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.
Use this link it shows the question you are solving , but it has slightly different parameters ie the equilibrium of the spring. It show that you need
GPE + EPE = KE + GPE + EPE
ie TOP=SIDE. You can then rearrange this to get everything on one side except v (your speed).
Use this link it shows the question you are solving , but it has slightly different parameters ie the equilibrium of the spring. It show that you need
GPE + EPE = KE + GPE + EPE
ie TOP=SIDE. You can then rearrange this to get everything on one side except v (your speed).
I cant post link so search on google for part b) What is the magnitude of the force the hoop exerts on the bead and look for the mit link
Part b) is working forces. So you have centripetal acceleration = N-kx where kx is the spring force.
cent acc = mv^2/r so us a)to derive this.
hey man can you tell the answer,, plzz help
To solve this problem, we will use the principles of energy conservation and Newton's laws of motion.
(a) To find the speed v of the bead when θ = 90∘, we can use the principle of energy conservation. At the topmost point of the hoop, the potential energy is at its maximum and the kinetic energy is at its minimum. As the bead slides downward, the potential energy decreases and is converted into kinetic energy. This allows us to write the energy conservation equation:
Potential Energy + Spring Energy = Kinetic Energy
1/2*m*v^2 + 1/2*k*(R - R)^2 = m*g*R
The potential energy at the topmost point is m*g*R, as the mass m is at a height R above the bottom of the hoop. The spring energy is 1/2*k*(R - R)^2 = 0, as the spring is in its equilibrium (relaxed) length and has no potential energy. We can cancel out the common factors of m and solve for v:
1/2*v^2 = g*R
v^2 = 2*g*R
v = √(2*g*R)
So, the speed v of the bead when θ = 90∘ is √(2*g*R).
(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we can consider the forces acting on the bead at this point. There are two forces: gravity and the spring force.
The force of gravity is given by F_gravity = m*g, where m is the mass of the bead and g is the acceleration due to gravity. This force is acting vertically downward.
The spring force, F_spring, is given by Hooke's law: F_spring = -k*(x - R), where k is the spring constant and x is the displacement of the bead from its equilibrium position. At θ = 90∘, the bead is at its equilibrium position, so x - R = 0 and the spring force is zero. Therefore, the only force acting on the bead is the force of gravity.
The magnitude of the force the hoop exerts on the bead is equal in magnitude but opposite in direction to the force of gravity, so it is F_repulsive = -F_gravity = -m*g. Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is m*g.