A solid cylinder of mass 2 kg and radius 0.2 | vedclass

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(C) According to the principle of conservation of angular momentum:$L_{initial} = L_{final}$$I\omega + mvR = (I + mR^2)\omega'$Where $I$ is the moment

Vedclass · Accepted Answer

$10.3$

Vedclass · Answer

(C) According to the principle of conservation of angular momentum:$L_{initial} = L_{final}$$I\omega + mvR = (I + mR^2)\omega'$Where $I$ is the moment of inertia of the solid cylinder,$I = \frac{1}{2}MR^2$.$I = \frac{1}{2} 	imes 2 	imes (0.2)^2 = 0.04 \ kg \cdot m^2$.Initial angular momentum $L_i = I\omega + mvR = (0.04 	imes 3) + (0.5 	imes 5 	imes 0.2) = 0.12 + 0.5 = 0.62 \ kg \cdot m^2/s$.Final moment of inertia $I_f = I + mR^2 = 0.04 + (0.5 	imes (0.2)^2) = 0.04 + 0.02 = 0.06 \ kg \cdot m^2$.Using $L_i = I_f \omega'$:$0.62 = 0.06 	imes \omega'$$\omega' = \frac{0.62}{0.06} = \frac{62}{6} \approx 10.33 \ rad/s$.Thus,the angular velocity is $10.3 \ rad/s$.

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