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.
Consider a double star system under the influence of the gravitational force between the stars. Star 1 has mass m1 = 2.42 × 1031 kg and Star 2 has mass m2 = 2.32 × 1031 kg. Assume that each star undergoes uniform circular motion about the center of mass of the system (cm). In the figure below r1 is the distance between Star 1 and cm, and r2 is the distance between Star 2 and cm.
1)If the stars are always a fixed distance s=r1+r2 = 3.12 × 1018 m apart, what is the period of the orbit (in s)?
To solve this problem, we can apply the principles of conservation of energy and forces.
(a) To find the speed v of the bead when θ = 90∘, we can use the conservation of mechanical energy. At any point along the path, the total mechanical energy remains constant. The total mechanical energy at the starting point (at θ = 0) is given as:
E1 = mgh + (1/2)mv^2 + (1/2)k(R - R)^2
where m is the mass of the bead, g is the acceleration due to gravity, h is the height above the reference point, v is the velocity of the bead, k is the spring constant, and R is the radius of the hoop.
At the point when θ = 90∘, the bead is at the highest point of the hoop. At this point, the gravitational potential energy is zero and the spring energy is also zero (since the spring is in its relaxed state). Therefore, the only energy present is the kinetic energy:
E2 = (1/2)mv^2
Since the total mechanical energy is conserved, we can equate E1 and E2:
mgh + (1/2)mv^2 + (1/2)k(R - R)^2 = (1/2)mv^2
Since the height h is equal to R, the equation simplifies to:
mgR = (1/2)mv^2
Solving for v:
v = sqrt(2gR)
Therefore, the speed v of the bead when θ = 90∘ is sqrt(2gR).
(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 that point. The only forces acting on the bead are gravity and the tension in the spring.
The force of gravity acting on the bead is mg, directed vertically downwards.
The force exerted by the spring is given by Hooke's Law:
F = -k(R - R)
Since the spring is in its relaxed state, the force exerted by the spring is zero:
F = 0
Therefore, the only force acting on the bead when θ = 90∘ is the force of gravity. The magnitude of this force is mg.
So, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is mg.
To determine the speed of the bead when θ = 90∘, we can use the conservation of mechanical energy. Initially, the bead is at rest, so its initial kinetic energy is zero. The only energy present is potential energy due to gravity and the spring potential energy.
At θ = 90∘, the height of the bead is at its maximum, which is equal to the total length of the spring, 2R. At this point, all of the potential energy is converted into kinetic energy.
(a) To find the speed v of the bead when θ = 90∘, we can equate the initial potential energy to the final kinetic energy:
mg(2R) + 1/2k(R)^2 = 1/2mv^2
Simplifying the equation and rearranging for v, we get:
v = √(2gR + kR^2/m)
So, the speed of the bead when θ = 90∘ is given by the above equation.
(b) To find the magnitude of the force the hoop exerts on the bead at θ = 90∘, we can use Newton's second law. The net force acting on the bead is the sum of the gravitational force and the force from the spring:
F_net = mg - k(R - R)
Since θ = 90∘, the tension in the spring is balanced by the weight of the bead. Thus, the force exerted by the hoop is equal to the weight of the bead:
F_hoop = mg
So, the magnitude of the force the hoop exerts on the bead when θ = 90∘ is equal to mg.