Laws of Motion

JEE Main level, numerical answer, medium

199 of 210

Question 199 of 210

A particle is moving in a circle of radius 50cm50cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0t=0 is 4ms−14m{s}^{-1}, the time taken to complete the first revolution will be 1α[1−e−2π]s\dfrac{1}{\alpha }\left[1-{e}^{-2\pi }\right]s, where α=\alpha =______.

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Answer: 8

Given:
∣a⃗c∣=∣a⃗t∣\left|{\vec{a}}_{c}\right|=\left|{\vec{a}}_{t}\right|
⇒v2r=dvdt\Rightarrow \dfrac{{v}^{2}}{r}=\dfrac{dv}{dt}
⇒∫4vdvv2=∫0tdtr\Rightarrow {\int }_{4}^{v}\dfrac{dv}{{v}^{2}}={\int }_{0}^{t}\dfrac{dt}{r}
⇒[−1v]4v=tr\Rightarrow {\left[\dfrac{-1}{v}\right]}_{4}^{v}=\dfrac{t}{r}
⇒−1v+14=2t\Rightarrow \dfrac{-1}{v}+\dfrac{1}{4}=2t
⇒v=41−8t=dsdt\Rightarrow v=\dfrac{4}{1-8t}=\dfrac{ds}{dt}
⇒4∫0tdt1−8t=∫0sds\Rightarrow 4{\int }_{0}^{t}\dfrac{dt}{1-8t}={\int }_{0}^{s}ds
(r=0.5m,distancecovereds=2πr=π)\left(r=0.5m,\text{distance}\text{covered}s=2\pi r=\pi \right)
⇒4×[ln⁡(1−8t)]0t−8=π\Rightarrow 4\times \dfrac{[\ln (1-8t){]}_{0}^{t}}{-8}=\pi
⇒ℓn(1−8t)=−2π\Rightarrow \ell n(1-8t)=-2\pi
⇒1−8t=e−2π\Rightarrow 1-8t={e}^{-2\pi }
⇒t=(1−e−2π)18s\Rightarrow t=\left(1-{e}^{-2\pi }\right)\dfrac{1}{8}s
So, α=8\alpha =8.

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