MHT CET · Physics · Rotational Motion
A rotating body has angular momentum ' \(L\) '. If its frequency is doubled and kinetic energy is halved, its angular momentum will be
- A \(\frac{\mathrm{L}}{4}\)
- B \(\frac{\mathrm{L}}{2}\)
- C 2 L
- D 4 L
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{L}}{4}\)
Step-by-step Solution
Detailed explanation
Angular momentum of a particle performing UCM
\(\mathrm{L}=\mathrm{I} \omega...(i)\)
Kinetic energy, \(\mathrm{k}=\frac{1}{2} \mathrm{I} \omega^2\)...(ii)
\(\therefore \quad \mathrm{L}=\frac{2 \mathrm{k}}{\omega}\)
\(\ldots[\) From (i) and (ii)]
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{k}_1}{\mathrm{k}_2} \times \frac{\omega_2}{\omega_1} \\
& \frac{\mathrm{~L}}{\mathrm{~L}_2}=\frac{1}{\frac{1}{2}} \times \frac{2}{1} \\
& \therefore \quad \mathrm{~L}_2=\frac{\mathrm{L}}{4}
\end{aligned}\)
...(given)
\(\mathrm{L}=\mathrm{I} \omega...(i)\)
Kinetic energy, \(\mathrm{k}=\frac{1}{2} \mathrm{I} \omega^2\)...(ii)
\(\therefore \quad \mathrm{L}=\frac{2 \mathrm{k}}{\omega}\)
\(\ldots[\) From (i) and (ii)]
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{k}_1}{\mathrm{k}_2} \times \frac{\omega_2}{\omega_1} \\
& \frac{\mathrm{~L}}{\mathrm{~L}_2}=\frac{1}{\frac{1}{2}} \times \frac{2}{1} \\
& \therefore \quad \mathrm{~L}_2=\frac{\mathrm{L}}{4}
\end{aligned}\)
...(given)
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