A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes [Kerala CEE 2013]
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a
L2
b
L
c
L3
d
L4
answer is A.
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Detailed Solution
We know that, L=Iω=2πmr2fNow, ω′=ω/2Hence, L′=Iω′=mr2ω2=πmr2f∴ LL′=2mπr2fπmr2f⇒L′=L2