Question

What will be the force acting on a particle moving along a circle of radius R, if its kinetic energy k depends on the distance covered as $k=a{s}^2$?

• A. $2a\frac{R^2}{s}$
• B. $2as{\left[\begin{array}{c}1+\frac{s^2}{R^2}\end{array}\right]}^{1/2}$
• C. $\frac{2a{s}^2}{R}$
• D. $2as$

Answer 1

Answer: (B)

$2as{\left[\begin{array}{c}1+\frac{s^2}{R^2}\end{array}\right]}^{1/2}$

Given, $\frac{1}{2}m{v}^2=a{s}^2$

$\Rightarrow v=s\sqrt{\frac{2a}{m}}$

$\therefore a_R=\frac{v^2}{R}=\frac{2a{s}^2}{mR}$ ...(i)

$a_t=\frac{dv}{dt}=\frac{dv}{ds}.\frac{ds}{dt}=v\frac{dv}{dt}$ ...(ii)

From Eq. (i),

$a_t=\left[\begin{array}{c}s\sqrt{\frac{2a}{m}}\end{array}\right]\left[\begin{array}{c}\sqrt{\frac{2a}{m}}\end{array}\right]=\frac{2as}{m}$ ...(iii)

So, net accelerations $a_0=\sqrt{a_R^2+a_t^2}$

$=\sqrt{{\left[\begin{array}{c}\frac{2a{s}^2}{mR}\end{array}\right]}^2+{\left[\begin{array}{c}\frac{2as}{m}\end{array}\right]}^2}$

Hence, $a_0=\frac{2as}{m}\sqrt{1+{\left[\begin{array}{c}\frac{s}{R}\end{array}\right]}^2}$

$\therefore F=m{a}_0=2as\sqrt{1+{\left(\begin{array}{c}\frac{s}{R}\end{array}\right)}^2}$