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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]
detailed solution
Correct option is A
We know that, L=Iω=2πmr2fNow, ω′=ω/2Hence, L′=Iω′=mr2ω2=πmr2f∴ LL′=2mπr2fπmr2f⇒L′=L2Talk to our academic expert!
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A particle of mass m is moving in YZ-plane with a uniform velocity v with its trajectory running parallel to + ve Y-axis and intersecting Z-axis at z = a as shown in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is [NCERT Exemplar)
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