Question

A particle is moving in a circle of radius R 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 $v_o$ the time taken to complete the first revolution is

• A. $R/v_o$
• B. $v_o/R$
• C. $\left(R/v_o\right)\times (1-e^{-2\pi })$
• D. $\left(R/v_o\right)\times \left(e^{-2\pi }\right)$

Answer 1

The correct option is C $\left(R/v_o\right)\times (1-e^{-2\pi })$

According to question, $a_T=a_R$

$R\frac{dw}{dt}=R{w}^2$

$\frac{dw}{d{w}^2}=dt$

On integrating, ${\int }_{w_0}^w\frac{dw}{w^2}={\int }_0^{t}dt$

$\frac{1}{w_0}-\frac{1}{w}=t+c$

$w=\frac{w_0}{1-w_{0}t}$

${\int }_0^{2\pi }d\theta ={\int }_0^T\frac{w_{0}dt}{1-w_{0}t}$

$=\frac{1}{w_0}(1-e^{-2\pi })$

$=\frac{R}{v_0}(1-e^{-2\pi })(\because w_0=\frac{v_0}{R})$