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.

4 answers

sqrt(((2*(sqrt(2)-1)*k*R*R)/m)+2*g*R)
(N + hooke's Force*cos(pi/4))= m*v^2/R => N = ?
N=mv^2/R -((sqrt(2)-1)/sqrt(2))kR
anyone got all the questions??
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