A particle of mass ' $m$ ' is performing uniform circular motion along a circular path of radius ' $r$ '.…
A particle of mass ' $m$ ' is performing uniform circular motion along a circular path of radius ' $r$ '. Its angular momentum about the axis passing through the centre and perpendicular to the plane is ' $L$ '. The kinetic energy of the particle is
$\frac{\mathrm{L}^2}{2 \mathrm{mr}^2}$
$\frac{2 \mathrm{~L}^2}{\mathrm{mr}^2}$
$\frac{\mathrm{L}^2}{\mathrm{mr}^2}$
$\frac{2 \mathrm{~L}^2}{3 \mathrm{mr}^2}$
Solution
$\begin{aligned} & E=\frac{1}{2} I \omega^2 \\ & L=I \omega \Rightarrow L^2=I^2 \omega^2 \\ \therefore \quad & E=\frac{1}{2} \cdot \frac{L^2}{I} \\ & B u t I=M r^2 \\ \therefore \quad & E=\frac{1}{2} \frac{L^2}{M r^2}=\frac{L^2}{2 \mathrm{Mr}^2}\end{aligned}$