A fan of moment of inertia 0.6 , kg cdot m^2 | vedclass

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Question

(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

$0.6\pi \, kg \cdot m^2/s$

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.Given: Moment of inertia $I = 0.6 \, kg \cdot m^2$.Frequency of rotation $f = 0.5 \, rev/s$.The angular velocity $\omega$ is related to the frequency by $\omega = 2\pi f$.Substituting the values: $\omega = 2 	imes \pi 	imes 0.5 = \pi \, rad/s$.Now,calculate the angular momentum: $L = I\omega = 0.6 	imes \pi = 0.6\pi \, kg \cdot m^2/s$.Therefore,the correct option is $A$.

$A$ fan of moment of inertia $0.6 \, kg \cdot m^2$ is turned up to a working speed of $0.5$ revolutions per second. The angular momentum of the fan is:

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