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.
To find the answers to these questions, we can use the principle of conservation of mechanical energy and the forces acting on the bead at different positions.
(a) To find the speed of the bead when θ = 90∘, we need to find the total mechanical energy of the system at that position.
The mechanical energy of the system is the sum of the kinetic energy and the potential energy. At θ = 90∘, the potential energy will be at its maximum.
The potential energy at θ = 90∘ is given by the gravitational potential energy and the potential energy stored in the spring.
Gravitational potential energy: U_gravity = m * g * h
At θ = 90∘, the height h is equal to R (the radius of the hoop), so U_gravity = m * g * R.
The potential energy stored in the spring at θ = 90∘ is given by: U_spring = (1/2) * k * (x - R)^2
At θ = 90∘, the extension of the spring x is equal to 2R, so U_spring = (1/2) * k * (2R - R)^2 = (1/2) * k * R^2.
The total potential energy at θ = 90∘ is the sum of the gravitational potential energy and the spring potential energy:
U_total = U_gravity + U_spring = m * g * R + (1/2) * k * R^2.
The initial mechanical energy of the system is zero since the bead is released from rest.
At θ = 90∘, the total mechanical energy is equal to the kinetic energy of the bead:
E = (1/2) * m * v^2.
Setting the total mechanical energy equal to the kinetic energy, we have:
(1/2) * m * v^2 = m * g * R + (1/2) * k * R^2.
Simplifying and rearranging the equation, we can solve for v:
v^2 = 2 * g * R + k * R^2 / m.
Taking the square root of both sides, we get:
v = sqrt(2 * g * R + k * R^2 / m).
Therefore, the speed of the bead when θ = 90∘ is v = sqrt(2 * g * R + k * R^2 / m).
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we need to consider the forces acting on the bead at that position.
At θ = 90∘, the only force acting on the bead is the force of gravity.
The force of gravity is given by: F_gravity = m * g.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is equal to the force of gravity, which is F_gravity = m * g.