MCQ
A particle performs uniform circular motion with an angular momentum $L.$ If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes
- A$4L$
- B$2L$
- C$L/2$
- ✓$L/4$
✓
Answer
Correct option: D.
$L/4$
d
$\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^{2}$
$\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^{2}$
$\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2} \Rightarrow 2=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)\left(\frac{1}{2}\right)^{2} \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=8$
Also, $\mathrm{L}^{2}=2 \mathrm{I}(\mathrm{KE})$
$\left(\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right)^{2}=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)\left(\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}\right)=8 \times 2=16$
$\Rightarrow \frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}=4 \quad \Rightarrow \mathrm{L}_{2}=\frac{\mathrm{L}_{1}}{4}$
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