Question

The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends on the distance covered as $K=a{s}^2,$ where $a$ is a constant. The force acting on the particle is

• A. $2a\frac{s^2}{R}$
• B. $2as{(1+\frac{s^2}{R^2})}^{\frac{1}{2}}$
• C. $2as$
• D. $2a\frac{R^2}{s}$

Answer 1

The correct option is B $2as{(1+\frac{s^2}{R^2})}^{\frac{1}{2}}$

According to given problem, $K=\frac{1}{2}M{v}^2=a{s}^2$ (Here, $M=$ mass)
$\Rightarrow v=s\sqrt{\frac{2a}{M}}...\left(i\right)$
So, radial acceleration $a_R=\frac{v^2}{R}=\frac{2a{s}^2}{MR}$

Tangential acceleration
$a_t=\frac{dv}{dt}=\frac{dv}{ds}.\frac{ds}{dt}=v\frac{dv}{ds}$ (By chain rule)

Using $v=s\sqrt{\frac{2a}{M}}$ yields,
$a_t=\left[s\sqrt{\frac{2a}{M}}\right]\left[\sqrt{\frac{2a}{M}}\right]=\frac{2as}{M}$

Net acceleration $a=\sqrt{a_R^2+a_t^2}$
$a=\sqrt{{\left(\frac{2a{s}^2}{MR}\right)}^2+{\left(\frac{2as}{M}\right)}^2}=\frac{2as}{M}\left(\sqrt{1+{\left(\frac{s}{R}\right)}^2}\right)$

Force on the particle
$\therefore F=Ma=2as\sqrt{1+(s/R{)}^2}$