Thekinetic energyofaparticle movingalong a circle of radius R depends on the distance covered S as K `=alphaS^(2),` where `alpha` is a constant. Find theaorce acting on the particle as a function of S.

Explanation

In circular motion, thefore acting is tangentialas well as fadial.The two forces are given by the expressions
`F_("tangential")=ma_(t)m(dv)/(dt)and F_("radial")=me_(c)=mv^(2)R`
Now `K=alphaS^(2)`
Also `K=1/2mv^(2)`
So, we have
`1/2mv^(2)=alphaS^(2)`
`v^(2)=(2alphaS^(2))/(m)`
Differentiating w.r.t time, we get
`2vxx(dv)/(dt)=((2x)/(m))2S(dS)/(dt)`
Now `(dS)/(dt)=v` (speed)
So we have
`(dv)/(dt)=(2alphsS)/(m)`
Now `F_("tangential")=(mdv)/(dt)=2alphaS`
Also, `F_("radial")=ma_(c)=(mv^(2))/(R)=(2aS^(2))/(R)`
Net force `=sqrt(F_("tangential")^(2)+F_("radial")^(2))=2alphaSsqrt(1+(S^(2))/(R^(2)))`