The angular momentum of a rotating body is L. | vedclass

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

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$\frac{1}{9} L$

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(A) The angular momentum $L$ of a rotating body is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. The kinetic energy $K$ is given by $K = \frac{1}{2} I\omega^2 = \frac{L^2}{2I}$.Given: Initial state is $L_1 = L$,$K_1 = K$,$\omega_1 = \omega$,$I_1 = I$.New state: Frequency $f_2 = 3f_1$,which implies $\omega_2 = 3\omega_1 = 3\omega$. Kinetic energy $K_2 = \frac{1}{3} K_1 = \frac{K}{3}$.Using $K = \frac{L^2}{2I}$,we have $I = \frac{L^2}{2K}$.For the new state: $I_2 = \frac{L_2^2}{2K_2}$.Also,$L_2 = I_2 \omega_2 = I_2 (3\omega)$.From $K_2 = \frac{1}{2} I_2 \omega_2^2$,we have $\frac{K}{3} = \frac{1}{2} I_2 (3\omega)^2 = \frac{9}{2} I_2 \omega^2$.Since $K = \frac{1}{2} I \omega^2$,we substitute this: $\frac{1}{3} (\frac{1}{2} I \omega^2) = \frac{9}{2} I_2 \omega^2$.Solving for $I_2$: $I_2 = \frac{I}{27}$.Now,calculate the new angular momentum $L_2 = I_2 \omega_2 = (\frac{I}{27}) (3\omega) = \frac{I\omega}{9} = \frac{L}{9}$.

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third,the new angular momentum becomes:

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