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

Options

A. `( R)/(v_(0)) (1 - e^(-2pi))`
B. `(2R)/(v_(0)) (1- e^(-2pi))`
C. `(3R)/(v_(0)) ( 1 - e^(-2pi))`
D. None of these

Answer

A

Explanation

Centripetal acceleration is given by
`a_N =(dv)/(dt)=(v^2)/(R)`
`int_0^t(dt)/(R)=int_(v_0)^v(dv)/(v^2)`
`or " "t =-R[1/v]_(v_0)^vrArrv=(v_0R)/(R-v_0t)`
Also `(dr)/(dt)=(v_0R)/((R-v_0t))`
`int_0^(2piR)dr =v_0Rint_0^T(dt)/(R-v_0t)`
`rArr" "T=R/v_0(1-e^(-2pi))`