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

Options

A. `(R )/(v_0)`
B. `(R )/(v_0) (1-e^(-2pi ))`
C. `(R )/(v_0 ) e^(-2pi)`
D. `(2 pi R )/(v_0)`

Answer

B

Explanation

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) or v= (v_0R )/( (R-v_0 t))`
As ` v=(dr)/(dt) = (v_0 R )/((R-v_0t)) therefore therefore (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 ))`