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Kinetic energy of a particle moving along a circle of radisu R depends on the distance covered as `K =as^(2)` where a is a constant . Find the force acting on the particle as a function of s.A. `(2a)/(s)sqrt(1+((s)/(R))^(2))`B. `2assqrt(1+((R)/(s))^(2))`C. `2assqrt(1+((s)/(R))^(2))`D. `(2s)/(a)sqrt(1+((R)/(s))^(2))` |
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Answer» (c) According to given problem `(1)/(2)Mv^(2)=as^(2)` " " [Hence, M =mass] `rArr" " v=ssqrt((2a)/(M))" " (i)` So, `" " a_(g) =(v^(2))/(R)=(2as^(2))/(MR)` Further more as `a_(t)=(dv)/(dt)=(dv)/(ds),(ds)/(dt)=v(dv)/(ds)` (By chain rule) from Eq. (i) we get `i,e" " v=sqrt((2a)/(M))` yeilds `:." " a_(t)=[ssqrt((2a)/(M))][sqrt((2a)/(M))]=(2as)/(M) " " ( :.v/s=sqrt((2a)/(m)))` so that," " `a_("net")=sqrt(a_(R)^(2)+a_(t)^(2))` `=sqrt(((2as^(2))/(MR))^(2)+((2a)/(M))^(2))=(2as)/(M)(sqrt(1+((s)/(R))^(2)))` `:. " " F-Ma_("net")=2assqrt(1+(s//R)^(2))` |
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