A particle is moving in a circle of radius `R` in such a way that at any instant the normal and tangential component 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))`
B. `(R)/(v_(0))e^(-2pi)`
C. `(R)/(v_(0))(1-e^(-2pi))`
D. `(R)/(v_(0))(1+e^(-2pi))`

Answer

C

Explanation

`a_(t)=a_(r)`
. `(dv)/(dt)=(v^(2))/(R)` , `(v=speed)`
:. `int_(vo)^(v)v^(-2)dv=(1)/(R)int_(0)^(t)dt`
`[(1)/(v_(0))-(1)/(v)]=(t)/(R)`
or `(1)/(v)=(1)/(v_(0))-(t)/(R)=(R-v_(0)t)/(Rv_(0))`
or `v=(Rv_(0))/(R-v_(0)t)`
:. `(dx)/(dt)=(Rv_(0))/(R-v_(0)t)(x="distance travelled")`
or `int_(0)^(x)dx=int_(0)^(t)(Rv_(0))/(R-v_(0)t)dt`
`x=Rv_(0)(-(1)/(v_(0)))[In(R-v_(0)t)]_(0)^(t)`
`=-RIn(1-(v_(0)t)/(R))` :. `1-(v_(0)t)/(R)=e^(-x//R)`
or `t=t=(R)/(v_(0))(1-e^(-2pi))`
Putting `x=2piR` , we get
`t=(R)/(v_(0))(1-e^(-2pi))`