To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,
[tex]v = r\omega \rightarrow \omega = \frac{v}{r}[/tex]
Here,
v = Lineal velocity
[tex]\omega[/tex]= Angular velocity
r = Radius
Our values are
[tex]v = 6/ms[/tex]
[tex]r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m[/tex]
Replacing to find the angular velocity we have,
[tex]\omega = \frac{6m/s}{0.06m}[/tex]
[tex]\omega = 100rad/s[/tex]
Convert the units to RPM we have that
[tex]\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})[/tex]
[tex]\omega = 955.41rpm[/tex]
Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm
