The centripetal force on a particle moving in a uniform circular motion on a horizontal circle of radius r is ` (-sigma)/(r^(2))` , then find the mechanical energy of this particle .

Explanation

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)`