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):

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

$v=1.57m/s$

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

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

$F_C=\frac{m{v}^2}{r}=2\times \frac{\frac{1}{2}m{v}^2}{r}=2\times \frac{K}{r}=2\times \frac{(0.616)}{(0.5)}=2.46N$

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

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

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.