A bead having mass m is projected along a stationary circular hoop as shown. The

Question

A bead having mass m is projected along a stationary circular hoop as shown. The bead has a hole through it so it is constrained to stay on the hoop, and there is no friction between the bead and the hoop. At the bottom of the hoop, the bead is given an initial speed v0= squre root of (8gR). (a) what is the speed of v of the bead at the top of the hoop?.(b) what are the magnitude and direction of the normal force and contact that the hoop exerts on the bead when the bead is at the top of the hoop ?

Explanation / Answer

a) mg*2r = 1/2m*8gr - 1/2mv^2

2mgr = 4mgr - 1/2mv^2

v = sqrt (4gr)

b) at top Normal will act downwards

N+mg = mv^2/r

N = mv^2/r - mg = 4mg - mg = 3mg

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