A particle is moving in a circle of radius R in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at t=0 is v0 , the time taken to complete the first revolution is

Question

A particle is moving in a circle of radius R in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at t=0 is v0 , the time taken to complete the first revolution is (A) (R/v0) (B) (R/v0)(.1-e- 2 π .) (C) (R/v0)e- 2 π  (D) (2 π R/v0)

Tardigrade Admin · Accepted Answer

When a particle moves in a circular motion, it is acted upon by centripetal force directed towards the centre. Hence, centripetal acceleration is

a_{N} = \frac{d v}{d t} = \frac{v ^{2}}{R}

or 0 \int t \frac{d t}{R} = v_{0} \int v \frac{d v}{v ^{2}}

or t = - R [\frac{1}{v}]_{v_{0}}^{v}

v = \frac{v _{0} R}{R - v _{0} t}

A l so \frac{d r}{d t} = \frac{v _{0} R}{( R - v _{0} t )}

0 \int 2 π R d r = v_{0} R 0 \int T \frac{d t}{R - v _{0} t}

⟹ T = \frac{R}{v _{0}} (1 - e^{- 2 π})

Q. A particle is moving in a circle of radius $R$ in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at $t = 0$ is $v_{0}$ , the time taken to complete the first revolution is

$\frac{R}{v _{0}}$

$\frac{R}{v _{0}} (1 - e^{- 2 π})$

$\frac{R}{v _{0}} e^{- 2 π}$

$\frac{2 π R}{v _{0}}$

Q. A particle is moving in a circle of radius R in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at t=0 is v0​ , the time taken to complete the first revolution is

v0​R​

v0​R​(1−e−2π)

v0​R​e−2π

v0​2πR​

Q. A particle is moving in a circle of radius $R$ in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at $t = 0$ is $v_{0}$ , the time taken to complete the first revolution is

$\frac{R}{v _{0}}$

$\frac{R}{v _{0}} (1 - e^{- 2 π})$

$\frac{R}{v _{0}} e^{- 2 π}$

$\frac{2 π R}{v _{0}}$