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.
unanswered
(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.
Got the first one:
sqrt((0.83*k*R^2)/m+2*g*R)
Can't seem to find the second (b)
me 2,,unable to get the second! but trying!! plzz help someone
(2*(sqrt(2)-1)*k*R)+(2*m*g)-((sqrt(2)-1)*(k*R)*(1/sqrt(2)))
a) Fx(x)= -8*x*(x^2-1)
b). x=1
c) v(x=0)= 1.5
d) v(x=-1)= sqrt(6)
To answer these questions, we need to analyze the forces acting on the bead as it moves in the vertical hoop.
(a) To find the speed v of the bead at θ = 90∘, we can consider the conservation of energy. Initially, at θ = 0, the bead is at rest, so its initial kinetic energy is zero. At θ = 90∘, the bead reaches its maximum height and momentarily comes to rest again.
The total mechanical energy is conserved in this system. Initially, the energy is purely potential energy, given by:
Initial potential energy = m * g * R
At θ = 90∘, the total energy is the sum of potential energy and the energy stored in the spring, given by:
Final total energy = m * g * h + 0.5 * k * (R - h)^2
Where h is the height of the bead from the lowest point of the motion.
Since the speed v = 0 at θ = 90∘, the kinetic energy is zero. Equating the initial and final energies, we have:
m * g * R = m * g * h + 0.5 * k * (R - h)^2
Simplifying this equation will lead to an expression for h. Substituting this value of h back into the equation, we can solve for the speed v.
(b) To find the magnitude of the force the hoop exerts on the bead at θ = 90∘, we can consider the forces acting on the bead. At this point, the net force must be directed towards the center of the hoop to keep the bead in circular motion.
The forces acting on the bead are gravity (mg) and the force from the spring (k * (R-h)). The net force is given by:
Net force = mg - k * (R-h)
The magnitude of the force the hoop exerts on the bead is equal to the centripetal force required to keep it in circular motion:
For the bead to move in a horizontal circle, the net force must be inward and equal to mv^2/R, where v is the speed of the bead.
So, we can set the net force equal to mv^2/R:
mg - k * (R-h) = m * v^2 / R
Simplifying this equation will lead to an expression for the magnitude of the force the hoop exerts on the bead.
It's important to note that the expressions for both the speed v and the force magnitude depend on the height h, which needs to be calculated using the equation derived in part (a).