Question

A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential component of its acceleration are equal. If its speed at $t=0$ is $v_0.$ The time taken to complete the first revolution is

• A. $\frac{R}{v_0}$
• B. $\frac{R}{v_0}{e}^{-2\pi }$
• C. $\frac{R}{v_0}(1-e^{-2\pi })$
• D. $\frac{R}{v_0}(1+e^{-2\pi })$

Answer 1

Answer: (C)

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

Given; $a_c=a_t$

$R{w}^2=\frac{dv}{dt}$ where $v=Rw$

$⟹w^2=\frac{dw}{dt}$

Integrating, ${\int }_{w_o}^w\frac{1}{w^2}dw={\int }_0^{t}dt$

$\frac{1}{w}=\frac{1}{w_o}-t$

$\frac{w}{w_o}=\frac{1}{1-w_{o}t}$ where $w=\frac{d\theta }{dt}$

Integrating again, ${\int }_0^{2\pi }d\theta =w_o{\int }_0^t\frac{1}{1-w_{o}t}dt$

$⟹2\pi =-ln(1-w_{o}t)$

OR $t=\frac{1}{w_o}(1-e^{2\pi })$ $as(v_o=R{w}_o)$