A particle of mass 'm' is performing U.C.M. along a circle of radius 'r'. The relation between centripetal…
A particle of mass 'm' is performing U.C.M. along a circle of radius 'r'. The relation between centripetal acceleration 'a' and kinetic energy 'E' is given by
$a=\frac{2 \mathrm{E}}{m r}$
$a=\left(\frac{2 E}{m r}\right)^{2}$
$a=\frac{E}{m r}$
$a=2 \mathrm{Em}$
Solution
$E=\frac{1}{2} m \omega^{2} r^{2}$
$\therefore 2 E=m \omega^{2} r^{2}$
$\therefore r \omega^{2}=\frac{2 E}{m r}$
$a=\frac{v^{2}}{r}=r \omega^{2}$
$a=\frac{2 E}{m r}$