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A particle performs rotational motion with an angular momentum ' $L$ '. if frequency of rotation is doubled and its kinetic energy becomes one fourth, the angular momentum becomes.
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Verified Answer
The correct answer is:
$\frac{L}{8}$
Kinetic energy $\mathrm{k}=\frac{1}{2} \mathrm{I} \omega^2$
$$
\begin{aligned}
& \therefore \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{I}_2 \omega_2^2}{\mathrm{I}_1 \omega_1^2} \\
& \therefore \frac{1}{4}=\frac{\mathrm{I}_2}{\mathrm{I}_1} \cdot 4 \\
& \therefore \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{16} \\
& \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\frac{\mathrm{I}_2 \omega_2}{\mathrm{I}_1 \omega_1}=\frac{1}{16} \times 2=\frac{1}{8} \\
& \therefore \mathrm{L}_2=\frac{\mathrm{L}_1}{8}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{I}_2 \omega_2^2}{\mathrm{I}_1 \omega_1^2} \\
& \therefore \frac{1}{4}=\frac{\mathrm{I}_2}{\mathrm{I}_1} \cdot 4 \\
& \therefore \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{16} \\
& \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\frac{\mathrm{I}_2 \omega_2}{\mathrm{I}_1 \omega_1}=\frac{1}{16} \times 2=\frac{1}{8} \\
& \therefore \mathrm{L}_2=\frac{\mathrm{L}_1}{8}
\end{aligned}
$$
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