Question

A particle is moving in a circle of radius $R$ with initial speed $v$. It starts retarding with constant retardation $\frac{v^2}{4\pi R}$. The number of the revolution it makes in time $\frac{8\pi R}{v}$ is

• A. 3
• B. 4
• C. 5
• D. 2

Answer 1

The correct option is D 2

Accelaration $a=\frac{-v^2}{4\pi R}$
Initial speed = $v$
Time $t=\frac{8\pi R}{v}$
We know $a=\frac{dv}{dt}$
$\frac{v^2}{4\pi R}=\frac{dv}{dt}$

${\int }_v^{v^{\prime }}\frac{dv}{v^2}={\int }_0^{\frac{8\pi R}{v}}\frac{dt}{4\pi R}$ velocity at $\frac{8\pi R}{v}is{v}^{\prime }$
$\frac{1}{v}-\frac{1}{v^{\prime }}=\frac{-2}{v}$
$\Rightarrow v^{\prime }=\frac{v}{3}$
Now,
We know $a=\frac{vdv}{ds}$
$\frac{-v}{4\pi R}=\frac{vdv}{ds}$
${\int }_v^{\frac{v}{3}}={\int }_0^s\frac{ds}{4\pi R}=\frac{dt}{4\pi R}$

$ln\left(\frac{1}{3}\right)=\frac{-s}{4\pi R}$
$s=4\pi R\times ln3$

So,number of revolution $n=\frac{4\pi R\times ln3}{2\pi R}$
$n\times ln3=2\times 1.098$
$n=2.198$
So, total number of revolution is $2$.