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

Answer 1

The correct option is B

$T\propto R^2$

Step 1. Given data

Central fictitious force, $F_c\propto \frac{1}{R^3}$ [Where, $R$ is the radius.]

Step 2. Finding the time period of revolution, $T$

$$\begin{gathered} F_c\propto \frac{1}{R^3} \\ {F}_c=\frac{k}{R^3} \end{gathered}$$ [Where, $k$ is constant]

Also, the formula of the central fictitious force,

$F_c=m{\omega }^{2}R$

Equating both the values of central fictitious force, we get

$$\begin{gathered} \frac{k}{R^3}=m{\omega }^{2}R \\ {\omega }^2=\frac{k}{m{R}^4} \\ {\left(\frac{2\mathrm{\pi }}{T}\right)}^2=\frac{k}{m{R}^4} \\ {T}^2\propto R^4 \\ T\propto R^2 \end{gathered}$$ [Where, $\omega =\frac{2\mathrm{\pi }}{T}$ is angular frequency, $m$ is mass]

Hence, the option $\left(B\right)$ is correct.