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.

Question

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.

Answer 1

sqrt(((2(sqrt(2)-1)kRR)/m)+2gR)

Answer 2

(N + hooke's Forcecos(pi/4))= mv^2/R => N = ?

Answer 3

N=mv^2/R -((sqrt(2)-1)/sqrt(2))kR

Answer 4

anyone got all the questions??

Answer 5

To solve this problem, we need to consider the forces acting on the bead at different positions along the hoop's path.

(a) To find the speed of the bead when θ = 90∘, we first need to determine the potential energy when the bead is at the starting position (θ = 0∘) and when it reaches θ = 90∘.

At θ = 0∘, the spring is at its relaxed length, so the potential energy of the spring is zero. The total potential energy at this position is given by the gravitational potential energy:

U(θ=0) = mgh

where h is the height of the bead above the lowest position (h = R).

At θ = 90∘, the spring is stretched to its maximum length (2R). The potential energy at this position is the sum of the gravitational potential energy and the potential energy stored in the stretched spring. The spring potential energy is given by:

Uspring = (1/2)kx^2

where x is the displacement of the bead from the equilibrium position (x = 2R - R = R). The total potential energy at θ = 90∘ is:

U(θ=90∘) = mgh + (1/2)kx^2

To find the speed v of the bead at θ = 90∘, we can use the conservation of mechanical energy:

E = KE + U

where E is the total mechanical energy, KE is the kinetic energy, and U is the potential energy.

Since the bead is released from rest, its initial kinetic energy is zero:

E(θ=0) = mgh

At θ = 90∘, the total mechanical energy is converted entirely into kinetic energy:

E(θ=90∘) = KE(θ=90∘) = (1/2)mv^2

Equating the initial and final energies, we get:

mgh = (1/2)mv^2 + mgh + (1/2)kx^2

The mass cancels out, and we can solve for v:

v^2 = 2gh + kx^2
v = √(2gh + kx^2)

Substituting h = R and x = R, we get:

v = √(2gR + kR^2)

So, the speed v of the bead when θ = 90∘ is √(2gR + kR^2).

(b) To find the magnitude of the force the hoop exerts on the bead when θ = 90∘, we need to consider the net force acting on the bead along the radial direction.

At any position on the hoop, the net force acting on the bead is the sum of the gravitational force and the spring force. At θ = 90∘, the gravitational force is directed downward and the spring force is directed upward.

The gravitational force is given by:

Fg = mg

The spring force is given by Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:

Fs = -kx

Since the displacement x when θ = 90∘ is R, the spring force is:

Fs = -kR

The net force is the sum of the gravitational force and the spring force:

Fnet = Fg + Fs = mg - kR

So, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is |Fnet| = |mg - kR|.