Two rigid bodies A and B rotate with rotation | vedclass

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Question

(C) The rotational kinetic energy $E$ is given by the formula $E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L = I \omega$ is the angular moment

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The rotational kinetic energy $E$ is given by the formula $E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L = I \omega$ is the angular momentum.From this,we can express the angular momentum as $L^2 = 2EI$,or $L = \sqrt{2EI}$.For body $A$,$L_A = \sqrt{2 E_A I_A}$.For body $B$,$L_B = \sqrt{2 E_B I_B}$.The ratio of angular momenta is $\frac{L_A}{L_B} = \sqrt{\frac{2 E_A I_A}{2 E_B I_B}} = \sqrt{\frac{E_A}{E_B} \cdot \frac{I_A}{I_B}}$.Given $I_A = I_B/4$,we have $\frac{I_A}{I_B} = \frac{1}{4}$.Given $E_A = 100 E_B$,we have $\frac{E_A}{E_B} = 100$.Substituting these values: $\frac{L_A}{L_B} = \sqrt{100 \cdot \frac{1}{4}} = \sqrt{25} = 5$.

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