A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is Vo, the time taken to complete the first revolution is

A particle is moving in a circle of radius R in such a way that at any

1.	A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is Vo, the time taken to complete the first revolution is
Answer» `(R )/(v_0)` `(R )/(v_0) (1-e^(-2pi ))` `(R )/(v_0 ) e^(-2pi)` `(2 pi R )/(v_0)` Solution :NORMAL acceleration, `a_n =(v^2)/(R)` TANGENTIAL acceleration,`a_t = (dv)/(dt)` As` a_t-a_n `( given ) ` therefore(dv )/( dt ) = (v^2)/(R )or (dt)/(R )= (dv )/(v^2)` Integrating the above EQUATION, we get `int_(0)^(1)(dt)/(R )= int_(v_0)^(v)(dv )/(v^2) or(t)/(R )=- [ (1)/(v)]_(v_0)^(v) orv= (v_0R )/( (R-v_0 t))` As `v=(dr)/(dt) = (v_0 R )/((R-v_0t)) thereforetherefore(dr )/(R )= (v_0dt)/( (R-v_0 t))` INTEGRATE the above equation, we get` int_(0)^(2pi R )(dr )/(R )=int_(0)^(T)(v_0 dt)/(R-v0 t))` on simplification, we get `T= (R )/(v_0)(1-e^(-2pi ))`