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  • A particle performs uniform circular motion with an angular momentum . If the frequency of particle&#3

A particle performs uniform circular motion with an angular momentum \mathrm{L}. If the frequency of particle's motion is double and its kinetic energy is halved then determine value of angular momentum.
 

Option: 1

\mathrm{L/2}


Option: 2

\mathrm{2L}


Option: 3

\mathrm{L/4}


Option: 4

\mathrm{L/3}


Answers (1)

best_answer

use, \mathrm{L= mvr }...........(1)

but \mathrm{v=r \omega \: and\: r=2 \pi f }

Now, \mathrm{L=m v \times\left[\frac{v}{2 \pi f}\right] =\frac{1}{2} m v^2 \times \frac{1}{\pi f} }

                                           \mathrm{=\frac{k E}{\pi f} \quad\quad\left[\because k=\frac{1}{2} m v^2\right]}

if the frequency is doubled and K.E. is halved then, New angular momentum,

\mathrm{L^{\prime}=\frac{K E^{\prime}}{\pi f^{\prime}} =\frac{\frac{K E}{2}}{\pi(2 f)} }

                     \mathrm{ =\frac{L}{4} }

Hence option 3 is correct.


 

Posted by

Rakesh

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