Question bank/JEE Main/Physics

JEE MainPhysicsEasyRigid Body Dynamics

A particle performs uniform circular motion with an angular
momentum L. If thefrequency of particle's motion is doubled and its
kinetic energy is halved, the angularmomentum becomes

Show answer and solution
L=IωL = I\omega
ω′=2ω\omega^{'} = 2\omega
12(12Iω2)=12I′ω′2\frac{1}{2}\left( \frac{1}{2}I\omega^{2} \right) = \frac{1}{2}I^{'}\omega^{'2}
Iω22=I′4ω2\frac{I\omega^{2}}{2} = I^{'}4\omega^{2}
I′=(I8)⇒L′=I′ω′=I82ω=Iω4=(L/4)I^{'} = \left( \frac{I}{8} \right) \Rightarrow L^{'} = I^{'}\omega^{'} = \frac{I}{8}2\omega = \frac{I\omega}{4} = (L/4)

Source: Grass-G / P-07 Physics paper 17 October.docx

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