The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as KE = as 2 , where a is a constant. Find the force acting on the particle as a function of s .

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TestHub · Accepted Answer

There are two forces acting on the body as it moves along the circle: i) a tangential force that tangentially accelerates the body as it moves, given by $m \frac{d v}{d t}$ , and ii) the centripetal force that causes the body to turn, given by $\frac{m v ^{2}}{R}$ . Thus, we have:

The kinetic energy is $K E = \frac{1}{2} m v^{2} = a s^{2}$ , so $v^{2} = \frac{2 a s ^{2}}{m}$ .

The tangential force is $F_{t} = \frac{d ( K E )}{d s} = \frac{d ( a s ^{2} )}{d s} = 2 a s$ .

The centripetal force is $F_{c} = \frac{m v ^{2}}{R} = \frac{m}{R} (\frac{2 a s ^{2}}{m}) = \frac{2 a s ^{2}}{R}$ .

The net force is $F = F_{t}^{2} + F_{c}^{2} = (2 a s)^{2} + (\frac{2 a s ^{2}}{R})^{2} = 2 a s 1 + \frac{s ^{2}}{R ^{2}}$ .

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