A smooth ring of mass m and radius R = 1 m is pulled at P with a constant acceleration a = 4 m s-2 on a horizontal surface such that the plane of the ring lies on the surface. Find the angular acceleration of the ring at the given position. (in rad/s2)
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answer is 2.
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Detailed Solution
The pseudo force acting at the CM of the ring is F→ps=−ma i^ which produces a torque about P given as τ→ps=−(ma)(R)k^This produces an angular acceleration α→=τ→psIP=−maR k^2mR2 ∵IP=mR2+mR2=2mR2=−a2Rk^α=42×1=2 rad/s2