Question

The kinetic energy of a particle moving along a circle of radius $R$ depends on distance $x$ as $K=a{x}^2,$ where $a$ is a constant. Then

• A. Centripetal force is acting on the particle.
• B. Tangential force is acting on the particle.
• C. Resultant force (F) makes angle $(\theta ={30}^{\circ })$ with Centripetal force $F_C$, when the magnitude of $F_C$ and Tangential force $F_T$ are equal.
• D. The angle made by $F$ (Net force) with centripetal force will be $\text{tan}\theta =\frac{R}{x}.$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Centripetal force is acting on the particle.
B Tangential force is acting on the particle.
D The angle made by $F$ (Net force) with centripetal force will be $\text{tan}\theta =\frac{R}{x}.$

Kinetic Energy $K=a{x}^2$
$\therefore \frac{1}{2}m{v}^2=a{x}^2\Rightarrow v^2=\frac{2}{m}a\times x^2$
$\therefore v=\sqrt{\frac{2a}{m}}x\Rightarrow \int \frac{dx}{x}=\int \sqrt{\frac{2a}{m}}dt$
$\Rightarrow \text{ln x}=\sqrt{\frac{2a}{m}}t+C$
$\Rightarrow x=e^{kt+C}\text{where}(k=\sqrt{\frac{2a}{m}})$
$\therefore v=k{e}^{kt+C}\text{Tangential acceleration}\Rightarrow a_T=\frac{dv}{dt}=k^2{e}^{kt}$
hence $F_T=m{k}^2{e}^{kt+C}$
Centripetal force $F_C=\frac{m{v}^2}{R}=\frac{2a{x}^2}{R}$
Angle made by net force with centripetal force is
$\therefore \text{tan}\theta =\frac{F_T}{F_C}=\frac{m{k}^2{e}^{kt+C}}{\frac{2a{x}^2}{R}}=\frac{m{k}^{2}x}{\frac{2a{x}^2}{R}}=\frac{2ax}{\frac{2a{x}^2}{R}}=\frac{xR}{x^2}=\frac{R}{x}$