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A hoop of radius r and mass m rotating with an angular velocity omegha is placedona rough horizontal surface.The initial velocity of the center of hoop is

zero.What will be the velocity of the center of hoop when it cease to slip?
Answer is r omegha/2
plz solve the problem from easy method



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 initial kinetic energy = 1/2 I ω²  = 1/2 m r² ω²
         I = moment of inertia of  hoop of radius r and mass m. = m r²
         ω = angular velocity.
 initially the center of mass is not moving.

let the hoop rotate and roll on the surface with an angular velocity = ω₁
  So its kinetic energy when it rolls = 1/2 I ω₁²  + 1/2 m (rω₁)²
         1/2 m r² ω² =  1/2 m r² ω₁²  + 1/2 m r² ω₁²
                 ω² = 2 ω₁²  =>
    ω₁ = ω /√2
         so velocity  v = r ω₁ = r ω/√2

2 5 2
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