JEE MAIN - Physics (2003 - No. 28)
A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is
$${L \over 4}$$
$$2L$$
$$4L$$
$${L \over 2}$$
Magyarázat
We know Rotational Kinetic Energy$$={1 \over 2}I{\omega ^2},$$
Angular Momentum $$L = I\omega \Rightarrow I = {L \over \omega }$$
$$\therefore$$ Initial $$K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega $$
Final $$K.E'$$ = $${{K.E} \over 2}$$ = $${1 \over 2}{L'} \times 2\omega $$
$$\therefore$$ $${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}} $$
$$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$$
$$\therefore$$ $$L' = {L \over 4}$$
Angular Momentum $$L = I\omega \Rightarrow I = {L \over \omega }$$
$$\therefore$$ Initial $$K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega $$
Final $$K.E'$$ = $${{K.E} \over 2}$$ = $${1 \over 2}{L'} \times 2\omega $$
$$\therefore$$ $${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}} $$
$$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$$
$$\therefore$$ $$L' = {L \over 4}$$