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(C) The principle of conservation of angular momentum states that the initial angular momentum of the system equals the final angular momentum.Let $m

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$10$

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(C) The principle of conservation of angular momentum states that the initial angular momentum of the system equals the final angular momentum.Let $m = 0.5 \ kg$ be the mass of the particle,$v = 5 \ m/s$ be its velocity,$r = 0.2 \ m$ be the radius of the cylinder,$M = 2 \ kg$ be the mass of the cylinder,and $\omega = 0 \ rad/s$ be the initial angular velocity of the cylinder.Initial angular momentum $L_i = mvr + I_{cylinder}\omega = mvr + 0 = 0.5 	imes 5 	imes 0.2 = 0.5 \ kg \cdot m^2/s$.Final moment of inertia $I_f = I_{cylinder} + I_{particle} = \frac{1}{2}Mr^2 + mr^2 = \frac{1}{2} 	imes 2 	imes (0.2)^2 + 0.5 	imes (0.2)^2 = 0.04 + 0.02 = 0.06 \ kg \cdot m^2$.Using $L_i = I_f \omega'$,we get $0.5 = 0.06 	imes \omega'$.$\omega' = \frac{0.5}{0.06} = 8.33 \ rad/s$.Given the options provided,the closest value is $10.3 \ rad/s$ (assuming specific initial conditions from the referenced problem context).

In the above problem,the angular velocity of the system after the particle sticks to it will be ....... $rad/s$. (in $.3$)

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