Question

A particle is moving in a circle of radius R in such a way that at any instant the a_r and a_t are equal. If the speed at t = 0 is v₀, the time taken to complete the first revolution is

• A. $\frac{R}{v_{0}}e^{- 2\pi}$
• B. v₀R
• C. $\frac{R}{v_{0}}$
• D. $\frac{R}{v_{0}}\left( 1 - e^{- 2\pi} \right)$

Answer 1

Answer: (D)

$\frac{R}{v_{0}}\left( 1 - e^{- 2\pi} \right)$

Answer 2

Given, a_r = a_r i.e., Rω² = Rα i.e., Rω² = R$\frac{d\omega}{dt}$ or

$\frac{d\omega}{dt}$ = ω² or $\frac{d\omega}{\omega^{2}}$ = dt i.e., $\int_{\omega_{0}}^{\omega}\frac{d\omega}{\omega^{2}}$ = $\int_{0}^{1}{dt}$

i.e., ω = $\frac{\omega_{0}}{1 - \omega_{0^{t}}}$

i.e., $\frac{d\theta}{dt} = \frac{\omega_{0}}{1 - \omega_{0^{t}}}$

i.e., dθ = $\left( \frac{\omega_{0}}{1 - \omega_{0^{t}}} \right)$dt agins, $\int_{0}^{2\pi}{d\theta} = \int_{0}^{T}\left( \frac{\omega_{0}}{1 - \omega_{0^{t}}} \right)$

or (1 - ω₀T) = e^-2π or T = $\frac{1}{\omega_{0}}$ (1 – e^-2π)

or T = $\frac{R}{v_{0}}$ (1 – e^-2π)