A particle is moving on a circle of radius `R` such that at every instant the tangential and radial accelerations are equal in magnitude. If the velocity of the particle be `v_(0)` at `t=0`, the time for the completion of the half of the first revolution will be

Options

A. `R//v_(0)`
B. `(R//v_(0))(1-e^(-pi))`
C. `(R//v_(0))e^(-pi)`
D. `(R//v_(0))(1-e^(-2pi))`

Answer

B

Explanation

Speed just before collision = `sqrt(2gl(1-costheta_(0)))`
Speed just after collision = `e sqrt(2gl(1-cos theta_(0)))`
From COE,
`(1)/(2)me^(2)[2gl(1-costheta_(0))]=mgl(1-costheta)`
`therefore theta = cos^(-1){1-e^(2)(1-costheta_(0))}`