In circular motion, assuming `barv=barwxxbar r`, obtain an expressio for the resultantj acceleration of a particle in terms of tangential and radial component.

Explanation

Let the particle be moving around a circular path of constant radiuss r. If it speeds up or slows down, speed `(omega)` and the linear speed (v) change with time. Change with time. Thus, at any instant, `omega` v and r are related as :
`vecv=vecomegaxxvecr`
On differentiating, we get
`(dvecv)/("dt")=d/("dt")(vecr+vecomega)`
`(dvecv)/("dt")=d/("dt")xx(vecr+vecomega)xx(dvecr)/("dt")`
`(dvecomega)/("dt")=vecalpha"and"(dvecr)/("dt")=vecr`
`:." "(dvecv)/("dt") =vecalphaxxvecr+vecomegaxxvecv`
`vecalphaxxvecr` is along the tangent of circumference of circular path, it si tangential acceleration `veca_(T)`.
`vecomegaxxvecv` is along the radius of circle, pointing towards centre, it is radial accearation `veca_(r)`.
` veca=veca_(T)+veca_(r)`