Question

A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^3.$ Its time period of revolution will be given by:

• A. $T\propto R^{\frac{5}{2}}$
• B. $T\propto R^2$
• C. $T\propto R^{\frac{4}{3}}$
• D. $T\propto R^{\frac{3}{2}}$

Answer 1

The correct option is B $T\propto R^2$

Given that,
$F$ $\propto \frac{1}{R^3}$

$F=$ $\frac{K}{R^3}$

$\frac{m{v}^2}{R}=\frac{K}{R^3}$

$m(\omega R{)}^2=\frac{K}{R^2}$

${\omega }^2=\frac{K}{m}\left(\frac{1}{R^4}\right)$

As $\omega =\frac{2\pi }{T}$

${\left(\frac{2\pi }{T}\right)}^2\propto \frac{1}{R^4}$

$\frac{4{\pi }^2}{T^2}\propto \frac{1}{R^4}$

$T$ $\propto R^2$