Rotational Motion Question 7
Question: A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is
[AIEEE 2003]
Options:
A) $ \frac{L}{4} $
B) 2L
C) 4L
D) $ \frac{L}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
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[a] Angular momentum $ L=l\omega $ (i)
Kinetic energy $ K=\frac{1}{2}l{{\omega }^{2}}=\frac{1}{2}L\omega $ [from Eq.(i)]
$ \therefore $ $ L=\frac{2K}{\omega } $
Now, the new angular momentum
$ L’=\frac{2( \frac{K}{2} )}{2\omega } $
$ ( \because K’=\frac{K}{2}and\omega ‘=2\omega ) $
$ \Rightarrow $ $ L’=\frac{L}{4} $
BETA
