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Question: A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal t...

A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to −(k/r2)- \left( k / r ^ { 2 } \right)where k is constant. The total energy of the particle is:

A

−kr- \frac { k } { r }

B

−k2r- \frac { k } { 2 r }

C

k2r\frac { k } { 2 r }

D

2kr\frac { 2 k } { r }

Answer

−k2r- \frac { k } { 2 r }

Explanation

Solution

Since the particle is moving in horizontal circle, centripetal force,

F=mv2r=kr2\mathrm { F } = \frac { \mathrm { mv } ^ { 2 } } { \mathrm { r } } = \frac { \mathrm { k } } { \mathrm { r } ^ { 2 } }

…..(i)

Kinetic energy of the particle,

(using (i))

As F=−dudrF = \frac { - \mathrm { du } } { \mathrm { dr } }

potential energy,

U=∫∞rFdr=−∫∞r(−kr2)dr=k∫∞rr−2dr=−kr\mathrm { U } = \int _ { \infty } ^ { \mathrm { r } } \mathrm { Fdr } = - \int _ { \infty } ^ { \mathrm { r } } \left( \frac { - \mathrm { k } } { \mathrm { r } ^ { 2 } } \right) \mathrm { dr } = \mathrm { k } \int _ { \infty } ^ { \mathrm { r } } \mathrm { r } ^ { - 2 } \mathrm { dr } = \frac { - \mathrm { k } } { \mathrm { r } }

total energy =