A boy is pushing a ring of mass 2 kg and rad | vedclass

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(C) Let $M = 2 \ kg$,$R = 0.5 \ m$,$a = 0.3 \ m/s^2$,and $N = 2 \ N$ (force applied by the stick).For a ring,the moment of inertia about the center is

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$4$

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(C) Let $M = 2 \ kg$,$R = 0.5 \ m$,$a = 0.3 \ m/s^2$,and $N = 2 \ N$ (force applied by the stick).For a ring,the moment of inertia about the center is $I = MR^2$.Since it rolls without slipping,the angular acceleration is $\alpha = a/R$.Let $f_s$ be the friction from the ground and $f_a$ be the friction from the stick.Applying Newton's second law for translational motion: $N - f_s = Ma$.$2 - f_s = 2 	imes 0.3 = 0.6 \implies f_s = 1.4 \ N$.Applying torque equation about the center of the ring: $(f_s - f_a)R = I\alpha$.$(1.4 - f_a)R = (MR^2)(a/R) = MaR$.$1.4 - f_a = Ma = 2 	imes 0.3 = 0.6$.$f_a = 1.4 - 0.6 = 0.8 \ N$.The friction from the stick is $f_a = \mu N$,where $\mu = P/10$.$0.8 = (P/10) 	imes 2$.$0.8 = P/5 \implies P = 4$.

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