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A ring of diameter 1m is rotating about a central axis perpendicular to its diameter. It is rotating with a speed of 10rad/s. What force should be applied on it tangentially to stop it in exactly 3 rotations? Let the mass of the ring be 2kg.

Correct Answer: 25/3π N
Let w be the angular velocity & a be the angular acceleration. Moment of inertia of the ring = MR2 = 2*0.25 = 0.5kgm2. Angular distance = 3 rotations = 3*2π rad = 6π rad. Using, w2 = w02 + 2aθ, we get: 0 = 100 – 2a*6π. a = 100/12π. Torque = Ia = rF ∴ F = Ia/r = 0.5*100/(12π*0.5) = 25/3π N.

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