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A thin ring of radius R and mass 2m lies on a horizontal frictionless surface against a vertical wall. Two beads, each of mass m, are connected by a spring of stiffness k and can slide on the ring. Initially the beads are in unstable equilibrium at the ends of the ring's diameter pointing toward the wall, and the ring is held fixed. After a small push the beads slide away from the wall symmetrically. When the distance of each bead from the wall reaches the maximum value of 1.6R, the ring is released. (Gravity neglected.) Which of the following statements are correct? (A) The natural (undeformed) length of the spring is 1.8R. (B) The maximum speed v of each bead as they move away from the wall is 0.2R*sqrt(k/(2m)). (C) The distance of the beads from the wall when they next lie along the ring's diameter is 1.3R. (D) The distance of the beads from the wall when they next lie along the ring's diameter is 1.4R.
- (A) Natural length = 1.8R
- (B) Max bead speed = 0.2R*sqrt(k/(2m))
- (C) Distance from wall at next diameter position = 1.3R
- (D) Distance from wall at next diameter position = 1.4R
Correct answer: (A) Natural length = 1.8R
Solution
Set x-axis toward wall. Ring centre at (R,0) from wall. Beads initially at phi=pi/2 (along y-diameter, spring length=2R). Wait - actually beads start at phi=0 direction (one at wall, one at 2R). After symmetric push to angle phi from y-axis, spring length = 2R*sin(phi). Energy conservation: (1/2)k(2R-L)² = (1/2)k(1.6R-L)² gives L=1.8R (A correct). Maximum speed when spring at natural length: 2*(1/2)*m*v² = (1/2)k(0.2R)² => v = 0.2R*sqrt(k/(2m)) (B correct). After ring released, COM fixed at 1.3R from wall. When beads return to diameter (phi=pi/2), ring centre = COM = 1.3R; bead distance from wall = 1.3R (C correct, D wrong).
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