed v of the bead when θ = 90∘, we can use energy conservation. At θ = 0, the bead has gravitational potential energy but no kinetic energy since it is at rest. As the bead moves to θ = 90∘, it loses some of its potential energy but gains kinetic energy.
Let's denote the initial gravitational potential energy as U(0) and the spring potential energy as Uspring(0). The total initial potential energy is the sum of these two:
U(0) = m*g*R
The final kinetic energy at θ = 90∘ is equal to the initial gravitational potential energy:
K(90∘) = U(0)
The final spring potential energy at θ = 90∘ is 0 since the spring is at its relaxed length.
Uspring(90∘) = 0
Using energy conservation, we have:
U(0) + Uspring(0) = K(90∘) + Uspring(90∘)
Simplifying, we get:
m*g*R + 0 = 1/2 * m * v^2 + 0
Solving for v, we find:
v = √(2*g*R)
So the speed v of the bead when θ = 90∘ is √(2*g*R).
Now let's move to part (b) and find the magnitude of the force the hoop exerts on the bead when θ = 90∘.
The force exerted by the hoop on the bead is a centripetal force that keeps the bead moving in a circular path. This force is provided by both gravity and the spring.
At θ = 90∘, the gravitational force is pointing downwards and has a magnitude of m*g.
The spring force acts along the direction of the spring, which is tangent to the circular path at the bottom of the hoop. Its magnitude is given by Hooke's Law:
Fspring = k * (x - L)
Where x is the displacement of the bead from its equilibrium position and L is the relaxed length of the spring (which is equal to R in this case).
At θ = 90∘, the displacement x is equal to R, so:
Fspring = k * (R - R) = 0
Therefore, the spring force is 0 when θ = 90∘.
The magnitude of the force the hoop exerts on the bead is equal to the gravitational force:
Fhoop = m * g
So the magnitude of the force the hoop exerts on the bead when θ = 90∘ is m*g.