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 speed v of the bead when θ = 90 degrees, we need to use the conservation of mechanical energy.
First, let's consider the different forms of energy involved:
1. Gravitational potential energy (Ug): Given by Ug = mgh, where m is the mass of the bead, g is the acceleration due to gravity, and h is the height of the bead from the equilibrium position of the spring (R-Rcosθ).
2. Elastic potential energy (Us): Given by Us = (1/2)kx^2, where k is the spring constant and x is the compression or extension of the spring from its equilibrium position (R-Rcosθ-R = -Rcosθ).
3. Kinetic energy (K): Given by K = (1/2)mv^2, where v is the speed of the bead.
The total mechanical energy (E) is the sum of the gravitational potential energy and the elastic potential energy:
E = Ug + Us.
At θ = 0 degrees, the bead is at the equilibrium position, so the total mechanical energy is simply the potential energy stored in the spring when it is extended by -R:
E = (1/2)k(-R)^2 = (1/2)kR^2.
At θ = 90 degrees, the bead is at the highest point on the hoop. At this point, the bead is at its maximum potential energy, and all of its energy is in the form of kinetic energy:
E = (1/2)mv^2.
Setting the expressions for E equal to each other and solving for v, we get:
(1/2)kR^2 = (1/2)mv^2.
Simplifying, we find:
v^2 = kR^2/m.
Taking the square root of both sides, we get:
v = sqrt(kR^2/m).
Therefore, the speed v of the bead when θ = 90 degrees is given by v = sqrt(kR^2/m).
For part (b), to find the magnitude of the force the hoop exerts on the bead when θ = 90 degrees, we need to consider the net force acting on the bead.
At θ = 90 degrees, the bead is at the highest point on the hoop. The net force acting on the bead is the sum of the gravitational force (Fg) and the spring force (Fs) in the vertical direction.
1. Gravitational force: Fg = mg, where m is the mass of the bead and g is the acceleration due to gravity.
2. Spring force: Fs = -kx, where k is the spring constant and x is the compression or extension of the spring from its equilibrium position (R - Rcosθ - R = -Rcosθ).
At the highest point, the net vertical force should be zero since the bead is in equilibrium. Therefore, we have:
-Fg + Fs = 0.
Substituting the expressions for Fg and Fs, we get:
-mg - k(-Rcosθ) = 0.
Simplifying, we find:
mg = kRcosθ.
At θ = 90 degrees, cosθ = 0, so the equation becomes:
mg = kR(0).
Since the magnitude of the force the hoop exerts on the bead is equal to mg, we have:
|F| = mg = 0.
Therefore, the magnitude of the force the hoop exerts on the bead when θ = 90 degrees is zero.