LOOP, SPRING AND BEAD (14 points possible)
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 answer these questions, we need to analyze the forces acting on the bead at the given position. Let's break down the problem step by step.
(a) What is the speed v of the bead when θ = 90∘?
At θ = 90∘, the bead is at the highest point of the loop. Since there is no friction and the bead is released from rest, the only forces acting on the bead are gravity and the spring force.
The gravitational force can be calculated as F_gravity = m * g, where m is the mass of the bead and g is the acceleration due to gravity.
The spring force is proportional to the displacement of the spring from its relaxed length. In this case, the displacement is the difference between the equilibrium length of the spring and the current position of the bead. Since the equilibrium length of the spring is R and the current position of the bead is at the highest point of the loop, the displacement is -R.
We can use Hooke's law to calculate the spring force:
F_spring = -k * displacement = -k * (-R) = k * R
The net force acting on the bead at this point is the sum of the gravitational force and the spring force:
Net force = F_gravity + F_spring = m * g + k * R
According to Newton's second law, the net force is equal to the mass times the acceleration of the bead:
Net force = m * a
Setting these two expressions equal to each other, we can solve for the acceleration, which is related to the speed:
m * g + k * R = m * a
Since the bead is at the highest point of the loop, the acceleration is equal to the centripetal acceleration, which is v^2 / R, where v is the speed of the bead.
m * g + k * R = m * (v^2 / R)
Rearranging the equation, we can solve for v:
v^2 = (m * g * R + k * R^2) / m
Taking the square root of both sides, we find:
v = sqrt((m * g * R + k * R^2) / m)
Therefore, the speed v of the bead when θ = 90∘ is sqrt((m * g * R + k * R^2) / m).
(b) What is the magnitude of the force the hoop exerts on the bead when θ = 90∘?
The force the hoop exerts on the bead is the normal force, which is equal to the gravitational force acting on the bead at this point.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is F_gravity = m * g.