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(C) Let $m = 0.1 \ kg$ be the mass of the sleeve,$R = 5 \ m$ be the radius,and $\omega = 2 \ rad/s$ be the angular velocity.For the sleeve to be in eq

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$2.0$

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(C) Let $m = 0.1 \ kg$ be the mass of the sleeve,$R = 5 \ m$ be the radius,and $\omega = 2 \ rad/s$ be the angular velocity.For the sleeve to be in equilibrium relative to the rotating frame,the forces acting on it are gravity ($mg$ downwards) and the normal force $(N)$ from the wire.The horizontal component of the normal force provides the centripetal acceleration: $N \sin 	heta = m \omega^2 (R \sin 	heta)$.Since $\sin 	heta 
eq 0$ (not at top or bottom),we can cancel $\sin 	heta$ to get $N = m \omega^2 R$.Substituting the values: $N = (0.1 \ kg) 	imes (2 \ rad/s)^2 	imes (5 \ m) = 0.1 	imes 4 	imes 5 = 2.0 \ N$.

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