Newtons Laws EX 43

The rotor rotates around the axis shown. The man in the rotor has the radial acceleration which points towards the center of his trajectory. The center of his trajectory lies on the axis and therefore, in this position, the man has an acceleration towards the right. We can determine the magnitude of his acceleration from the radius and the period.

Since the radius $r=2m$ and the period $T=2s$ we find the speed $v={2\pi r \over T}={2\pi (2) \over (2)}=2\pi =6.28m/s$

We can also determine the magnitude of his acceleration $a={v^{2} \over r}={(6.28)^{2} \over 2}=19.7m/s^{2}$.

The free body diagram below also shows the forces exerted on the man: $F_{G}$, $F_{N}$ and $F_{f}$. Note that the force of friction points upward because it is the force which will prevent the man from sliding down.

Now we can compute the components of all the forces:

Forces	x-component	y-component
$F_{G}$	0	-700 N
$F_{N}$	$F_{N}\cos(0^{\circ })=F_{N}$	0
$F_{f}$	0	$F_{f}$

Now we will use the Newton's Second Law:

$F_{N}=ma=(70)(19.7)=1380N$

$F_{f}-700=0$. From this equation we find that $F_{f}=700N$.