A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be , where

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Solution

Q.
A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be $\frac{1}{α} [1 - e^{- 2 π}] s$ , where $α =$ _____________ . [2024]

Ans.
(8) $| {\vec{a}}_{c} | = | {\vec{a}}_{t} |$

   $\frac{v^{2}}{r} = \frac{d v}{d t} \Rightarrow \int_{4}^{v} \frac{d v}{v^{2}} = \int_{0}^{t} \frac{d t}{r}$

   $\Rightarrow {[\frac{- 1}{v}]}_{4}^{v} = \frac{t}{r} \Rightarrow \frac{- 1}{v} + \frac{1}{4} = 2 t$

   $\Rightarrow v = \frac{4}{1 - 8 t} = \frac{d s}{d t}$

$4 \int_{0}^{t} \frac{d t}{1 - 8 t} = \int_{0}^{s} d s$

   $(r = 0.5 m s = 2 π r = π)$

   $4 \times \frac{{[l n (1 - 8 t)]}_{0}^{t}}{- 8} = π$

$l \ln (1 - 8 t) = - 2 π$

$1 - 8 t = e^{- 2 π} \Rightarrow t = \frac{1}{8} (1 - e^{- 2 π})$

   $S o, α = 8$

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