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A particle executes uniform circular motion with angular momentum 'L'. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its
new angular momentum is
Options:
new angular momentum is
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Verified Answer
The correct answer is:
$\frac{L}{4}$
(B)
$\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}, \quad \mathrm{~L}=\mathrm{I} \omega$
$\mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime} \omega^{\prime 2}, \quad \omega^{\prime}=2 \omega$
$\therefore \mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}, \quad \mathrm{~K}^{\prime}=\frac{\mathrm{K}}{2}$
$\therefore \frac{1}{2}\left(\frac{1}{2} \mathrm{I} \omega^{2}\right)=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}$
$\therefore \mathrm{I}^{\prime}=\frac{\mathrm{I}}{8}$
$\mathrm{~L}^{\prime}=\mathrm{I}^{\prime} \omega^{\prime}=\frac{\mathrm{I}}{8} \times 2 \omega=\frac{\mathrm{I} \omega}{4}=\frac{\mathrm{L}}{4}$
$\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}, \quad \mathrm{~L}=\mathrm{I} \omega$
$\mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime} \omega^{\prime 2}, \quad \omega^{\prime}=2 \omega$
$\therefore \mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}, \quad \mathrm{~K}^{\prime}=\frac{\mathrm{K}}{2}$
$\therefore \frac{1}{2}\left(\frac{1}{2} \mathrm{I} \omega^{2}\right)=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}$
$\therefore \mathrm{I}^{\prime}=\frac{\mathrm{I}}{8}$
$\mathrm{~L}^{\prime}=\mathrm{I}^{\prime} \omega^{\prime}=\frac{\mathrm{I}}{8} \times 2 \omega=\frac{\mathrm{I} \omega}{4}=\frac{\mathrm{L}}{4}$
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