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. The bead starts on the top of the circle opposing gravitational pull of the earth
(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 speed of the bead when θ = 90∘, we need to consider the forces acting on the bead at that point. Since the bead is on the top of the circle opposing the gravitational pull of the Earth, the forces acting on the bead are its weight and the spring force.
Let's break it down step by step:
(a) Finding the speed v of the bead when θ = 90∘:
1. The weight of the bead is given by the equation W = mg, where m is the mass of the bead and g is the acceleration due to gravity.
2. The net force acting on the bead is the sum of the weight and the spring force. At θ = 90∘, the spring is compressed to its maximum extent.
3. The force exerted by the spring is given by Hooke's Law: F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.
4. At θ = 90∘, the spring is compressed by an amount equal to the diameter of the circle (2R) because the equilibrium length of the spring is R.
5. Using the laws of motion, we can write the equation for the net force acting on the bead as: F_net = mg - k(2R).
6. Since the net force is equal to the mass times acceleration (F_net = ma), we have: ma = mg - k(2R).
7. Solving for a, we get: a = (mg - k(2R)) / m.
8. Now we can find the speed v of the bead when θ = 90∘ by using the equation for acceleration: v^2 = u^2 + 2as, where u is the initial velocity (which is negligible in this case) and s is the displacement.
9. At θ = 90∘, the bead has moved a distance equal to the circumference of the circle (2πR).
10. Plugging in the values, we get: v^2 = 0 + 2[(mg - k(2R)) / m] * (2πR).
11. Simplifying, we get: v^2 = 4πRg - 4πkR^2 / m.
12. Taking the square root of both sides, we can express the speed v in terms of m, R, k, and g: v = √(4πRg - 4πkR^2 / m).
(b) Finding the magnitude of the force the hoop exerts on the bead when θ = 90∘:
1. At θ = 90∘, the bead is moving in a circular path with constant speed.
2. By considering the centripetal force required to keep the bead moving in a circle, we can find the magnitude of the force the hoop exerts on the bead.
3. The centripetal force is given by the equation F_c = mv^2 / R, where m is the mass of the bead, v is the velocity, and R is the radius of the circular path.
4. Plugging in the expression for v from part (a), we get: F_c = m[√(4πRg - 4πkR^2 / m)]^2 / R.
5. Simplifying, we get: F_c = 4πRmg - 4πkR^2.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is 4πRmg - 4πkR^2, expressed in terms of m, R, k, and g.