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

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(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular v

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.For a solid cylinder rotating about its own axis,the moment of inertia is $I = \frac{1}{2}MR^2$.Given: Mass $M = 2\ kg$,Radius $R = 0.2\ m$,Angular velocity $\omega = 3\ rad/s$.Substituting the values:$I = \frac{1}{2} 	imes 2\ kg 	imes (0.2\ m)^2 = 1 	imes 0.04 = 0.04\ kg\cdot m^2$.Now,calculate the angular momentum:$L = I\omega = 0.04\ kg\cdot m^2 	imes 3\ rad/s = 0.12\ J-s$.Therefore,the angular momentum of the cylinder before collision is $0.12\ J-s$.

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