Question

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

• A. 3
• B. 4
• C. 2
• D. none of these

Answer 1

Answer: (C)

2

Given,

$a=-\frac{u^2}{4\pi R}$

$t=\frac{8\pi R}{u}$

$v=0m/s$

From 3rd equation of motion,

$2as=v^2-u^2$

$s=\frac{v^2-u^2}{2a}$

$s=\frac{0^2-u^2}{-2u^2/4\pi R}=2\pi R$

Angle, $\varphi =\frac{2\pi R}{R}$

$\varphi =2\pi$ which is equal to one revolution.

Time taken to stop the particle,

$v=u+at$

$0=u-\frac{u^2}{4\pi R}t$

$t^{\prime }=\frac{4\pi R}{u}$

Time left to move the particle,

$t^{\prime \prime }=t-t^{\prime }$

$t^{\prime \prime }=\frac{8\pi R}{u}-\frac{4\pi R}{u}=\frac{4\pi R}{u}$

From 2nd equation of motion, particle starts from the rest,

$s=\frac{1}{2}a{t}^2=\frac{1}{2}\times \frac{u^2}{4\pi R}(4\pi R/u{)}^2$

$s=2\pi R$

Angle, ${\varphi }^{\prime }=\frac{2\pi R}{R}=2\pi$ one revolution.

Hence, the total revolution is $2$.

The correct option is C.