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 $v_0$, the time taken to complete the first revolution is:

• A. $\frac{R}{v_0}{e}^{-2\pi }$
• B. $v_{0}R$
• C. $\frac{R}{v_0}$
• D. $\frac{R}{v_0}(1-e^{-2\pi })$

Answer 1

The correct option is D $\frac{R}{v_0}(1-e^{-2\pi })$

Given, $a_r=a_t$ i.e $R{\omega }^2=R\alpha$
i.e., $R{\omega }^2=R\frac{d\omega }{dt}$ or $\frac{d\omega }{dt}={\omega }^2$

or, $\frac{d\omega }{{\omega }^2}=dt$
$\Rightarrow$ ${\int }_{{\omega }_0}^{\omega }\frac{d\omega }{{\omega }^2}={\int }_0^{t}dt$
or, $\omega =\frac{{\omega }_0}{1-{\omega }_{0}t}$
or, $d\theta =\left(\frac{{\omega }_0}{1-{\omega }_{0}t}\right)dt$
Again, ${\int }_0^{2\pi }d\theta ={\int }_0^T\left(\frac{{\omega }_0}{1-{\omega }_{0}t}\right)dt$
or, $(1-{\omega }_{0}T)=e^{-2\pi }$ or $T=\frac{1}{{\omega }_0}(1-e^{-2\pi })$
or, $T=\frac{R}{v_0}(1-e^{-2\pi })$