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.
where is theta=0?
on the top part of the circle opposite the standard direction gravitational force
To solve this problem, we need to consider the forces acting on the bead.
Let's start by analyzing the forces when the bead is at θ = 90 degrees.
(a) The speed v of the bead when θ = 90 degrees:
At θ = 90 degrees, the bead is at the highest point of the hoop. The forces acting on the bead are gravity and the spring force.
Gravity is pulling the bead downward with a force of mg, where g is the acceleration due to gravity.
The spring force is determined by Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position and the spring constant. In this case, the displacement is R - R = 0, since the bead is at the equilibrium position. Therefore, the spring force is zero.
Since there are no forces acting horizontally, the bead just falls vertically due to gravity. The vertical component of velocity can be obtained using the conservation of mechanical energy:
Initial potential energy = Final kinetic energy
mgh = (1/2)mv^2
Since the bead is released from rest, the initial velocity is zero and h = 2R (height of the hoop).
mgh = (1/2)mv^2
mg * 2R = (1/2)mv^2
2mgR = (1/2)mv^2
2gR = (1/2)v^2
v^2 = 4gR
Therefore, the speed v of the bead when θ = 90 degrees is given by:
v = sqrt(4gR) = 2sqrt(gR)
(b) The magnitude of the force the hoop exerts on the bead when θ = 90 degrees:
At θ = 90 degrees, the bead is at the highest point of the hoop. Here, the hoop exerts a normal force on the bead, providing the required centripetal force for circular motion.
The centripetal force is given by:
Fc = mv^2 / R
Substituting the value of v from part (a), we have:
Fc = m(2sqrt(gR))^2 / R
Fc = 4mgR / R
Fc = 4mg
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90 degrees is 4mg.