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a small particle of mass m atached with a light inextensible spring L is moving in a vertical circle .minimum velocity required for the particle to go around complete vertical circle at highest ponit is √gl.in the given case the particle is moving in complete vertical circle & ratio of it's maximum to minimum velocity is 2:1What is the minimum velocity a)4√gl/3b)2√gl/3c)√gl/3d)3√gl/3ans (B) 2√gl/3
In this problem, the given velocity(√gl ) is the minimum velocity required for the particle at its highest point to complete vertical circle.
Let, u & v be the velocity of the particle at its lowest and highest point respectively.
u : v = 2:1 => u = 2v.----------->(1)
according to work-energy theorem,
mg(2l) = (1/2)mu2 - (1/2)mv2 --------------->(2)
from eqns (1)& (2), we get v = (2/3)(√gl).
This is the minimum velocity of the particle in its complete circular motion in this present case.
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