Work EX 5

(a) The speed of the particle in uniform circular motion can be found from the period (T = 2 s) and the radius (r = 0.5 m):

$v={2\pi r \over T}$

$v=1.57m/s$

(b) $K={\frac {1}{2}}mv^{2}={\frac {1}{2}}(0.5)(1.57)^{2}=0.616J$

(c) The force acting on the particle acts towards the center (i.e.: is centripetal) and is given by:

$F_{C}={mv^{2} \over r}=2\times {{\frac {1}{2}}mv^{2} \over r}=2\times {K \over r}=2\times {(0.616) \over (0.5)}=2.46N$

(d) There is no work done by the force. The displacement $\Delta r$ is in the direction of the velocity. The centripetal force is perpendicular to the velocity and hence to $\Delta r$, so the dot product is zero:

$W=F\cdot \Delta r=Fr\cos(90^{\circ })=0J$

Since there is no work done on the particle, there is no change in kinetic energy and its speed stays constant even though there is a force acting on it.