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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
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Verified Answer
The correct answer is:
$a=\frac{2 \mathrm{E}}{m r}$
(C)
$E=\frac{1}{2} m \omega^{2} r^{2}$
$\therefore 2 E=m \omega^{2} r^{2}$
$\therefore r \omega^{2}=\frac{2 E}{m r}$
$a=\frac{v^{2}}{r}=r \omega^{2}$
$a=\frac{2 E}{m r}$
$E=\frac{1}{2} m \omega^{2} r^{2}$
$\therefore 2 E=m \omega^{2} r^{2}$
$\therefore r \omega^{2}=\frac{2 E}{m r}$
$a=\frac{v^{2}}{r}=r \omega^{2}$
$a=\frac{2 E}{m r}$
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