The centripetal force on a particlemoving in a uniform circular motion on a horizontal circle of radius r is(-sigma)/(r^(2)) , thenfind themechanical energy of this particle .

The centripetal force on a particlemoving in a uniform circular motion

1.	The centripetal force on a particlemoving in a uniform circular motion on a horizontal circle of radius r is(-sigma)/(r^(2)) , thenfind themechanical energy of this particle .
Answer» Solution :If the centripetal FORCE `F_(C)` is in the uniform circular motion , then ` \| F_(C )\| == SIGMA/(r^(2))` ` :. (mv^(2))/r = (sigma)/(r^(2))` ` :. Mv^(2) = siga/r` ` :. ` Kinetic energy `1/2 mv^(2) = sigma/(2r) ""…(1)` and for potential energy , `F = - (dV)/(dr)` ` :. dV = 0 Fdr` ` :. dV = -Fdr` ` :. int dV = int - Fdr ""( :. F = F_(C))` ` :. V = int -(-sigma/(r^(2)))dr` ` = sigma int 1/(r^(2)) dr` ` = sigma [-1/r]` ` :. V = -sigma/r "" .....(2)` MECHANICAL energy `E = K+V` ` = sigma/(2r) -sigma/r [ :. " From equation (1) and (2) "] ` ` :. E = - sigma/(2r)`