a particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at `t=0` is `v_(0)` then time it takes to complete the first revolution is `R/(alphav_(0))(1-e^(-betapi))`. Find the value of `(alpha+beta)`.

Answer

3

Explanation

`(dv)/(dt)=v^(2)/RrArr underset(v_(0))overset(v)(int)(dv)/v^(2)=1/Runderset(0)overset(t)(int)dt rArr (-1/v)_(v_(0))^(v)=t/RrArr v=v_(0)/(1-v_(0)/Rt)rArr (ds)/(dt)=v_(0)/(1-v_(0)/Rt)rArr underset(0)overset(2piR)(int)ds=underset(0)overset(t)(int)v_(0)/((1-v_(0)/Rt))dt`
`rArr 2pi R=-R[ln(1-v_(0)/Rt)]_(0)^(t)rArr 2pi=-ln(1-v_(0)/Rt)rArr 1-v_(0)/Rt=e^(-2pi)rArr t=R/v_(0)(1-e^(-2pi))`
`rArr alpha=1, beta=2 rArr (alpha+beta)=(1+2)=3`