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 \(\mathrm{t}=0\) is \(4 \mathrm{~m} / \mathrm{s}\), the time taken to complete the first revolution will be \(\frac{1}{\alpha}\left[1-\mathrm{e}^{-2 \pi}\right] \mathrm{s}\), where \(\alpha=\) ______.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Kinematics

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

4 m / s

, the time taken to complete the first revolution will be

\frac{1}{α} [1 - e^{- 2 π}] s

, where

α =

______.

Answer: 8

Solution

Equal normal and tangential acceleration means

\frac{v ^{2}}{r} = \frac{d v}{d t}

Separating variables and integrating from

v = 4

m/s at

t = 0

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

[- \frac{1}{v}]_{4}^{v} = \frac{t}{r}

- \frac{1}{v} + \frac{1}{4} = \frac{t}{0.5} = 2 t

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

For one complete revolution, the arc length is

s = 2 π r = π

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

- \frac{1}{2} [ln (1 - 8 t)]_{0}^{t} = π

ln (1 - 8 t) = - 2 π

1 - 8 t = e^{- 2 π}

t = \frac{1}{8} (1 - e^{- 2 π})

Comparing with the given form

t = \frac{1}{α} (1 - e^{- 2 π})

α = 8

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Kinematics →

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Kinematics

Solution

More JEE Main questions on Kinematics