Q. A particle is undergoing uniform circular motion with angular momentum $L$. While moving on the same path if its kinetic energy becomes four times, then its angular momentum will be
J & K CETJ & K CET 2015System of Particles and Rotational Motion
Solution:
Angular momentum during uniform circular motion $L=m v r$
Kinetic energy $K=\frac{1}{2} m v^{2}$
$=\frac{1}{2} m \times\left(\frac{L}{m r}\right)^{2}=\frac{L^{2}}{2 m r^{2}}$
According to the question,
$K'=\frac{\left(L'\right)^{2}}{2 m r^{2}}=4\, K$
$\Rightarrow \frac{\left(L'\right)^{2}}{2 m r^{2}}=\frac{4 \times L^{2}}{2 m r^{2}}$
$\Rightarrow L'=2\, L$
