Answer to Question #137951 in Mechanics | Relativity for Sneha

Question

The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = as2, where a is a constant. Find the force acting on the particle as a function of s.

Answer 1

Explanations & Calculations

• If the mentioned formula $\small T = as^2$ shows the kinetic energy of the particle as $\small E_k = as^2$ then by applying the Newton's second law towards the center of the particle's path the force acting on it could easily be calculated.
• When a particle moves on a circular path, a particular force is needed in order to stick on to that path or it moves on a straight line & that force is called the centripetal force ($\small F_c$).
• Therefore,

$\qquad\qquad \begin{aligned} \small F_c &= ma\\ &= \small m\frac{v^2}{R}\\ &= \small \frac{2}{R}\times \frac{1}{2}mv^2 : \text{rearrange to form the expression for} E_k\\ &= \small \frac{2}{R}\times as^2\\ F_c&= \small \bold{\frac{2as^2}{R}} \end{aligned}$

• But note that it's said that the kinetic is not a constant as it is subjected to change as the covered distance changes meaning that there may be it's speed changes over time hence the velocity change both in radial & tangential directions.
• Therefore, a tangential acceleration is present hence a tangential force therefore a resultant force to be calculated.
• Tangential acceleration

$\qquad\qquad \begin{aligned} \small a_t &= \small \frac{dv}{dt} \end{aligned}$

• As its known from the given relationship

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2&= \small as^2\\ \small v^2&= \small \frac{2as^2}{m}\\ \small 2v\frac{dv}{dt}&= \small \frac{4as}{m}\frac{ds}{dt}=\frac{4as}{m}\times v\\ \therefore \small \frac{dv}{dt} = a_t&= \small \frac{2as}{m} \end{aligned}$

• Therefore, the tangential force $\small F_t = ma_t= \bold{2as}$
• These forces are normal to each other hence the resultant is

$\qquad\qquad \begin{aligned} \small F_{r} &= \small \sqrt{F_r^2+F_t^2}\\ &= \small \sqrt{\frac{4a^2s^4}{R^2}+4a^2s^2}\\ F_r&= \small \bold{\frac{2as}{R}\sqrt{R^2+s^2}} \end{aligned}$