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A hoop of radius r and mass m rotating with angular velocity ω₀ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. The velocity of the centre of the hoop when it ceases to slip is

Asked in JEE Main Online 2013 · Rolling dynamics and slipping

Answer: (4) (rω₀)/2

Step-by-step solution

Friction reduces ω and builds v until v=ω r. Angular momentum about the contact line is conserved: Iω₀=Iω+mvr with I=mr².

mr²ω₀=mr²v/r+mvr=2mvr, so v=(rω₀)/2.

Why the other options are wrong

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