A particle is whirled in a circular path by a string with a constant angular velocity and its angular momentum is L. If the length of the string is now halved keeping the angular velocity same, the angular momentum is:

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$\frac{L}{4}$

$\frac{L}{2}$

answer is A.

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Detailed Solution

L = ${mr}^{2} ω, Now Given r^{'} = \frac{r}{2} and ω^{'} = ω$

$L^{'} = mω \frac{r^{2}}{4} = \frac{L}{4}$

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