Question

The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as K.E. = as² where $a$ is a constant. The force acting on the particle is

• A. $2a\frac{s^{2}}{R}$
• B. $2as\left( 1 + \frac{s^{2}}{R^{2}} \right)^{1/2}$
• C. $2as$
• D. $2a\mspace{6mu}\frac{R^{2}}{s}$

Answer 1

Answer: (B)

$2as\left( 1 + \frac{s^{2}}{R^{2}} \right)^{1/2}$

Answer 2

In non-uniform circular motion two forces will work on a particle $F_{c}\text{and }F_{t}$

So the net force $F_{Net} = \sqrt{F_{c}^{2} + F_{t}^{2}}$ ….(i)

Centripetal force $F_{c} = \frac{mv^{2}}{R} = \frac{2as^{2}}{R}$ ….(ii)

[As kinetic energy $\frac{1}{2}mv^{2} = as^{2}$ given]

Again from : $\frac{1}{2}mv^{2} = as^{2} \Rightarrow$ $v^{2} = \frac{2as^{2}}{m} \Rightarrow v = s\sqrt{\frac{2a}{m}}$

Tangential acceleration $a_{t} = \frac{dv}{dt} = \frac{dv}{ds}.\frac{ds}{dt}$ $\Rightarrow$

$a_{t} = \frac{d}{ds}\left\lbrack s\sqrt{\frac{2a}{m}} \right\rbrack.v$

$a_{t} = v\sqrt{\frac{2a}{m}} = s\sqrt{\frac{2a}{m}}\sqrt{\frac{2a}{m}}$ $= \frac{2as}{m}$

and $F_{t} = ma_{t} = 2as$ ….(iii)

Now substituting value of F_c and F_t in equation (i) $\therefore F_{Net} = \sqrt{\left( \frac{2as^{2}}{R} \right)^{2} + (2as)^{2}} = 2as\left\lbrack 1 + \frac{s^{2}}{R^{2}} \right\rbrack^{1/2}$