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.
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.
sqrt(( (2*sqrt(2)-2)*k*R^2)/m+2*g*R)
(2*(sqrt(2)-1)*k*R)+(2*m*g)-((sqrt(2)-1)*(k*R)*(1/sqrt(2)))
A satellite with a mass of ms = 8.00 × 103 kg is in a planet's equatorial plane in a circular "synchronous" orbit. This means that an observer at the equator will see the satellite being stationary overhead (see figure below). The planet has mass mp = 9.59 × 1025 kg and a day of length T = 0.5 earth days (1 earth day = 24 hours).
(a) How far from the center (in m) of the planet is the satellite?
(b) What is the escape velocity (in km/sec) of any object that is at the same distance from the center of the planet that you calculated in (a)?
A bungee jumper jumps (with no initial speed) from a tall bridge attached to a light elastic cord (bungee cord) of unstretched length L. The cord first straightens and then extends as the jumper falls. This prevents her from hitting the water! Suppose that the bungee cord behaves like a spring with spring constant k = 110 N/m. The bridge is h = 110 m high and the jumper's mass is m = 50 kg. Use g = 10 m/s2.
(a) What is the maximum allowed length L of the unstretched bungee cord (in m) to keep the jumper alive? (Assume that the spring constant doesn't depend on L).
(b) Before jumping, our jumper verified the spring constant of the cord. She lowered herself very slowly from the bridge to the full extent of the cord and when she is at rest she measured the distance to the water surface. What was the measured distance (in m)?
A planet has a single moon that is solely influenced by the gravitational interaction between the two bodies. We will assume that the moon is moving in a circular orbit around the planet and that the moon travels with a constant speed in that orbit. The mass of the planet is mp = 5.69 × 1025 kg. The mass of the moon is mm = 8.25 × 1022 kg. The radius of the orbit is R = 7.76 × 108 m.
1)What is the period of the moon's orbit around the planet in earth days (1 earth day = 24 hours).
help smeone plzz.!!
To find the magnitude of the force the hoop exerts on the bead at θ = 90∘, we need to consider the forces acting on the bead at that position.
First, we have the gravitational force acting on the bead due to its weight. This force can be calculated using Newton's second law:
F_gravity = m * g
where m is the mass of the bead and g is the acceleration due to gravity.
Next, we have the force exerted by the spring. The elongation of the spring at θ = 90∘ is given by R (since the equilibrium length of the spring is R and the bead is at the lowest point of the hoop). The force exerted by the spring can be calculated using Hooke's law:
F_spring = k * (R - R)
Since the bead is at rest, the net force on it must be zero. Therefore, the magnitude of the force the hoop exerts on the bead can be found by summing the gravitational force and the force exerted by the spring:
F_hoop = F_gravity + F_spring
Substituting the equations for F_gravity and F_spring, we have:
F_hoop = m * g + 0
Hence, the magnitude of the force the hoop exerts on the bead at θ = 90∘ is:
F_hoop = m * g
Therefore, the answer is expressed in terms of m, R, k, and g.