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