Answer
a) $v_T=1.25m/s$
b) $\omega=8rev/s$
Work Step by Step
a) At $r=0.0568m$, we have $\omega=3.5rev/s\times(\frac{2\pi rad}{1rev})=22rad/s$
The constant tangential speed is $v_T=r\omega=1.25m/s$
b) At $r=0.0249m$, the angular speed is $$\omega=\frac{v_T}{r}=50.2rad/s\times\Big(\frac{1rev}{2\pi rad}\Big)=8rev/s$$
