TS EAMCET · Physics · Rotational Motion
A particle performs uniform circular motion with an angular momentum L. If the frequency of the particle's motion is doubled and its kinetic energy is halved, then its angular momentum becomes
- A \(2 \mathrm{~L}\)
- B \(4 \mathrm{~L}\)
- C \(\frac{\mathrm{L}}{2}\)
- D \(\frac{\mathrm{L}}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{L}}{4}\)
Step-by-step Solution
Detailed explanation
Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) and rotational kinetic energy \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^2 \therefore \mathrm{K}=\frac{1}{2} \mathrm{~L} \omega\)…
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