InterviewSolution
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InterviewSolution
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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. |
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Answer» 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)))` |
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