A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to - $K / r^{2}$ , where K is a constant. The total energy of the particle is

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Solution

The correct option is B
Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}} ∴ K . E . = \frac{1}{2} m v^{2} = \frac{K}{2 r}$
$U = - \int_{\infty}^{r} F d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$
Total energy $E = K . E . + P . E . = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$

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A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to -K/r2 , where K is a constant. The total energy of the particle is

The correct option is B Here mv2r = Kr2 ∴K.E. = 12mv2 = K2r U = −∫r∞F dr = −∫r∞(−Kr2)dr = −Kr Total energy E = K.E. + P.E. = K2r − Kr = −K2r

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to - $K / r^{2}$ , where K is a constant. The total energy of the particle is

The correct option is B
Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}} ∴ K . E . = \frac{1}{2} m v^{2} = \frac{K}{2 r}$
$U = - \int_{\infty}^{r} F d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$
Total energy $E = K . E . + P . E . = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$