A particle of mass ‘m’ moves in circular path of radius ‘R’ such that tangential acceleration is equal to a normal acceleration At t=0, the velocity of particle is $V_{0}$

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Time taken to complete one revolution is $\frac{1}{ω_{0}} ({1-e}^{-π})$

Angular velocity of particle at time ‘t’ is $ω = \frac{ω_{0}}{1- ω_{0} t}$

Time taken to complete one revolution is $\frac{1}{ω_{0}} (1 - e^{- 2 π})$

Angular velocity of particle at time ‘t’ is $ω = \frac{ω_{0}}{1+ ω_{0} t}$

answer is B, D.

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Detailed Solution

$α_{n} = α_{t}$

$ω^{2} R = R \frac{d ω}{d t}$

$\int_{0}^{t} d t = \int_{ω_{0}}^{ω} ω^{- 2} d ω$

$t = \frac{- 1}{ω} + \frac{1}{ω_{0}}$

$ω = \frac{ω_{0}}{1 - ω_{0} t}$

$\frac{d θ}{d t} = \frac{ω_{0}}{1 - ω_{0} t}$

$\int_{0}^{2 π} d θ = ω_{0} \int_{0}^{T} \frac{d t}{1 - ω_{0} t}$

${[θ]}_{0}^{2 π} = \frac{ω_{0} {[\ln (1 - ω_{0} t)]}_{0}^{T}}{- ω_{0}}$

$2 π - 0 = - [\ln (1 - ω_{0} T) - \ln (1)]$

$2 π = \ln (1) - \ln (1 - ω_{0} T)$

$2 π = \ln \frac{1}{1 - ω_{0} T}$

$2 π = \log_{e}^{1 / 1 - ω_{0} T}$

$\frac{1}{1 - ω_{0} T} = e^{2 π} \Rightarrow 1 - ω_{0} T = e^{- 2 π}$

$T = \frac{1 - e^{- 2 π}}{ω_{0}}$

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