Question

The kinetic energy of a particle along a circle of radius R is given as K = As where A is a constant and s is the distance travelled. What is the net force acting on the particle –

• A. $\frac{A}{R}\sqrt{1 + \frac{R^{2}}{s^{2}}}$
• B. $A\sqrt{1 + \frac{4s^{2}}{R^{2}}}$
• C. $4As\sqrt{1 + \frac{s^{2}}{R^{2}}}$
• D. None of the above

Answer 1

Answer: (B)

$A\sqrt{1 + \frac{4s^{2}}{R^{2}}}$

Answer 2

K = $\frac{1}{2}$ mv² = As

$\frac{dK}{dt}$ = vm $\frac{dv}{dt}$ = A $\frac{ds}{dt}$ ̃ m $\frac{dv}{dt}$ = A

̃ dv = $\frac{A}{m}$dt ̃ v = $\frac{A}{m}$t

a_t = $\frac{A}{m}$, a_c = $\frac{v^{2}}{R} = \frac{A^{2}t^{2}}{m^{2}R}$

̃ a₀ = $\sqrt{a_{t}^{2} + a_{c}^{2}} = \sqrt{\frac{A^{2}}{m^{2}} + \left( \frac{A^{2}t^{2}}{m^{2}R} \right)^{2}}$Also

$\frac{ds}{dt} = \frac{A}{m}t \Rightarrow s = \frac{A}{2m}t^{2} \Rightarrow t^{2} = \frac{2ms}{A}$

Then, a = $\sqrt{\frac{A^{2}}{m^{2}} + \left( \frac{A^{2}}{m^{2}R} \times \frac{2ms}{A} \right)^{2}}$

= $\sqrt{\frac{A^{2}}{m^{2}} + \frac{4A^{2}s^{2}}{m^{2}R^{2}}} = \frac{A}{m}\sqrt{1 + \frac{4s^{2}}{R^{2}}}$

Thus net force on the particle = ma = A$\sqrt{1 + \frac{4s^{2}}{R^{2}}}$