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 it's acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be (1)/(α)[1 - e-2π] s, where α = _________. Numerical answer type. Work it out, then open the answer.

Answer: 8. dv/dt = v²/R ⇒ 1/v = 1/v₀ - t/R So v = ds/dt = (v₀ R)/(R - v₀ t) ⇒ s = -Rln (1 - (v₀ t)/(R)) One revolution, s = 2π R: 1 - (v₀ t)/(R) = e-2π ⇒ t = R/v₀(1 - e-2π) R/v₀ = 0.5/4 = 1/8, so α = 8

A particle is moving in a circle of radius 50 cm in such a way that… | JEE Main 2024 (29 Jan, Shift 2)

Q11JEE Main 2024 · 29 JAN · SHIFT 2Difficult

Non-Uniform Circular Motion

A particle is moving in a circle of radius $50\text{ cm}$ in such a way that at any instant the normal and tangential components of it's acceleration are equal. If its speed at $t = 0$ is $4\text{ m/s}$ , the time taken to complete the first revolution will be $\frac{1}{\alpha}\left[1 - e^{-2\pi}\right]\text{ s}$ , where $\alpha =$ _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

$\frac{dv}{dt} = \frac{v^2}{R} \Rightarrow \frac{1}{v} = \frac{1}{v_0} - \frac{t}{R}$
So $v = \frac{ds}{dt} = \frac{v_0 R}{R - v_0 t} \Rightarrow s = -R\ln\left(1 - \frac{v_0 t}{R}\right)$
One revolution, $s = 2\pi R$ : $1 - \frac{v_0 t}{R} = e^{-2\pi} \Rightarrow t = \frac{R}{v_0}\left(1 - e^{-2\pi}\right)$
$\frac{R}{v_0} = \frac{0.5}{4} = \frac{1}{8}$ , so $\alpha = 8$

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