A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/(2r^(2)) .Its energy is

A particle is moving in a circular path of radius a under the action o

1.	A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/(2r^(2)) .Its energy is
Answer» `k/(4a^(2))` `k/(2a^(2))` zero `-3/2 k(a^(2))` Solution :`U = k/(2r^(2)) "" ` ….(1) ` :. (dU)/(dr) = k/2 (-2r^(-3) (dr)/(dr))` ` :. F = k/(r^(3))` Now particle is moving in CIRCULAR path of radius a instead of RADIUSR , the CENTRIPETAL force `(mv^(2))/a = k/(a^(3)) "" [ :. r =a]` ` :. 1/2 mv^(2) =k/(2a^(2))` ` :. k = k/(2a^(2))""...(2)` ` :. "Total energy " = U+k` ` = - k/(2a^(2) +k/(2a^(2))` `[ :. " from equ .(1) and (2) r =1"] ` = 0